AG SRB U5 - 072656 - DR. D. Dambreville's Math...

1 2 3 4 5 6 7 8 9 10

ISBN 978-0-8251-7257-1

Copyright © 2013

J. Weston Walch, Publisher

Portland, ME 04103

www.walch.com

Printed in the United States of America

EDUCATIONWALCH

These materials may not be reproduced for any purpose.The reproduction of any part for an entire school or school system is strictly prohibited.

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iiiTable of Contents

Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . v

Unit 5: Quadratic FunctionsLesson 1: Interpreting Structure in Expressions . . . . . . . . . . . . . . . . . . . . . . . . . . . . U5-1Lesson 2: Creating and Solving Quadratic Equations in One Variable . . . . . . . . U5-18Lesson 3: Creating Quadratic Equations in Two or More Variables . . . . . . . . . . U5-58Lesson 4: Solving Systems of Equations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . U5-109Lesson 5: Interpreting Quadratic Functions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . U5-131Lesson 6: Analyzing Quadratic Functions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . U5-165Lesson 7: Building Functions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . U5-204Lesson 8: Transforming Functions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . U5-226Lesson 9: Fitting Quadratic Functions to Data . . . . . . . . . . . . . . . . . . . . . . . . . . . U5-259

Answer Key . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . AK-1

Table of Contents

vIntroduction

Welcome to the CCGPS Analytic Geometry Student Resource Book. This book will help you learn how to use algebra, geometry, data analysis, and probability to solve problems. Each lesson builds on what you have already learned. As you participate in classroom activities and use this book, you will master important concepts that will help to prepare you for the EOCT and for other mathematics assessments and courses.

This book is your resource as you work your way through the Analytic Geometry course. It includes explanations of the concepts you will learn in class; math vocabulary and definitions; formulas and rules; and exercises so you can practice the math you are learning. Most of your assignments will come from your teacher, but this book will allow you to review what was covered in class, including terms, formulas, and procedures.

• In Unit 1: Similarity, Congruence, and Proofs, you will learn about dilations, and you will construct lines, segments, angles, polygons, and triangles. You will explore congruence and then define, apply, and prove similarity. Finally, you will prove theorems about lines, angles, triangles, and parallelograms.

• In Unit 2: Right Triangle Trigonometry, you will begin by exploring trigonometric ratios. Then you will go on to apply trigonometric ratios.

• In Unit 3: Circles and Volume, you will be introduced to circles and their angles and tangents. Then you will learn about inscribed polygons and circumscribed triangles by constructing them and proving properties of inscribed quadrilaterals. You will construct tangent lines and find arc lengths and areas of sectors. Finally, you will explain and apply area and volume formulas.

• In Unit 4: Extending the Number System, you will start working with the number system and rational exponents. Then you will perform operations with complex numbers and polynomials.

• In Unit 5: Quadratic Functions, you will begin by identifying and interpreting structures in expressions. You will use this information as you learn to create and solve quadratic equations in one variable, including taking the square root of both sides, factoring, completing the square, applying the quadratic formula, and solving quadratic inequalities. You will move on to solving quadratic equations in two or more variables, and solving systems

Introduction

Introductionvi

of equations. You will learn to analyze quadratic functions and to build and transform them. Finally, you will solve problems by fitting quadratic functions to data.

• In Unit 6: Modeling Geometry, you will study the links between the two math disciplines, geometry and algebra, as you derive equations of a circle and a parabola. You will use coordinates to prove geometric theorems about circles and parabolas and solve systems of linear equations and circles.

• In Unit 7: Applications of Probability, you will explore the idea of events, including independent events, and conditional probability.

Each lesson is made up of short sections that explain important concepts, including some completed examples. Each of these sections is followed by a few problems to help you practice what you have learned. The “Words to Know” section at the beginning of each lesson includes important terms introduced in that lesson.

As you move through your Analytic Geometry course, you will become a more confident and skilled mathematician. We hope this book will serve as a useful resource as you learn.

U5-1

Lesson 1: Interpreting Structure in Expressions

UNIT 5 • QUADRATIC FUNCTIONS

Lesson 1: Interpreting Structure in Expressions

Common Core Georgia Performance Standards

MCC9–12.A.SSE.1a★

MCC9–12.A.SSE.1b★

WORDS TO KNOW

binomial a polynomial with two terms

coefficient the number multiplied by a variable in an algebraic expression

constant term a term whose value does not change

factor one of two or more numbers or expressions that when multiplied produce a given product

like terms terms that contain the same variables raised to the same power

monomial an expression with one term, consisting of a number, a variable, or the product of a number and variable(s)

polynomial a monomial or the sum of monomials

quadratic equation an equation that can be written in the form ax2 + bx + c = 0, where x is the variable, a, b, and c are constants, and a ≠ 0

Essential Questions

1. How are quadratic expressions and quadratic equations alike? How are they different?

2. How are verbal phrases translated into quadratic expressions?

3. How does changing the value of a coefficient, constant, or variable in an expression change the value of the expression?

4. How are quadratic functions different from linear functions?

U5-2Unit 5: Quadratic Functions

quadratic expression an algebraic expression that can be written in the form ax2 + bx + c, where x is the variable, a, b, and c are constants, and a ≠ 0

term a number, a variable, or the product of a number and variable(s)

trinomial a polynomial with three terms

variable a letter used to represent a value or unknown quantity that can change or vary

Recommended Resources• MathIsFun.com. “Algebra—Basic Definitions.”

http://www.walch.com/rr/00098

This website gives an overview of the important vocabulary for this lesson. Color-coded expressions help users visualize the differences between similar terms.

• Quia. “Rags to Riches: Combining Like Terms.”


Players combine like terms to simplify expressions in this multiple-choice game modeled on the TV show, Who Wants to Be a Millionaire? Players can use up to three hints on their quest to reach the million-dollar question.

• TheMathGames.com. “Rj Collects Like Terms.”


Players guide Rj the robot to drop blocks labeled with terms into pipes with like terms.

U5-3Lesson 1: Interpreting Structure in Expressions

IntroductionAlgebraic expressions are mathematical statements that include numbers, operations, and variables to represent a number or quantity. We know that a variable is a letter used to represent a value or unknown quantity that can change or vary. We have seen several linear expressions such as 2x + 1. In this example, the highest power of the variable x is the first power. In this lesson, we will look at expressions where the highest power of the variable is 2.

Key Concepts

• A quadratic expression is an expression where the highest power of the variable is the second power.

• A quadratic expression can be written in the form ax2 + bx + c, where x is the variable, and a, b, and c are constants.

• Both b and c can be any number, but a cannot be equal to 0 because quadratic expressions must contain a squared term.

• An example of a quadratic expression is 4x2 + 6x – 2. When a quadratic expression is set equal to 0, as in 4x2 + 6x – 2 = 0, the resulting equation is called a quadratic equation. A quadratic equation is an equation that can be written in the form ax2 + bx + c = 0, where x is the variable, a, b, and c are constants, and a ≠ 0.

• The quadratic expression 4x2 + 6x – 2 is made up of many component parts: terms, factors, coefficients, and constants.

• A term is a number, a variable, or the product of a number and variable(s).

• There are 3 terms in the given expression: 4x2, 6x, and –2.

• A factor is one of two or more numbers or expressions that when multiplied produce a given product. In the given expression, the factors of 4x2 are 4 and x2 and the factors of 6x are 6 and x.

• The number multiplied by a variable in an algebraic expression is called a coefficient. In the given expression, the coefficient of the term 4x2 is 4 and the coefficient of the term 6x is 6.

• When there is no number before a variable, the coefficient is either 1 or –1 because x means 1x and –x2 means –1x2.

Lesson 5.1.1: Identifying Terms, Factors, and Coefficients


• A term that does not contain a variable is called a constant term because the value of the term does not change. In the given expression, –2 is a constant.

• Two or more terms that contain the same variables raised to the same power are called like terms. Like terms can be combined by adding. Be sure to follow the order of operations when combining like terms.

• 4x2 + 6x – 2 has no like terms, so let’s use another expression as an example: 9x2 – 8x2 + 2x.

• In the expression 9x2 – 8x2 + 2x, 9x2 and –8x2 are like terms. After simplifying the expression by combining like terms (9x2 and –8x2), the result is x2 + 2x.

• A monomial is a number, a variable, or the product of a number and variable(s). We can also think of a monomial as an expression containing only one term. 5x2 is an example of a monomial.

• A polynomial is a monomial or the sum of monomials. A polynomial can have any number of terms.

• A binomial is a polynomial with two terms. 6x + 9 is an example of a binomial.

• A trinomial is a polynomial with three terms. 4x2 + 6x – 2 is an example of a trinomial.


Guided Practice 5.1.1Example 1

Identify each term, coefficient, and constant of 6(x – 1) – x(3 – 2x) + 12. Classify the expression as a monomial, binomial, or trinomial. Determine whether it is a quadratic expression.

1. Simplify the expression.

The expression can be simplified by following the order of operations and combining like terms.

6(x – 1) – x(3 – 2x) + 12 Original expression

6x – 6 – x(3 – 2x) + 12 Distribute 6 over x – 1.

6x – 6 – 3x + 2x2 + 12 Distribute –x over 3 – 2x.

3x + 6 + 2x2 Combine like terms: 6x and –3x; –6 and 12.

2x2 + 3x + 6 Rearrange terms so the powers are in descending order.

2. Identify all terms.

There are three terms in the expression: 2x2, 3x, and 6.

3. Identify all coefficients.

The number multiplied by a variable in the term 2x2 is 2; the number multiplied by a variable in the term 3x is 3; therefore, the coefficients are 2 and 3.

4. Identify any constants.

The quantity that does not change (is not multiplied by a variable) in the expression is 6; therefore, 6 is a constant.


5. Classify the expression as a monomial, binomial, or trinomial.

The polynomial is a trinomial because it has three terms.

6. Determine whether the expression is a quadratic expression.

It is a quadratic expression because it can be written in the form ax2 + bx + c, where a = 2, b = 3, and c = 6.

Example 2

Translate the verbal expression “take triple the difference of 12 and the square of x, then increase the result by the sum of 3 and x” into an algebraic expression. Identify the terms, coefficients, and constants of the given expression. Is the expression quadratic?

1. Translate the expression by breaking it down into pieces.

“The difference of 12 and the square of x” translates to (12 – x2).

Triple this expression is 3(12 – x2).

“The sum of 3 and x” translates to (3 + x).

Increasing the original expression by this sum translates to 3(12 – x2) + (3 + x).

2. Simplify the expression.

3(12 – x2) + (3 + x) Expression

36 – 3x2 + (3 + x) Distribute 3 over (12 – x2).

–3x2 + x + 39 Combine like terms.

The simplified expression is –3x2 + x + 39.


There are three terms in the expression: –3x2, x, and 39.



The number multiplied by a variable in the term –3x2 is –3; the number multiplied by a variable in the term x is 1; therefore, –3 and 1 are coefficients.


The number that is not multiplied by a variable in the expression is 39; therefore, 39 is a constant.


It is a quadratic expression because it can be written in the form ax2 + bx + c, where a = –3, b = 1, and c = 39.

Example 3

A fence surrounds a park in the shape of a pentagon. The side lengths of the park in feet are given by the expressions 2x2, 3x + 1, 3x + 2, 4x, and 5x – 3. Find an expression for the perimeter of the park. Identify the terms, coefficients, and constant in your expression. Is the expression quadratic?

1. Find an expression for the perimeter of the park.

Add like terms to find the perimeter, P.

P = 2x2 + (3x + 1) + (3x + 2) + 4x + (5x – 3) Set up the equation using the given expressions.

P = 2x2 + 3x + 3x + 4x + 5x + 1 + 2 – 3 Reorder like terms.

P = 2x2 + 15x Combine like terms.

The expression for the park’s perimeter is 2x2 + 15x.


There are two terms in this expression: 2x2 and 15x.



The number multiplied by a variable in the term 2x2 is 2; the number multiplied by a variable in the term 15x is 15; therefore, 2 and 15 are coefficients.


Every number in the expression is multiplied by a variable; therefore, there is no constant.


It is a quadratic expression because it can be written in the form ax2 + bx + c, where a = 2, b = 15, and c = 0.

UNIT 5 • QUADRATIC FUNCTIONSLesson 1: Interpreting Structure in Expressions

PRACTICE


Use what you know about the components of expressions to complete problems 1–3.

1. Identify the terms, coefficients, constant, and factors of 30x2 – 18x + 72.

2. Simplify the expression 5x + 4(5x – x) – 2x(6) and classify it as a monomial, binomial, or trinomial.

3. Write a quadratic expression that contains three terms, coefficients of –1 and 1, and a constant of 52.

For problems 4 and 5, determine whether each expression is a quadratic expression. Explain your reasoning.

4. 20x(4 – 5x) + 3(x – 8)

5. 12x(x2 – 4x) – 2(3 + x)

continued

Practice 5.1.1: Identifying Terms, Factors, and Coefficients


PRACTICE


For problems 6–10, write an algebraic expression. Identify the terms, coefficients, and constants of the given expressions. Determine whether the expression is quadratic and explain your reasoning.

6. the product of 9 and x, decreased by the sum of 8 and the square of x

7. double the sum of 2 and x increased by one-half x2

8. The area of a square is the square of its side, s.

9. The volume of a cube is the cube of the length of its side, s.

10. The surface area of a sphere with radius r is four times the product of π and the square of the radius.


Lesson 5.1.2: Interpreting Complicated Expressions

IntroductionVariables change, but constants remain the same. We need to understand how each term of an expression works in order to understand how changing the value of a variable will affect the value of the expression. In this lesson, we will explore these relationships.

Key Concepts

• We can translate verbal expressions into algebraic expressions to analyze how changing values of coefficients, constants, or variables will affect the value of the expression.

• Changing the value of a variable will not change the value of a constant, just as changing the value of a constant will not change terms containing variables.

• It is important to always follow the correct order of operations.

• Simplifying expressions lets us more easily see how different terms interact with one another.

• Use the Distributive Property to multiply binomials.

• When presented with a polynomial in factored form, multiply the factors to see if the polynomial is a quadratic. Quadratic expressions are of the form ax2 + bx + c, where a ≠ 0.

• Sometimes it is easier to work with the factored form of an expression. For instance, since we know that a positive number multiplied by a positive number is always a positive number, it is easier to determine values of a variable that make an expression positive by finding the values that make all of its factors positive.

• We can use similar logic to determine the values that make an expression negative or equal to 0.


Example 1

Show that (x + 2)(2x – 1) is a quadratic expression by writing it in the form ax2 + bx + c. Identify a, b, and c.

1. Multiply the factors using the Distributive Property.

(x + 2)(2x – 1) Original expression

x(2x) + x(–1) + 2(2x) + 2(–1) Distribute x over (2x – 1) and

2 over (2x – 1).

2x2 – x + 4x – 2 Combine like terms.

2x2 + 3x – 2

2. Compare the resulting polynomial to ax2 + bx + c.

Since 2x2 + 3x – 2 has the form ax2 + bx + c, where a ≠ 0, (x + 2)(2x – 1) is a quadratic expression.

3. Identify a, b, and c.

In the polynomial, a = 2, b = 3, and c = –2.

Guided Practice 5.1.2


Example 2

What values of x make the expression (x + 2)(x – 3) positive?

1. Determine the sign possibilities for each factor.

The expression will be positive when both factors are positive or both factors are negative.

2. Determine the values of the variable that make both factors positive.

x + 2 is positive when x > –2.

x – 3 is positive when x > 3.

Both factors are positive when x > –2 and x > 3, or when x > 3.

3. Determine the values of the variable that make both factors negative.

x + 2 is negative when x < –2.

x – 3 is negative when x < 3.

Both factors are negative when x < –2 and x < 3, or when x < –2.

4. Analyze the range of values of x at which the factors are either both positive or both negative to determine the outer limits of values for x.

The value of the expression is positive when x > 3 or x < –2.


Example 3

The length of each side of a square is increased by 2 centimeters. How does the perimeter change? How does the area change?

1. Find an expression for the perimeter of the original square.

Let x be the length of one side of the square in centimeters.

Multiply each side length by 4 since a square has four sides of equal length.

The perimeter of the square is 4x centimeters.

2. Find an expression for the perimeter of the new square and compare the new perimeter to the original perimeter.

One side of the new square is x + 2 centimeters.

The new perimeter is 4(x + 2) or 4x + 8 centimeters.

Next, subtract the old perimeter from the new perimeter to determine the difference.

4x + 8 – 4x = 8

The new perimeter is 8 centimeters longer than the original perimeter.

3. Find an expression for the area of the original square.

The area of the original square is x2 square centimeters.

4. Find an expression for the area of the new square and compare the new area to the original area.

The new area is (x + 2)2 or x2 + 4x + 4 square centimeters.

Subtract the old area from the new area to determine the difference.

x2 + 4x + 4 – x2 = 4x + 4

The new area is 4x + 4 square centimeters larger than the old area.


Example 4

A car’s total stopping distance in feet depends on many factors, but can be

approximated by the expression 11

10

1

192x x+ , for which x is the speed of the car in

miles per hour. Is this expression quadratic? What effect does doubling the car’s speed

from 10 mph to 20 mph have on the total stopping distance?

1. Determine whether the expression is quadratic.

This expression is quadratic because it can be written in the form

ax2 + bx + c, with a = 1

19, b =

11

10, and c = 0.

2. Determine the stopping distance of a car traveling 10 mph.

Substitute 10 for x to find the total stopping distance of a car traveling 10 mph.

11

1010

1

1910 2( ) ( )+ = 11 +

100

19 ≈ 16.26 feet

The stopping distance of a car traveling 10 mph is approximately 16.26 feet.

3. Determine the stopping distance of a car traveling 20 mph.

Substitute 20 for x to find the total stopping distance of a car traveling 20 mph.

+11

10(20)

1

19(20)2 = 22 +

400

19 ≈ 43.05 feet

The stopping distance of a car traveling 20 mph is approximately 43.05 feet.

4. Compare the stopping distances for a car traveling 10 mph and a car traveling 20 mph.

When the speed of the car doubles from 10 mph to 20 mph, the stopping distance almost triples from about 16 feet to about 43 feet.


PRACTICE


For problems 1 and 2, use what you know about expressions to answer the questions.

1. What values of x make the expression (x + 4)(x – 6) negative?

2. What values of x make the expression (5x + 7)(2x – 8) positive?

For problems 3 and 4, show that the expression is a quadratic expression by writing it in the form ax2 + bx + c. Identify a, b, and c.

3. (9x – 15)(2x + 1)

4. (x – 2)(x + 3) – (x – 4)(2x + 5)

For problems 5 and 6, determine whether each expression is a quadratic expression. Explain your reasoning.

5. (x – 1)2 + 10

continued

Practice 5.1.2: Interpreting Complicated Expressions


PRACTICE


6. (x + 4)(x + 1)(x – 1)

For problems 7–10, translate any verbal expressions to quadratic expressions, and then answer the questions.

7. The population of a city t years after 2000 is represented by the quadratic expression 100t2 + 300t + 50,000. How did the city’s population change from 2000 to 2012?

8. The surface area in feet of a cylinder with a radius of 1 foot can be found by adding the product of 2π and the square of the radius to the product of 2π and the radius. How does the surface area change when the radius is tripled from 1 foot to 3 feet?

9. How does the area of a square change when the side length is tripled?

10. The surface area of a sphere is the product of 4π and the square of the radius. How does the surface area change when the radius is halved?

U5-18

Lesson 2: Creating and Solving Quadratic Equations in One Variable


Unit 5: Quadratic Functions


MCC9–12.A.SSE.2

MCC9–12.N.CN.7

MCC9–12.A.CED.1★

MCC9–12.A.REI.4a

MCC9–12.A.REI.4b

WORDS TO KNOW

complex number a number in the form a + bi, where a and b are real numbers, and i is the imaginary unit

discriminant an expression whose solved value indicates the number and types of solutions for a quadratic. For a quadratic equation in standard form (ax2 + bx + c = 0), the discriminant is b2 – 4ac.

factor to write an expression as the product of its factors

greatest common factor (GCF)

the largest factor that two or more terms share

imaginary number any number of the form bi, where b is a real number, i = −1 , and b ≠ 0

imaginary unit, i the letter i, used to represent the non-real value, i = −1

Essential Questions

1. How can we solve a quadratic equation by taking square roots?

2. How can we solve a quadratic equation by factoring?

3. How can we solve a quadratic equation by completing the square?

4. How is the quadratic formula derived?

5. How is a quadratic inequality solved?

U5-19Lesson 2: Creating and Solving Quadratic Equations in One Variable

interval the set of all real numbers between two given numbers. The two numbers on the ends are the endpoints. The endpoints might or might not be included in the interval depending on whether the interval is open, closed, or half-open.

irrational number a number that cannot be written as m

n , where m and

n are integers and n ≠ 0; any number that cannot be

written as a decimal that ends or repeats

leading coefficient the coefficient of the term with the highest power. For a quadratic equation in standard form (ax2 + bx + c), the leading coefficient is a.

perfect square trinomial a trinomial of the form x bxb

2

2

2+ +

that can be

written as the square of a binomial

prime an expression that cannot be factored

quadratic equation an equation that can be written in the form ax2 + bx + c = 0, where x is the variable, a, b, and c are constants, and a ≠ 0

quadratic formula a formula that states the solutions of a quadratic

equation of the form ax2 + bx + c = 0 are given by

xb b ac

a=− ± −2 4

2. A quadratic equation in this

form can have no real solutions, one real solution,

or two real solutions.

quadratic inequality an inequality that can be written in the form ax2 + bx + c < 0, ax2 + bx + c ≤ 0, ax2 + bx + c > 0, or ax2 + bx + c ≥ 0

rational number any number that can be written as m

n, where both m

and n are integers and n ≠ 0; any number that can be

written as a decimal that ends or repeats

real numbers the set of all rational and irrational numbers

root(s) solution(s) of a quadratic equation


Zero Product Property if the product of two factors is 0, then at least one of the factors is 0

Recommended Resources• Cliffs Notes. “Solving Quadratic Inequalities.”


This website explains how to solve a quadratic inequality. Users are guided through several examples.

• Khan Academy. “Solving Quadratic Equations by Square Roots.”


This video explains how to solve quadratic equations by taking square roots.

• Math Portal. “Quadratic Equation Solver.”


This interactive website guides users through solving a quadratic equation in standard form. Users input an equation, and then the calculator shows the steps to solve the quadratic by completing the square or by using the quadratic formula.

• MathIsFun.com. “Completing the Square.”


This website offers the user a tutorial on completing the square.

• National Library of Virtual Manipulatives. “Algebra Tiles.”


This virtual manipulative shows users how to use algebra tiles to multiply binomials, multiply polynomials using the Distributive Property, factor polynomials, and square binomials. The manipulative requires Java software to run.


IntroductionYou can determine how far a ladder will extend from the base of a wall by creating a quadratic equation and then taking the square root. To find this length, you only need to find the positive square root because a negative distance would not make sense in this situation. For certain types of quadratics, we can solve by taking the square root, but in most problems, we need to take both the positive and negative square root.

Key Concepts

• The imaginary unit i represents the non-real value i = −1 . i is the number whose square is –1. We define i so that i = −1 and i 2 = –1.

• An imaginary number is any number of the form bi, where b is a real number, i = −1 , and b ≠ 0.

• A complex number is a number with a real component and an imaginary component. Complex numbers can be written in the form a + bi, where a and b are real numbers, and i is the imaginary unit. For example, 5 + 3i is a complex number. 5 is the real component and 3i is the imaginary component.

• Real numbers are the set of all rational and irrational numbers. Real numbers do not contain an imaginary component.

• Real numbers are rational numbers when they can be written as m

n, where

both m and n are integers and n ≠ 0. Rational numbers can also be written as a

decimal that ends or repeats. The real number 0.4 is a rational number because

it can be written as the fraction 2

5.

• Real numbers are irrational when they cannot be written as m

n , where m and

n are integers and n ≠ 0. Irrational numbers cannot be written as a decimal

that ends or repeats. The real number 3 is an irrational number because

it cannot be written as the ratio of two integers. Other examples of irrational

numbers include 2 and π.

• A quadratic equation is an equation that can be written in the form ax2 + bx + c = 0, where a ≠ 0.

Lesson 5.2.1: Taking the Square Root of Both Sides


• Quadratic equations can have no real solutions, one real solution, or two real solutions. When a quadratic has no real solutions, it has two complex solutions.

• Quadratic equations that contain only a squared term and a constant can be solved by taking the square root of both sides. These equations can be written in the form x2 = c, where c is a constant.

• When we take the square root of both sides, we need to remember that a

number and its opposite have the same square. Therefore, rather than simply

taking the positive square root, we need to take the positive and negative

square root. For x2 = c, we find that x c= ± .

• We can use a similar method to solve quadratic equations in the form (ax + b)2 = c.

• c tells us the number and type of solutions for the equation.

c Number and type of solutionsNegative Two complex solutions0 One real, rational solutionPositive and a perfect square Two real, rational solutionsPositive and not a perfect square Two real, irrational solutions


Example 1

Solve 2x2 – 5 = 195 for x.

1. Isolate x2.

2x2 – 5 = 195 Original equation

2x2 = 200 Add 5 to both sides.

x2 = 100 Divide both sides by 2.

2. Use a square root to find all possible solutions to the equation.

Take the square root of both sides. Remember that both 102 and (–10)2 equal 100.

x = ± = ±100 10

The equation 2x2 – 5 = 195 has two solutions, 10 and –10.



Example 2

Solve (x – 1)2 + 15 = –1 for x.

1. Isolate the squared binomial.

(x – 1)2 + 15 = –1 Original equation

(x – 1)2 = –16 Subtract 15 from both sides.

2. Use a square root to isolate the binomial.

Take the square root of both sides. Remember to use the ± sign.

x− = ± −1 16

3. Simplify the square root.

There is a negative number under the radical, so the answer will be a complex number.

x− = ± −1 16 Equation

x− = ± −1 1 16( )( ) Write –16 as a product of a perfect square and –1.

x− = ± − •1 1 16 Product Property of Square Roots

x – 1 = ±4i Simplify.

4. Isolate x.

x – 1 = ±4i Equation

x = 1 ± 4i Add 1 to both sides.

The equation (x – 1)2 + 15 = –1 has two solutions, 1 ± 4i.


Example 3

Solve 4(x + 3)2 – 10 = –6 for x.

1. Isolate the squared binomial.

4(x + 3)2 – 10 = –6 Original equation

4(x + 3)2 = 4 Add 10 to both sides.

(x + 3)2 = 1 Divide both sides by 4.

x+ = ±3 1 Take the square root of both sides.

x + 3 = ±1

2. Isolate x.

x + 3 = ±1 Equation

x = –3 ± 1 Subtract 3 from both sides.

3. Split the answer into two separate expressions and evaluate.

x = –3 + 1 = –2

x = –3 – 1 = –4

The equation 4(x + 3)2 – 10 = –6 has two solutions, –2 and –4.

UNIT 5 • QUADRATIC FUNCTIONSLesson 2: Creating and Solving Quadratic Equations in One Variable

PRACTICE


Solve each equation for x.

1. x2 = 4

2. x2 + 8 = 4

3. x2 + 5 = –3

4. (x – 4)2 = 40

5. 2(x + 6)2 – 6 = –6

6. 8(x – 5)2 = 56

Use what you know about square roots to complete problems 7–10.

7. When does a quadratic equation in the form ax2 + b = c have only one real solution?

8. The area of a square with sides of length s is given by s2. The area of a square is 49 square inches. What is the length of one side of the square?

9. The area of a circle with radius r is given by πr2. The area of a circle is 20π square units. What is the radius of the circle?

10. The surface area of a sphere with radius r is given by 4πr2. If the surface area of a sphere is 20 square feet, what is the radius of the sphere?

Practice 5.2.1: Taking the Square Root of Both Sides


IntroductionRecall that a factor is one of two or more numbers or expressions that when multiplied produce a given product. We can factor certain expressions by writing them as the product of factors.

The Zero Product Property states that if the product of two factors is 0, then at least one of the factors is 0. After setting a quadratic equation equal to 0, we can sometimes factor the quadratic expression and solve the equation by setting each factor equal to 0.

Key Concepts

• The greatest common factor, or GCF, is the largest factor that two or more terms share.

• You should always check to see if the terms of an expression have a greatest common factor before attempting to factor further.

• The value of a for a quadratic expression in the form ax2 + bx + c is called the leading coefficient, or lead coefficient, because it is the coefficient of the term with the highest power.

• To factor a trinomial with a leading coefficient of 1 in the form x2 + bx + c, find two numbers d and e that have a product of c and a sum of b.

• The factored form of the expression will be (x + d)(x + e).

• When finding d and e, be careful with the signs.

• The table that follows shows what the signs of d and e will be based on the signs of b and c.

Signs of b, c, d, and e

b c d e+ + + +– + – –

+ –Opposite signs; the number with the larger absolute value is positive.

– –Opposite signs; the number with the larger absolute value is negative.

Lesson 5.2.2: Factoring


• You may be able to factor expressions with lead coefficients other than 1 in your head or by using guess-and-check.

• Expressions with lead coefficients other than 1 in the form ax2 + bx + c can sometimes be factored by grouping.

• If you struggle to factor expressions in your head or by using guess-and-check, factoring by grouping is a more structured alternative.

Factoring by Grouping1. Begin by finding two numbers d and e whose product is ac and whose

sum is b.

2. Rewrite the expression by replacing bx with dx + ex: ax2 + dx + ex + c.

3. Factor the greatest common factor from ax2 + dx.

4. Factor the greatest common factor from ex + c.

5. Factor the greatest common factor from the resulting expression.

• A quadratic expression in the form (ax)2 – b2 is called a difference of squares.

• The difference of squares (ax)2 – b2 can be written in factored form as (ax + b)(ax – b).

• Some expressions cannot be factored. These expressions are said to be prime.

• Although the difference of squares is factorable, the sum of squares is prime.

• For example, (3x)2 – 52 = 9x2 – 25 = (3x + 5)(3x – 5), but (3x)2 + 52 = 9x2 + 30x + 25 and is not factorable.


Example 1

Factor x2 – 8x + 15.

1. The lead coefficient is 1, so we begin by finding two numbers whose product is 15 and whose sum is –8.

The numbers are –3 and –5 because (–3)(–5) = 15 and –3 + –5 = –8.

2. Find the factors.

Use –3 and –5 to find the factors.

The factors will be (x – 3) and (x – 5).

3. Write the expression as the product of its factors.

The factors are (x – 3) and (x – 5), so the product of the factors is (x – 3)(x – 5).

You can check your work by multiplying the factors, (x – 3)(x – 5). The product should be the original expression.

(x – 3)(x – 5) = x2 – 8x + 15


Example 2

Solve 8x2 – 8 = –x2 + 56 by factoring.

1. Rewrite the equation so that all terms are on one side.

8x2 – 8 = –x2 + 56 Original equation

9x2 – 8 = 56 Add x2 to both sides.

9x2 – 64 = 0 Subtract 56 from both sides.


2. Factor the difference of squares.

The expression on the left side can be rewritten in the form (3x)2 – 82.

We can use this form to rewrite the expression as the difference of squares to factor the expression.

(3x + 8)(3x – 8) = 0

3. Use the Zero Product Property to solve.

The expression will equal 0 only when one of the factors is equal to 0.

Set each factor equal to 0 and solve.

3x + 8 = 0 3x – 8 = 0

3x = –8 3x = 8

x = −8

3x =

8

3

8x2 – 8 = –x2 + 56 when x = −8

3 or

8

3.

Example 3

Solve x2 + 8x = 20 by factoring.

1. Rewrite the equation so all the terms are on one side of the equation.

x2 + 8x = 20 Original equation

x2 + 8x – 20 = 0 Subtract 20 from both sides.

2. Find the factors.

The lead coefficient of the expression is 1, so begin by finding two numbers whose product is –20 and whose sum is 8.

The numbers are –2 and 10 because (–2)(10) = –20 and –2 + 10 = 8.

Therefore, the factors are (x – 2) and (x + 10).


Example 4

Factor 3x2 – x – 10 by grouping.

1. Find two numbers whose product is ac and whose sum is b.

The expression is in the form ax2 + bx + c.

a = 3, b = –1, and c = –10

ac = 3(–10) = –30

We need to find two numbers whose product is –30 and whose sum is –1.

The numbers are –6 and 5 because (–6)(5) = –30 and –6 + 5 = –1.

2. Replace bx in the original expression.

The numbers we found have a sum of b, so we can replace bx with –6x + 5x.

3x2 – 6x + 5x – 10


(x – 2)(x + 10) = 0


The expression will equal 0 only when one of the factors is equal to 0. Set each factor equal to 0 and solve.

x – 2 = 0 x + 10 = 0

x = 2 x = –10

x2 + 8x = 20 when x = 2 or when x = –10.


Example 5

Solve 7x2 + 63x – 70 = 0.

1. Factor out the greatest common factor of the expression.

The greatest common factor is 7.

7(x2 + 9x – 10) = 0

2. Find two numbers whose product is –10 and whose sum is 9.

The numbers are –1 and 10 because (–1)(10) = –10 and –1 + 10 = 9.

3. Use these numbers to find the factors.

The factors are (x – 1) and (x + 10).

3. Factor the greatest common factor of the two left-hand terms. Do the same with the two right-hand terms.

The greatest common factor of the left-hand terms is 3x.

3x(x – 2) + 5x – 10

The greatest common factor of the right-hand terms is 5.

3x(x – 2) + 5(x – 2)

4. Factor the greatest common factor of the expression.

The greatest common factor of the expression is (x – 2).

3x2 – x – 10 = (x – 2)(3x + 5)

We have just factored by grouping. You can check your work by multiplying the factors. The product should be the original expression.


Example 6

Solve 8x2 + 18x = 5.

1. Rewrite the equation so that all the terms are on one side of the equation.

8x2 + 18x = 5 Original equation

8x2 + 18x – 5 = 0 Subtract 5 from both sides.


7(x – 1)(x + 10) = 0


The expression will equal 0 only when one of the factors is equal to 0. The factor of 7 will never equal 0. Set each of the remaining factors equal to 0 and solve.

x – 1 = 0 x + 10 = 0

x = 1 x = –10

7x2 + 63x – 70 = 0 when x = 1 or when x = –10.


2. Factor the equation.

The lead coefficient is not 1, so factor by grouping.

Begin by finding two numbers whose product is ac and whose sum is b.

a = 8, b = 18, and c = –5

ac = 8(–5) = –40

Find two numbers whose product is –40 and whose sum is 18.

The numbers are 20 and –2 because (20)(–2) = –40 and 20 + –2 = 18.

Replace bx in the original expression.

The numbers we found have a sum of b, so we can replace bx with 20x – 2x.

8x2 + 20x – 2x – 5 = 0

Factor out the greatest common factor of the left-hand terms. Do the same with the right-hand terms.

The greatest common factor of the left-hand terms is 4x.

4x(2x + 5) –2x – 5 = 0

The greatest common factor of the right-hand terms is –1.

4x(2x + 5) – 1(2x + 5) = 0

Factor out the greatest common factor of the expression, (2x + 5).

(2x + 5)(4x – 1) = 0

3. Set each factor equal to 0 and solve.

2x + 5 = 0 4x – 1 = 0

2x = –5 4x = 1

x = −5

2x =

1

4

8x2 + 18x = 5 when x = −5

2 or when x =

1

4.


PRACTICE


Factor each expression completely.

1. 12b2 – 42b

2. x2 + x – 30

3. 16s2 + 1

For problems 4–7, solve each equation by factoring.

4. x2 – 2x + 1 = 0

5. 3a2 – 6a = 0

6. 7x2 + 10x + 3 = 0

7. 4x2 + 6x – 10 = 0

Use the given information to solve problems 8–10. Determine whether your answers are reasonable and explain why or why not.

8. The income in dollars for a charity fund-raiser can be expressed by 360p – 12p2, where p is the ticket price. What ticket price(s) will result in an income of $0?

9. A rectangular garden shed has an area of x2 – 11x + 18 square feet. Find the width of the carpet if the length is x – 2 feet.

10. The length of a rectangle is 4 centimeters longer than its width. The area of the rectangle is 320 square centimeters. Find the width of the rectangle.

Practice 5.2.2: Factoring


Lesson 5.2.3: Completing the Square

Introduction

A trinomial of the form x bxb

2

2

2+ +

that can be written as the square of a

binomial is called a perfect square trinomial. We can solve quadratic equations by

transforming the left side of the equation into a perfect square trinomial and using

square roots to solve.

Key Concepts

• When the binomial (x + a) is squared, the resulting perfect square trinomial is x2 + 2ax + a2.

• When the binomial (ax + b) is squared, the resulting perfect square trinomial is a2x2 + 2abx + b2.

Completing the Square to Solve Quadratics1. Make sure the equation is in standard form, ax2 + bx + c.

2. Subtract c from both sides.

3. Divide each term by a to get a leading coefficient of 1.

4. Add the square of half of the coefficient of the x-term to both sides to complete the square.

5. Express the perfect square trinomial as the square of a binomial.

6. Solve by using square roots.


Example 1

Solve x2 – 8x + 16 = 4 by completing the square.

1. Determine if x2 – 8x + 16 is a perfect square trinomial.

Take half of the value of b and then square the result. If this is equal to the value of c, then the expression is a perfect square trinomial.

b

2

8

216

2 2

=−

=

x2 – 8x + 16 is a perfect square trinomial because the square of half of –8 is 16.

2. Write the left side of the equation as a binomial squared.

Half of b is –4, so the left side of the equation can be written as (x – 4)2.

(x – 4)2 = 4

3. Isolate x.

(x – 4)2 = 4 Perfect square trinomial

x – 4 = ± 2 Take the square root of both sides.

x = 4 ± 2 Add 4 to both sides.

x = 4 + 2 = 6 or x = 4 – 2 = 2Split the answer into two separate equations and solve for x.

4. Determine the solution(s).

The equation has two solutions, x = 2 or x = 6.



Example 2

Solve x2 + 6x + 4 = 0 by completing the square.

1. Determine if x2 + 6x + 4 is a perfect square trinomial.

Take half of the value of b and then square the result. If this is equal to the value of c, then the expression is a perfect square trinomial.

b

2

6

29

2 2

=

=

x2 + 6x + 4 is not a perfect square trinomial because the square of half of 6 is not 4.

2. Complete the square.

x2 + 6x + 4 = 0 Original equation

x2 + 6x = –4 Subtract 4 from both sides.

x2 + 6x + 32 = –4 + 32 Add the square of half of the coefficient of the x-term to both sides to complete the square.

x2 + 6x + 9 = 5 Simplify.


Half of b is 3, so the left side of the equation can be written as (x + 3)2.

(x + 3)2 = 5


4. Isolate x.

(x + 3)2 = 5 Equation

x+ = ±3 5 Take the square root of both sides.

x = − ±3 5 Subtract 3 from both sides.


The equation x2 + 6x + 4 = 0 has two solutions, x = − ±3 5 .

Example 3

Solve 5x2 – 50x – 120 = 0 by completing the square.

1. Determine if 5x2 – 50x – 120 = 0 is a perfect square trinomial.

The leading coefficient is not 1.

First divide both sides of the equation by 5 so that a = 1.

5x2 – 50x – 120 = 0 Original equation

x2 – 10x – 24 = 0 Divide both sides by 5.

Now that the leading coefficient is 1, take half of the value of b and then square the result. If the expression is equal to the value of c, then it is a perfect square trinomial.

b

2

10

225

2 2

=−

=

5x2 – 50x – 120 = 0 is not a perfect square trinomial because the square of half of –10 is not –24.



x2 – 10x – 24 = 0 Equation

x2 – 10x = 24 Add 24 to both sides.

x2 – 10x + (–5)2 = 24 + (–5)2 Add the square of half of the coefficient of the x-term to both sides to complete the square.

x2 – 10x + 25 = 49 Simplify.


Half of b is –5, so the left side of the equation can be written as (x – 5)2.

(x – 5)2 = 49

4. Isolate x.

(x – 5)2 = 49 Equation

x− = ± = ±5 49 7 Take the square root of both sides.

x = 5 ± 7 Add 5 to both sides.

x = 5 + 7 = 12 or x = 5 – 7 = –2Split the answer into two separate equations and solve for x.


The equation 5x2 – 50x – 120 = 0 has two solutions, x = –2 or x = 12.


PRACTICE


For problems 1–4, find c so that the expression is a perfect square trinomial.

1. x2 + 22x + c

2. x2 + 100x + c

3. x2 – 9x + c

4. x x c24

5− +

Solve problems 5–7 by completing the square.

5. x2 – 8x + 2 = 0

6. 2x2 + 2x = 5

7. x2 + 4x = 21

Use what you know about squares and factoring to solve problems 8–10. Determine whether your answers are reasonable and explain why or why not.

8. A dog pen has an area of 60 square feet. The width of the pen is 2 feet shorter than its length. Find the length of the pen.

9. A student kicks a ball during gym class. The ball’s height in feet x seconds after being kicked is given by –16x2 + 40x. When will the ball hit the ground?

10. The fuel economy in miles per gallon of a certain truck is given by the expression –0.02x2 + 1.5x + 3.4, where x is the truck’s speed in miles per hour. For what speed(s) does the truck have a fuel economy of 20 miles per gallon?

Practice 5.2.3: Completing the Square


Lesson 5.2.4: Applying the Quadratic Formula

IntroductionCompleting the square can be a long process, and not all quadratic expressions can be factored. Rather than completing the square or factoring, we can use a formula that can be derived from the process of completing the square. This formula, called the quadratic formula, can be used to solve any quadratic equation in standard form, ax2 + bx + c = 0.

Key Concepts

• A quadratic equation in standard form, ax2 + bx + c = 0, can be solved for x by

using the quadratic formula: xb b ac

a=− ± −2 4

2.

• Solutions of quadratic equations are also called roots.

• The expression under the radical, b2 – 4ac, is called the discriminant.

• The discriminant tells us the number and type of solutions for the equation.

Discriminant Number and type of solutionsNegative Two complex solutions0 One real, rational solutionPositive and a perfect square Two real, rational solutionsPositive and not a perfect square Two real, irrational solutions



Given the standard form of a quadratic equation, ax2 + bx + c, derive the quadratic formula by completing the square.

1. Begin with a quadratic equation in standard form.

ax2 + bx + c = 0

2. Subtract c from both sides.

ax2 + bx = –c

3. Divide both sides by a.

xb

ax

c

a2 + =

−


Add the square of half of the coefficient of the x-term to both sides to complete the square.

Add b

a2

2

to both sides to complete the square.

xb

ax

b

a

c

a

b

a2

2 2

2 2+ +

=−

+


5. Write the left side as a binomial squared and simplify the right side.

xb

a

c

a

a b+

=−

+2 4

2 2 2

xb

a

ac

a

b

a

xb

a

b ac

a

+

=−

+

+

=−

2

4

4 4

2

4

4

2

2

2

2

2 2

22

6. Take the square root of both sides and simplify the right side.

xb

a

b ac

a

xb

a

b ac

a

+ = ±−

+ = ±−

2

4

4

2

4

2

2

2

2

7. Subtract b

a2 from both sides to solve for x.

xb

a

b ac

a=−

±−

2

4

2

2

8. The quadratic formula can be written as two fractions, as in step 7. However, the fractions are often combined.

xb b ac

a=− ± −2 4

2


Example 2

Use the discriminant of 3x2 – 5x + 1 = 0 to identify the number and type of solutions.

1. Determine a, b, and c.

a = 3, b = –5, and c = 1

2. Substitute the values for a, b, and c into the formula for the discriminant, b2 – 4ac.

b2 – 4ac = (–5)2 – 4(3)(1) = 25 – 12 = 13

The discriminant of 3x2 – 5x + 1 = 0 is 13.

3. Use what you know about the discriminant to determine the number and type of solutions for the quadratic equation.

The discriminant, 13, is positive, but it is not a perfect square. Therefore, there will be two real, irrational solutions.


Example 3

Solve 2x2 – 5x = 12 using the quadratic formula.

1. Write the quadratic in standard form.

2x2 – 5x = 12 Original equation

2x2 – 5x – 12 = 0 Subtract 12 from both sides.

2. Determine the values of a, b, and c.

a = 2, b = –5, and c = –12

3. Substitute the values of a, b, and c into the quadratic formula.

xb b ac

a=− ± −2 4

2Quadratic formula

x =− − ± − − −( ) ( ) ( )( )

( )

5 5 4 2 12

2 2

2

Substitute values for a, b, and c.

x =± +5 25 96

4Simplify.

x =±5 121

4

x =±5 11

4


Since the discriminant, 121, is positive and a perfect square, there are two real, rational solutions.

Write the fraction from step 3 as two fractions and simplify.

x =+

= =5 11

4

16

44 x =

−=−

=−5 11

4

6

4

3

2

x = 4 x =−32

The solutions to the equation 2x2 – 5x = 12 are x = 4 or x = −3

2.


Example 4

Solve x2 = 2x – 1 using the quadratic formula.

1. Put the quadratic in standard form.

x2 = 2x – 1 Original equation

x2 – 2x = –1 Subtract 2x from both sides.

x2 – 2x + 1 = 0 Add 1 to both sides.

2. Determine the values of a, b, and c.

a = 1, b = –2, and c = 1

3. Substitute the values of a, b, and c into the quadratic formula.

xb b ac

a=− ± −2 4

2Quadratic formula

x =− − ± − −( ) ( ) ( )( )

( )

2 2 4 1 1

2 1

2


x =± −2 4 4

2Simplify.

x =±2 0

2

x =±2 0

2

4. Determine the solutions(s).

Since the discriminant is 0, there is one real, rational solution.

x =±

= =2 0

2

2

21

x = 1

The solution to the equation x2 = 2x – 1 is x = 1.


Example 5

Solve 5x2 + 2x + 3 = 0 using the quadratic formula.

1. The quadratic is already in standard form. Determine the values of a, b, and c.

a = 5, b = 2, and c = 3

2. Substitute these values into the quadratic formula.

xb b ac

a=− ± −2 4

2Quadratic formula

x =− ± −( ) ( ) ( )( )

( )

2 2 4 5 3

2 5

2


x =− ± −2 4 60

10Simplify.

x =− ± −2 56

10


Since the discriminant is negative, there are two complex solutions.

Simplify to determine the solutions.

x =− ± −2 56

10

x =− ± −2 4 1 14

10

xi

=− ±2 2 14

10

xi

=− ±1 14

5

The solutions to the equation 5x2 + 2x + 3 = 0 are xi

=− ±1 14

5.


PRACTICE


For problems 1 and 2, find the discriminant. Write the number and type of roots of the equation.

1. 3x2 – 5x + 1 = 0

2. –2x2 – 4x = 12

For problems 3–6, solve using the quadratic formula.

3. x2 + 2x + 1 = 0

4. 2x2 + 16x + 40 = 0

5. 3x2 – 7x + 14 = 0

6. –6x = 7x2 – x – 12

Read each scenario and use your knowledge of the quadratic formula to answer the questions.

7. The height of a golf ball in meters x seconds after it has been hit is given by –4.9x2 + 42x. When does the ball hit the ground?

8. A girl downloads about 24x – x2 songs each month, where x is the price of one song. For what price(s) does the girl download 100 songs?

9. An apple falls from a tall branch. Its height in feet x seconds after it falls is given by 40 – 16x2. After how many seconds will the apple hit the ground?

10. Can a quadratic equation have one real solution and one complex solution?

Practice 5.2.4: Applying the Quadratic Formula


Lesson 5.2.5: Solving Quadratic InequalitiesIntroductionWe have looked at a variety of methods to solve quadratic equations. The solutions to quadratic equations are points. Quadratic inequalities can be written in the form ax2 + bx + c < 0, ax2 + bx + c ≤ 0, ax2 + bx + c > 0, or ax2 + bx + c ≥ 0. The solutions to quadratic inequalities are written as intervals. An interval is the set of all real numbers between two given numbers. The two numbers on the ends are the endpoints. The endpoints might or might not be included in the interval depending on whether the interval is open, closed, or half-open.

Key Concepts

• If the right side of an inequality is 0, we can use logic to determine the sign possibilities of the factors.

• Another method of solving quadratic inequalities is to set the right side equal to 0 if it isn’t already, replace the inequality sign with an equal sign, and solve the equation.

• Recall that multiplying or dividing both sides of an inequality by a negative number requires reversing the inequality symbol.

• Use these solutions to create regions on a number line and test points in each region to solve the inequality.

• The solutions to a quadratic inequality can be one interval or two intervals.

• If the quadratic equation has only complex solutions, the expression is either always positive or always negative. In these cases, the inequality will have no solution or infinitely many solutions.

• Solutions of quadratic inequalities are often graphed on number lines.

• Recall that the solutions to an inequality graphed on a number line are represented by a darkened line.

• The endpoints of the solution interval are represented by either an open dot or a closed dot.

• Graph the endpoints as an open dot if the original inequality symbol is < or >.

• Graph endpoints as a closed dot if the original inequality symbol is ≤ or ≥.


Example 1

For what values of x is (x – 2)(x + 10) > 0?

1. Determine the sign possibilities for each factor.

The expression will be positive when both factors are positive or both factors are negative.

2. Determine when both factors are positive.

x – 2 is positive when x > 2.

x + 10 is positive when x > –10.

Both factors are positive when x > 2 and x > –10, or when x > 2.

3. Determine when both factors are negative.

x – 2 is negative when x < 2.

x + 10 is negative when x < –10.

Both factors are negative when x < 2 and x < –10, or when x < –10.

(x – 2)(x + 10) > 0 when x > 2 or x < –10.



Example 2

Solve x2 + 8x + 7 ≤ 0. Graph the solutions on a number line.

1. Replace the inequality sign with an equal sign.

x2 + 8x + 7 = 0

2. Solve the equation.

The equation factors easily.

Find two numbers whose product is 7 and whose sum is 8.

The numbers are 1 and 7 because (1)(7) = 7 and 1 + 7 = 8.

Use these numbers to find the factors.

The factors are (x + 1) and (x + 7).

Write the expression as the product of factors.

(x + 1)(x + 7) = 0

Use the Zero Product Property to solve.

x + 1 = 0 x + 7 = 0

x = –1 x = –7

3. Use a number line and test points to solve the inequality.

The sign of the original inequality was ≤. Therefore, we are looking for solutions that are ≤ 0. Plot the solutions to the equation on a number line. This divides the number line into three regions: numbers less than –7 and –1; numbers in between –7 and –1; and numbers greater than –7 and –1.

– 10 – 8 – 6 – 4 – 2 20

Region 3

–7 –1

Region 1 Region 2 Region 3

Test one point from each region of our number line. Pick any point from the region that is easy to work with.

(continued)


Test x = –10 from Region 1:

(–10)2 + 8(–10) + 7 = 27

Set up an inequality that compares the result of your test x-value with 0. Refer back to the sign used in the original inequality—in this case, ≤.

27 ≤ 0 is not true. Therefore, points in Region 1 (x < –7) are not solutions.


(–2)2 + 8(–2) + 7 = 4 – 16 + 7 = –5

Set up an inequality that compares the result of your test x-value with 0.

–5 ≤ 0 is true. Therefore, points in Region 2 (–7 ≤ x ≤ –1) are solutions.

Test x = 0 from Region 3:

(0)2 + 8(0) + 7 = 7


7 ≤ 0 is not true. Therefore, points in Region 3 (x > –1) are not solutions.

From the test points, we can see that the expression is less than or equal to 0 when x is between –7 and –1.

– 8– 10 – 6 – 4 – 2 20–7 –1

Notsolutions Solutions

Notsolutions

4. Solve the inequality and graph the solutions on a number line.

– 8– 10 – 6 – 4 – 2 20–7 –1

x2 + 8x + 7 ≤ 0 when –7 ≤ x ≤ –1.


Example 3

Solve 4x – 1 > 8 – x2. Graph the solutions on a number line.

1. Move all terms to the left side of the inequality.

4x – 1 > 8 – x2 Original inequality

x2 + 4x – 1 > 8 Add x2 to both sides.

x2 + 4x – 9 > 0 Subtract 8 from both sides.

2. Replace the inequality sign with an equal sign.

x2 + 4x – 9 = 0

3. Solve the equation.

The equation cannot be factored; it is prime.

Solve the equation by completing the square.

x2 + 4x – 9 = 0 Equation

x2 + 4x = 9 Add 9 to both sides.

x2 + 4x + 22 = 9 + 22 Add the square of half of the coefficient of the x-term to both sides to complete the square.

x2 + 4x + 4 = 13 Simplify.

(x + 2)2 = 13Express the left side as a perfect square trinomial.

x+ = ±2 13Take the square root of both sides to isolate the binomial.

x = − ±2 13 Isolate x by subtracting 2 from both sides.

x ≈ –5.61 or x ≈ 1.61


4. Use a number line and test points to solve the inequality.

The sign of the original inequality was >. Therefore, we are looking for solutions that are > 0. Plot the solutions to the equation on a number line.

– 8– 10 – 6 – 4 – 2 20 4

1.61–5.61

Region 1 Region 2 Region 3

Test one point from each region of our number line. Pick any point from the region that is easy to work with.


(–10)2 + 4(–10) – 9 = 51

Set up an inequality that compares the result of your test x-value with 0. Refer back to the sign used in the original inequality, in this case >.

51 > 0 is true. Therefore, points in Region 1 (x < –5.61) are solutions.


(0)2 + 4(0) – 9 = –9


–9 > 0 is not true. Therefore, points in Region 2 (–5.61 ≤ x ≤ 1.61) are not solutions.


(2)2 + 4(2) – 9 = 3


3 > 0 is true. Therefore, points in Region 3 (x > 1.61) are solutions.

– 8– 10 – 6 – 4 – 2 0 2 4

1.61–5.61

Not solutions SolutionsSolutions

From the test points, we can see that the expression is greater than 0 when x is less than –5.61 or when x is greater than 1.61.


5. Graph the solutions on a number line.

– 8– 10 – 6 – 4 – 2 20 4

1.61–5.61

4x – 1 > 8 – x2 when x < –5.61 or when x > 1.61.

Example 4

For what values of x is 3x2 + 2x + 2 < 0?

1. Set the inequality equal to 0 and then find the discriminant.

Find the discriminant of 3x2 + 2x + 2 = 0.

a = 3, b = 2, and c = 2

b2 – 4ac Formula for discriminant

= 22 – 4(3)(2) Substitute values for a, b, and c.

= 4 – 24 Simplify.

= –20

The discriminant is negative; therefore, the equation has no real solutions.

2. Use the discriminant to solve the inequality.

Since the equation has no real solutions, the expression will either always be positive or always be negative. Test any point.

Test x = 0:

3(0)2 + 2(0) + 2 = 2

The expression will always be positive. Therefore, 3x2 + 2x + 2 < 0 has no real solutions.


PRACTICE


For problems 1–7, solve each quadratic inequality. Graph the solution(s), if any, on a number line.

1. (2x + 1)(x – 1) ≥ 0

2. (x – 2)(5x + 7) ≤ 0

3. x2 + 4 < 0

4. x2 + 13x + 22 ≥ 0

5. 2x2 + 16x – 3 > 0

6. x2 + 10x – 3 < 7x

7. 4x + 7 ≤ 4x2

For problems 8–10, solve each inequality using the given information.

8. The height of a helicopter in meters x seconds after it takes off is given by –4.9x2 + 42x. When is the helicopter more than 5 meters above the ground?

9. Livia drops a water balloon from her apartment window onto the ground below. The balloon’s height above the ground in feet x seconds into the drop is given by the expression –5x2 + 5x + 4. When is the balloon more than 6 feet above the ground?

10. The length of a pendulum in centimeters is given by the expression 9 8

4

2

2

. x

π,

where x is the time in seconds for the pendulum to swing from one side to the

other. When the length of the pendulum is greater than 5 centimeters, what

can you say about the time it takes for the pendulum to swing from one side to

the other?

Practice 5.2.5: Solving Quadratic Inequalities

U5-58

Lesson 3: Creating Quadratic Equations in Two or More Variables





MCC9–12.A.SSE.3a★

MCC9–12.A.SSE.3b★


Essential Questions

1. How are the graphs of quadratics related to their functions?

2. How do you rearrange a formula when solving for a squared term?

3. How do you find the vertex of a quadratic from its three different forms?

4. What types of real-world situations are modeled by quadratic equations?

5. What do the solutions to quadratic equations represent in terms of real-world situations?

WORDS TO KNOW

axis of symmetry of a parabola

the line through the vertex of a parabola about which

the parabola is symmetric. The equation of the axis of

symmetry is =−

xb

a2.

intercept form the factored form of a quadratic equation, written as f(x) = a(x – p)(x – q), where p and q are the zeros of the function

key features of a quadratic the x-intercepts, y-intercept, where the function is increasing and decreasing, where the function is positive and negative, relative minimums and maximums, symmetries, and end behavior of the function used to describe, draw, and compare quadratic functions

literal equation an equation that involves two or more variables

U5-59Lesson 3: Creating Quadratic Equations in Two or More Variables

maximum the largest y-value of a quadratic

minimum the smallest y-value of a quadratic

parabola the U-shaped graph of a quadratic function

quadratic function a function that can be written in the form f(x) = ax2 + bx + c, where a ≠ 0. The graph of any quadratic function is a parabola.

standard form of a quadratic function

a quadratic function written as f(x) = ax2 + bx + c, where a is the coefficient of the quadratic term, b is the coefficient of the linear term, and c is the constant term

vertex form a quadratic function written as f(x) = a(x – h)2 + k, where the vertex of the parabola is the point (h, k); the form of a quadratic equation where the vertex can be read directly from the equation

vertex of a parabola the point at which the curve changes direction; the maximum or minimum

x-intercept the point at which the graph crosses the x-axis; written as (x, 0)

y-intercept the point at which the graph crosses the y-axis; written as (0, y)


Recommended Resources• IXL Learning. “Functions: Write linear, quadratic, and exponential functions.”


This website provides practice and instant feedback on determining linear, exponential, or quadratic equations given a table of data.

• IXL Learning. “Quadratic equations: characteristics of quadratic functions.”


Users can practice determining the key features of graphs of quadratic functions, including vertices, y-intercepts, maximums, and more. Instant feedback is provided along with explanations for incorrect answers.

• Massey University. “Mathematical Formulae.”


This website offers basic instructions on rearranging various formulas, including those with squares.


IntroductionImagine the path of a basketball as it leaves a player’s hand and swooshes through the net. Or, imagine the path of an Olympic diver as she arcs away from the diving board, slips deep into the water, and resurfaces. Both traveling paths make a U-shaped curve known as a parabola. A parabola is the U-shaped graph of a quadratic function, which is an equation with a degree of 2. Formally, a quadratic function is a function that can be written in the form f(x) = ax2 + bx + c, where a ≠ 0. Quadratics can be used to model varying situations. In this lesson, you will learn about and practice identifying the important points on a parabola and use those points to sketch the corresponding graph.

Key Concepts

• The standard form of a quadratic function is f(x) = ax2 + bx + c, where a is the coefficient of the quadratic term, b is the coefficient of the linear term, and c is the constant term.

• The x-intercepts of the graph of a quadratic function are the points at which the graph crosses the x-axis, and are written as (x, 0).

• The y-intercept of the graph of a quadratic function is the point at which the graph crosses the y-axis and is written as (0, y).

• The graph on the following page shows the location of the parabola’s y-intercept.

Lesson 5.3.1: Creating and Graphing Equations Using Standard Form


–2 –1 0 1 2 3 4 5

–6

–5

–4

–3

–2

–1

1

2

3

4

y-intercept

• The equation of the parabola is y = x2 – 2x – 4.

• Note that the y-intercept of this equation is –4 and the constant term of the quadratic equation is –4.

• The y-intercept of a quadratic is the c value of the quadratic equation when written in standard form.

• The axis of symmetry of a parabola is the line through the vertex of a parabola about which the parabola is symmetric.


–2 –1 0 1 2 3 4 5

–6

–5

–4

–3

–2

–1

1

2

3

4

Axis of symmetry

Vertex (1, –5)

• The axis of symmetry extends through the vertex of the graph.

• The vertex of a parabola is the point on a parabola where the graph changes direction, (h, k), where h is the x-coordinate and k is the y-coordinate.

• The equation of the axis of symmetry is =−

xb

a2.

• In the example above, the equation of the axis of symmetry is x = 1 because the vertical line through 1 is the line that cuts the parabola in half.

• The axis of symmetry is sometimes written as =−

hb

a2, where h is the

x-coordinate of the vertex.

• The vertex of a quadratic lies on the axis of symmetry, as shown in the graph on the next page.


–2 –1 0 1 2 3 4 5

–6

–5

–4

–3

–2

–1

1

2

3

4

Axis of symmetryx = 1

Vertex (1, –5)

• The formula =−

xb

a2 is not only used to find the equation of the axis of

symmetry, but also to find the x-coordinate of the vertex.

• To find the y-coordinate, substitute the value of x into the original function,

(h, k) = − −

b

af

b

a2,

2.

• The vertex gives information about the maximum or minimum value of a quadratic.

• The maximum is the largest y-value of a quadratic and the minimum is the smallest y-value.

• One might see these terms in real-world situations such as maximizing profit and minimizing cost, among others.

• From the equation of a function in standard form, you can determine if you have a maximum or minimum based on the sign of the coefficient of the quadratic term, a.


• If a > 0, the parabola opens up and therefore has a minimum value.

• If a < 0, the parabola opens down and therefore has a maximum value.

Minimum Maximumy = 3x2 y = –3x2

a = 3 and a > 0 a = –3 and a < 0

x0Minimum

y

–1 1x

0Maximum

y

–1 1

• To create the equation of a quadratic in standard form, one must know some combination of the following key features: y-intercept, axis of symmetry or the vertex, and maximum or minimum.

• The key features of a quadratic are the x-intercepts, y-intercept, where the function is increasing and decreasing, where the function is positive and negative, relative minimums and maximums, symmetries, and end behavior of the function.

• To create the equation of a quadratic given the vertex and y-intercept, first set

the x-coordinate of the vertex equal to −ba2

and isolate b.

• Second, substitute the expression that is equal to b, the y-intercept, and the coordinates of the vertex into the standard form of a quadratic equation, y = ax2 + bx + c.

• Solve the equation for a and then substitute the value of a into the equation you created in the first step to determine the value of b.



Find the y-intercept and vertex of the function f(x) = –2x2 + 4x + 3. Determine whether the vertex is a minimum or maximum point on the graph.

1. Determine the y-intercept.

The function f(x) = –2x2 + 4x + 3 is written in standard form, f(x) = ax2 + bx + c.

When x = 0, the y-intercept equals c, which is 3.

Verify that 3 is the y-intercept.

f(x) = –2x2 + 4x + 3 Original equation

f(0) = –2(0)2 + 4(0) + 3 Substitute 0 for x.

f(0) = 3 Simplify.

The y-intercept is the point (0, 3).

2. Determine the vertex of the function by using (h, k) = − −

b

af

b

a2,

2,

where h is the x-coordinate of the vertex and k is the function, f(x),

evaluated for h.

f(x) = –2x2 + 4x + 3 is in standard form; therefore, a = –2 and b = 4.

=−

xb

a2Equation to determine the vertex

( )=−−

x4

2 2Substitute –2 for a and 4 for b.

x = 1 Simplify.

The vertex has an x-value of 1.

(continued)


Since h is the x-coordinate of the vertex, or the input value of the vertex, we can find the output value, k, by evaluating the function for f(1).

f(x) = –2x2 + 4x + 3 Original equation

f(1) = –2(1)2 + 4(1) + 3 Substitute 1 for x.

f(1) = 5 Simplify.

The y-value of the vertex is 5.

The vertex is the point (1, 5).

3. Use the value of a to determine if the vertex is a maximum or a minimum value.

Given that f(x) = –2x2 + 4x + 3 is written in standard form, a = –2.

Since a < 0, the parabola opens down.

As such, the vertex (1, 5) is a maximum point.

Example 2

h(x) = 2x2 – 11x + 5 is a quadratic function. Determine the direction in which the function opens, the vertex, the axis of symmetry, the x-intercept(s), and the y-intercept. Use this information to sketch the graph.

1. Determine whether the graph opens up or down.

h(x) = 2x2 – 11x + 5 is in standard form; therefore, a = 2.

Since a > 0, the parabola opens up.


2. Find the vertex and the equation of the axis of symmetry.

h(x) = 2x2 – 11x + 5 is in standard form; therefore, a = 2 and b = –11.

=−

xb


( )( )=

− −x

11

2 2Substitute 2 for a and –11 for b.

x = 2.75 Simplify.

The vertex has an x-value of 2.75.

Since the input value is 2.75, find the output value by evaluating the function for h(2.75).

h(x) = 2x2 – 11x + 5 Original equation

h(2.75) = 2(2.75)2 – 11(2.75) + 5 Substitute 2.75 for x.

h(2.75) = –10.125 Simplify.

The y-value of the vertex is –10.125.

The vertex is the point (2.75, –10.125).

Since the axis of symmetry is the vertical line through the vertex, the equation of the axis of symmetry is x = 2.75.

3. Find the y-intercept.

h(x) = 2x2 – 11x + 5 is in standard form, so the y-intercept is the constant c, which is 5.

The y-intercept is (0, 5).


4. Find the x-intercepts, if any exist.

The x-intercepts occur when y = 0.

Substitute 0 for the output, h(x), and solve.

This equation is factorable, but if we cannot easily identify the factors, the quadratic formula always works.

Note both methods.

Solved by factoring: Solved using the quadratic formula:

h(x) = 2x2 – 11x + 50 = 2x2 – 11x + 50 = (2x – 1)(x – 5)

0 = 2x – 1 or 0 = x – 5x = 0.5 or x = 5

h(x) = 2x2 – 11x + 5

0 = 2x2 – 11x + 5

a = 2, b = –11, and c = 5

xb b ac

a=− ± −2 4

2

x =− −( )± −( ) − ( )( )

( )11 11 4 2 5

2 2

2

x =±11 81

4

x =±11 9

4

=−

=+

x x11 9

4or

11 9

4

= =x x2

4or

20

4x = 0.5 or x = 5

The x-intercepts are (0.5, 0) and (5, 0).


5. Plot the points from steps 2–4 and their symmetric points over the axis of symmetry.

Connect the points with a smooth curve.

x

y

0 2 4 6 8

–10

–8

–6

–4

–2

2

4

6

Axis of symmetry

2.25 units 2.25 units(0, 5) (5.5, 5)

(0.5, 0) (5, 0)

(2.75, –10.125)

Example 3

R(x) = 2x2 + 8x + 8 is a quadratic function. Determine the direction in which the function opens, the vertex, the axis of symmetry, the x-intercept(s), and the y-intercept. Use this information to sketch the graph.


R(x) = 2x2 + 8x + 8 is written in standard form; therefore, a = 2.

Since a > 0, the parabola opens up.



R(x) = 2x2 + 8x + 8 is in standard form; therefore, a = 2 and b = 8.

=−

xb


( )( )=

−x

8

2 2Substitute 2 for a and 8 for b.

x = –2 Simplify.

The vertex has an x-value of –2.

Since the input value is –2, find the output value by evaluating the function for R(–2).

R(x) = 2x2 + 8x + 8 Original equation

R(–2) = 2(–2)2 + 8(–2) + 8 Substitute –2 for x.

R(–2) = 0 Simplify.

The y-value of the vertex is 0.

The vertex is the point (–2, 0).

Since the axis of symmetry is the vertical line through the vertex, the equation of the axis of symmetry is x = –2.


The function, R(x) = 2x2 + 8x + 8, is in standard form, so the y-intercept is the constant c, which is 8.





Substitute 0 for the output, R(x), and solve using the quadratic formula or by factoring.

Solved by factoring: Solved using the quadratic formula:

R(x) = 2x2 + 8x + 8

0 = 2x2 + 8x + 8

0 = 2(x2 + 4x + 4)

0 = 2(x + 2)2

0 = (x + 2)

x = –2

R(x) = 2x2 + 8x + 8

0 = 2x2 + 8x + 8

a = 2, b = 8, and c = 8

xb b ac

a=− ± −2 4

2

x =−( )± ( ) − ( )( )

( )8 8 4 2 8

2 2

2

x =− ±8 0

4x = –2

The x-intercept is (–2, 0). This is a special case where the vertex is also the x-intercept.



x

y

–5 –4 –3 –2 –1 0 1

2

4

6

8

Axis of symmetry

2 units 2 units(–4, 8) (0, 8)

(–2, 0)

For a more accurate graph, determine an additional pair of symmetric points.

Choose any x-coordinate on the left or right of the axis of symmetry.

Evaluate the function for the chosen value of x to determine the output value.

Let’s choose x = –1.

R(x) = 2x2 + 8x + 8 Original equation

R(–1) = 2(–1)2 + 8(–1) + 8 Substitute –1 for x.

R(–1) = 2 Simplify.

(–1, 2) is an additional point.

Plot (–1, 2) on the same graph.

(–1, 2) is 1 unit from the axis of symmetry.

Locate the point that is symmetric to the point (–1, 2).(continued)


(–3, 2) is also 1 unit from the axis of symmetry and is symmetrical to the original point (–1, 2).

You can verify that (0, 8) and (–4, 8) are the same distance from the axis of symmetry and are also symmetrical by referring to the graph.

x

y

–5 –4 –3 –2 –1 0 10

2

4

6

8

(–2, 0)

(–3, 2) (–1, 2)

(–4, 8) (0, 8)2 units

1 unit 1 unit

2 units

Axis of symmetry

Example 4

g(x) = –x2 + 8x – 17 is a quadratic function. Determine the direction in which the function opens, the vertex, the axis of symmetry, the x-intercept(s), and the y-intercept. Use this information to sketch the graph.


g(x) = –x2 + 8x – 17 is written in standard form; therefore, a = –1.

Since a < 0, the parabola opens down.



g(x) = –x2 + 8x – 17 is written in standard form; therefore, a = –1 and b = 8.

=−

xb


( )( )=−−

x8

2 1Substitute –1 for a and 8 for b.

x = 4 Simplify.

The vertex has an x-value of 4.

Since the input value is 4, find the output value by evaluating the function for g(4).

g(x) = –x2 + 8x – 17 Original equation

g(4) = –(4)2 + 8(4) – 17 Substitute 4 for x.

g(4) = –1 Simplify.

The y-value of the vertex is –1.

The vertex is the point (4, –1).

Since the axis of symmetry is the vertical line through the vertex, the equation of the axis of symmetry is x = 4.


The function g(x) = –x2 + 8x – 17 is in standard form, so the y-intercept is the constant c, which equals –17.

The y-intercept is (0, –17).




Substitute 0 for the output, g(x), and solve using the quadratic formula since the function is not factorable.


–x2 + 8x – 17 = 0 Set the equation equal to 0.

Determine the values of a, b, and c.

a = –1, b = 8, and c = –17

xb b ac

a=− ± −2 4

2Quadratic formula

x =−( )± ( ) − −( ) −( )

−( )8 8 4 1 17

2 1

2Substitute –1 for a, 8 for b, and –17 for c.

x =− ± −

−8 4

2Simplify.

In this case, the discriminant, –4, is negative, which means there are no real solutions.

This also means that there are no x-intercepts.



xy

0–1 1 2 3 5 7 94 6 8

–20

–14

–16

–18

–10

–2

–4

–6

–8

–12

Axis of symmetry

(0, –17) (8, –17)

(4, –1)

4 units 4 units

For a more accurate graph, determine an additional pair of symmetric points.

Choose any x-coordinate on the left or right of the axis of symmetry.

Evaluate the function for the chosen value of x to determine the output value.

Let’s choose x = 1.


g(1) = –(1)2 + 8(1) – 17 Substitute 1 for x.

g(1) = –10 Simplify.

(1, –10) is an additional point.

Plot (1, –10) on the same graph.

(1, –10) is 3 units from the axis of symmetry.

Locate the point that is symmetric to the point (1, –10).(continued)


(7, –10) is also 3 units from the axis of symmetry and is symmetrical to the original point (1, –10).

You can verify that (1, –10) and (7, –10) are the same distance from the axis of symmetry and are also symmetrical by referring to the graph.

xy

0–1 1 2 3 5 7 94 6 8

–20

–14

–16

–18

–10

–2

–4

–6

–8

–12

Axis of symmetry

(0, –17) (8, –17)

(4, –1)

4 units 4 units

3 units 3 units(1, –10) (7, –10)


Example 5

Create the equation of a quadratic function given a vertex of (2, –4) and a y-intercept of 4.

1. Write an equation for b in terms of a.

Set the x-coordinate of the vertex equal to −ba2

.

=−ba

22

Substitute 2 for x.

4a = –b Multiply both sides by 2a.

–4a = b Multiply both sides by –1.

b = –4a

2. Substitute the expression for b from step 1, the coordinates of the vertex, and the y-intercept into the standard form of a quadratic equation.

y = ax2 + bx + c Standard form of a quadratic equation

y = ax2 + (–4a)x + c Substitute –4a for b.

(–4) = a(2)2 + (–4a)(2) + c Substitute the vertex (2, –4) for x and y.

(–4) = a(2)2 + (–4a)(2) + 4 Substitute the y-intercept of 4 for c.

–4 = 4a – 8a + 4 Simplify, then solve for a.

–8 = –4a

a = 2


3. Substitute the value of a into the equation for b from step 1.

b = –4a Equation from step 1

b = –4(2) Substitute 2 for a.

b = –8 Simplify.

4. Substitute a, b, and c into the standard form of a quadratic equation.

y = ax2 + bx + c Standard form of a quadratic equation

y = 2x2 – 8x + 4 Substitute 2 for a, –8 for b, and 4 for c.

The equation of the quadratic function with a vertex of (2, –4) and a y-intercept of 4 is y = 2x2 – 8x + 4.

UNIT 5 • QUADRATIC FUNCTIONSLesson 3: Creating Quadratic Equations in Two or More Variables

PRACTICE


Sketch a graph for each of the following quadratic functions.

1. a(x) = 2x2 – 6x + 4

2. e(x) = x2

3. f (x) = x 2 + 2

Find the y-intercept and vertex of the following functions. State whether the vertex is a minimum or maximum point on the graph and explain your reasoning.

4. n(h) = –2h2 – 7h

5. l(r) = 4r 2 + 40r + 7

6. f(x) = –2x2 + 4x + 3

Could the following graph represent the given function? Explain your reasoning.

7. d(t) = t2 – 3t – 5

x

y

–2 0 2 4 6

–15

–10

–5

5

continued

Practice 5.3.1: Creating and Graphing Equations Using Standard Form


PRACTICE


Use your knowledge of quadratic functions to complete the problems that follow.

8. Create the equation of a quadratic function with a vertex of (5, 6) and a y-intercept of –69.

9. The path of a ball shot from a slingshot in the air can be modeled by the

function h t t t( )= − +5

2202 , where h is the height, in feet, of the ball above

ground t seconds after it is released. Graph the function to approximate the

maximum height the ball reaches.

10. A sock manufacturing company’s profit p (in hundreds of dollars) after selling x thousand pairs of socks can be modeled by the function p(x) = –4x2 + 40x – 2. How many pairs of socks must be sold in order to maximize profits?


Lesson 5.3.2: Creating and Graphing Equations Using the x-intercepts

IntroductionThe equation of a quadratic function can be written in several different forms. We have practiced using the standard form of a quadratic function to identify key information and graph the function. In this lesson, we will explore writing quadratic functions in intercept form or factored form, graphing a quadratic function from intercept form, and creating the equation of a quadratic function given its graph. An equation written in intercept form allows you to easily identify the x-intercepts of a quadratic function.

Key Concepts

• A quadratic function in standard form can be created from the intercept form.

• The intercept form of a quadratic function is f(x) = a(x – p)(x – q), where p and q are the zeros of the function.

• To write a quadratic in intercept form, distribute the terms of the first set of parentheses over the terms of the second set of parentheses and simplify.

• For example, f(x) = 3(x – 4)(x – 1) becomes f(x) = 3x2 – 15x + 12.

• The y-intercept of a quadratic function written in intercept form can be found by substituting 0 for x.

• The x-intercept(s) of a quadratic function written in intercept form can be found by substituting 0 for y.

• The x-intercepts, also known as the zeros, roots, or solutions of the quadratic, are often identified as p and q.

• The axis of symmetry is located halfway between the zeros. To determine its

equation, use the formula xp q

=+2

.

• Recall that the axis of symmetry extends through the vertex of a quadratic function.

• As such, the x-coordinate of the vertex is the value of the equation of the axis of symmetry.

• If the vertex is the point (h, k), then the equation of the axis of symmetry is x = h.

• The y-value can be found by substituting the x-value into the equation.



Determine the equation of a quadratic in standard form, given the zeros x = 2 and x = –2, and the point (0, 3).

1. Write the zeros as expressions equal to 0 to determine the factors of the quadratic.

Recall that when solving a quadratic in factored form, you take all factors and set them equal to 0 to determine the solution, thus the name “zeros.”

To work backward, we must undo the solving process to find the factors of the equation.

x = 2 x = –2

x – 2 = 0 x + 2 = 0

The factors of the equation are (x – 2) and (x + 2), so the equation is f(x) = a(x – 2)(x + 2).

2. Use the point (0, 3) to find a, the coefficient of the x2 term.

The equation f(x) = a(x – 2)(x + 2) is written in intercept form, or f(x) = a(x – p)(x – q).

Though a could have a value of 1, thus not affecting the equation we just found, this step is important for the cases when a ≠ 1.

Substitute the point (0, 3) into the equation to find a.

f(x) = a(x – 2)(x + 2) Equation

3 = a(0 – 2)(0 + 2) Substitute the point (0, 3) for x and y.

3 = a(–2)(2) Simplify, then solve for a.

3 = –4a

− =3

4a

The coefficient of x2 is −3

4.


3. Write the equation in standard form.

Insert the value found for a into the equation from step 1 and solve.

f(x) = a(x – 2)(x + 2) Equation

f x x x( ) ( )( )= − +3

42 2– Substitute −

3

4 for a.

f x x( )= − −( )3

442 Simplify by multiplying the factors.

f x x( )= − +3

432

The equation of a quadratic in standard form, with zeros x = 2

and x = –2, and the point (0, 3), is f x x( )= − +3

432 .

Example 2

Identify the x-intercepts, the axis of symmetry, and the vertex of the quadratic f(x) = (x + 5)(x + 2). Use this information to graph the function.

1. Identify the x-intercepts and plot the points.

The x-intercepts of the quadratic are the zeros or solutions of the quadratic.

Start by setting the equation equal to 0 and solving to determine the zeros.

f(x) = (x + 5)(x + 2) Original equation

0 = (x + 5)(x + 2) Set the equation equal to 0.

0 = x + 5 or 0 = x + 2 Set each factor equal to 0.

–5 = x or –2 = x Solve each equation for x.


2. Determine the axis of symmetry to find the vertex.

The axis of symmetry is the line that divides the parabola in half.

To determine its equation, find the midpoint of the two zeros and draw the vertical line through that point.

Insert the values of x into the formula xp q

=+2

to find the midpoint.

Let p = –5 and q = –2.

x =− + −( )

=−

= −5 2

2

7

23 5.

The equation of the axis of symmetry is x = –3.5.

The axis of symmetry extends through the vertex of the quadratic. This means that the vertex (h, k) of the graph has an x-value of –3.5.

Substitute –3.5 for x in the original equation to determine the y-value of the vertex.

f(x) = (x + 5)(x + 2) Original equation

f(–3.5) = (–3.5 + 5)(–3.5 + 2) Substitute –3.5 for x.

f(–3.5) = (1.5)(–1.5) Simplify.

f(–3.5) = –2.25

The y-value of the vertex is –2.25.

The vertex is the point (–3.5, –2.25).


3. Graph the zeros, the vertex, and the axis of symmetry of the given equation.

x

y

–8 –6 –4 –2 0 2 4 6 8

–6

–4

–2

0

2

4

6

(–5, 0) (–2, 0)

(–3.5, –2.25)

4. Sketch the function based on the zeros and the vertex.

x

y

–8 –6 –4 –2 0 2 4 6 8

–10

–8

–6

–4

–2

0

2

4

6

8

10

(–5, 0) (–2, 0)

(–3.5, –2.25)


Example 3

Use the intercepts and a point on the graph below to write the equation of the function in standard form.

x

y

–8 –6 –4 –2 0 2 4 6 8

–8

–6

–4

–2

0

2

4

6

8

(4, –3)

1. Find the x-intercepts of the graph and write them as zeros of the equation.

Identify the two points of intersection and write them as zeros to prepare to write the factors of the equation.

The x-intercepts are (–2, 0) and (3, 0).

The zeros are x = –2 and x = 3.


2. Write the zeros as expressions equal to 0 to determine the factors of the quadratic.

x = –2 x = 3

x + 2 = 0 x – 3 = 0

The factors of the equation are (x + 2) and (x – 3), so the equation is f(x) = a(x + 2)(x – 3).

3. Use the point (4, –3) to find a, the coefficient of the x2 term.

The equation f(x) = a(x + 2)(x – 3) is written in intercept form, or f(x) = a(x – p)(x – q).

Substitute (4, –3) into the equation to find a.

f(x) = a(x + 2)(x – 3) Equation

–3 = a(4 + 2)(4 – 3) Substitute (4, –3) for x and f(x).

–3 = a(6)(1) Simplify.

–3 = 6a

− =1

2a

The coefficient of x2 is −1

2.

4. Write the equation in standard form.

Insert the value found for a into the equation from step 1 and solve.

f(x) = a(x + 2)(x – 3) Equation

f x x x( )1

22 3( )( )= − + − Substitute −

1

2 for a.

f x x x( )1

262( )= − − − Simplify by multiplying the factors.

f x x x( )1

2

1

232=− + + Distribute −

1

2 over the parentheses.

The standard form of the equation is f x x x( )1

2

1

232=− + + .


PRACTICE


continued

Identify the x-intercepts of the following quadratic functions. Determine the axis of symmetry for each.

1. h(t) = (–16t + 1)(t – 7)

2. y x x= −

+

23

4

7

2

Determine the equation of each quadratic in standard form, given the zeros and a point.

3. x = –4, x = –2; (–3, –1)

4. x = 15, x = 5; (0, 75)

Sketch a graph for each of the following quadratic functions.

5. f(x) = (x – 3)(x – 4)

6. g(x) = (x – 3)(x – 2)

Practice 5.3.2: Creating and Graphing Equations Using the x-intercepts


PRACTICE


Given the graph of a quadratic function, use the intercepts and a point to write the equation of the function in standard form.

7.

x

y

–6 –4 –2 0 2 4

–40

–30

–20

–10

0

10

20

(–3, –36)

continued


PRACTICE


8.

x

y

–1 0 1 2 3 4 5 6 7

–8

–6

–4

–2

0

2

9. A walkway is being installed around a rectangular playground. The playground is 30 feet by 12 feet, and the total area of the playground and the walkway is 1,288 ft2. What is the width of the walkway?

10. A high school senior vacationing in Negril, Jamaica, for her senior trip jumped

off a 20-foot cliff into a pool of water. The height of the senior above the water

is modeled by the function h t t t( )= − + +21

4

5

4, where h(t) is the height of the

senior above the water in feet t seconds after jumping off the cliff. How many

seconds will it take for the senior to reach the water?


IntroductionQuadratic functions are used to model various situations. Some situations are literal, such as determining the shape of a parabola, and some situations involve applying the key features of quadratics to real-life situations. For example, an investor might want to predict the behavior of a particular mutual fund over time, or an NFL scout might want to determine the maximum height of a ball kicked by a potential football punter.

In this lesson, we will look specifically at the vertex form of a quadratic, f(x) = a(x – h)2 + k, where the vertex is the point (h, k). The vertex can be read directly from the equation.

Key Concepts

• Standard form, intercept form, and vertex form are equivalent expressions written in different forms.

• Standard form: f(x) = ax2 + bx + c, where a is the coefficient of the quadratic term, b is the coefficient of the linear term, and c is the constant term

• Intercept form: f(x) = a(x – p)(x – q), where p and q are the zeros of the function

• Vertex form: f(x) = a(x – h)2 + k, where the vertex of the parabola is the point (h, k)

• To identify the vertex directly from an equation in vertex form, identify h (the x-coordinate of the vertex) and k (the y-coordinate of the vertex).

• Note that the original equation in vertex form has the quantity x – h, so if the equation has a subtraction sign then the value of h is h.

• This is true because x – (–h) simplifies to x + h.

• However, if the quantity is written as x + h, the value of h is –h. A quadratic function in standard form can be created from vertex form, f(x) = a(x – h)2 + k, where (h, k) is the vertex of the quadratic.

• To do so, distribute and simplify by combining like terms.

• For example, f(x) = 3(x – 2)2 + 4 becomes f(x) = 3x2 – 12x + 16.

• A quadratic function in vertex form can be created from standard form, f(x) = ax2 + bx + c.

Lesson 5.3.3: Creating and Graphing Equations Using Vertex Form


• To do so, complete the square, or determine the value of c that would make ax2 + bx + c a perfect square trinomial.

• To complete the square, take the coefficient of the linear term, divide by the product of 2 and the coefficient of the quadratic term, and square the quotient.

ax2 + bx + c

a xb

ax

b

a

b

ac2

2 2

2 2+ +

−

+

a xb

aa

b

ac+

−

+2 2

2 2

• Since the quotient of b and 2a is a constant term, we can combine it with the

constant c to get the equation a xb

ak+

+2

2

, where k ab

ac= −

+2

2

.

• For example, f(x) = 2x2 – 12x + 22 becomes f(x) = 2(x – 3)2 + 4.

• When graphing a quadratic using vertex form, if the vertex is the y-intercept, choose two pairs of symmetric points to plot in order to sketch the most accurate graph.



Given the quadratic function f x x( )= +( ) −1

26 22 , identify the vertex and determine

whether it is a minimum or maximum.

1. Identify the vertex.

The vertex form of a quadratic is f(x) = a(x – h)2 + k, with the vertex (h, k).

f x x( )= +( ) −1

26 22 is in vertex form; therefore, the vertex of this

function is (–6, –2).

2. Determine if the vertex is a minimum or maximum.

If a > 0, the quadratic opens up and thus has a lowest point or a minimum.

If a < 0, the quadratic opens down and thus has a highest point or a maximum.

f x x( )= +( ) −1

26 22 is in vertex form, so a =

1

2.

1

20> , so the vertex of the function is a minimum.


Example 2

Determine the equation of a quadratic function that has a minimum at (–4, –8) and passes through the point (–2, –5).

1. Substitute the vertex into f(x) = a(x – h)2 + k.

f(x) = a(x – h)2 + k Vertex form

f(x) = a[x – (–4)]2 + (–8) Substitute (–4, –8) for h and k.

f(x) = a(x + 4)2 – 8 Simplify.

2. Substitute the point (–2, –5) into the equation from step 1 and solve for a.

f(x) = a(x + 4)2 – 8 Equation

–5 = a[(–2) + 4]2 – 8 Substitute (–2, –5) for x and f(x).

–5 = a(2)2 – 8 Simplify.

–5 = 4a – 8

3 = 4a

3

4= a

3. Substitute a into the equation from step 1.

f(x) = a(x + 4)2 – 8

f x x( )= +( ) −3

44 82

The equation of the quadratic function with a minimum at (–4, –8)

and passing through the point (–2, –5) is f x x( )= +( ) −3

44 82 .


Example 4

Sketch a graph of the quadratic function y = (x + 3)2 – 8. Label the vertex, the axis of symmetry, the y-intercept, and one pair of symmetric points.

1. Identify the vertex and the equation of the axis of symmetry.

Given the vertex form of a quadratic function, f(x) = a(x – h)2 + k, the vertex is the point (h, k).

The vertex of the quadratic y = (x + 3)2 – 8 is (–3, –8).

The axis of symmetry extends through the vertex.

The equation of the axis of symmetry is x = –3.

Example 3

Convert the function g(x) = –7x2 + 14x to vertex form.

1. Complete the square to rewrite the function in vertex form.

g(x) = –7x2 + 14x Original function

g(x) = –7(x2 – 2x) Factor out –7.

g(x) = –7(x2 – 2x + 1 – 1) Complete the square.

g(x) = –7(x2 – 2x + 1) + 7 Rewrite the equation as a perfect square trinomial.

g(x) = –7(x – 1)2 + 7 Rewrite the equation as a binomial squared.

2. Summarize your result.

The function g(x) = –7x2 + 14x written in vertex form is g(x) = –7(x – 1)2 + 7.



The parabola crosses the y-axis when x = 0.

Substitute 0 for x to find y.

y = (x + 3)2 – 8 Original equation

y = (0 + 3)2 – 8 Substitute 0 for x.

y = 32 – 8 Simplify.

y = 1

The y-intercept is the point (0, 1).

3. Find an extra point to the left or right of the axis of symmetry.

Choose an x-value and substitute it into the equation to find the corresponding y-value.

Typically, choosing x = 1 or x = –1 is simplest arithmetically, if these numbers aren’t already a part of the vertex or axis of symmetry.

In this case, let’s use x = 1.

y = (x + 3)2 – 8 Original equation

y = (1 + 3)2 – 8 Substitute 1 for x.

y = 42 – 8 Simplify.

y = 8

The parabola passes through the point (1, 8).

x = 1 is 4 units to the right of the axis of symmetry, x = –3.

4 units to the left of the axis of symmetry and horizontal to (1, 8) is the symmetric point (–7, 8).


4. Plot the points you found in steps 2 and 3 and their symmetric points over the axis of symmetry.

x

y

–8 –6 –4 –2 0 2 4 6 8

–10

–8

–6

–4

–2

0

2

4

6

10

8Axis of symmetryx = –3(–7, 8) (1, 8)

(–6, 1) (0, 1)

V (–3, –8)


PRACTICE


Given the following quadratic functions, identify the vertices, state whether the function has a minimum or maximum, and explain your reasoning.

1. y = –3(x + 4)2 + 1

2. f(x) = 4(x – 2)2

Determine the equation of a quadratic function that satisfies the given criteria.

3. The function’s minimum is at (5.5, –6) and it passes through the point (3, 0).

4. The function’s minimum is at (2, 0) and it passes through the point (0, 20).

Convert each quadratic function given in standard form to vertex form.

5. f(x) = x2 – 2x – 2

6. g(x) = 0.3x2 + 1.2x + 1.2

Sketch a graph of each quadratic function. Label the vertex, the axis of symmetry, and one pair of symmetric points.

7. y = (x + 4)2 + 2

8. u(x) = –(x – 5)2 – 2.5

Create the equation of a quadratic with the given characteristics.

9. The vertex is at (6, 72) and the x-intercept is at (12, 0).

10. The vertex is at (10.5, 385) and the x-intercept is at (21, 0).

Practice 5.3.3: Creating and Graphing Equations Using Vertex Form


Introduction

The formula for finding the area of a trapezoid, A b b h= +( )1

2 1 2 , is a formula because

it involves two or more variables that have specific and special relationships with

the other variables. In this case, the formula involves four variables: b1, a base of the

trapezoid; b2, the other base of the trapezoid; h, the height of the trapezoid; and A,

the area of the trapezoid. When you rearrange the formula, you are simply changing

the focus of the formula. In its given form, the focus is A, the area. However, when it

is necessary to find the height, h becomes the focus of the formula and thus is isolated

using proper algebraic properties. In this lesson, you will rearrange literal equations

(equations that involve two or more variables) and formulas with a degree of 2.

Key Concepts

• Literal equations and formulas contain equal signs.

• Just as in any other equation, we must apply proper algebraic properties to maintain a balance when changing the focus of an equation or formula. That is, if you subtract a value from one side of the equation, you must do the same to the other side, and so on.

• When you change the focus of a literal equation or formula, you are isolating the variable in question.

• To isolate a variable that is squared, perform the inverse operation by taking the square root of both sides of the equation.

• When you take the square root of a real number there are two solutions: one is positive and the other is negative.

• Take (a)2, for example. When you square a, the result is a2. The same, however, is true for (–a)2.

• When you square –a, the result is still a2 because the product of two negatives is a positive. In other words, (–a)2 = a2.

• Since taking a square root involves finding a number that you can multiply by itself to result in the square, we must take into account both the positive and negative.

Lesson 5.3.4: Rearranging Formulas


• That is, a a2 = ± .

• When solving for a squared term of a formula, the focus is important in determining whether to use two solutions.

• If the focus is any quantity that would never be negative in real life, such as distance, time, or population, it is appropriate to ignore the negative.

• When solving for a variable in a multi-step equation, first isolate the term containing the variable using subtraction or addition. Then determine which operations are applied to the variable, and undo them in reverse order.


Example 2

Solve y = 3(x – 7)2 + 8 for x.

1. Isolate x.

y = 3(x – 7)2 + 8 Given equation

y – 8 = 3(x – 7)2 Subtract 8 from both sides.

y x−=

−( )8

3

3 7

3

2

Divide both sides by 3.

yx

−= −( )8

37 2 Simplify.

yx

−= −( )8

37 2

Take the square root of both sides.

±−

= −y

x8

37 Simplify.

78

3±

−=

yx Add 7 to both sides.


Solve the equation x2 + y2 = 100 for y.

1. Isolate y.

Begin by subtracting x2 from both sides.

x2 + y2 = 100 Original equation

y2 = 100 – x2 Subtract x2 from both sides.

y x2 2100= − Take the square root of both sides.

y x= ± −100 2 Simplify, remembering that the result could be positive or negative.


The formula x2 + y2 = 100 can be rewritten as y x= ± −100 2 .



The equation y = 3(x – 7)2 + 8 solved for x is xy

= ±−

78

3.

Example 3

The formula for the area of a square is A = s2, where s is the length of a side of the square. Solve the formula for s.

1. Isolate s.

A = s2 Formula for the area of a square

A s= 2 Take the square root of both sides.

± =A s Simplify.


In this case, the length of a square cannot be negative (or it would not exist), so the negative is disregarded.

The equation A = s2 solved for s is s A= .


Example 4

The equation to graph an ellipse is x h

a

y k

b

−( )+

−( )=

2

2

2

2 1 , where (h, k) is the center of

the ellipse, a is the number of units right and left of the center on the ellipse, and b is the

number of units up and down from the center on the ellipse. Solve the formula for a.

b

b

a a

(h, k)

x

y

1. Isolate a.

x h

a

y k

b

−( )+

−( )=

2

2

2

2 1 Equation of an ellipse

x h

a

y k

b

−( )= −

−( )2

2

2

21 Subtract the fraction y k

b

−( )2

2 from both sides.

ax h

aa

y k

b2

2

22

2

21−( )

= −

−( )

Multiply both sides by a2.

x h ay k

b−( ) = −

−( )

2 2

2

21

Simplify.

(continued)


x h

y k

b

ay k

b

y k

b

−( )

−−( )

=

−−( )

−−( )

2

2

2

2

2

2

2

21

1

1

Divide each side by 12

2−−( )

y k

b.

x h

y k

b

a−( )

−−( )

=2

2

2

2

1Simplify.

x h

y k

b

a−( )

−−( )

=2

2

2

2

1Take the square root of both sides.

±−( )

− −( )=

x h

b y k

b

a2

2 2

2

Simplify.

±−( )

− −( )=

b x h

b y ka

2 2

2 2


Since a is the horizontal distance from the center of the ellipse, this is another situation where we disregard the negative.

The equation x h

a

y k

b

−( )+

−( )=

2

2

2

2 1 solved for a is

ab x h

b y k=

−( )− −( )

2 2

2 2 .


PRACTICE


Practice 5.3.4: Rearranging FormulasFor problems 1–4, solve each equation for x.

1. x2 + 4y2 = 360

2. y = 16x2 – 9

3. y = 4(x + 7)2

4. 2.5(x – 6.5)2 – 20 = y

Problems 5–10 each describe a different mathematical formula. Solve each problem for the requested variable.

5. The formula to determine the power of an electrical charge is PV

R=

2

, where

P is the power, V is the electric potential difference (a value that can be negative),

and R is the resistance. Solve this equation for the electric potential difference.

6. The area of a sector of a circle is represented by A r=1

22θ , where A is the area of

the sector, r is the radius, and θ is the angle created by the sides of the sector and

the center of the circle. Solve for the radius.

continued


PRACTICE


7. The formula a2 + b2 = c2 is the Pythagorean Theorem, where c is the length of the hypotenuse of a right triangle and a and b are the lengths of the other two legs. Rearrange this formula so that the focus is c.

8. In physics, the circular motion of an object can be defined by the formula av

r=

2

,

where a is the centripetal acceleration (directed towards the center of the circle),

v is the tangential velocity, and r is the radius. Solve for the tangential velocity.

9. The formula for a circle drawn in the coordinate plane is (x – h)2 + (y – k) 2 = r 2, where (h, k) is the center of the circle and r is the radius. Solve for y.

10. The formula to find the surface area of a sphere is SA = 4πr2, where SA is the surface area and r is the radius of the sphere. Solve for the radius.

U5-109

Lesson 4: Solving Systems of Equations



Common Core Georgia Performance Standard

MCC9–12.A.REI.7

WORDS TO KNOW

point(s) of intersection the ordered pair(s) where graphed functions intersect on a coordinate plane; these are also the solutions to systems of equations

quadratic formula a formula that states the solutions of a quadratic

equation of the form ax2 + bx + c = 0 are given by

xb b ac

a=− ± −2 4

2. A quadratic equation in this form

can have no real solutions, one real solution, or two

real solutions.

quadratic-linear system a system of equations in which one equation is quadratic and one is linear

substitution the replacement of a term of an equation by another term that is known to have the same value

system of equations a set of equations with the same unknowns

Essential Questions

1. How are points of intersection related to the solution of a system of equations?

2. At most, how many solutions can a quadratic-linear system have?

3. Is it possible for a quadratic-linear system to have no real solutions?


Recommended Resources• MathBits.com. “Solving a Linear-Quadratic System.”


This site describes linear-quadratic systems and demonstrates how to graph them using a TI-83/84 graphing calculator.

• Math Warehouse. “Systems of Linear and Quadratic Equations.”


This site offers instruction on identifying the systems of linear and quadratic equations graphically, and includes practice problems.

• Oswego City School District Regents Exam Prep Center. “Linear-Quadratic Systems.”


This site offers a quick and effective review of systems of equations, and also provides practice problems with hidden but accessible solutions.

U5-111Lesson 4: Solving Systems of Equations

Lesson 5.4.1: Solving Systems Graphically

IntroductionA system of equations is a set of equations with the same unknowns. A quadratic-linear system is a system of equations in which one equation is quadratic and one is linear. We learned previously how to solve systems of linear equations. In this lesson, we will learn to solve a quadratic-linear system by graphing.

Key Concepts

• A quadratic-linear system may have one real solution, two real solutions, or no real solutions.

• The solutions of a quadratic-linear system can be found by graphing.

• The points of intersection of the graphed quadratic and linear equation are the ordered pair(s) where graphed functions intersect on a coordinate plane. These are also the solutions to the system.

Graphed Quadratic-Linear Systems

x

y

–8 –6 –4 –2 0 2 4 6 8

–10

–8

–6

–4

–2

0

2

4

6

8

10

x

y

-8 -6 -4 -2 0 2 4 6 8

-10

-8

-6

-4

-2

0

2

4

6

8

10

x

y

-8 -6 -4 -2 0 2 4 6 8

-10

-8

-6

-4

-2

0

2

4

6

8

10

If the equations intersect in two places, they have two real solutions.

If the equations intersect in one place, they have one real solution.

If the equations do not intersect, they have no real solutions.

• Real-world problems may appear to have two solutions when graphed, but in fact only have one because of the context of the problem. Traditionally, negative values are ignored. For example, if graphing the distance a car traveled in relation to time, if one algebraic solution had a negative value, it would not be a true solution since distance and time are always positive.


Example 1

Graph the system of equations below to determine the real solution(s), if any exist.

y x x

y x

= + −= −

2 2 2

2 2

1. Graph the quadratic function, y = x2 + 2x – 2.

If graphing by hand, first plot the y-intercept, (0, –2).

Next, use the equation for the axis of symmetry to determine the x-coordinate of the vertex of the parabola.

xb

a=−

=−

=−

= −2

2

2 1

2

21

( )

Draw the axis of symmetry, x = –1.

Substitute the x-coordinate to determine the corresponding y-coordinate of the vertex.

y = (–1)2 + 2(–1) – 2 = 1 – 2 – 2 = –3

Plot the vertex, (–1, –3).

Finally, since the y-intercept is 1 unit right of the axis of symmetry, an additional point on the parabola is 1 unit left of the axis of symmetry at (–2, –2). Sketch the curve.

If using a graphing calculator, follow the directions appropriate to your model.

On a TI-83/84:

Step 1: Press [Y=].

Step 2: At Y1, type [X, T, θ, n][x2][+][2][X, T, θ, n][–][2].

Step 3: Press [GRAPH].

On a TI-Nspire:

Step 1: Press [menu] and select 3: Graph Type, then select 1: Function.

(continued)



Step 2: At f1(x), enter [x], hit the [x2] key, then type [+][2][x][–][2] and press [enter] to graph the quadratic function.

x

y

–8 –6 –4 –2 0 2 4 6 8

–8

–6

–4

–2

0

2

4

6

8

10

2. Graph the linear function, y = 2x – 2, on the same grid.

If graphing by hand, first plot the y-intercept, (0, –2).

Note: Since the two functions have the same y-intercept, we have already determined one solution.

Next, use the slope of 2 to graph additional points in either direction.

Sketch the line.

On a TI-83/84:

Step 1: Press [Y=].

Step 2: At Y2, type [2][X, T, θ, n][–][2].


On a TI-Nspire:

Step 1: Press [menu] and select 3: Graph Type, then select 1: Function.

(continued)


Step 2: At f 2(x), enter [2][x][–][2] and press [enter] to graph the linear function.

x

y

–8 –6 –4 –2 0 2 4 6 8

–8

–6

–4

–2

0

2

4

6

8

10

(0, –2)

3. Note any intersections that exist as the solution or solutions.

On a TI-83/84:

Step 1: To approximate the intersections, press [2nd][TRACE] and select intersect.

Step 2: Press [ENTER].

Step 3: At the prompt that reads “First curve?”, press [ENTER].

Step 4: At the prompt that reads “Second curve?”, press [ENTER].

Step 5: At the prompt that reads “Guess?”, arrow as close as possible to the first point of intersection and press [ENTER]. Write down the approximate ordered pair.

On a TI-Nspire:

Step 1: Press [menu] and select 7: Points & Lines, then select 3: Intersection Point(s).

Step 2: Select the graphs of both functions.

The only point of intersection is (0, –2).

The solution of y x x

y x

= + −= −

2 2 2

2 2 is (0, –2).


Example 2


y x

y x x

=

= + −

3

4 22

1. Graph the quadratic function, y = x2 + 4x – 2, using the methods demonstrated in Example 1.

x

y

–8 –6 –4 –2 0 2 4 6 8

–8

–6

–4

–2

0

2

4

6

8

10


2. Graph the linear function, y = 3x, on the same grid.

x

y

–8 –6 –4 –2 0 2 4 6 8

–8

–6

–4

–2

0

2

4

6

8

10

(–2, –6)

(1, 3)


The two points of intersection are (–2, –6) and (1, 3).

The solutions of y x

y x x

=

= + −

3

4 22 are (–2, –6) and (1, 3).


Example 3


4 –10

4 62

=−

= − −

y x

y x x

1. Graph the quadratic function, y = x2 – 4x – 6, using the methods demonstrated in Example 1.

x

y

–8 –6 –4 –2 0 2 4 6 8

–14

–12

–10

–8

–6

–4

–2

0

2

4


2. Graph the linear function, y = –4x +10, on the same grid.

x

y

–8 –6 –4 –2 0 2 4 6 8

–14

–12

–10

–8

–6

–4

–2

0

2

4


The two functions do not intersect.

The system 4 –10

4 62

=−

= − −

y x

y x x has no real solutions.

UNIT 5 • QUADRATIC FUNCTIONSLesson 4: Solving Systems of Equations

PRACTICE


For problems 1 and 2, use the graphs to determine the solution(s) to the system of equations, if any exist.

1.

x

y

–8 –6 –4 –2 0 2 4 6 8

–8

–6

–4

–2

0

2

4

6

8

10

continued

Practice 5.4.1: Solving Systems Graphically


PRACTICE


2.

x

y

–8 –6 –4 –2 0 2 4 6 8

–8

–6

–4

–2

0

2

4

6

8

10

For problems 3–8, graph each system. Determine the real solution(s), if any exist.

3. y x

y x

= − −= − +

2 4

5 7

4. y x x

y

= + −= −

3 9 12

12

2

5. y x x

y x

= + −

= − −

5 11 30

52

370

2

6. y x x

y

= + −= −

7 15 20

30

2

continued


PRACTICE


7. y x x

y x

= − − −= −

3 8 640

22 5 862 5

2

. .

8. y x x

y x

= + += +

2 6 25

12 5 15.

For problems 9 and 10, use the given information to create a graph. Then use the graph to answer the questions.

9. The price of a particular stock, P(x), over a 1-year period can be modeled by P(x) = 0.75x2 – 6x + 20, for which x is the number of months. The price of a different stock, C(x), can be modeled by C(x) = 2.75x + 2 during the same year. Did these two stocks have the same value at any time during the year? If so, at what time(s) and at what value(s)?

10. A classic rock band, the Ailing Jones, released two versions of their latest album: a stand-alone CD, and a deluxe boxed set that includes the CD and a coffee table book. A record store manager compared sales of the stand-alone CD with sales of the deluxe boxed set. His figures showed that the total number of CDs sold followed the function c(x) = 108x – 6x2, where c(x) is the total number of CDs sold x days after the album was released. Sales of the deluxe boxed set followed the function d(x) = 100x. After how many days was the number of CDs sold equal to the number of deluxe boxed sets sold? How many of each version of the album were sold in that time?


Lesson 5.4.2: Solving Systems Algebraically

IntroductionPreviously, we learned how to solve quadratic-linear systems by graphing and identifying points of intersection. In this lesson, we will focus on solving a quadratic-linear system algebraically. When doing so, substitution is often the best choice. Substitution is the replacement of a term of an equation by another that is known to have the same value.

Key Concepts

• When solving a quadratic-linear system, if both functions are written in function form such as “y =” or “f(x) =”, set the equations equal to each other.

• When you set the equations equal to each other, you are replacing y in each equation with an equivalent expression, thus using the substitution method.

• You can solve by factoring the equation or by using the quadratic formula, a formula that states the solutions of a quadratic equation of the form ax2 + bx +

c = 0 are given by xb b ac

a=− ± −2 4

2.



Solve the given system of equations algebraically.

y x x

y x

= − += − +

2 11 28

3 12

1. Since both equations are equal to y, substitute by setting the equations equal to each other.

–3x + 12 = x2 – 11x + 28Substitute –3x + 12 for y in the first equation.

2. Solve the equation either by factoring or by using the quadratic formula.

Since a (the coefficient of the squared term) is 1, it’s simplest to solve by factoring.

–3x + 12 = x2 – 11x + 28 Equation

0 = x2 – 8x + 16Set the equation equal to 0 by adding 3x to both sides, and subtracting 12 from both sides.

0 = (x – 4)2 Factor.

x – 4 = 0 Set each factor equal to 0 and solve.

x = 4

Substitute the value of x into the second equation of the system to find the corresponding y-value.

y = –3(4) + 12 Substitute 4 for x.

y = 0

For x = 4, y = 0. Therefore, (4, 0) is the solution.


3. Check your solution(s) by graphing.

x

y

-8 -6 -4 -2 0 2 4 6 8

-5

0

5

(4, 0)

The equations do indeed intersect at (4, 0); therefore, (4, 0) checks out as the solution to this system.

Example 2


y x x

y x

= + += −

2 13 15

1

2


x – 1 = 2x2 + 13x + 15 Substitute x – 1 for y in the first equation.



Since the equation can be factored easily, choose this method.

x – 1 = 2x2 + 13x + 15 Equation

0 = 2x2 + 12x + 16Set the equation equal to 0 by subtracting x from both sides and then adding 1 to both sides.

0 = 2(x + 2)(x + 4) Factor.

Next, set the factors equal to 0 and solve.

x + 2 = 0 x + 4 = 0

x = –2 x = –4

Substitute each of the values you found for x into the second equation of the system to find the corresponding y-value.

y = (–2) – 1 Substitute –2 for x.

y = –3

y = (–4) – 1 Substitute –4 for x.

y = –5

For x = –2, y = –3, and for x = –4, y = –5. Therefore, (–2, –3) and (–4, –5) are the solutions to the system.



x

y

-8 -6 -4 -2 0 2 4 6 8

-5

0

5

(–2, –3)

(–4, –5)

The equations do indeed intersect at (–2, –3) and (–4, –5);

therefore, (–2, –3) and (–4, –5) check out as the solutions to

this system.

Example 3


y x x

y x

= − −= − −

7 25 12

3 30

2


–3x – 30 = 7x2 – 25x – 12Substitute –3x – 30 for y in the first equation.



The quadratic equation cannot easily be factored, so use the quadratic formula.

The quadratic formula is xb b ac

a=− ± −2 4

2.

–3x – 30 = 7x2 – 25x – 12 Equation

0 = 7x2 – 22x + 18Set the equation equal to 0 by adding 3x and 30 to both sides; then, apply the quadratic formula.

a = 7, b = –22, c = 18

x =− −( )± −( ) − ( )( )

( )22 22 4 7 18

2 7

2Substitute values for a, b, and c into the quadratic formula.

x =± −22 20

14Simplify.

Since the value under the square root is negative, there are no real solutions to the quadratic. Simplify to find the complex solutions.

xi

=±22 2 5

14

xi

=±11 5

7



xy

-15 -10 -5 0 5 10 15

-30

-20

-10

Since the functions do not intersect, a complex solution is feasible.


PRACTICE


In problems 1–8, solve each system of equations algebraically. Check your solutions graphically.

1. y x

y

= − −= −

2 5

5

2

2. y x x

y x

= − + += − +

2 4 6

2 11

3. y x x

y x

= + −

= −

2 2 6

1

27

4. y x

y x

= − += +

2 3

4 5

2

5. y x x

y x

= − +

=

1

422

6. y x x

y x

= + += +

2 12 5

16 1

2

7. y x x

y x

= += −

2

3 1

8. y x x

y x

= + −= −

2 3 9

5 8continued

Practice 5.4.2: Solving Systems Algebraically


PRACTICE


For problems 9 and 10, use the given information to write a system of equations, and then solve the system algebraically to answer the questions.

9. A soccer ball is kicked so that its height in feet t seconds after it is kicked can be modeled by the function h(t) = –16t2 + 56t + 4. A hawk flies from its nest 45 feet above the ground at the same time that the player kicks the ball. The hawk’s flight can be modeled by the function h

2(t) = 45 – 12t. After how many

seconds will the hawk and ball reach the same height above the ground?

10. Janetta is a hairstylist who accepts tips. Her profit P each week can be modeled by the function P(c) = –200c2 + 2400c – 4700, where c is the charge per customer. Bertram is the manager at the salon. He is paid a flat rate and cannot accept tips. His profit each week can be modeled by the function P = 500. What must Janetta charge in order for her profit to match Bertram’s profit? Explain.

U5-131

Lesson 5: Interpreting Quadratic Functions




MCC9–12.F.IF.4★

MCC9–12.F.IF.5★

MCC9–12.F.IF.6★

Essential Questions

1. What information can be gathered by analyzing the key features of a quadratic function?

2. What properties must be true for a function to be identified as odd, even, or neither?

3. What are the applications of the domain of a quadratic function?

4. Why is it important to find the average rate of change when calculating the slope of a quadratic function?

WORDS TO KNOW

average rate of change the ratio of the difference of output values to the

difference of the corresponding input values: f b f a

b a

( )− ( )−

; a measure of how a quantity changes

over some interval

concave down a graph of a curve that is bent downward, such as a quadratic function with a maximum value

concave up a graph of a curve that is bent upward, such as a quadratic function with a minimum value


concavity with respect to a curve, the property of being arched upward or downward. A quadratic with positive concavity will increase on either side of the vertex, meaning that the vertex is the minimum or lowest point of the curve. A quadratic with negative concavity will decrease on either side of the vertex, meaning that the vertex is the maximum or highest point of the curve.

decreasing the interval of a function for which the output values are becoming smaller as the input values are becoming larger

domain the set of all input values (x-values) that satisfy the given function without restriction

end behavior the behavior of the graph as x becomes larger or smaller

even function a function that, when evaluated for –x, results in a function that is the same as the original function; f(–x) = f(x)

extrema the minima or maxima of a function

increasing the interval of a function for which the output values are becoming larger as the input values are becoming larger

inflection point a point on a curve at which the sign of the curvature (i.e., the concavity) changes

key features of a quadratic the x-intercepts, y-intercept, where the function is increasing and decreasing, where the function is positive and negative, relative minimums and maximums, symmetries, and end behavior of the function used to describe, draw, and compare quadratic functions

neither describes a function that, when evaluated for –x, does not result in the opposite of the original function (odd) or the original function (even)

odd function a function that, when evaluated for –x, results in a function that is the opposite of the original function; f(–x) = –f(x)

slope the measure of the rate of change of one variable with

respect to another variable; slope = rise

run2 1

2 1

�

�

−−

= =y y

x x

y

x;

the slope in the equation y = mx + b is m.

U5-133Lesson 5: Interpreting Quadratic Functions

Recommended Resources• ChiliMath. “Finding the Domain and Range of a Function.”


This website provides a summary of finding the domain and range for various types of functions as well as practice problems. Illustrated examples walk through finding the domain and range for different situations.

• JamesRahn.com. “Rate of Change.”


This website provides a summary, practice, and an answer key for problems related to average rate of change. The intended audience is precalculus and calculus students; however, the summary is written so students of all levels feel comfortable exploring the concept.

• MathIsFun.com. “Even and Odd Functions.”


This website provides a summary and practice problems for even functions, odd functions, and functions that are neither odd nor even. The site also illustrates the differences in behavior for each type of function.


IntroductionThe tourism industry thrives on being able to provide travelers with an amazing travel experience. Specifically, in areas known for having tropical weather, tour planners want to maximize profit each month by identifying the warmest and coolest months, and then plan tours accordingly. Tour planners might use quadratic models to determine when profits are increasing or decreasing, when they maximized, and/or how profits change in the earlier months versus the later months by looking at the key features of the quadratic functions. In this lesson, you will learn about the key features of a quadratic function and how to use graphs, tables, and verbal descriptions to identify and apply the key features.

Key Concepts

• The key features of a quadratic function are distinguishing characteristics used to describe, draw, and compare quadratic functions. These key features include the x-intercepts, y-intercept, where the function is increasing and decreasing, where the function is positive and negative, relative minimums and maximums, symmetries, and end behavior of the function.

• The x-intercepts of a quadratic function occur when the parabola intersects the x-axis. In the graph that follows, the x-intercepts occur when x = 2 and when x = –2.

Lesson 5.5.1: Interpreting Key Features of Quadratic Functions


x

y

–3 –2 –1 0 1 2 3

–4

–2

2

4

6

8

x-intercepts

• The equation of the x-axis is y = 0; therefore, the x-intercepts can also be found in a table by identifying when the y-value is 0.

• The table of values below corresponds to the parabola we just saw. Notice that the same x-intercepts can be found where the table shows y is equal to 0.

x y

–4 12

–2 0

0 –4

2 0

4 12

• The ordered pair that corresponds to an x-intercept is always of the form (x, 0). The x-intercepts are also the solutions of a quadratic function.

• The y-intercepts of a quadratic function occur when the parabola intersects the y-axis. In the next graph, the y-intercept occurs when y = –4.


x

y

–3 –2 –1 0 1 2 3

–4

–2

2

4

6

8

y-intercept

• The equation of the y-axis is x = 0; therefore, the y-intercept can also be found in a table by identifying when the x-value is 0.

• Notice in the table of values that corresponds to the parabola above, the same y-intercept can be found where x is 0.

x y

–4 12

–2 0

0 –4

2 0

4 12

• The ordered pair that corresponds to a y-intercept is always of the form (0, y).

• Recall that the vertex is the point on a parabola where the graph changes direction.

• The maximum or minimum of the function occurs at the vertex of the parabola.

• The vertex is also the point where the parabola changes from increasing to decreasing.


• Increasing refers to the interval of a function for which the output values are becoming larger as the input values are becoming larger.

• Decreasing refers to the interval of a function for which the output values are becoming smaller as the input values are becoming larger.

• Recall that parabolas are symmetric to a line that extends through the vertex, called the axis of symmetry.

• Any point to the right or left of the parabola is equidistant to another point on the other side of the parabola.

• A parabola only increases or decreases as x becomes larger or smaller.

• Read the graph from left to right to determine when the function is increasing or decreasing.

• Trace the path of the graph with a pencil tip. If your pencil tip goes down as you move toward increasing values of x, then f(x) is decreasing.

• If your pencil tip goes up as you move toward increasing values of x, then f(x) is increasing.

• For a quadratic, if the graph has a minimum value, then the quadratic will start by decreasing toward the vertex, and then it will increase.

• If the graph has a maximum value, then the quadratic will start by increasing toward the vertex, and then it will decrease.

• The vertex is called an extremum. Extrema are the maxima or minima of a function.

• The concavity of a parabola is the property of being arched upward or downward.

• A quadratic with positive concavity will increase on either side of the vertex, meaning that the vertex is the minimum or lowest point of the curve.

• A quadratic with negative concavity will decrease on either side of the vertex, meaning that the vertex is the maximum or highest point of the curve.

• A quadratic that has a minimum value is concave up because the graph of the function is bent upward.

• A quadratic that has a maximum value is concave down because the graph of the function is bent downward.

• The graphs that follow demonstrate examples of parabolas as they decrease and then increase, and vice versa. Trace the path of each parabola from left to right with your pencil to see the difference.


Decreasing then Increasing Increasing then Decreasing

Vertex: (0, –4); minimum

x < 0 = decreasing

x > 0 = increasing

Direction: concave up

Vertex: (0, 4); maximum

x < 0 = increasing

x > 0 = decreasing

Direction: concave down

x

y

–4 –2 0 2 4

–4

–2

2

4

x

y

–4 –2 0 2 4

–4

–2

2

4

• The inflection point of a graph is a point on a curve at which the sign of the curvature (i.e., the concavity) changes. In the following graph, the curvature starts out as concave down, but then switches to concave up at (–1, 1). The point (–1, 1) is the point of inflection.

• The vertex of a quadratic function is also the point of inflection.

x

y

–10 –9 –8 –7 –6 –5 –4 –3 –2 –1 0 1 2 3 4 5 6 7 8 9 10

–10

–9

–8

–7

–6

–5

–4

–3

–2

–1

1

2

3

4

5

6

7

8

9

10


• End behavior is the behavior of the graph as x becomes larger or smaller.

• If the highest exponent of a function is even, and the coefficient of the same term is positive, then the function is approaching positive infinity as x approaches both positive and negative infinity.

• If the highest exponent of a function is even, but the coefficient of the same term is negative, then the function is approaching negative infinity as x approaches both positive and negative infinity.

Even and Positive Even and Negativef(x) = x2 – 4

Highest exponent: 2

Coefficient of x2: positive

As x approaches positive infinity, f(x) approaches positive infinity.

As x approaches negative infinity, f(x) approaches positive infinity.

x

y

–4 –2 0 2 4

–4

–2

2

4

f(x) = –x2 + 4

Highest exponent: 2

Coefficient of x2: negative

As x approaches positive infinity, f(x) approaches negative infinity.

As x approaches negative infinity, f(x) approaches negative infinity.

x

y

–4 –2 0 2 4

–4

–2

2

4

• Functions can be defined as odd or even based on the output yielded when evaluating the function for –x.

• For an odd function, f(–x) = –f(x). That is, if you evaluate a function for –x, the resulting function is the opposite of the original function.

• For an even function, f(–x) = f(x). That is, if you evaluate a function for –x, the resulting function is the same as the original function.


• If evaluating the function for –x does not result in the opposite of the original function or the original function, then the function is neither odd nor even.

• Though all quadratics have an even power, not all quadratics are even functions.

• It is important to evaluate the function for –x when the quadratic includes both a linear and a constant term.


Example 1

A local store’s monthly revenue from T-shirt sales is modeled by the function f(x) = –5x2 + 150x – 7. Use the equation and graph to answer the following questions: At what prices is the revenue increasing? Decreasing? What is the maximum revenue? What prices yield no revenue? Is the function even, odd, or neither?

x

y

0 5 10 15 20 25 30

200

400

600

800

1000

1200

Mon

thly

T-s

hirt

reve

nue

(dol

lars

)

T-shirt price (dollars)

1. Determine when the function is increasing and decreasing.

Use your pencil to determine when the function is increasing and decreasing.

Moving from left to right, trace your pencil along the function.

The function increases until it reaches the vertex, then decreases.

The revenue is increasing when the price per shirt is less than $15 or when x < 15.

The vertex of this function has an x-value of 15.

The revenue is decreasing when the price per shirt is more than $15 or when x > 15.



2. Determine the maximum revenue.

Use the vertex of the function to determine the maximum revenue.

The T-shirt price that maximizes revenue is x = 15.

The maximum is the corresponding y-value.

Since it is difficult to estimate accurately from this graph, substitute x into the function to solve.

f(x) = –5x2 + 150x – 7 Original function

f(15) = –5(15)2 + 150(15) – 7 Substitute 15 for x.

f(15) = 1118 Simplify.

The maximum revenue is $1,118.

3. Determine the prices that don’t yield revenue.

Identify the x-intercepts.

The x-intercepts are 0 and 30, so the store has no revenue when the shirts cost $0 and $30.

4. Determine if the function is even, odd, or neither.

Evaluate the function for –x.

f(x) = –5x2 + 150x – 7 Original function

f(–x) = –5(–x)2 + 150(–x) – 7 Substitute –x for x.

f(–x) = –5x2 – 150x – 7 Simplify.

Since f(–x) is neither the original function nor the opposite of the original function, the function is not even or odd; it is neither.

5. Use the graph of the function to verify that the function is neither odd nor even.

Since the function is not symmetric over the y-axis or the origin, the function is neither even nor odd.


Example 2

A function has a minimum value of –5 and x-intercepts of –8 and 4. What is the value of x that minimizes the function? For what values of x is the function increasing? Decreasing?

1. Determine the x-value that minimizes the function.

Quadratics are symmetric functions about the vertex and the axis of symmetry, the line that divides the parabola in half and extends through the vertex.

The x-value that minimizes the function is the midpoint of the two x-intercepts.

Find the midpoint of the two points by taking the average of the two x-coordinates.

x =− +

= −8 4

22

The value of x that minimizes the function is –2.

2. Determine when the function is increasing and decreasing.

Use the vertex to determine when the function is increasing and when it is decreasing.

The minimum value is –5 and the vertex of the function is (–2, –5).

From left to right, the function decreases as it approaches the minimum and then increases.

The function is increasing when x > –2 and decreasing when x < –2.


Example 3

The table below shows the predicted temperatures for a summer day in Woodland, California. At what times is the temperature increasing? Decreasing?

Time Temperature (ºF)

8 a.m. 52

10 a.m. 64

12 p.m. 72

2 p.m. 78

4 p.m. 81

6 p.m. 76

1. Use the table to determine approximate intervals of increasing and decreasing temperatures.

Examine what is happening to the temperatures as the day progresses from morning to evening.

The values are increasing when y2 – y

1 is positive, or when the

subsequent output value is larger than the preceding value. In this case, the temperature starts at 52º at 8 a.m. and increases to 81º at 4 p.m.

The values are decreasing when y2 – y

1 is negative or when the

subsequent output value is less than the preceding value. At 81º, the temperature decreases to 76º.

The temperatures in Woodland on this summer day appear to be increasing from about 8 a.m. to 4 p.m. The temperatures are decreasing from 4 p.m. to 6 p.m.


2. Use graphing technology to verify the information that is assumed from the table.

On a TI-83/84:

Step 1: Press [STAT].

Step 2: Press [ENTER] to select Edit.

Step 3: Enter x-values into L1. Enter times based on a 24-hour clock for times after 12 p.m. For example, 1 p.m. should be entered as hour 13.

Step 4: Enter y-values into L2.

Step 5: Press [2nd][Y=].

Step 6: Press [ENTER] twice to turn on the Stat Plot.

Step 7: Press [ZOOM][9] to select ZoomStat and show the scatter plot.


Step 9: Arrow to the right to select Calc.

Step 10: Press [5] to select QuadReg.

Step 11: Enter [L1][,][L2], Y1. To enter Y

1, press [VARS] and arrow

over to the right to “Y-VARS.” Select 1: Function. Select 1: Y1.

Step 12: Press [ENTER] to see the graph of the data and the quadratic equation.

On a TI-Nspire:

Step 1: Press the [home] key and select the Lists & Spreadsheet icon.

Step 2: Name Column A “time” and Column B “temperature.”

Step 3: Enter x-values under Column A. Enter times based on a 24-hour clock for times after 12 p.m. For example, 1 p.m. should be entered as hour 13.

Step 4: Enter y-values under Column B.

Step 5: Select Menu, then 3: Data, and then 6: Quick Graph.

Step 6: Press [enter].

Step 7: Move the cursor to the x-axis and choose “time.”(continued)


Step 8: Move the cursor to the y-axis and choose “temperature.”

Step 9: Select Menu, then 4: Analyze, then 6: Regression, and then 4: Show Quadratic.

Step 10: Move the cursor over the equation and press the center key in the navigation pad to drag the equation for viewing, if necessary.

5 10 15

20

0

40

60

80

Time (24-hour clock)

Tem

pera

ture

(ºF)

y

x

3. State your conclusion.

The highest temperature (the maximum) in the table will occur at the point of inflection, or in this case, the time at which the temperature goes from increasing to decreasing.

The highest temperature is 81º, and this occurs at 4 p.m.

The maximum temperature appears to happen at hour 16, according to the quadratic model, or at around 4 p.m.

The high is slightly less than 81º, the predicted temperature for that hour.

UNIT 5 • QUADRATIC FUNCTIONSLesson 5: Interpreting Quadratic Functions

PRACTICE


For each of the functions below, use graphing technology to answer the following questions: What are the x-values for which the function is increasing? Decreasing? What is the maximum or minimum value of the function? What are the x-intercepts? Is the function even, odd, or neither?

1. f(x) = x2 – 3x – 6

2. g(x) = –x2 – 4x + 7

3. y = –4x2 + 8x + 12

4. h(x) = 5x2

Given the descriptions of the quadratic functions below, answer the following questions: What is the value of x that minimizes or maximizes the function? For what values of x is the function increasing? Decreasing?

5. A function has a minimum value of –16.3 and x-intercepts of 9.3 and –41.9.

6. A function has a minimum value of –8.125 and x-intercepts of 4.27 and 0.23.

7. A function has a maximum value of 0.417 and x-intercepts of –1.618 and 0.618.

8. A function has a minimum value of –1.02 and x-intercepts of –0.165 and 0.124.

continued

Practice 5.5.1: Interpreting Key Features of Quadratic Functions


PRACTICE


Use the tables and scenarios that follow to complete the remaining problems.

9. You are practicing punting the football before football tryouts. You kick the ball from the ground represented by the point (0, 0), and the path of the ball is parabolic. The table below represents the height of the ball seconds after being kicked. Use a quadratic model to determine at what times the height of the ball is increasing and decreasing.

Time (seconds) Ball height (feet)

0 0

0.5 12

1 16

1.5 12

2 0

10. The table below shows the height of a signal flare seconds after it is shot from the deck of a ship. Signal flares explode when they reach their highest point. Use a quadratic model to determine how high the flare will be when it explodes.

Time (seconds) Height of flare (feet)

0 112

1 192

2 240

3 256

4 240

5 192


Lesson 5.5.2: Identifying the Domain of a Quadratic Function

IntroductionThe domain of a function is all input values that satisfy the function without restriction. Typically the domain is all real numbers; however, in the case of an application problem, one must determine the most reasonable input values that satisfy the given situation. In this lesson, you will practice identifying the domain of a quadratic function given an equation, a graph, and a real-world context.

Key Concepts

• When a quadratic function does not have a specified interval and is not an application of a real-world situation, its domain is all real numbers.

• This is expressed by showing that the input values exist from negative infinity to infinity as −∞< <∞x .



Describe the domain of the quadratic function g(x) = 1.5x2.

1. Sketch a graph of the function.

x

y

–3 –2 –1 0 1 2 3

2

4

6

8

2. Describe what will happen if the function continues.

Looking at the function, you can see that the function will continue to increase upward and the function will continue to grow wider.

Growing wider without end means that the domain of this function is all real numbers as x increases to infinity and decreases to negative infinity, or −∞< <∞x .


Example 2

Describe the domain of the following function.

x

y

–3 –2 –1 0 1 2 3

–8

–6

–4

–2

1. Describe what is happening to the width of the function as x approaches positive and negative infinity.

The function will continue to open down as x approaches both positive and negative infinity.

2. Determine the domain of the function.

Growing wider without end means that the domain of this function is all real numbers as x increases to infinity and decreases to negative infinity. The domain of the function is all real values, −∞< <∞x .


Example 3

Amit is a diver on the swim team. Today he’s practicing by jumping off a 14-foot platform into the pool. Amit’s height in feet above the water is modeled by f(x) = –16x2 + 14, where x is the time in seconds after he leaves the platform. About how long will it take Amit to reach the water? Describe the domain of this function.

1. Sketch a graph of the situation.

x

y

–1.5 –1 –0.5 0 0.5 1 1.5

2

4

6

8

10

12

14

2. Identify the x-intercepts of the function.

Using graphing technology, the x-intercepts occur when x ≈ –0.94 and x ≈ 0.94.


3. Use the x-intercepts to provide a reasonable domain for this context.

In the context of diving into the swimming pool, the domain is not reasonable until Amit jumps off the platform.

Amit jumps from 14 feet and at that height, the time is 0 seconds.

The positive x-intercept is the moment Amit makes contact with the water.

It will take him 0.94 second to reach the water.

The reasonable domain of this context is 0 to 0.94 seconds or 0 < x < 0.94.


PRACTICE


Use graphing technology to describe the domain of each quadratic function.

1. y = 3x2 – 4x + 2

2. y x x= − − −7

432 52

3. f(x) = 6x2 + 9x – 1

4. g(x) = 2x2 – 12x – 9

Describe the domain of the following functions in words and as an inequality.

5.

x

y

–8 –6 –4 –2 0

2

4

6

8

10

12

14

16

continued

Practice 5.5.2: Identifying the Domain of a Quadratic Function


PRACTICE


6.

x

y

2 4 6 8

–14

–16

–12

–10

–8

–6

–4

–2

2

0

7.

x

y

–4 –2 0 2 4

–10

–8

–6

–4

–2

2

continued


PRACTICE


Use the given information to solve the following problems.

8. A kickball is kicked from the ground and travels a parabolic path. The path can be modeled by the function h(t) = –2t2 + 20t, where h(t) is the height of the kickball in feet above the ground t seconds after being kicked. Assuming the ball lands on level ground, about how long is the ball in the air?

9. The height of baseballs thrown by an automatic baseball-pitching machine can be modeled by the function h(t) = –16t2 + 48t + 3.5, where h(t) is the height of the ball t seconds after being released. If the batter misses the ball, how long does it take the ball to hit the ground? Assume there is no net or catcher behind the plate to stop the ball.

10. A movie theater manager believes that the theater loses money as ticket prices go up. The theater’s average weekly sales can be modeled by the quadratic function R(x) = –700x2 + 7700x + 245,000, where R(x) is the weekly revenue in dollars and x is the number of $0.50 increases in price. For what number of $0.50 increases will the theater continue to produce revenue? After how many $0.50 increases will the theater receive the greatest revenue?


Lesson 5.5.3: Identifying the Average Rate of Change

IntroductionThe average running speed of a human being is 5–8 mph. Imagine you set out on an 8-mile run that takes you no longer than 1 hour. You run often, so you run consistently and at the same speed for the entire hour. The rate of change of your position for your runs is always 8 mph and can be modeled linearly, because the rate of change is constant. However, a friend of yours, averaging the same distance in an hour, hasn’t built up the endurance needed to run 8 miles consistently. Sometimes your friend runs 9 mph, sometimes he stops to rest, sometimes he walks, and then he resumes running at 8 mph. He might do this several times in the hour, sometimes running faster than you, sometimes slower, and sometimes not running at all. The rate of change of your friend’s speed is not constant and cannot be modeled linearly. The average rate of change for your friend’s speed, the ratio of the change in the output of a function to the change in the input for specific intervals, is inconsistent. In this lesson, you will practice calculating the average rate of change of quadratic functions over various intervals.

Key Concepts

• The average rate of change of a function is the rate of change between any two points of a function; it is a measure of how a quantity changes over some interval.

• The average can be found by calculating the ratio of the difference of output

values to the difference of the corresponding input values, f b f a

b a

( )− ( )−

, from

x = a to x = b. This formula is often referred to as the average rate of change

formula.

• Recall that the slope of a linear function is found using the formula rise

run2 1

2 1

�

�

−−

= =y y

x x

y

x.

• Although the formula for calculating the average rate of change looks quite different from the formula used to find the slope of a linear function, they are actually quite similar.

• Both formulas are used to find the rate of change between two specific points.

• The rate of change of a linear function is always constant, whereas the average rate of change of a quadratic function is not constant.

• Choosing different x-values and their corresponding y-values will result in different rates of change.



Calculate the average rate of change for the function f(x) = x2 + 6x + 9 between x = 1 and x = 3.

1. Evaluate the function for x = 3.

f(x) = x2 + 6x + 9 Original function

f(3) = (3)2 + 6(3) + 9 Substitute 3 for x.

f(3) = 36 Simplify.

2. Evaluate the function for x = 1.

f(x) = x2 + 6x + 9 Original function


f(1) = 16 Simplify.

3. Use the average rate of change formula to determine the average rate of change between x = 1 and x = 3.

Average rate of change = f b f a

b a

( )− ( )−

Average rate of change formula

Average rate of change =( )− ( )

−f f3 1


Average rate of change =−36 16

2Substitute the values for f(3) and f(1).

Average rate of change = 10 Simplify.

The average rate of change of f(x) = x2 + 6x + 9 between x = 1 and x = 3 is 10.


Example 2

Use the graph of the function to calculate the average rate of change between x = –3 and x = –2.

x

y

–4 –3 –2 –1 0

2

4

6

1. Use the graph to identify f(–2).

According to the graph, f(–2) = –1.

2. Use the graph to identify f(–3).

According to the graph, f(–3) = 2.

3. Use the average rate of change formula to calculate the average rate of change between x = –3 and x = –2.

Average rate of change = f b f a

b a

( )− ( )−


Average rate of change = =−( )− −( )−( )− −( )

f f2 3

2 3 Substitute –3 for a and –2 for b.

Average rate of change =− −1 2

1Substitute the values for f(–3) and f(–2) found from the graph.

Average rate of change = –3 Simplify.

The average rate of change of the function between x = –3 and x = –2 is –3.


Example 3

For the function g(x) = (x – 3)2 – 2, is the average rate of change greater between x = –1 and x = 0 or between x = 1 and x = 2?

1. Calculate the average rate of change between x = –1 and x = 0.

Evaluate the function at x = –1 and x = 0.

For x = –1: For x = 0:

g(x) = (x – 3)2 – 2 g(x) = (x – 3)2 – 2

g(–1) = [(–1) – 3]2 – 2 g(0) = [(0) – 3]2 – 2

g(–1) = 14 g(0) = 7

Average rate of change = g b g a

b a

( )− ( )−


Average rate of change =( )− −( )− −( )

g g0 1

0 1 Substitute –1 for a and 0 for b.

Average rate of change =−7 14

1Substitute the values for g(–1) and g(0).


The average rate of change between x = –1 and x = 0 is –7.


2. Calculate the average rate of change between x = 1 and x = 2.

Evaluate the function at x = 1 and x = 2.

For x = 1: For x = 2:

g(x) = (x – 3)2 – 2 g(x) = (x – 3)2 – 2

g(1) = [(1) – 3]2 – 2 g(2) = [(2) – 3]2 – 2

g(1) = 2 g(2) = –1


b a

( )− ( )−


Average rate of change =( )− ( )

−g g2 1


Average rate of change =− −1 2

1Substitute the values for g(1) and g(2).


The average rate of change between x = 1 and x = 2 is –3.

3. Compare the averages.

Since –3 > –7, the average rate of change of g(x) = (x – 3)2 – 2 is greater between x = 1 and x = 2 than it is between x = –1 and x = 0.


Example 4

Find the average rate of change between x = –0.75 and x = –0.25 for the following function.

x y

–1 0

–0.75 3.44

–0.5 6.25

–0.25 8.44

0 10

0.25 10.94

1. Identify the output values when x = –0.75 and x = –0.25.

Refer to the table.

When x = –0.75, y = 3.44.

When x = –0.25, y = 8.44.

2. Calculate the average rate of change between x = –0.75 and x = –0.25 by applying the average rate of change formula.


b a

( )− ( )−


Average rate of change =( )− ( )− − −

g g– . – .

. ( . )

0 25 0 75

0 25 0 75Substitute –0.75 for a and –0.25 for b.

Average rate of change =−

− − −( )8 44 3 44

0 25 0 75

. .

. .Substitute the values for g(–0.75) and g(–0.25).

Average rate of change =5

0 5.Simplify.

Average rate of change = 10

The average rate of change of the function is 10.


PRACTICE


Calculate the average rate of change for the functions below between x = –4 and x = –2.

1. f(x) = 2(x – 1)2 – 4

2. g(x) = 12 – 2(x + 1)2

3. h x x( )= −1

412

4. x y

–6 110–5 77–4 50–3 29–2 14–1 50 51 5

5. x y

–6 162–5 116–4 78–3 48–2 26–1 120 61 8

6.

x

y

–6 –4 –2 0 2

–2

2

4

6

continued

Practice 5.5.3: Identifying the Average Rate of Change


PRACTICE


For the following functions, is the average rate of change greater between x = –2 and x = –1 or between x = –1 and x = 0?

7. y x x= − −1

282

8. a(x) = –2x2 – 2x – 5

9. y = 6x2 + 31x – 12

Read the scenario and use the information in it to answer the question.

10. A mother drops an apartment key down to her son from several floors above. The function h(t) = –16t2 + 60 is used to model the key’s height, h(t), in feet t seconds after being released. What is the key’s average rate of change 1.5 to 2 seconds after being dropped?

U5-165

Lesson 6: Analyzing Quadratic Functions




MCC9–12.F.IF.7a★

MCC9–12.F.IF.8a

MCC9–12.F.IF.9

MCC9–12.F.LE.3★

Essential Questions

1. Howisaquadraticequationsimilartoalinearequation?Howisitdifferent?

2. Howisaquadraticequationsimilartoanexponentialequation?Howisitdifferent?

3. Inwhatsituationsisitappropriatetouseaquadraticmodel?

WORDS TO KNOW

axis of symmetry of a parabola

thelinethroughthevertexofaparabolaaboutwhich

theparabolaissymmetric.Theequationoftheaxisof

symmetryis xb

a2=−

.

extrema theminimaormaximaofafunction

first difference inasetofdata,thechangeinthey-valuewhenthex-valueisincreasedby1

intercept thepointatwhichalineinterceptsthex-ory-axis

intercept form thefactoredformofaquadraticequation,writtenasf(x)=a(x–p)(x–q),wherepandqarethezerosofthefunction

maximum thelargesty-valueofaquadratic

minimum thesmallesty-valueofaquadratic

parabola theU-shapedgraphofaquadraticfunction

root(s) solution(s)ofaquadraticequation


second difference in a set of data, the change in successive first differences

standard form of a quadratic function

a quadratic function written as f(x) = ax2 + bx + c, where a is the coefficient of the quadratic term, b is the coefficient of the linear term, and c is the constant term

symmetrical having two identical parts when reflected over a line or rotated around a point

vertex form a quadratic function written as f(x) = a(x – h)2 + k, where the vertex of the parabola is the point (h, k); the form of a quadratic equation where the vertex can be read directly from the equation


x-intercept the point at which the graph crosses the x-axis; written as (x, 0)

y-intercept the point at which the graph crosses the y-axis; written as (0, y)

zeros the x-values of a function for which the function value is 0

Recommended Resources• IXL Learning. “Solve an Equation Using the Zero Product Property.”


This interactive website gives a series of problems and scores them immediately. If the user submits a wrong answer, a description and process for arriving at the correct answer are provided. Users solve quadratic equations by setting factors equal to 0. This activity is meant as a review.

• PhET Interactive Simulations. “Equation Grapher 2.02.”


This website allows users to compare the graphs of various self-created equations.

• West Texas A&M University Virtual Math Lab. “Graphs of Quadratic Functions.”


This tutorial offers a review and worked examples for writing and graphing quadratic functions in different forms, as well as practice problems with worked solutions for reference.

U5-167Lesson 6: Analyzing Quadratic Functions

IntroductionWe have studied the key features of the graph of a parabola, such as the vertex and x-intercepts. In this lesson, we will review the definitions of key features and learn to identify them from the equation of a quadratic function. Once you identify the key features, you can make a sketch of the graph of a quadratic equation.

Key Concepts

• A quadratic function in standard form, or general form, is written as f(x) = ax2 + bx + c, where a is the coefficient of the quadratic term, b is the coefficient of the linear term, and c is the constant term.

• The graph of a quadratic is U-shaped and called a parabola.

• The extremum of a graph is the function value that achieves either a maximum or a minimum.

• The maximum is the largest y-value of a quadratic, and the minimum is the smallest y-value of a quadratic.

• If a > 0, the parabola is concave up and the quadratic has a minimum.

• If a < 0, the parabola is concave down and the quadratic has a maximum.

• The extreme values of a quadratic occur at the vertex, the point at which the curve changes direction.

• If the quadratic equation is given in standard form, the vertex can be found by

identifying the x-coordinate of the vertex using b

a2

−.

• Substitute the value of the x-coordinate into the quadratic equation to find the y-coordinate of the vertex.

• The vertex of a quadratic function is b

af

b

a2,

2

− −

.

• Another way to find the vertex is to complete the square on the standard form of the parabola in order to convert it to vertex form.

• The vertex form of a quadratic function is f(x) = a(x – h)2 + k, where the coordinate pair (h, k) is the location of the vertex.

• The intercept of a graph is the point at which a line intercepts the x- or y-axis.

• The y-intercept of a function is the point at which the graph crosses the y-axis. This occurs when x = 0. The y-intercept is written as (0, y).

Lesson 5.6.1: Graphing Quadratic Functions


• The x-intercepts of a function are the points at which the graph crosses the x-axis. This occurs when y = 0. The x-intercept is written as (x, 0).

• The zeros of a function are the x-values for which the function value is 0.

• The intercept form of the quadratic function, written as f(x) = a(x – p)(x – q), where p and q are the zeros of the function, can be used to identify the x-intercepts. Set the factored form equal to 0. Then set each factor equal to 0. As long as the coefficients of x are 1, the x-intercepts are located at (r, 0) and (s, 0).

• These values for x are the roots, or the solutions, of the quadratic equation.

• Parabolas are symmetrical; that is, they have two identical parts when rotated around a point or reflected over a line.

• This line is the axis of symmetry, the line through the vertex of a parabola about which the parabola is symmetric. The equation of the axis of symmetry

is xb

a2=−

.

• Symmetry can be used to find the vertex of a parabola if the vertex is not known.

• If you know the x-intercepts of the graph or any two points on the graph with the same y-value, the x-coordinate of the vertex is the point halfway between the values of the x-coordinates.

• For x-intercepts (r, 0) and (s, 0), the x-coordinate of the vertex is r s+

2.

16

14

12

10

6

4

2

2

4

6

8

10

10 5 5 10

Axis of symmetry

x-intercept

Vertex

y-intercept

x-intercept0


• To graph a function using a graphing calculator, follow these general steps for your calculator model.

On a TI-83/84:

Step 1: Press the [Y=] button.

Step 2: Type the function into Y1, or any available equation. Use the [X, T, θ, n] button for the variable x. Use the [x2] button for a square.

Step 3: Press [WINDOW]. Enter values for Xmin, Xmax, Ymin, and Ymax. The Xscl and Yscl are arbitrary. Leave Xres = 1.


On a TI-Nspire:

Step 1: Press the [home] key.

Step 2: Arrow over to the graphing icon and press [enter].

Step 3: Type the function next to f1(x), or any available equation, and press [enter]. Use the [X] button for the letter x. Use the [x2] button for a square.

Step 4: To change the viewing window, press [menu]. Select 4: Window/Zoom and select A: Zoom – Fit.


Example 1

Given the function f(x) = –2x2 + 16x – 30, identify the key features of the graph: the extremum, vertex, x-intercept(s), and y-intercept. Then sketch the graph.

1. Determine the extremum of the graph.

The extreme value is a minimum when a > 0. It is a maximum when a < 0.

Because a = –2, the graph opens downward and the quadratic has a maximum.

2. Determine the vertex of the graph.

The maximum value occurs at the vertex.

The vertex is of the form b

af

b

a2,

2

− −

.

Use the original equation f (x) = –2x2 + 16x – 30 to find the values of a and b in order to find the x-value of the vertex.

xb

a2=− Formula to find the x-coordinate of

the vertex of a quadratic

x16

2( 2)

( )=−−

Substitute –2 for a and 16 for b.

x = 4 Simplify.

The x-coordinate of the vertex is 4.

Substitute 4 into the original equation to find the y-coordinate.

f(x) = –2x2 + 16x – 30 Original equation


f(4) = 2 Simplify.

The y-coordinate of the vertex is 2.

The vertex is located at (4, 2).



3. Determine the x-intercept(s) of the graph.

Since the vertex is above the x-axis and the graph opens downward, there will be two x-intercepts.

Factor the quadratic and set each factor equal to 0.


f(x) = –2(x2 – 8x + 15) Factor out the greatest common factor.

f(x) = –2(x – 3)(x – 5) Factor the trinomial.

0 = –2(x – 3)(x – 5)Set the factored form equal to 0 to find the intercepts.

x – 3 = 0 or x – 5 = 0 Set each factor equal to 0 and solve for x.

x = 3 or x = 5 Simplify.

The x-intercepts are (3, 0) and (5, 0).

4. Determine the y-intercept of the graph.

The y-intercept occurs when x = 0.

Substitute 0 for x in the original equation.



f(0) = –30 Simplify.


When the quadratic equation is written in standard form, the y-intercept is c.


5. Graph the function.

Use symmetry to identify additional points on the graph.

The axis of symmetry goes through the vertex, so the axis of symmetry is x = 4.

For each point to the left of the axis of symmetry, there is another point the same distance on the right side of the axis and vice versa.

The point (0, –30) is on the graph, and 0 is 4 units to the left of the axis of symmetry.

The point that is 4 units to the right of the axis is 8, so the point (8, –30) is also on the graph.

Determine two additional points on the graph.

Choose an x-value to the left or right of the vertex and find the corresponding y-value.



f(1) = –16 Simplify.

An additional point is (1, –16).

(1, –16) is 3 units to the left of the axis of symmetry.

The point that is 3 units to the right of the axis is 7, so the point (7, –16) is also on the graph.

Plot the points and join them with a smooth curve.

35

30

25

20

15

5

5

10

15

20

25

30

35

40

45

50

55

60

2 2 4 6 8 1

–

(0, –30) (8, –30)

(3, 0)

f(x) = –2x2 + 16x – 30

(4, 2)

(5, 0)

(1, –16) (7, –16)

0


Example 2

Given the function f(x) = x2 + 6x + 9, identify the key features of its graph: the extremum, vertex, x-intercept(s), and y-intercept. Then sketch the graph.


The extreme value is a minimum when a > 0. It is a or maximum when a < 0.

Because a = 1, the graph opens upward and the quadratic has a minimum.


The minimum value occurs at the vertex.

Find the vertex by completing the square.

f(x) = x2 + 6x + 9 Original equation

f(x) = (x + 3)2 Write the perfect square trinomial as a binomial squared.

The vertex is located at (–3, 0).

3. Determine the x-intercept(s) of the graph.

Since the vertex is on the x-axis, there will be one x-intercept.

The squared binomial is the factored form of the equation.

Set the squared binomial equal to 0.

y = (x + 3)2 Squared binomial

0 = (x + 3)(x + 3) Factored binomials

x + 3 = 0 or x + 3 = 0Set each factor equal to 0 and solve for x.

x = –3 or x = –3 Simplify.

The x-intercept is (–3, 0).







f(0) = 9 Simplify.



Use symmetry to identify an additional point on the graph.

The axis of symmetry goes through the vertex, so the axis of symmetry is x = –3.

For each point to the left of the axis of symmetry, there is another point the same distance on the right side of the axis and vice versa.

The point (0, 9) is on the graph, and 0 is 3 units to the right of the axis of symmetry.

The point that is 3 units to the left of the axis is –6, so the point (–6, 9) is also on the graph.




f(–1) = (–1)2 + 6(–1) + 9 Substitute –1 for x.

f(–1) = 4 Simplify.

An additional point is (–1, 4).

(–1, 4) is 2 units to right of the axis of symmetry.

The point that is 2 units to the left of the axis is –5, so the point (–5, 4) is also on the graph.

(continued)



14

13

12

10

9

8

7

6

5

4

3

2

1

1

3

4

5

6

7

12 10 8 6 4 2 2 4

(–6, 9) (0, 9)

f(x) = x2 + 6x + 9

(–3, 0)

(–5, 4) (–1, 4)

0

Example 3

Given the function f(x) = –2(x + 1)(x + 5), identify the key features of its graph: the extremum, vertex, x-intercept(s), and y-intercept. Then sketch the graph.


The extreme value is either a minimum, when a > 0, or a maximum, when a < 0.

Because a = –2, the graph opens down and the quadratic has a maximum.


2. Determine the x-intercepts of the graph.

The equation is in factored form.

The zeros occur when y = 0.

Find the zeros by setting the equation equal to 0.

Then set each factor equal to 0.

f(x) = –2(x + 1)(x + 5) Original equation

0 = –2(x + 1)(x + 5) Set the equation equal to 0.

x + 1 = 0 or x + 5 = 0

x = –1 or x = –5

The x-intercepts are (–1, 0) and (–5, 0).


Use the axis of symmetry to identify the vertex.

The axis of symmetry is halfway between the x-intercepts.

Find the midpoint between the x-intercepts.

x x1 2

2

+

Midpoint formula

x( 1) ( 5)

2=

− + −

Substitute –1 for x

1 and –5 for x

2.

x = –3 Simplify.

The axis of symmetry is x = –3, so the x-coordinate of the vertex is –3.

To find the y-coordinate, substitute –3 into the original equation.


f(–3) = –2[(–3) + 1][(–3) + 5] Substitute –3 for x.

f(–3) = –2(–2)(2) Simplify.

f(–3) = 8

The vertex is (–3, 8).






f(0) = –2[(0) + 1][(0) + 5] Substitute 0 for x.

f(0) = –2(1)(5) Simplify.

f(0) = –10



Use symmetry to identify another point on the graph.

Because 0 is 3 units to the right of the axis of symmetry, the point 3 units to the left of the axis will have the same value, so (–6, –10) is also on the graph.




f(–2) = –2[(–2) + 1][(–2) + 5] Substitute –2 for x.

f(–2) = 6 Simplify.

An additional point is (–2, 6).

(–2, 6) is 1 unit to right of the axis of symmetry.

The point that is 1 unit to the left of the axis is –4, so the point (–4, 6) is also on the graph.

(continued)



24

22

20

18

16

14

10

8

6

4

2

2

4

6

8

10

14

16

18

20

22

24

26

28

2 10 8 6 4 2 2 4

f(x) = –2(x + 1)(x + 5) (–3, 8)

(–5, 0)

(–4, 6) (–2, 6)

(–1, 0)

(0, –10)(–6, –10)

0

UNIT 5 • QUADRATIC FUNCTIONSLesson 6: Analyzing Quadratic Functions

PRACTICE


For each function that follows, identify the intercepts, vertex, and maximum or minimum. Then, sketch the graph of the function.

1. y = x2 + 6x – 7

2. y = –x2 – 8x – 12

3. y = x2 + 4x + 3

4. y x x= − −1

222

5. y = (x + 7)(x + 3)

6. y = 3x2 + 6x + 3

7. y = –2x2 – 12x – 16

For each problem that follows, determine whether the function has a minimum or maximum, identify the maximum or minimum, and identify the intercepts.

8. A golfer’s ball lands in a sand trap 4 feet below the playing green. The path of the ball on her next shot is given by the equation y = –16x2 + 20x – 4, where y represents the height of the ball after x seconds.

9. The revenue, R(x), generated by an increase in price of x dollars for an item is represented by the equation R(x) = –10x2 + 100x + 750.

10. The flight of a rubber band follows the quadratic equation H(x) = –x2 + 6x + 7, where H(x) represents the height of the rubber band in inches and x is the horizontal distance the rubber band travels in inches after launch.

Practice 5.6.1: Graphing Quadratic Functions


Lesson 5.6.2: Writing Equivalent Forms of Quadratic Functions

IntroductionQuadratic equations can be written in standard form, factored form, and vertex form. While each form is equivalent, certain forms easily reveal different features of the graph of the quadratic function. In this lesson, you will learn to use the processes of factoring and completing the square to show key features of the graph of a quadratic function and determine how these key features relate to the characteristics of a real-world situation.

Key Concepts

• Recall that the standard form, or general form, of a quadratic function is written as f(x) = ax2 + bx + c, where a is the coefficient of the quadratic term, b is the coefficient of the linear term, and c is the constant term.

• The process of completing the square can be used to transform a quadratic equation from standard form to vertex form, f(x) = a(x – h)2 + k.

• Vertex form can be used to identify the key features of a function’s graph.

• The vertex of a parabola is the point where the graph changes from increasing to decreasing, or vice versa.

• In vertex form, the extremum of the graph of a quadratic equation is easily identified using the vertex, (h, k).

• If a < 0, the function achieves a maximum, where k is the y-coordinate of the maximum and h is the x-coordinate of the maximum.

• If a > 0, the function has a minimum, where k is the y-coordinate of the minimum and h is the x-coordinate of the minimum.

• Because the axis of symmetry goes through the vertex, the axis of symmetry is easily identified from vertex form as x = h.

• The process of factoring can be used to transform a quadratic equation in standard form to factored form, f(x) = a(x – r)(x – s).

• The zeros of a function are the x-values where the function value is 0.

• Setting the factored form equal to 0, 0 = a(x – r)(x – s), the zeros are easily identified as r and s.


• As long as the coefficients of x are 1, the x-intercepts can be identified as (r, 0) and (s, 0).

• The axis of symmetry is easily identified from the factored form as the axis

of symmetry occurs at the midpoint between the zeros. Therefore, the axis of

symmetry is xr s

=+2

.


Example 1

Suppose that the flight of a launched bottle rocket can be modeled by the equation y = –x2 + 6x, where y measures the rocket’s height above the ground in meters and x represents the rocket’s horizontal distance in meters from the launching spot at x = 0. How far does the bottle rocket travel in the horizontal direction from launch to landing? What is the maximum height the bottle rocket reaches? How far has the bottle rocket traveled horizontally when it reaches its maximum height?

1. Identify the zeros of the function.

In the original equation, y represents the height of the bottle rocket. At launch and landing, the height of the bottle rocket is 0.

Write the original equation in factored form. Set it equal to 0 to identify the zeros of the function.

y = –x2 + 6x Original equation

0 = –x2 + 6x Set the equation equal to 0.

0 = –(x2 – 6x) Factor out –1.

0 = –x(x – 6) Factor the binomial.

Solve for x by setting each factor equal to 0.

–x = 0 or x – 6 = 0

x = 0 or x = 6

The x-intercepts are at x = 0 and x = 6. Find the distance between the two points to determine how far the bottle rocket travels in the horizontal direction.

6 – 0 = 6

The bottle rocket travels 6 meters in the horizontal direction from launch to landing.



2. Determine the maximum height of the bottle rocket.

The maximum height occurs at the vertex.

Write the equation in vertex form by completing the square.

y = –x2 + 6x Original equation

y = –(x2 – 6x)Factor out the common factor, –1, from the variable terms.

y = –(x2 – 6x + 9) + 9Add and subtract the square of

1

2 of the x-term.

Be sure to multiply the subtracted term by a, –1.

y = –(x – 3)2 + 9Write the trinomial as a binomial squared and simplify the constant term.

The vertex form is y = –(x – 3)2 + 9. The vertex is (3, 9). The maximum value is the y-coordinate of the vertex, 9.

The bottle rocket reaches a maximum height of 9 meters.

3. Determine the horizontal distance from the launch point to the maximum height of the bottle rocket.

We know that the bottle rocket is launched from the point (0, 0) and reaches a maximum height at (3, 9). Subtract the x-values of the two points to find the distance traveled horizontally.

3 – 0 = 3

Another method is to take the total distance traveled horizontally from launch to landing and divide it by 2 to find the same answer. This is because the maximum value occurs halfway between the zeros of the function.

6

23=

The bottle rocket has traveled 3 meters horizontally when it reaches its maximum height.


Example 2

Decreasing the cost of an item can result in a greater number of sales. The revenue function that predicts the revenue in dollars, R(x), for each $1 change in price, x, for a particular item is R(x) = –100x2 + 1400x + 24,000. What is the maximum value of the function? What does the maximum value mean in the context of the problem? What price increase maximizes the revenue and what does it mean in the context of the problem? What are the zeros of the function and what do they represent? What increase in price results in the same revenue as not increasing the price at all?

1. Determine the maximum value of the function.

The maximum value of the function occurs at the vertex. Complete the square to find the vertex.

R(x) = –100x2 + 1400x + 24,000 Original equation

R(x) = –100(x2 – 14x) + 24,000Factor out the common factor, –100, from the variable terms.

R(x) = –100(x2 – 14x + 49) + 24,000 + 4900

Add and subtract the square of 1

2 of

the x-term. Be sure to multiply the

subtracted term by a, –100.

R(x) = –100(x – 7)2 + 28,900Write the trinomial as a binomial squared and add the constants.

The vertex is (7, 28,900); therefore, the maximum value is 28,900.

2. Determine what the maximum value means in the context of the problem.

The maximum value of 28,900 means that the maximum revenue resulting from increasing the price by x dollars is $28,900.


3. Determine the price increase that will maximize the revenue and what it means in the context of the problem.

The maximum value occurs at the vertex, (7, 28,900).

This means an increase in price of $7 will result in the maximum revenue.

4. Determine the zeros of the function and what they represent in the context of the problem.

The zeros of the function occur when the function, R(x) = –100x2 + 1400x + 24,000, equals 0.

Write the equation in factored form to identify the zeros.

Set the equation equal to 0 and solve for the zeros.

R(x) = –100x2 + 1400x + 24,000 Original equation

R(x) = –100(x2 – 14x – 240) Factor out the common factor, –100.

R(x) = –100(x – 24)(x + 10) Factor the trinomial.

Set each factor equal to 0 and solve for x.

x – 24 = 0 or x + 10 = 0

x = 24 or x = –10

The zeros of the quadratic are 24 and –10.

To interpret the zeros, recall that x represents each dollar change in the price of the product. An increase of –10 would not make sense in the context of this problem. An increase of $24 would result in zero revenue.


5. Determine the increase in price that results in the same revenue as not increasing the price at all.

A zero increase in price results when x = 0.

Recall that the function is symmetric.

The axis of symmetry can be found using the x-value of the vertex, 7.

Zero is 7 units to the left of the vertex, so the point that is symmetric is 7 units to the right of the vertex. The x-value that is 7 units to the right of 7 is 14.

Substitute 0 and 14 for x to confirm.

When x = 0: When x = 14:

R(x) = –100x2 + 1400x + 24,000 R(x) = –100x2 + 1400x + 24,000

R(0) = –100(0)2 + 1400(0) + 24,000

R(14) = –100(14)2 + 1400(14) + 24,000

R(0) = 24,000 R(14) = –19,600 + 19,600 + 24,000

R(14) = 24,000

An increase in price of $14 would result in revenue of $24,000, which is the same revenue as not increasing the price at all.


Example 3

A football is kicked and follows a path given by y = –0.03(x – 30)2 + 27, where y represents the height of the ball in feet and x represents the ball’s horizontal distance in feet. What is the maximum height the ball reaches? What horizontal distance maximizes the height? What are the zeros of the function? What do the zeros represent in the context of the problem? What is the total horizontal distance the ball travels? If the ball reaches a height of 20.25 feet after traveling 15 feet horizontally, will the ball make it over a 10-foot-tall goal post that is 45 feet from the kicker?

1. Determine the maximum height of the ball.

The maximum occurs at the vertex.

The maximum value can be identified from the vertex form of the quadratic.

The quadratic, y = –0.03(x – 30)2 + 27, is already in vertex form, f(x) = a(x – h)2 + k, where the vertex is (h, k).

The vertex is (30, 27) and the maximum value is 27.

The maximum height the ball reaches is 27 feet.

2. Determine the horizontal distance of the ball when it reaches its maximum height.

The x-coordinate of the vertex maximizes the quadratic.

The vertex is (30, 27).

The ball will have traveled 30 feet in the horizontal direction when it reaches its maximum height.


3. Determine the zeros of the function.

The zeros of the function occur when the function value is 0.

The factored form of the quadratic equation can be used to identify the zeros of the function.

First write the function in standard form.

y = –0.03(x – 30)2 + 27 Original equation

y = –0.03(x2 – 60x + 900) + 27 Square the binomial.

y = –0.03x2 + 1.8x – 27 + 27Distribute –0.03 over the equation in parentheses.

y = –0.03x2 + 1.8x Simplify.

Factor the quadratic and set it equal to 0.

y = –0.03x2 + 1.8x Standard form of the function

y = –0.03x(x – 60) Factor out the common factor, –0.03x.

0 = –0.03x(x – 60) Set the equation in factored form equal to 0.

–0.03x = 0 or x – 60 = 0 Set each factor equal to 0 and solve for x.

x = 0 or x = 60

The zeros are 0 and 60.

4. Determine what the zeros represent in the context of the problem.

The zeros represent where the ball was kicked from at a horizontal distance of 0 feet, and where the ball lands at a horizontal distance of 60 feet.

5. Determine the total horizontal distance the ball travels.

The distance between the two zeros gives the total horizontal distance traveled, 60 feet.


6. Determine whether the ball will clear the goal post.

The scenario asked if the ball would make it over a 10-foot-tall goal post that is 45 feet from the kicker if the ball reached a height of 20.25 feet after traveling 15 feet horizontally.

The distance between the two zeros is 60 feet.

The axis of symmetry is half the distance between the zeros at x = 30.

Fifteen feet is 15 feet to the left of the axis of symmetry.

The height of the ball 15 units to the right of the axis of symmetry will be the same as the height 15 units to the left of the axis of symmetry, or 20.25 feet.

This is above the goal-post height of 10 feet.

The ball will go over the 10-foot-tall goal post.


PRACTICE


For problems 1–3 below, complete each of the following tasks:

a. Write the equation in vertex form.

b. Identify the maximum or minimum value of the function.

c. Write the equation of the axis of symmetry.

d. Write the equation in factored form.

e. Identify the zeros of the function.

1. y = x2 – 8x + 12

2. y x x= − − +1

242

3. y = –16t2 + 48t

Use the given information in each scenario that follows to complete the remaining problems.

4. A butterfly descends toward the ground and then flies back up. The butterfly’s descent can be modeled by the equation h(t) = t2 – 10t + 26, where h(t) is the butterfly’s height above the ground in feet and t is the time in seconds since you saw the butterfly. Write the equation in vertex form and identify the vertex. What is the meaning of the vertex in the context of the problem?

5. A cliff diver jumps upward from the edge of a cliff then begins to descend, so that his path follows a parabola. The diver’s height, h(t), above the water in feet is given by h(t) = –2t2 + 4t + 50, where t represents the time in seconds. Write the function in vertex form. What is the vertex and what does it represent in the context of the problem? How many seconds after the start of the dive does the diver reach the initial height?

continued

Practice 5.6.2: Writing Equivalent Forms of Quadratic Functions


PRACTICE


6. The revenue of producing and selling widgets is given by R(w) = –0.8w2 + 40w, where w is the number of widgets produced and R(w) is the amount of revenue in dollars. Write the function in factored form. What are the zeros and what do they represent in the context of the problem? What number of widgets maximizes the revenue?

7. A football is kicked and follows a path given by y = –0.03x2 + 1.2x, where y represents the height of the ball in feet and x represents the horizontal distance in feet. Write the equation in vertex form. What is the vertex and what does it mean in the context of the problem? How far does the ball travel in the horizontal direction?

8. A frog hops from the bank of a creek onto a lily pad. The path of the jump can

be modeled by the equation h x x x( )= − + +1

22 22 , where h(x) is the frog’s height

in feet above the water and x is the number of seconds since the frog jumped.

Write the equation in vertex form and identify the vertex. What does the vertex

represent in the context of the problem? What is the axis of symmetry? After

how many seconds does the height of the frog reach the initial height?

9. The revenue, R(x), generated by an increase in price of x dollars for an item is represented by the equation R(x) = –2x2 + 20x + 150. Write the equation in vertex form and identify the vertex. What does the vertex represent in the context of the problem? What is the axis of symmetry? What increase in price results in the same revenue as not increasing the price at all?

10. Decreasing the cost of an item can result in a greater number of sales. The revenue function that predicts the revenue in dollars, R(x), for each $1 decrease in price, x, for a certain item is R(x) = –x2 + 16x + 260. Write the equation in factored form. Identify the zeros. What do the zeros represent in the context of the problem? What is the axis of symmetry? What increase in price results in the same revenue as not increasing the price at all?


Lesson 5.6.3: Comparing Properties of Functions Given in Different Forms

IntroductionRelationships can be presented through different forms: tables, equations, graphs, and words. This section explores how to compare functions presented in all four forms.

Key Concepts

• Recall that for a quadratic function in standard form, f(x) = ax2 + bx + c, the value of a determines whether the quadratic has a maximum or a minimum.

• If a is negative, the quadratic function will have a maximum.

• If a is positive, the quadratic function will have a minimum.

• Recall that linear functions have a constant slope.

• When x increases by 1, y increases by a constant value.

• A linear function will either increase, decrease, or remain constant.

• The change in y when x increases by 1 is called a first difference.

• Recall that exponential functions either increase or decrease.

• There is a constant multiple in the rate of change between the y-values.

• A growing exponential function will eventually exceed a linear or quadratic function.

• A quadratic function changes from increasing to decreasing, or vice versa.

• There is neither a constant rate of change nor a constant multiple in a quadratic function.

• In a quadratic model, the change in the first differences is constant.

• The change in first differences is called a second difference.

• Compare the y-values of the vertices of quadratic functions to determine the greatest maximum or least minimum values between two or more quadratic functions.


Example 1

Through comparison shopping, you have obtained the yearly insurance rate quotes from three car insurance companies. The insurance rate quote is given as a function of the age of the customer. Using the information presented below, which company has the lowest rate for the youngest drivers?

• Wreck-for-Less Insurance: R(x) = 0.87x2 – 91.48x + 3185.16, where R(x) is the yearly premium in dollars and x is the age in years.

• Careless Insurance: A 51-year-old customer pays a yearly premium of $900. For every 5-year change in age, younger or older, the customer is charged an additional $300, then $500, then $700, and so on.

• Fender and Bender Insurance:

Age of customer Under 24 25–29 30–39 40–49 50–59 Over 59

Insurance rate ($) 1,750 1,500 1,125 916 875 916

1. Verify that the vertex for each insurance company rate is the minimum value.

For Wreck-for-Less, the value of a is positive, implying that it is an upward-opening parabola.

The rates for Careless Insurance increase as the customer’s age moves farther away from age 51, so the value at age 51 must be a minimum value. Therefore, the parabola opens upward.

The table for Fender and Bender Insurance decreases then increases, so it is also an upward-opening parabola.



2. Find the vertex for each insurance company.

Find the vertex of the given Wreck-for-Less function using b

af

b

a2,

2

− −

.

xb

a2=−

Formula to find the x-value of the vertex

x( 91.48)

2(0.87)=− −

Substitute 0.87 for a and –91.48 for b.

x =91 48

1 74

.

. Simplify.

x ≈ 52.57

Use the given function to find fb

a2

−

.

R(x) = 0.87x2 – 91.48x + 3185.16 Original equation

R(52.57) = 0.87(52.57)2 – 91.48(52.57) + 3185.16Substitute 52.57 for x.

R(52.57) ≈ 780.39 Simplify.

The y-value of the vertex is about 780.39.

The vertex for Wreck-for-Less is approximately (52.57, 780.39).

The table for Fender and Bender shows that the vertex is found within the 50–59 age group at a rate of $875.

The minimum value for Careless Insurance is (51, 900), since the rate increases as the age moves away from 51 in either direction.

3. Determine the company with the cheapest rate.

Compare the y-values of each vertex.

780.39 < 875 < 900

The minimum premium for Wreck-for-Less is the cheapest, at about $780.39.


Example 2

Three students are shooting wads of paper with a rubber band, aiming for a trash can in the front of the room. The height of each student’s paper wad in feet is given as a function of the time in seconds. Which student’s paper wad flies the highest?

• The path of Alejandro’s paper wad is modeled by the equation f(x) = –x2 + 2x + 7.

• Melissa’s paper wad is estimated to reach the heights shown in the table below.

x 0 2 3 4y 3 6 7 6

• After 3 seconds, Connor’s paper wad achieves a maximum height of 6.5 feet above the floor.

1. Determine if each function represents a quadratic.

Alejandro’s path is represented by a quadratic function written in standard form.

Melissa’s table has symmetry about x = 3, which implies a quadratic relationship.

In general, a projectile follows a parabolic path, gaining height until reaching a maximum, then descending. We can assume Connor’s paper wad will follow a parabolic path.

2. Verify that the extremum for each function is the maximum value of each function.

The value of a in the equation for Alejandro’s paper wad is a negative value. Therefore, the function has a maximum value.

The table for Melissa’s paper wad increases then decreases, so its path will have a maximum value.

The maximum height of the path of Connor’s paper wad is given.


3. Determine the vertex of each function.

Find the vertex of Alejandro’s function using b

af

b

a2,

2

− −

.

x = b

a2

−

Formula to find the x-coordinate of the vertex

x = (2)

2( 1)

−−

Substitute –1 for a and 2 for b.

x = 1 Simplify.

Use the given function to find fb

a2

−

= f(1).

f(x) = –x2 + 2x + 7 Original equation

f(1) = –(1)2 + 2(1) + 7 Substitute 1 for x.

f(1) = –1 + 2 + 7 Simplify.

f(1) = 8

The vertex is at (1, 8).

Find the vertex for Melissa’s paper wad using symmetry. Her paper wad’s function is symmetric about the vertex at (3, 7).

The vertex for Connor’s paper wad is given as (3, 6.5).

4. Use the vertices to determine whose paper wad goes the highest.

Compare the y-values of the vertices.

8 > 7 > 6.5

Alejandro’s paper wad flies the highest, at 8 feet.


Example 3

Which of the following quadratic functions has a vertex with a larger y-value: f(x) = 2x2 – 12x + 25, or g(x) as presented in the table?

x –4 –3 –2 0 2

g(x) 7 8 7 –1 –17

1. Verify for each function whether the vertex is a minimum or maximum.

For f(x), a is positive, which means that the vertex is a minimum.

For g(x), as x increases, the y-values first increase and then decrease. This indicates the vertex is a maximum.

2. Find the vertex for each function.

For f(x), complete the square of f(x) = 2x2 – 12x + 25.

f(x) = 2(x2 – 6x) + 25Factor 2 out of each term containing an x.

f(x) = 2(x2 – 6x + 9) + 25 – 2(9) Add and subtract 2(9).

f(x) = 2(x – 3)2 + 25 – 2(9)Rewrite the perfect square trinomial as a squared binomial.

f(x) = 2(x – 3)2 + 7 Simplify.

From this form, we can see that the vertex of f(x) is (3, 7).

For g(x), the points are symmetric about x = –3. Therefore, the line of symmetry contains (–3, 8), which must be the vertex.

3. Determine which vertex has a larger y-value.

Compare the y-values of each function.

7 < 8

The function g(x) has a vertex with a larger y-value.


Example 4

You are considering investing $5,000 in one of two mutual funds. The first fund will pay $500 each year. The second fund is predicted to have end-of-year balances as shown in the following table. Which fund should you choose if you want to withdraw your money after 5 years? Which fund should you choose if you want to invest the money for 10 years?

x (year) 0 1 2 3 4 5I(x) ($) 5,000 5,200 5,500 5,900 6,400 7,000

1. Determine if the first fund is quadratic.

The first fund has a constant rate of change of $500 each year; therefore, it is linear.

2. Determine if the second fund is quadratic.

There is no constant rate of change; therefore, the function is not linear.

In the first year, the fund increases by $100, then $200, then $300, and so on.

There is no constant multiple from one end-of-year balance to the next.

This means it is not exponential.

If we find the second differences by subtracting $100 from $200, $200 from $300, and so on, we find that the second differences are constant.

Therefore, the function is quadratic.

3. Determine which fund you should choose if you want to withdraw your money after 5 years.

At the end of the fifth year, the first fund would contain $5000 + $500(5) = $7500.

The second fund contains only $7,000 after 5 years.

If you want to withdraw your money after 5 years, you should choose the first fund.


4. Determine which fund should you choose if you want to invest the money for 10 years.

The first fund increases by $500 each year, but the second fund increases by more and more each year.

At the end of the fifth year, the second account holds only $500 less than the first fund, but the balance will increase by $600 to exceed the first fund by the end of the sixth year.

Based on this information, the second fund would have more money after 10 years. This can be verified. At the end of the tenth year, the first fund would contain $5000 + $500(10) = $10,000, and the second fund would continue to increase following the same pattern and contain $11,500.

Example 5

Suppose that you have been offered a position at a prestigious company. You may choose how your salary is paid. Option 1 is described by the quadratic equation S(x) = 2500x2 + 2500x + 60,000, where x is the number of years you are with the company and S(x) is the yearly salary in dollars. Option 2 has a starting yearly salary of $35,000, but you will get a 25% raise each year. Make a table of values for each salary and graph both functions on a coordinate plane. If you plan to work for this company for 5 years, which option should you choose? If you plan to work for this company until you retire at age 70, which option should you choose?

1. Write an equation for Option 2.

Since you have been given the equation for Option 1 already, start by determining the equation for Option 2. Option 2 is an exponential function because it increases by the same factor (125%) each year. We can write the function by multiplying the initial salary by 1.25 raised to the number of years.

T(x) = 35,000(1.25)x


2. Make a table of values for each salary.

Salaries listed in the table that follows are in thousands of dollars.

Year 1 2 3 4 5 6 7 8 9 10 11Option 1 65 75 90 110 135 165 200 240 285 335 390Option 2 43.8 54.6 68.4 85.4 106.8 133.5 166.9 208.6 260.8 326.0 407.5

3. Graph both functions on the same coordinate plane.

Plot the points from the table and connect them with a smooth curve. Label each function clearly.

Years of employment

20 4 6 8 10 12 14 16 18

Sala

ry in

dol

lars

50,000

100,000

150,000

200,000

250,000

300,000

350,000

400,000

450,000

500,000

550,000

y = 35,000

y = 60,000

Option 1

Option 2

(10.4, 356,410)

y

x

4. Determine which option you should choose if you plan to work for this company for only 5 years.

The table of values shows us that Option 1 yields a salary of $135,000 in the fifth year, compared to $106,800 for Option 2.

If you plan to work for this company for only 5 years, you should select Option 1.


5. Determine which option you should choose if you plan to work for this company until retirement at age 70.

We can see from the graph that although Option 2 pays less for the first 10 years, from year 11 on it is the better choice.

Exponential functions eventually exceed quadratic functions.

If you plan to work for this company until retirement, you should select Option 2.


PRACTICE


Use the following information to solve problems 1–3.

Three parts suppliers have different rates for bulk purchases. You need to buy bolts from one of the suppliers. The prices at Nuts and Bolts Company tend to follow the equation P(x) = 0.00015x2 + 0.005x. At Loose Screws, customers pay $10.00 for the first 100 pieces, then $0.05 for each piece over 100. The prices for Nuts for You are shown in the following table.

Number of bolts 100 200 300 400

Price ($) 20 24 26 27

1. Write an expression for the revenue generated at Loose Screws.

2. If you would like to buy 200 bolts, which company would offer the best deal? Why?

3. If you would like to buy 600 bolts, which company would offer the best deal? Why?

Use the following information to solve problems 4–7.

Three turtles are running a race. They are free to roam in any direction. The location of the first turtle, Agatha, can be given by the equation A(t) = 2(t – 2)2 – 9, where A(t) is the distance in feet from the starting line and t is the number of seconds since the race started. The location of the second turtle, Bertha, is given in the table that follows. The location of the third turtle, Charlotte, is given by the equation C(t) = 3t2 – 12t – 6.

t 1 2 3 4 5B(t) –15 –18 –15 –6 9

4. Which turtle is winning the race at t = 3?

continued

Practice 5.6.3: Comparing Properties of Functions Given in Different Forms


PRACTICE


5. Which turtle is winning the race at t = 6?

6. Is there a point at which any of the turtles are tied?

7. If the finish line were at 40 feet, which turtle would you predict to win? Why?

Use the given information to answer questions 8–10.

8. Which of the following parabolas has the vertex with the greater y-value: a parabola with two x-intercepts with a > 0, or a parabola with two x-intercepts with a < 0?

9. Which function would achieve a lower minimum value: a parabola with no x-intercepts and a > 0, or a parabola with two x-intercepts and a > 0?

10. One city’s population can be modeled by the quadratic equation p(x) = 2x2 + 34x + 120, where x is the number of years after the year 2000 and p is the population. A second city’s population starts at 10,000 people but grows by 10% each year and is given by the equation f(x) = 10(1.1)x. Each equation yields the population in thousands and x is the number of years since 2000. In what year does the population of the second city exceed the population of the first city?

Unit 5: Quadratic FunctionsU5-204

Lesson 7: Building FunctionsUNIT 5 • QUADRATIC FUNCTIONS


MCC9–12.F.BF.1a★

MCC9–12.F.BF.1b★

WORDS TO KNOW

concavity with respect to a curve, the property of being arched upward or downward. A quadratic with positive concavity will increase on either side of the vertex, meaning that the vertex is the minimum or lowest point of the curve. A quadratic with negative concavity will decrease on either side of the vertex, meaning that the vertex is the maximum or highest point of the curve.

curve the graphical representation of the solution set for y = f(x). In the special case of a linear equation, the curve will be a line.

function a relation in which every element of the domain is paired with exactly one element of the range; that is, for every value of x, there is exactly one value of y.

leading coefficient the coefficient of the term with the highest power


Essential Questions

1. What is the connection between the vertex of a parabola and maximum/minimum values?

2. What is the connection between the x2 coefficient and a maximum or minimum?

3. What is concavity?

4. How can mathematical expressions be used to describe real-world problems?

U5-205Lesson 7: Building Functions

Recommended Resources• Khan Academy. “Quadratics.”


This tutorial offers lessons and videos showing how to factor quadratic equations.

• LivePhysics.com. “Quadratic Equation Graph.”


This parabolic shape simulator allows the user to manipulate the values of a, b, and c in the standard form of a quadratic, y = ax2 + bx + c, to see an interactive animation of the resulting curve.

• WolframAlpha. “Mathematical Functions.”


Users can choose from several different types of functions and input values. The site will show how the values look as an equation, provide results of the equation, and graph it.


IntroductionA function is a relation in which every element of the domain is paired with exactly one element of the range; that is, for every value of x, there is exactly one value of y. When writing functions based on real-world information, first we must carefully analyze the words describing the situation. For example, if we say an unknown length is 2 more than twice another length, we must be able to translate the given information into an algebraic expression. In this case there are two lengths, with one length dependent on the other. If we let x represent the second length, then the first length is 2 more (+) than twice (2 ×) x, or 2 + 2x.

In this lesson, the problems will involve different contexts that quadratic equations describe. Notice that every problem will either require you to identify the two distinct linear expressions whose product will build the quadratic, or to identify the vertex of the parabola along with one other point on the parabola (which will allow you to build the quadratic in the vertex form of a quadratic equation).

Key Concepts

• A quadratic expression is the product of two linear factors.

• The vertex of a parabola represents either the maximum or minimum y-value for the equation.

• The vertex of a parabola can be found using 2

,2

− −

b

af

b

a from the general

form of the quadratic equation, where f(x) = ax2 + bx + c.

• The vertex of a parabola can be found using (h, k) in vertex form, where f(x) = a(x – h)2 + k.

• The curve of a parabola is the graphical representation of the solution set for the equation of the parabola.

• A quadratic equation can be built using the vertex and any other point on the parabola.

• Concavity refers to the direction the parabola faces.

• The vertex of an upward-facing (or concave up) parabola will be at the bottom or minimum of the curve.

• Conversely, the vertex of a downward-facing (or concave down) parabola will be at the top or maximum of the parabola.

Lesson 5.7.1: Building Functions from Context


Concave Up Concave Down

The vertex is a minimum. The vertex is a maximum.

• The leading coefficient is the coefficient of the term with the highest power.

• In a quadratic equation, the leading coefficient is the number that is being multiplied by the x2 term.

• The leading coefficient of the equation of a parabola determines its concavity.

• If the leading coefficient is positive, the parabola is concave up and the graph has a minimum.

• If the leading coefficient is negative, the parabola is concave down and the graph has a maximum.


Example 1

A farmer is building a rectangular pen using 100 feet of electric fencing and the side of a barn. In addition to fencing, there will be a 4-foot gate also requiring the electric fencing on either side of the pen. The farmer wants to maximize the area of the pen. How long should he make each side of the fence in order to create the maximum area?

1. Write the expressions that describe the length of each side of the pen.

Starting with one of the sides that is perpendicular to the barn, we can say the length is an unknown amount, x, plus the 4-foot gate, or x + 4.

The length of the second side of the fence that is perpendicular to the barn will be the same length as the first.

The side that is parallel to the barn is whatever amount of fence is left over after creating the two perpendicular sides.

The total amount of fencing is 100 feet, and there are two sides of length (x + 4), so the length of the side that is parallel to the barn is 100 – 2(x + 4).

Simplifying the expression, we get 100 – 2x – 8, or 92 – 2x.

2. Build the equation that describes the area of the pen.

Remember that area equals length times width, or A = l • w.

Let l = x + 4 and w = 92 – 2x.

A(x) = (x + 4)(92 – 2x) Substitute values for length and width.

A(x) = 92x – 2x2 + 368 – 8x Multiply.

A(x) = –2x2 + 84x + 368 Reorder and simplify.



3. To find the maximum area, use the vertex.

Note that the leading coefficient of the equation is negative. This means the graph of the equation will be a downward-facing parabola. Therefore, the vertex of the parabola will describe the greatest amount of area.

A(x) = –2x2 + 84x + 368

Given that the general form of a quadratic function is y = ax2 + bx + c, we can determine that a = –2, b = 84, and c = 368.

Find the x-coordinate of the vertex by substituting in a and b values

from the quadratic function into the expression −ba2

:

−=−−

=−−

=b

a2

84

2 2

84

421

( )

( )


Find the y-coordinate of the vertex by substituting the x-coordinate from the vertex into the quadratic function.

A(x) = –2x2 + 84x + 368 Quadratic function

A(21) = –2(21)2 + 84(21) + 368 Substitute 21 for x.

A(21) = –882 + 1764 + 368 Simplify, then solve.

A(21) = 1250

The maximum area of the pen is 1,250 ft2.

4. Finally, use the x-value from the vertex to find the lengths of each side of the pen.

x = 21

Each side that is perpendicular to the barn is equal to x + 4.

(x + 4) = (21 + 4) = 25 feet

The side that is parallel to the barn is equal to 92 – 2x.

(92 – 2x) = [92 – 2(21)] = 50 feet


Example 2

An amusement park has commissioned the design of a steel roller coaster with a drop section that is modeled by a parabola. Part of the roller coaster’s track will go through an underground tunnel. In this section, the roller coaster will dip 12 feet below ground level. The roller coaster will dip below ground level at a horizontal distance of 24 feet from the peak just before the drop and reemerge to ground level at a horizontal distance of 36 feet from the peak just before the drop. Find an equation of the parabola that describes the drop and the height of the roller coaster at the peak.

1. Derive the coordinates of the points on the parabola.

Let the x-axis represent ground level and the y-intercept represent the peak of this section of the roller coaster.

Since the horizontal distances are given as 24 feet and 36 feet, it can be determined that the roller coaster dips below ground level at (24, 0) and reemerges at (36, 0).

2. Establish the linear factors and the equation of the parabola.

From the x-intercepts, the linear factors of the equation are (x – 24) and (x – 36). Therefore, the equation of the parabola is y = a(x – 24)(x – 36).

3. Find the vertex of the parabola.

Expand the factored form of the equation of the parabola into the general form, y = ax2 + bx + c.

y = a(x – 24)(x – 36) Equation of the parabola

y = a(x2 – 36x – 24x + 864) Multiply the factors.

y = a(x2 – 60x + 864) Simplify.

y = ax2 – 60ax + 864a Distribute a.

The x-coordinate for the vertex of a quadratic equation is −ba2

.

Substitute the values from the quadratic function to determine the

x-coordinate.

(continued)


−=− −

= =b

a

a

a

a

a2

60

2

3030

( )

( )

The x-coordinate is 30.

Combine this information with the given minimum of the drop, 12 feet below ground level, to find the vertex of the equation: (30, –12).

4. Find the value of a.

Substitute the point (30, –12) into the factored form of the quadratic equation and solve for a.

y = a(x – 24)(x – 36) Factored form of the quadratic equation

–12 = a(30 – 24)(30 – 36) Substitute 30 and –12 for x and y.

–12 = a(6)(–6) Simplify, then solve for a.

–12 = –36a

a =1

3

After solving for a, the equation of the parabola is y x x= − −1

324 36( )( ).

5. Find the height of the peak of this section of the roller coaster.

The height of the peak is the roller coaster’s vertical distance above ground level. Since the peak occurs at the y-intercept, where the x-coordinate equals 0, set x = 0 in the quadratic equation and solve for y.

y

y

y

y

= − −

= − −

=

=

1

30 24 0 36

1

324 36

1

3864

288

( )( )

( )( )

( )

The height of the roller coaster’s peak just before the drop is 288 feet.


Example 3

A suspension bridge has two cables secured at either end of the span by two supporting towers. The cables are attached to the tops of the towers. In the section between the two towers, the cables form a parabolic curve. At their lowest point, the cables are 15 feet from the surface of the bridge. The towers are 400 feet apart, and the vertical distance from the surface of the bridge to the top of each tower is 415 feet. What is a quadratic equation that describes the curve of the cables between the towers?

y

x

1. Determine the vertex of the parabola.

Since the cables are attached to the tops of the towers, this means the cables start 415 feet up from the surface of the bridge. The cables dip down to a height of 15 feet above the bridge. If we allow the y-axis to represent the midpoint between the towers, then we can say the vertex of the parabola is (0, 15).


2. Derive the equation of the parabola.

Substitute the vertex (0, 15) into the general equation for the vertex form of a parabola, y = a(x – h)2 + k, so that h = 0 and k = 15.

y = a(x – h)2 + k General equation

y = a(x – 0)2 + 15 Substitute 0 for h and 15 for k.

y = ax2 + 15 Simplify.

The equation of the parabola formed by the bridge cables is y = ax2 + 15.

3. Find another point on the parabola.

To solve for a, we need any other point on the parabola to substitute into the equation for x and y. In this case, we have been given the height of the supporting towers (415 feet) and their distance apart (400 feet). Since we let the axis of symmetry of the parabola be the y-axis, we can determine where the top of each tower lies on the parabola.

The y-coordinate for the top of each tower is 415, the height.

To find the x-coordinate, divide the horizontal distance between the towers, 400 feet, by 2.

400

2200=

Since the y-axis is the axis of symmetry of the parabola, and each y-coordinate is positive, we can determine that one tower is in Quadrant I, (x, y). The other tower is in Quadrant II, (–x, y). Substitute the x- and y-values into these forms to determine where the tops of the towers are located.

Tower in Quadrant I: (x, y) = (200, 415)

Tower in Quadrant II: (–x, y) = (–200, 415)

The tops of the towers lie at (–200, 415) and (200, 415). Therefore, we know the points (–200, 415) and (200, 415) lie on the parabola.

(continued)


(–200, 415) (200, 415)

(0, 15)

y

x

4. Solve for a.

Take one of the known points that lies on the parabola and substitute the x- and y-values into the equation of the parabola. Either point will do. Let’s use (200, 415).

y = ax2 + 15 Equation of the parabola

415 = a(200)2 + 15 Substitute 200 for x and 415 for y.

415 = a(40,000) + 15 Simplify.

400 = a(40,000) Subtract 15 from both sides.

400

40 000,= a

Divide both sides by 40,000.

a =1

100 Simplify.


5. Complete the equation of the curvature of the cable.

Substitute 1

100 for a in the equation of the parabola.

y = ax2 + 15, where a =1

100

y x= +1

100152

UNIT 5 • QUADRATIC FUNCTIONSLesson 7: Building Functions

PRACTICE


Use your knowledge of quadratic functions to complete each problem that follows.

1. Expand the linear factors of f(x), where f(x) = (3x + 4)(x + 3).

2. Let g(x) = 2x2 – x + 18. What is g(–3)?

3. The product of two consecutive integers is 5,402. Build a function that can be used to solve for the integers. What are the two integers?

Use the following scenario to complete problems 4 and 5.

A suspension bridge has two supporting towers with a cable secured at either end of the span and then draped off the towers. In the section between the two supporting towers, the cable forms a parabolic curve. At the lowest point, the cable is 25 feet from the surface of the bridge. The supporting towers are 350 feet apart, and rise 300 feet from the surface of the bridge to the top of the tower. Use the height of the right tower as the y-intercept.

4. What are the coordinates of the vertex and y-intercept?

5. What is the complete equation in vertex form of the suspension cable?

continued

Practice 5.7.1: Building Functions from Context


PRACTICE



Sabrina is a graphic artist. She wants to create a border around a rectangular image such that the borders on either side are equal but the borders on the top and bottom are twice as thick as the side borders. The image is 300 pixels tall and 700 pixels wide.

6. If the width of the side borders is represented by x, what is the function for the area of the completed image, borders included?

7. If the width of the side border were determined to be 50 pixels, what would be the area of the complete image, borders included?


An amusement park is building a steel roller coaster with a section so steep that when the roller coaster descends the riders feel almost weightless. The section is modeled by a concave down parabola. In this section, the roller coaster will start its ascent at (0, 0), reach the peak at 40 feet above the starting point, and return to its starting height at a horizontal distance of 120 feet from the start.

8. What are the coordinates of the vertex and the x-intercepts?

9. What is the equation of the parabolic curve of the roller coaster?

Use the given information to write a function for problem 10.

10. An equestrian center is building a new paddock or exercise area using the side of the stable as one of the borders. The stable master has purchased 200 feet of electrical fencing. What is the function that models the area of a rectangular-shaped paddock?


IntroductionAs is true with linear and exponential functions, we can perform operations on quadratic functions. Such operations include addition, subtraction, multiplication, and division. This lesson will focus on adding, subtracting, multiplying, and dividing functions to create new functions. The lesson will also explore the effects of dividing a quadratic by one of its linear factors.

Key Concepts

Operations with Functions

• Functions can be added, subtracted, multiplied, and divided.

• For two functions f(x) and g(x), the addition of the functions is represented as follows: ( )( ) ( ) ( )f g x f x g x+ = + .

• For two functions f(x) and g(x), the subtraction of the functions is represented as follows: ( )( ) ( ) ( )f g x f x g x− = − .

• For two functions f(x) and g(x), the multiplication of the functions is represented as follows: f g x f x g x( )( ) ( ) ( )• = • .

• For two functions f(x) and g(x), the division of the functions is represented as

follows: f

gx

f x

g x

=( )( )

( ).

• Adding and subtracting linear expressions from a quadratic will yield a quadratic.

• Multiplying and dividing a quadratic by anything other than a constant will not yield a quadratic.

Restricted Domains

• When considering the division of a quadratic by a linear factor, it is possible to create a linear expression with a restricted domain. For example:

• For f(x) = x2 + 5x + 6 and g(x) = x + 3, f

gx

( ) can be found such that

f

gx

f x

g x

x x

x

x x

x

= =+ ++

=+ +

+=( )

( )

( )

( )( ) (2 5 6

3

2 3

3

xx x

xx

+ ++

= +2 3

32

) ( ).

• In simpler terms, f

gx x

= +( ) 2 .

Lesson 5.7.2: Operating on Functions


• Remember that the denominator of a fraction cannot equal 0.

• Set the denominator equal to 0 and solve for x to find the restricted value(s) in the domain: x + 3 = 0, so x ≠ –3.

• Given the similar function h(x) = x + 2, the domain is all real numbers, and

the range is the same. However, since f(x) is divided by g(x), the domain of f

gx x

= +( ) 2 from the preceding example is all real numbers except for x =

–3 and the range is all real numbers except for y = –1.

• This is because when the restricted value of the domain (–3) is substituted into

the simplified form of f

gx x

= +( ) 2 and solved for y, we get:

• f

g

− = − +( ) ( )3 3 2

• f

g

− = −( )3 1

• Therefore, since x ≠ –3, then y ≠ –1.



Let f(x) = x2 – 3x + 4 and g(x) = x2 + 6x – 3. Build a new function, h(x), for which h(x) = (f + g)(x).

1. Expand the new function, h(x), into a form where substitution can be used.

h(x) = (f + g)(x) = f(x) + g(x)

The new function is expanded as h(x) = f(x) + g(x).

2. Add the functions.

f(x) = x2 – 3x + 4 and g(x) = x2 + 6x – 3 Given functions from problem statement

h(x) = f(x) + g(x) Expanded notation

h(x) = (x2 – 3x + 4) + (x2 + 6x – 3) Substitute values for f(x) and g(x).

h(x) = 2x2 + 3x + 1 Combine like terms.

The new function is h(x) = 2x2 + 3x + 1.

Example 2

Let f(x) = 3x + 4 and g(x) = 5x – 2. Build a new function, h(x), for which h(x) = (f • g)(x).

1. Expand the new function, h(x), into a form where substitution can be used.

h(x) = (f • g)(x) = f(x) • g(x)

The new function is expanded as h(x) = f(x) • g(x).

2. Multiply the functions.

f(x) = 3x + 4 and g(x) = 5x – 2 Given functions from problem statement

h(x) = f(x) • g(x) Expanded notation

h(x) = (3x + 4)(5x – 2) Substitute values for f(x) and g(x).

h(x) = 15x2 – 6x + 20x – 8 Multiply.

h(x) = 15x2 + 14x – 8 Combine like terms.

The new function is h(x) = 15x2 + 14x – 8.


Example 3

For f(x) = 3x2 + 13x – 10 and g(x) = x + 5, find f

gx

( ) . What type of function is the

quotient of f

gx

( ) ? Are there restrictions on the domain and range of the function

f

gx

( ) ?

1. Since the functions are being divided, write the functions f(x) and g(x) as a fraction.

f

gx

f x

g x

x x

x

= =+ −+

( )( )

( )

3 13 10

5

2

2. Factor the quadratic function, f(x).

3 13 10

5

3 2 5

5

2x x

x

x x

x

+ −+

=− ++

( )( )

3. Simplify the equation and define the type of equation of the simplified form.

Divide away the monomial (x + 5) from the top and bottom of the fraction:

( )( ) ( ) ( )3 2 5

5

3 2 5

53 2

x x

x

x x

xx

− ++

=− +

+= −

The function f

gx

( ) is a linear equation that graphs the line y = 3x – 2.

4. Look at the original fraction to see if there are restricted values on the domain.

In this case, x ≠ –5 because (–5) + 5 = 0 and division by 0 is undefined. Next, substitute x = –5 into the final equation to determine the restricted value(s) of y.

3x – 2 = 3(–5) – 2 = –17

Since x ≠ –5, then y ≠ –17.


Example 4

Zane is a textiles designer. His latest project is to design a rectangular area rug for a hotel lobby. The dimensions of the lobby are such that one set of walls is twice the length of the other set of walls. The rug must lay centered in the lobby, with each edge of the rug exactly 3 feet from each wall. What is the function in terms of x that describes the area of the lobby? What is the function in terms of x that describes the area of the rug? What is the function that describes the area of the lobby left uncovered by the rug?

1. Define the function that describes the area of the lobby.

Using x for the unknown quantity, the dimensions of the lobby are x feet wide by 2x feet long.

Since area = length • width, the area of the lobby can be described by the function f(x).

f(x) = l • w

f(x) = (2x)(x)

f(x) = 2x2

2. Define the function that describes the area of the rug.

The dimensions of the rug are 3 less than the length of the lobby on either side, so the shorter side is x – 2(3) = x – 6 and the longer side is 2x – 2(3) = 2x – 6.

Therefore, the area of the rug can be described by substituting values for the length (the longer side) and width (the shorter side) into the function g(x).

g(x) = l • w

g(x) = (2x – 6)(x – 6)

g(x) = 2x2 – 18x + 36


3. Define the function that represents the area of the lobby left uncovered by the rug.

The area of the lobby left uncovered by the rug can be described by subtracting the area of the rug, g(x), from the area of the lobby, f(x).

(f – g)(x) = f(x) – g(x)

(f – g)(x) = (2x2) – (2x2 – 18x + 36)

(f – g)(x) = 18x – 36


PRACTICE


Use your knowledge of operations on functions to complete the problems that follow.

1. Let f(x) = x2 + 8x + 11 and g(x) = 2x + 14. Build a new function h(x), for which h(x) = (f + g)(x).

2. Let s(x) = x2 – 7x – 14 and t(x) = 2x – 26. Build a new function u(x), for which u(x) = (s – t)(x).

3. Let j(x) = x + 5 and k(x) = 7x – 5. Build a new function m(x), for which m(x) = (f • g)(x).

4. Let f(x) = x2 + 13x + 42 and g(x) = x + 6. Build a new function h(x), for which

h(x) = f

gx

. State any restrictions on the domain and range.

5. Let g(x) = –2x2 – 9x + 17 and h(x) = x2 – 3. Build a new function, k(x), for which k(x) = (h – g)(x).

continued

Practice 5.7.2: Operating on Functions


PRACTICE


6. Let r(x) = x2 + 3x – 10 and s(x) = x – 2. Build a new function t(x), for which

t xr

sx( )=

. State any restrictions on the domain and range.

7. A homeowner is having a rectangular area rug custom designed for a room. The dimensions of the room are such that one set of walls is 4 feet longer than the other set of walls. The rug must sit centered in the room with each edge 3 feet from each wall. What is the function in terms of x that describes the area of the room? What is the function in terms of x that describes the area of the rug? What is the function that describes the area of the room left uncovered by the rug?

8. The altitude of a triangular sign is 18 inches less than 3 times its base. Define the functions that describe the altitude and base of the triangle, and then use these functions to build an area function of the triangle.

9. A zoo increased the lengths of both sides of its monkey park by the same amount. As a result, the monkey park is 200 feet by 300 feet. Define a function for each side length of the original park before it was enlarged and use those functions to build an area function.

10. The surface area of a cone is found by adding the area of the cone’s base to the cone’s lateral surface area. The base of a cone is a circle with an area of πr2, where r is the radius of the circle. The lateral surface of a cone is given by the equation πrs, where s is the slant height, or the distance from top of the cone to the edge of the base. Define a function f(r) that describes the area of the base of a cone. Define a function g(r) that describes the lateral surface area of a cone. Then simplify the function (f + g)(r).

U5-226

Lesson 8: Transforming FunctionsUNIT 5 • QUADRATIC FUNCTIONS



MCC9–12.F.BF.3

WORDS TO KNOW

horizontal compression squeezing of the parabola toward the y-axis

horizontal stretch pulling of the parabola and stretching it away from the y-axis

transformation adding or multiplying a constant to a function that changes the function’s position and/or shape

translation transforming a function where the shape and size of the function remain the same but the function moves horizontally and/or vertically; adding a constant to the independent or dependent variable

vertical compression squeezing of the parabola toward the x-axis

vertical stretch pulling of the parabola and stretching it away from the x-axis

Essential Questions

1. How is a function f(x) affected when multiplied by a constant k?

2. How is a function f(x) affected when x is multiplied by a constant k?

3. What are the effects of k on the graph of f(x + k) compared to f(x)?

4. What are the effects of k on the graph of f(x) + k compared to f(x)?

U5-227Lesson 8: Transforming Functions

Recommended Resources• MathIsFun.com. “Function Transformations.”


This site offers a simple, illustrated guide to how changing values in a function transforms the related graph. At the bottom of the page, users will find interactive review questions.

• Michael Mays and Laura Pyzdrowski. “Quadratic Functions Applet.”


This interactive graph allows the user to manipulate a, b, and c in y = ax2 + bx + c and then view both the resulting graph and a table of values. The applet requires Java to run.

• Seeing Math. “Quadratic Transformer Applet.”


This applet allows users to flip, skew, and otherwise transform a graph, and then view the equation of their graph in polynomial, vertex, or root form. The “Transform Function” button lets users create a new function by transforming an existing function. The applet requires Java to run.


IntroductionYou can change a function’s position or shape by adding or multiplying a constant to that function. This is called a transformation. When adding a constant, you can transform a function in two distinct ways. The first is a transformation on the independent variable of the function; that is, given a function f(x), we add some constant k to x: f(x) becomes f(x + k). The second is a transformation on the dependent variable; given a function f(x), we add some constant k to f(x): f(x) becomes f(x) + k.

In this lesson, we consider the transformation on a function by a constant k, either when k is added to the independent variable, x, or when k is added to the dependent variable, f(x). Given f(x) and a constant k, we will observe the transformations f(x) + k and f(x + k), and examine how transformations affect the vertex of a quadratic equation.

Key Concepts

• To determine the effects of the constant on a graph, compare the vertex of the original function to the vertex of the transformed function.

• Neither f(x + k) nor f(x) + k will change the shape of the function so long as k is a constant.

• Transformations that do not change the shape or size of the function but move it horizontally and/or vertically are called translations.

• Translations are performed by adding a constant to the independent or dependent variable.

Vertical Translations—Adding a Constant to the Dependent Variable, f(x) + k

• f(x) + k moves the graph of the function k units up or down depending on whether k is greater than or less than 0.

• If k is positive in f(x) + k, the graph of the function will be moved up.

• If k is negative in f(x) + k, the graph of the function will be moved down.

Lesson 5.8.1: Replacing f (x ) with f (x ) + k and f (x + k)


Vertical translations: f(x) + k

When k is positive, k > 0, the graph moves up:

f(x) + k

f(x)

When k is negative, k < 0, the graph moves down:

f(x) + k

f(x)

Horizontal Translations—Adding a Constant to the Independent Variable, f(x + k)

• f(x + k) moves the graph of the function k units to the right or left depending on whether k is greater than or less than 0.

• If k is positive in f(x + k), the function will be moved to the left.

• If k is negative in f(x + k), the function will be moved to the right.

Horizontal translations: f(x + k)

When k is positive, k > 0, the graph moves left:

f(x + k) f(x)

When k is negative, k < 0, the graph moves right:

f(x + k)f(x)


Example 1

Consider the function f(x) = x2 and the constant k = 2. What is f(x) + k? How are the graphs of f(x) and f(x) + k different?

1. Substitute the value of k into the function.

If f(x) = x2 and k = 2, then f(x) + k = x2 + 2.

2. Use a table of values to graph the functions on the same coordinate plane.

x f (x) f (x) + 2–2 4 6–1 1 30 0 21 1 32 4 6

-10 -9 -8 -7 -6 -5 -4 -3 -2 -1 0 1 2 3 4 5 6 7 8 9 10

-10-9-8-7-6-5-4-3-2-1

123456789

10 f(x) + 2

f(x)



3. Compare the graphs of the functions.

Notice the shape and horizontal position of the two graphs are the same. The only difference between the two graphs is that the value of f(x) + 2 is 2 more than f(x) for all values of x. In other words, the transformed graph is 2 units up from the original graph.

Example 2

Consider the function f(x) = x2 and the constant k = –3. What is f(x) + k? How are the graphs of f(x) and f(x) + k different?


If f(x) = x2 and k = –3, then f(x) + k = x2 – 3.

2. Use a table of values to graph the two functions on the same coordinate plane.

x f (x) f (x) – 3–4 16 13–2 4 10 0 –32 4 14 16 13

(continued)


-10 -9 -8 -7 -6 -5 -4 -3 -2 -1 0 1 2 3 4 5 6 7 8 9 10

-10-9-8-7-6-5-4-3-2-1

123456789

10

f(x) – 3

f(x)


Notice both the shape and horizontal position of the two graphs are the same. The only difference between the two graphs is that the value of f(x) – 3 is 3 less than f(x) for all values of x.

The easiest point to analyze on the graphs is the vertex. For f(x), the vertex occurs at (0, 0). For the graph of f(x) – 3, the vertex occurs at (0, –3). The transformed graph has moved down 3 units.


Example 3

Consider the function f(x) = x2, its graph, and the constant k = 4. What is f(x + k)? How are the graphs of f(x) and f(x + k) different?


If f(x) = x2 and k = 4, then f(x + k) = f(x + 4) = (x + 4)2.

2. Use a table of values to graph the two functions on the same coordinate plane.

x f (x) f (x + 4)–6 36 4–4 16 0–2 4 40 0 162 4 364 16 64

-10 -9 -8 -7 -6 -5 -4 -3 -2 -1 0 1 2 3 4 5 6 7 8 9 10

-10-9-8-7-6-5-4-3-2-1

123456789

10f(x + 4) f(x)



Notice the shape and vertical position of the two graphs are the same. The only difference between the two graphs is that every point on the curve of f(x) has been shifted 4 units to the left in the graph of f(x + 4).

Example 4

Consider the function f(x) = x2 and the constant k = –1. What is f(x + k)? How are the graphs of f(x) and f(x + k) different?


If f(x) = x2 and k = –1, then f(x + k) = f(x – 1) = (x – 1)2.

2. Using vertex or standard form, graph the two functions on the same coordinate plane.

Recall from Example 2 that the vertex of the parent function, f(x) = x2, is (0, 0).

The vertex of the transformed function, f(x – 1) = (x – 1)2, is (1, 0). This

can be verified by using − −

b

af

b

a2 2, .

To use this formula, expand the function so that it’s in the form of a quadratic, then find a, b, and c.

f(x – 1) = (x – 1)2 = x2 – 2x + 1

a = 1, b = –2, and c = 1

The x-coordinate of the vertex is given by 2

( 2)

2(1)1=

−=− −

=xb

a.

Substitute the x-value of the vertex into the function.

f(x – 1) = (x – 1)2

f(1) = [(1) – 1]2

f(1) = 0

Therefore, the vertex is (1, 0).(continued)


Graph the functions.

-5 -4 -3 -2 -1 0 1 2 3 4 5

-10-9-8-7-6-5-4-3-2-1

123456789

10

f(x – 1)f(x)


Notice both the shape and vertical position of the two graphs are the same. The only difference between the two graphs is that every point on the curve of f(x) has been shifted 1 unit to the right in the graph of f(x – 1).


Example 5

The revenue function for a model helicopter company is modeled by the curve f(x) = –5x2 + 400x, where x is the number of helicopters built per month and f(x) is the revenue. The owner wants to include rent in the revenue equation to determine the company’s profit per month. The company pays $2,250 per month to rent its warehouse. In terms of f(x), what equation now describes the company’s profit per month? Compare the vertices of the original function and the transformed function.

1. Build the new function.

The rent is being subtracted from the entire function. Therefore, the function follows the format f(x) + k, where k is negative.

The company makes f(x) = –5x2 + 400x dollars per month. The company also pays $2,250 dollars per month in rent. Therefore, the new function that describes the company’s profit, P(x), is as follows:

P(x) = f(x) – rent = f(x) – 2250 = –5x2 + 400x – 2250

2. Determine the vertex of f(x).

The vertex of f(x) has the x-coordinate −

=−

−=−−

=b

a2

400

2 5

400

1040

( )

( ).

Evaluate f(x) at the x-coordinate of the vertex to find the y-coordinate.

f(x) = –5x2 + 400x

f(40) = –5(40)2 + 400(40)

f(40) = –8000 + 16,000

f(40) = 8000

The vertex of f(x) is (40, 8000).


3. Determine the vertex of P(x).

The vertex of P(x) has the same x-coordinate as f(x) since the values of a and b are the same. The x-coordinate of the vertex of P(x) is 40.

Evaluate P(x) at the x-coordinate of the vertex to find the y-coordinate.

P(x) = –5x2 + 400x – 2250

P(40) = –5(40)2 + 400(40) – 2250

P(40) = 8000 – 2250

P(40) = 5750

The vertex of P(x) is (40, 5750).

4. Compare the vertices.

The vertex of P(x) is 2,250 units lower than the vertex of f(x). The model was shifted down 2,250 units.

UNIT 5 • QUADRATIC FUNCTIONSLesson 8: Transforming Functions

PRACTICE


For problems 1–3, let f(x) = x2. Write a function that translates f as described.

1. 3 units to the right

2. 4 units down

3. 6 units to the left and 1 unit down

For problems 4–6, let f(x) = x2. Graph g(x) by translating the graph of f.

4. g(x) = (x + 4)2

5. g(x) = x2 + 1

6. g(x) = (x – 2)2 – 5

Use what you know about translations of functions to solve each problem.

7. The graph shown below is a translation of f(x) = –x2. Write an equation for the graph. State the value of k that was used to transform the function horizontally and the value of k used to transform the function vertically.

-9 -8 -7 -6 -5 -4 -3 -2 -1 0 1 2 3 4 5 6 7 8 9

-10-9-8-7-6-5-4-3-2-1

123456789

10

continued

Practice 5.8.1: Replacing f (x ) with f (x ) + k and f (x + k)


PRACTICE


8. A mother and her son went golfing. The mother hit first. Her ball followed the path modeled by the equation f(x) = –0.0008x2 + 0.24x in the direction of the hole, and landed 12 yards short of the hole. The son teed off 12 yards closer to the hole because he is a beginner. He realized that if he could hit the ball on the same trajectory as his mother, his ball would land right by the hole. What is the equation that describes the path that the son’s ball should follow?

9. A paper wad is thrown from a height of 4 feet so that its path is modeled by the function f(x) = –0.05x2 + x + 4. If the exact same shot is taken from a balcony that is 15 feet above where the original shooter was standing, how far away will the paper wad hit the ground? What is the equation that models this shot?

10. Suzanne has a toy that launches rubber bands. The rubber bands always follow a path modeled by the function f(x) = –0.4(x – 5)2 + 10 when the launcher is at the “origin.” If the launcher is lifted up 3 feet and moved forward 4 feet, will a launched rubber band hit a painted target on a 10-foot-tall tree branch that is 12 feet from the origin? What is the function that models this new launcher position?


IntroductionTransformations can be made to functions in two distinct ways: by transforming the core variable of the function (multiplying the independent variable by k), and by transforming the function as a whole (multiplying the dependent variable by k). Previously, we saw how adding some constant k to the variable of a function or to the entire function affected the graph of the function.

In this lesson, we will see how multiplying by a constant affects the graph of a function. Given f(x) and a constant k, we will observe the transformations f(k • x) and k • f(x).

Key Concepts

Graphing and Points of Interest

• In the graph of a function, there are key points of interest that define the graph and represent the characteristics of the function.

• When a function is transformed, the key points of the graph define the transformation.

• The key points in the graph of a quadratic equation are the vertex and the roots, or x-intercepts.

Multiplying the Dependent Variable by a Constant, k: k • f(x)

• In general, multiplying a function by a constant will stretch or shrink (compress) the graph of f vertically.

• If k > 1, the graph of f(x) will stretch vertically by a factor of k (so the parabola will appear narrower).

• A vertical stretch pulls the parabola and stretches it away from the x-axis.

• If 0 < k < 1, the graph of f(x) will shrink or compress vertically by a factor of k (so the parabola will appear wider).

• A vertical compression squeezes the parabola toward the x-axis.

• If k < 0, the parabola will be first stretched or compressed and then reflected over the x-axis.

• The x-intercepts (roots) will remain the same, as will the x-coordinate of the vertex (the axis of symmetry).

• While k • f(x) = f(k • x) can be true, generally k • f(x) ≠ f(k • x).

Lesson 5.8.2: Replacing f (x ) with k • f (x ) and f (k • x )


Vertical stretches: when k > 1 in k • f(x)

k • f(x)f(x)

• The graph is stretched vertically by a factor of k.

• The x-coordinate of the vertex remains the same.

• The y-coordinate of the vertex changes.

• The x-intercepts remain the same.

Vertical compressions: when 0 < k < 1 in k • f(x)

k • f(x) f(x)

• The graph is compressed vertically by a factor of k.





Reflections over the x-axis: when k = –1 in k • f(x)

k • f(x)

f(x)

• The parabola is reflected over the x-axis.




• When k < 0, first perform the vertical stretch or compression, and then reflect the function over the x-axis.

Multiplying the Independent Variable by a Constant, k: f(k • x)

• In general, multiplying the independent variable in a function by a constant will stretch or shrink the graph of f horizontally.

• If k > 1, the graph of f(x) will shrink or compress horizontally by a factor of 1

k

(so the parabola will appear narrower).

• A horizontal compression squeezes the parabola toward the y-axis.

• If 0 < k < 1, the graph of f(x) will stretch horizontally by a factor of 1

k (so the

parabola will appear wider).

• A horizontal stretch pulls the parabola and stretches it away from the y-axis.

• If k < 0, the graph is first horizontally stretched or compressed and then reflected over the y-axis.

• The y-intercept remains the same, as does the y-coordinate of the vertex.

• When a constant k is multiplied by the variable x of a function f(x), the interval of the intercepts of the function is increased or decreased depending on the value of k.

• The roots of the equation ax2 + bx + c = 0 are given by the quadratic formula,

xb b ac

a=− ± −2 4

2.

• Remember that in the standard form of an equation, ax2 + bx + c, the only variable is x; a, b, and c represent constants.


• If we were to multiply x in the equation ax2 + bx + c by a constant k, we would arrive at the following:

a kx b kx c ak x bk x c( ) ( )2 2 2+ + =( ) +( ) +

• Use the quadratic formula to find the roots of ak x bk x c2 2( ) +( ) + :

xb b ac

a=− ± −2 4

2Quadratic equation

xbk bk k a c

k a

bk b k ack

ak=− ± ( ) − ( )

( ) =− ± −

2 2

2

2 2 2

2

4

2

4

2

Substitute ak2 for a and bk for b based on the equation in standard form.

xbk k b ac

ak

bk k b ac

ak=− ± −( )

=− ± −2 2

2

2

2

4

2

4

2Simplify.

xb b ac

ak

b b ac

a k=− ± −

=− ± −

•2 24

2

4

2

1

Horizontal compressions: when k > 1 in f(k • x)

f(k • x)f(x)

• The graph is compressed

horizontally by a factor of 1

k.

• The y-coordinate of the vertex remains the same.

• The x-coordinate of the vertex changes.


Horizontal stretches: when 0 < k < 1 in f(k • x)

f(k • x)

f(x)

• The graph is stretched horizontally

by a factor of 1

k.



Reflections over the y-axis: when k = –1 in f(k • x)

f(k • x) f(x)

• The parabola is reflected over the y-axis.



• The y-intercept remains the same.

• When k < 0, first perform the horizontal compression or stretch, and then reflect the function over the y-axis.


Example 1

Consider the function f(x) = x2, its graph, and the constant k = 2. What is k • f(x)? How are the graphs of f(x) and k • f(x) different? How are they the same?


If f(x) = x2 and k = 2, then k • f(x) = 2 • f(x) = 2x2.

2. Use a table of values to graph the functions.

x f (x) k • f (x)–2 4 8–1 1 20 0 01 1 22 4 8

-10 -9 -8 -7 -6 -5 -4 -3 -2 -1 0 1 2 3 4 5 6 7 8 9 10

-10-9-8-7-6-5-4-3-2-1

123456789

10f(x) k • f(x)



3. Compare the graphs.

Notice the position of the vertex has not changed in the transformation of f(x). Therefore, both equations have same root, x = 0. However, notice the inner graph, 2x2, is more narrow than x2 because the value of 2 • f(x) is increasing twice as fast as the value of f(x). Since k > 1, the graph of f(x) will stretch vertically by a factor of 2. The parabola appears narrower.

Example 2

Consider the function f(x) = x2 – 81, its graph, and the constant k = 3. What is f(k • x)? How do the vertices and the zeros of f(x) and f(k • x) compare?


If f(x) = x2 – 81 and k = 3, then f(k • x) = f(3x) = (3x)2 – 81 = 9x2 – 81.

2. Use the zeros and the vertex of f(x) to graph the function.

To find the zeros of f(x), set the function equal to 0 and factor.

x2 – 81 = 0 Set the function equal to 0.

(x + 9)(x – 9) = 0 Factor using the difference of two squares.

x + 9 = 0 or x – 9 = 0 Use the Zero Product Property.

x = –9 or x = 9 Solve for x.

The zeros are –9 and 9.

(continued)


The vertex of f(x) is (0, –81). This can be seen as the translation of the parent function f(x) = x2. The parent function is translated down 81 units.

-10 -9 -8 -7 -6 -5 -4 -3 -2 -1 0 1 2 3 4 5 6 7 8 9 10

-80

-70

-60

-50

-40

-30

-20

-10

10

20

30

40

f(x)

3. Using the zeros and the vertex of the transformed function, graph the new function on the same coordinate plane.

Set the transformed function equal to 0 and factor to find the zeros.

9x2 – 81 = 0 Set the function equal to 0.

9(x2 – 9) = 0 Use the GCF to factor out 9.

9(x + 3)(x – 3) = 0 Use the difference of two squares.

x + 3 = 0 or x – 3 = 0 Use the Zero Product Property.

x = –3 or x = 3 Solve for x.

The zeros are –3 and 3.(continued)


The vertex is the same as the original function, (0, –81). This again can be seen as the translation of 9x2 down 81 units.

-10 -9 -8 -7 -6 -5 -4 -3 -2 -1 0 1 2 3 4 5 6 7 8 9 10

-80

-70

-60

-50

-40

-30

-20

-10

10

20

30

40

f(x)

f(3x)


Notice the position of the vertex has not changed in the transformation

of f(x). However, notice the inner graph, f(3x) = 9x2 – 81, is narrower

than f(x). Specifically, the roots are x = –9 and 9 for f(x) and x = –3 and

3 for f(3x). This is because the roots of the function f(k • x) are 1

k times

the roots of f(x) in a quadratic equation. Since k > 1, the graph of f(x)

will shrink horizontally by a factor of 1

k, so the parabola

appears narrower.


Example 3

Consider the function f(x) = x2 – 6x + 8, its graph, and the constant k = –1. What is k • f(x)? How do the vertices and the zeros of f(x) and k • f(x) compare?


If f(x) = x2 – 6x + 8 and k = –1, then k • f(x) = –f(x) = –(x2 – 6x + 8) = –x2 + 6x – 8.

2. Use the roots and the vertex of the original function, f(x), to graph the function.

Set the original function equal to 0 and factor.

x2 – 6x + 8 = 0 Set the original function equal to 0.

(x – 4)(x – 2) = 0 Factor the equation.

x – 4 = 0 or x – 2 = 0 Set each factor equal to 0.

x = 4 or x = 2 Solve for x.

The roots are 4 and 2.

The vertex can be found using − −

b

af

b

a2 2, .


=−

xb

a.

a = 1, b = –6, and c = 8

2

( 6)

2(1)3=

−=− −

=xb

a


Evaluate f(x) for x = 3.

f(x) = x2 – 6x + 8

f(3) = (3)2 – 6(3) + 8

f(3) = 9 – 18 + 8

f(3) = –1

The vertex of the function is (3, –1).(continued)


-10 -9 -8 -7 -6 -5 -4 -3 -2 -1 0 1 2 3 4 5 6 7 8 9 10

-10-9-8-7-6-5-4-3-2-1

123456789

10

f(x)

3. Use the roots and the vertex of the transformed function to graph the function.

Set the transformed function equal to 0 and factor.

–x2 + 6x – 8 = 0 Set the transformed function equal to 0.

–(x2 – 6x + 8) = 0 Factor out –1.

–(x – 4)(x – 2) = 0 Factor the trinomial.

(x – 4) = 0 or (x – 2) = 0 Set each factor in x equal to 0.

x = 4 or x = 2 Solve for x.

The roots are 4 and 2.


b

af

b

a2 2, .

(continued)



=−

xb

a.

a = –1, b = 6, and c = –8

2

6

2( 1)3=

−=

−−

=xb

a

Evaluate the transformed function at f(3).

–f(x) = –x2 + 6x – 8

–f(3) = –(3)2 + 6(3) – 8

–f(3) = –9 + 18 – 8

–f(3) = 1

The vertex of the transformed function is (3, 1).

Graph the transformed function on the same coordinate plane as the original function.

-10 -9 -8 -7 -6 -5 -4 -3 -2 -1 0 1 2 3 4 5 6 7 8 9 10

-10-9-8-7-6-5-4-3-2-1

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10

f(x)

–f(x)



Notice that f(x) is reflected over the x-axis for k • f(x) where k is negative. Also notice the roots of the equations are the same. This is because k = –1.

Example 4

Consider the function f(x) = x2 – 6x + 8, its graph, and the constant k = –1. What is f(k • x)? How do the vertices and zeros of f(x) and f(k • x) compare?


If f(x) = x2 – 6x + 8 and k = –1, then f(k • x) = f(–x) = (–x)2 – 6(–x) + 8 = x2 + 6x + 8.

2. Use the roots and the vertex of the original function to graph the function.

The equation f(x) = x2 – 6x + 8 was explored in Example 3, where we found that the roots are 4 and 2, and the vertex is (3, –1).

-10 -9 -8 -7 -6 -5 -4 -3 -2 -1 0 1 2 3 4 5 6 7 8 9 10

-10-9-8-7-6-5-4-3-2-1

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10

f(x)


3. Use the roots and the vertex of the transformed function to graph the function.

Set the transformed function equal to 0 and factor.

x2 + 6x + 8 = 0 Set the transformed function equal to 0.

(x + 4)(x + 2) = 0 Factor the function.

(x + 4) = 0 or (x + 2) = 0 Set each factor equal to 0.

x = –4 or x = –2 Solve for x.

The roots are –4 and –2.


b

af

b

a2 2, .


=−

xb

a.

a = 1, b = 6, and c = 8

2

(6)

2(1)–3=

−=−

=xb

a

Evaluate the transformed function at f(–3) to find the y-coordinate of the vertex.

f(–x) = x2 + 6x + 8

f(–3) = (–3)2 + 6(–3) + 8

f(–3) = 9 – 18 + 8

f(–3) = –1

The vertex of the transformed function is (–3, –1).

(continued)


Graph the transformed function on the same coordinate plane as the original function.

-10 -9 -8 -7 -6 -5 -4 -3 -2 -1 0 1 2 3 4 5 6 7 8 9 10

-10-9-8-7-6-5-4-3-2-1

123456789

10

f(x)f(–x)


Notice that f(x) is reflected over the y-axis for f(k • x) where k is negative. Also notice the roots of the equations have the negative value of the x-coordinate compared to the original function.


Example 5

The dimensions of a rectangular garden edged with wood are such that the longer sides are 3 times the length of the shorter sides. Keeping the same ratio of side lengths, which would result in having a larger garden area: making the existing garden 5 times larger, or building 4 more gardens identical in size to the first?

1. From the description, build an equation that models the area of the garden.

Area is length times width, and the length of the garden is 3 times longer than its width.

Therefore, the area, a(x), is x • 3x = 3x2, where x is the width (the garden’s shorter side).

2. Use the equation to determine the area of the garden when the garden is made 5 times larger.

The garden’s width would now be 5x, and the area of the garden would be a(5x) = 3(5x)2 = 3(25x2) = 75x2.

3. Use the equation to determine the area of the garden when 4 more gardens equal in size to the original garden are built, for a total of 5 gardens.

If the area of one garden is a(x) = 3x2, then the area of 5 gardens is 5a(x) = 5(3x2) = 15x2.

4. Compare the results of the two scenarios.

The area of the garden in the first scenario is 75x2 units2.

The area of the 5 gardens in the second scenario is 15x2 units2.

The area of the garden in the first scenario is 5 times the area of the gardens in the second scenario; therefore, making the existing garden 5 times larger would result in a larger area.


PRACTICE


Use what you have learned about transformations of functions to solve problems 1 and 2.

1. For the function f(x) = x2 – 3x – 4, find 3 • f(x), and describe the changes that occur to the graph of f as a result of multiplying the function by 3. Check your answers by comparing the two functions on your graphing calculator.

2. For the function f(x) = x2 – 3x + 2, find f(2x), and describe the changes that occur to the graph of f as a result of multiplying the variable x by 2. Check your answers by comparing the two functions on your graphing calculator.

Use the graphs and the given information to complete problems 3 and 4.

3. Consider the graphs of the functions f(x) and g(x) shown below. The equation for f(x) is f(x) = x2 – 4x + 3. What could be the equation for g(x)?

-5 -4 -3 -2 -1 0 1 2 3 4 5

-10-9-8-7-6-5-4-3-2-1

123456789

10

f(x) g(x)

continued

Practice 5.8.2: Replacing f (x ) with k • f (x ) and f (k • x )


PRACTICE


4. Consider the graphs of the functions f(x) and g(x) shown below. The equation for f(x) is f(x) = x2 – x – 1. What could be the equation for g(x)?

-5 -4 -3 -2 -1 0 1 2 3 4 5

-10-9-8-7-6-5-4-3-2-1

123456789

10

f(x) g(x)

Complete each of the following tasks for the functions in problems 5–7.

• Graph f(x).

• List the key features you can use to graph g(x).

• Graph g(x).

• Check your graphs on your calculator.

5. f(x) = x2 + 2x – 3; g(x) = –2f(x)

6. f(x) = x2 + 2x – 3; g(x) = f(–2x)

7. f(x) = x2 – 4; g(x) = 1

2− • f(x)

continued


PRACTICE


Read each scenario and use the given information to solve problems 8–10.

8. A celebrity has a rectangular closet such that one side is 3 times as long as the other side. He would like to have more space for his coats, and he is deciding between two options. He could either triple the lengths of the sides of the existing closet, or he could build 2 more closets of the same size. Which option would give him the most area for his coats? Explain.

9. Dina manages a swimwear store in a beach town. She knows the equation that

models the store’s profit per month in the summer is f(x) = 3x2 + 300x, where

x is the average price charged per swimsuit. If swimsuit sales drop by half in

the winter, is the new model for the store’s profit f(2x), f x1

2

, 2 • f(x), or 1

2f x( ) ? Explain.

10. Zion and Zavier built a small catapult that launches beanbags for physics class.

The catapult can launch a beanbag so that the bag follows a path modeled by

the equation f(x) = –0.004x2 + 0.792x + 1.6. Zion says that the beanbag would go

farther if it followed the path g x f x( )=

1

2. Zavier says the beanbag would go

farther if it followed the path g(x) = 2 • f(x). Who is correct? Which equation for

g(x) would allow a launched beanbag to achieve the same height as a beanbag in

the original equation? Which equation for g(x) would allow a launched beanbag

to go the same distance as a beanbag in the original equation?

U5-259

Lesson 9: Fitting Quadratic Functions to Data




MCC9–12.S.ID.6a★

WORDS TO KNOW

extrapolation a prediction of the value of a dependent variable that is outside the range of the given data

first difference in a set of data, the change in the y-value when the x-value is increased by 1

interpolation a prediction of the value of a dependent variable that is within the range of the given data

quadratic regression the process of finding the equation of a parabola that fits a given set of data

quadratic regression equation

an equation that fits a parabola to the data

regression the mathematical process for determining an equation from a set of given data, used to make predictions for values of an independent variable

regression equation an equation that best represents a set of data; can be used to predict missing data or future data

second difference in a set of data, the change in successive first differences

Essential Questions

1. What is the relationship between a curve fitted to data and the actual data?

2. What are the benefits of using a curve fitted to data as opposed to the actual data?

3. What are the drawbacks to using a curve fitted to data as opposed to the actual data?


Recommended Resources• IXL Learning. “Identify linear, quadratic, and exponential functions from graphs.”


This interactive website gives a series of problems and scores them immediately. If the user submits a wrong answer, a description and process for arriving at the correct answer are provided. These problems show a graph and users are required to choose which function is depicted. This activity is meant as a review of distinguishing between the types of functions.

• IXL Learning. “Identify linear, quadratic, and exponential functions from tables.”


This interactive website provides practice problems for identifying types of functions from tables of values.

• mathispower4u. “Quadratic Regression on the TI-84—Example 1.”


This video demonstrates techniques for performing quadratic regression on the TI-84 graphing calculator.

U5-261Lesson 9: Fitting Quadratic Functions to Data

IntroductionData surrounds us in the real world. Every day, people are presented with numbers and are expected to make predictions about future events based upon that given data. A regression equation is an equation that best represents a set of data, and it can be used to predict missing data or future data. Different types of equations are suited to different types of data. Regression is the mathematical process for determining an equation from a set of given data. Regression is used to make predictions for values of an independent variable. Some data is best represented by linear or exponential equations, as you have seen previously. Quadratic regression is the process of finding the equation of a parabola that fits a given set of data. In this lesson, you will work with data sets that are best represented by quadratic equations, and you will learn how to write a quadratic regression equation, a regression equation that fits a parabola to data.

Key Concepts

• A linear equation describes a situation where there is a near-constant rate of change.

• An exponential equation describes a situation where the data changes by a constant multiple.

• A quadratic equation describes data that increases then decreases, or vice versa.

• If you are given a set of data and you are not sure whether the data is best modeled by a linear regression or a quadratic regression, you can look at the first and second differences.

• In a linear model, the y-value changes by a constant when the x-value increases by 1. The change in y when x increases by 1 is called a first difference. If your first differences are all about the same, then a linear model is appropriate.

• In a quadratic model, the first differences are not the same, but the change in the first differences is constant. The change in successive first differences is called a second difference.

• A quadratic regression equation fits a parabola to the data.

• The regression equation closely models the data but is not necessarily an exact fit. Actual data values and regression values might differ.

Lesson 5.9.1: Solving Problems Given Functions Fitted to Data


• Regression equations can be used to make predictions about the dependent variable for given values of the independent variable.

• Interpolation is when a regression equation is used to make predictions about a dependent variable that is within the range of the given data.

• Think of interpolation as finding “missing” data points within the given data.

• To interpolate, substitute the x-value into the given regression equation and solve for the y-value.

• Extrapolation is when a regression equation is used to make predictions about a dependent variable that is outside the range of the given data.

• Think of extrapolation as predicting data values based on the model outside of the given data.

• To extrapolate, substitute the x-value into the regression equation and solve for the y-value.

• The farther away you move from the given data to make predictions, the less accurate your predictions become.

• Be careful when extrapolating data and always make sure the predictions are reasonable.

• To write a regression model for a set of data without a calculator, first plot the given points. If the data has a basic parabolic shape (the values rise and then fall, or vice versa), you can write a quadratic equation to model the data.

• To write a quadratic regression equation, determine where the vertex would be for a parabola that models your data. Then, use either the two x-intercepts or the y-intercept to write the equation.

• Graph your regression equation to make sure that it approximates the data.


Example 1

The following data table shows a car’s speed in miles per hour and the car’s fuel efficiency in miles per gallon for each speed.

Speed (mph) 18.6 24.9 31.1 37.3 43.5 49.7 55.9 62.1

Fuel efficiency (mpg) 26.1 29.4 31.4 33.1 33.1 31.4 29.4 26.1

A quadratic regression equation that models this data is given by m(x) = –0.0146x2 + 1.1802x + 9.1356, where x is speed in mph and m(x) is fuel efficiency in mpg. A scatter plot of the data with the graph of this model is shown below.

–20 –10 0 10 20 30 40 50 60 70 80 90 100

–5

5

10

15

20

25

30

35

40

Speed (mph)

Fuel

e�

cien

cy (m

pg)

Use the given regression model to find the car’s fuel efficiency in miles per gallon when this car is traveling 31.1 mph. Compare your answer to the data in the table. Do these values match? Then use the graph to estimate the speed(s) that will result in fuel efficiencies of about 25 mpg and 40 mpg. Use the model to check your estimates.



1. Use the regression model to find the fuel efficiency in miles per gallon when this car is traveling 31.1 mph.

Substitute 31.1 for x in the given equation.

m(x) = –0.0146x2 + 1.1802x + 9.1356 Quadratic regression equation

m(x) = –0.0146(31.1)2 + 1.1802(31.1) + 9.1356 Substitute 31.1 for x.

m(x) ≈ 31.7186 Solve.

According to the regression model, a car traveling 31.1 mph would get approximately 31.7186 miles per gallon.

2. Compare your answer to the data in the table.

The data in the table indicates that a car traveling 31.1 mph gets 31.4 mpg.

These values do not match exactly, but they are very close. A regression model does not necessarily pass through every data point, so individual values may be slightly different when the model and the actual data are compared. However, it is still a good model for representing the data.


3. Use the graph to estimate the speed(s) that will result in fuel efficiency of about 25 mpg. Use the model to check your estimates.

Graph a horizontal line at y = 25.

–20 –10 0 10 20 30 40 50 60 70 80 90 100

–5

5

10

15

20

25

30

35

40

Speed (mph)

Fuel

e�

cien

cy (m

pg)

y = 25

The horizontal line indicates where the mpg is equal to 25. This line crosses the parabola twice. Those x-values appear to be about 17 mph and 64 mph.


4. Use the model equation to check your estimate.

Since 25 is the y-value, set the given equation equal to 25.

–0.0146x2 + 1.1802x + 9.1356 = 25

Next, subtract 25 from both sides to set the equation equal to 0.

–0.0146x2 + 1.1802x – 15.8644 = 0

Now that the quadratic equation is set up in the form ax2 + bx + c = 0, determine the values of a, b, and c.

a = –0.0146, b = 1.1802, and c = –15.8644

Substitute these values into the quadratic formula, xb b ac

a=− ± −2 4

2,

and solve the resulting equation for x.

x =− ± − − −

−( . ) ( . ) ( . )( . )

( .

1 1802 1 1802 4 0 0146 15 8644

2 0

2

00146)


x =− ±

−1 1802 0 46639108

0 0292

. .

.

Simplify, then solve.

x ≈ 17.0 or x ≈ 63.8

The quadratic formula yields x-values of approximately 17.0 mph and 63.8 mph. These values are a good match for the values estimated using the graph.

At speeds of about 17.0 mph and 63.8 mph, we can expect to attain a fuel efficiency of about 25 mpg.


5. Use the graph to estimate the speed(s) that will result in a fuel efficiency of about 40 mpg.

Graph a horizontal line at y = 40.

–20 –10 0 10 20 30 40 50 60 70 80 90 100

–5

5

10

15

20

25

30

35

40

Speed (mph)

Fuel

e�

cien

cy (m

pg)

y = 40

The horizontal line on the graph at y = 40 indicates at which speeds the car gets 40 miles per gallon. The maximum of the parabola is below this line. Therefore, this vehicle cannot achieve a fuel efficiency of 40 mpg at any speed.


6. Use the model equation to check your result.

Since 40 is the y-value, set the given equation equal to 40.

–0.0146x2 + 1.1802x + 9.1356 = 40

Subtract 40 from both sides to set the equation equal to 0.

–0.0146x2 + 1.1802x – 30.8644 = 0

Find the discriminant to see if a solution exists.

The formula for the discriminant is d = b2 – 4ac.

a = –0.0146, b = 1.1802, and c = –30.8644

d = (1.1802)2 – 4(–0.0146)(–30.8644) Substitute values for a, b, and c.

d = 1.39287204 – 1.80248096 Simplify, then solve.

d ≈ –0.41

The discriminant is negative. This means that, according to the model, it is not possible for this particular car to get 40 mpg at any speed.


Example 2

Use the regression model and graph from Example 1 to find the x- and y-intercepts of the graph. Interpret their meanings. Then, use the equation to predict the car’s fuel efficiency at the speeds of 20 mph, 65 mph, 75 mph, and 90 mph. Determine whether each of these predictions is an interpolation or an extrapolation, and whether any of the predictions seem unreasonable within the context of the problem.

The following data table from Example 1 shows a car’s speed in miles per hour and the car’s fuel efficiency in miles per gallon for each speed.

Speed (mph) 18.6 24.9 31.1 37.3 43.5 49.7 55.9 62.1

Fuel efficiency (mpg) 26.1 29.4 31.4 33.1 33.1 31.4 29.4 26.1

A quadratic regression equation that models this data is given by m(x) = –0.0146x2 + 1.1802x + 9.1356, where x is speed in mph and m(x) is fuel efficiency in mpg. A scatter plot of the data with the graph of this model is shown below.

–20 –10 0 10 20 30 40 50 60 70 80 90 100

–5

5

10

15

20

25

30

35

40

Speed (mph)

Fuel

e�

cien

cy (m

pg)


1. Find the x-intercepts of the graph.

The equation is –0.0146x2 + 1.1802x + 9.1356.

Solve this equation for x using the quadratic formula, xb b ac

a=− ± −2 4

2.

a = –0.0146, b = 1.1802, c = 9.1356

x =− ± − −

−( . ) ( . ) ( . )( . )

( .

1 1802 1 1802 4 0 0146 9 1356

2 0 01

2

446)


x =− ±

−1 1802 1 92639108

0 0292

. .

.

Simplify, then solve.

x ≈ –7.11 or x ≈ 87.95

The quadratic formula yields x-values of approximately –7.11 and 87.95 mph.

2. Interpret the meaning of the x-intercepts.

The x-intercepts are x ≈ –7.11 and x ≈ 87.95. According to the model, these are the speeds that have a fuel efficiency of 0 mpg. A car cannot achieve negative miles per hour; therefore, –7.11 does not make sense in the context of the problem. At 87.95 mph, a car is still moving, so it would still be using gas; therefore, a fuel efficiency of 0 mpg does not make sense within the context of the problem. The model appears to be valid only for data points near the values given in the data table.


Substitute 0 for x to find the y-intercept.

m(x) = –0.0146x2 + 1.1802x + 9.1356 Quadratic regression equation

m(0) = –0.0146(0)2 + 1.1802(0) + 9.1356 Substitute 0 for x.

m(0) = 9.1356 Solve.

The y-intercept is 9.1356.


4. Interpret the meaning of the y-intercept.

The y-intercept is 9.1356. This means that the car gets about 9.1 mpg when it is not moving. The only way for this to make sense is if the car is idling.

5. Use the equation to predict the fuel efficiency at the speeds of 20 mph, 65 mph, 75 mph, and 90 mph. Determine whether each of these predictions is an interpolation or an extrapolation, and if any of the predictions seem unreasonable within the context of the problem.

Predict the fuel efficiency at 20 mph and determine if the prediction seems to be reasonable.

Substitute 20 into the equation and solve.

m(20) = –0.0146(20)2 + 1.1802(20) + 9.1356

m(20) ≈ 26.9

A fuel efficiency of 26.9 mpg at 20 mph is reasonable. This is an interpolation because it falls within the range of the given data. The given data goes from 18.6 to 62.1 mph and 20 falls within that range.



m(65) = –0.0146(65)2 + 1.1802(65) + 9.1356

m(65) ≈ 24.2

A fuel efficiency of 24.2 mpg at 65 mph is reasonable. This is an extrapolation because it falls outside of the range of the given data. The given data goes from 18.6 to 62.1 mph and 65 falls outside of that range. However, it is close to the range of the given data, so the prediction is likely to be close.

(continued)




m(75) = –0.0146(75)2 + 1.1802(75) + 9.1356

m(75) ≈ 15.5

A fuel efficiency of 15.5 mpg at 75 mph might not be reasonable. This is an extrapolation because it falls outside of the range of the given data. Points from outside of the data set may be more unreliable than those from within the data or close to the endpoints.



m(90) = –0.0146(90)2 + 1.1802(90) + 9.1356

m(90) ≈ –2.9

A negative fuel efficiency is not only unreasonable, but also impossible. This is an extrapolation because it falls outside of the range of the given data. We must take care when extrapolating data to not use predictions that are unreasonable or impossible.


Example 3

The table below shows the height in feet of a children’s roller coaster at different times throughout the ride.

Time (seconds) 0 1.5 3 4.5 6

Height (feet) 0 6.9 9 6.5 0

Create a scatter plot of the data. Should a quadratic regression model be used to model the height of the roller coaster? If so, find a quadratic equation that fits this data.

1. Create a scatter plot of the data.

Plot the five given points on a coordinate plane with time in seconds along the x-axis and height in feet along the y-axis.

0 1 2 3 4 5 6 7

1

2

3

4

5

6

7

8

9

10

Time (seconds)

Hei

ght (

feet

)

Determine whether a quadratic regression model should be used to model the height of the roller coaster.

The scatter plot shows that the data is parabolic. The data rises, then falls. A quadratic regression model is appropriate.


2. Estimate the vertex of the parabola.

Since the x-intercepts are given, the line of symmetry will contain the midpoint of the segment determined by the x-intercepts, the vertical line x = 3. The roller coaster achieves a maximum height of 9 feet; this must occur at the vertex. Therefore, the vertex is at (3, 9).

3. Find the equation of the parabola.

The x-intercepts occur at 0 and 6, so the equation is y = a(x – 0)(x – 6), or y = ax(x – 6).

We can solve for a by substituting the vertex into the equation.

y = ax(x – 6)

9 = a(3)(3 – 6) Substitute (3, 9) for x and y.

9 = a(3)(–3) Simplify, then solve.

9 = –9a

–1 = a

(continued)


A regression equation that models this situation is y = –x(x – 6). Graph this parabola to verify.

0 1 2 3 4 5 6 7

1

2

3

4

5

6

7

8

9

10

Time (seconds)

Hei

ght (

feet

)

The graphed regression equation fits the data given in the table; therefore, the regression equation y = –x(x – 6) models the data.


Example 4

Look at the data given in the table that follows. What is the most appropriate regression for the data: linear, exponential, or quadratic? Create a scatter plot of the data and confirm the appropriateness of the model chosen.

x 1 2 3 4 5 6

y –1 10 25 44 67 94

1. Find the first differences by finding differences between consecutive y-values.

Start by subtracting the first y-value from the second y-value. The difference between the first two consecutive y-values is 10 – (–1) = 11. The table below shows the remaining differences.

x 1 2 3 4 5 6

y –1 10 25 44 67 94

11 15 19 23 27 First differences

2. Find the second differences by finding the differences between the first differences.

x 1 2 3 4 5 6

y –1 10 25 44 67 94

11 15 19 23 27 First differences

4 4 4 4 Second differences


3. Analyze the first and second differences.

The first differences are not constant, so a linear model will not be appropriate.

Since the second differences are constant, a quadratic model is appropriate.

4. Create a scatter plot of the data.

0 1 2 3 4 5 6 7

10

20

30

40

50

60

70

80

90

100

As the x-values increase by 1, the difference in y-values becomes increasingly greater. This is also true for exponential regressions.


5. Verify that an exponential model is not appropriate.

Already, first differences have shown that a linear regression is not appropriate. Second differences have shown that a quadratic fit is appropriate, but the scatter plot looks as if it could potentially be exponential.

To determine if an exponential fit is appropriate, find the ratios between the y-values. If the ratios are common, then an exponential fit is appropriate. If the ratios are not common, then an exponential fit is not appropriate.

For example, the ratio between the first two consecutive y-values is 10 ÷ (–1) = –10. The table below shows the remaining ratios.

x 1 2 3 4 5 6

y –1 10 25 44 67 94

–10 2.5 1.76 1.5 1.4 Ratios

The ratios are not similar; therefore, a common ratio does not exist. Thus, an exponential fit is not appropriate. The quadratic regression model is the most appropriate model to fit to the data.

UNIT 5 • QUADRATIC FUNCTIONSLesson 9: Fitting Quadratic Functions to Data

PRACTICE


Use the data and given information to solve problems 1–6.

The height of the water stream from a drinking fountain in inches above the ground is measured in inches to the right of the spout. The table below shows this data.

Distance from spout in inches 1 3 5 7

Height above ground in inches 38.57 41.07 39.46 33.93

1. Create a scatter plot of the data. Is a quadratic regression model appropriate? Explain.

2. Use the scatter plot to estimate the distance from the spout when the height of the water is 37 inches.

3. A quadratic regression model for the data is y = –0.5x2 + 3.2x + 36. Use this model to check your estimate from problem 2.

4. Using the model given in problem 3, find and interpret the x-intercepts.

5. Using the model given in problem 3, find and interpret the y-intercept.

6. Use the model given in problem 3 to estimate the height of the water 1 foot, or 12 inches, to the right of the spout. Is your answer reasonable? Why or why not?

continued

Practice 5.9.1: Solving Problems Given Functions Fitted to Data


PRACTICE


Use the data and the given model to answer questions 7 and 8.

A ball is dropped out of a second-story window. The height of the ball in feet above the ground is measured for the first second. The data is shown in the table below. A quadratic regression model for the data is y = –16.99x2 + 0.92x + 17.10.

Time (seconds) 0 0.2 0.4 0.6 0.8 1

Height (feet) 17.21 16.36 14.84 11.57 7.01 0.98

7. After how many seconds do you expect the ball to be at a height of 10 feet?

8. Use the model to predict the height of the ball after 0.1 second, 0.75 second, and 2 seconds. Are your answers reasonable? Explain.

Use what you know about functions fitted to data to complete problems 9 and 10.

9. Given the points (–3, 6), (–2, 2.5), (–1, 0), (0, –1.5), and (1, –2), create a scatter plot. Find a quadratic equation that describes these points without using a graphing calculator. Verify that a quadratic model is appropriate using second differences.

10. Evaluate the function that you found for problem 9 at x = 0 and x = –4.


IntroductionAs we saw in the previous lesson, we can use a process called regression to fit the best equation with the data in an attempt to draw conclusions and make predictions. We have learned how to interpret and analyze regression models and how to come up with a regression model using what we know about quadratic equations. In this lesson, we will learn how to easily find a quadratic regression with a graphing calculator.

Key Concepts

• A graphing calculator can graph a scatter plot of given data and find a regression equation that models the given data.

• Begin by entering the data into lists on your calculator, as outlined in the following steps.

• Decide whether the data would best be modeled with a quadratic regression, linear regression, or exponential regression. In this lesson, we focus on quadratic regressions, but you can choose the type of regression that is most appropriate for the situation.

• To decide which regression model is best, look at the scatter plot of the data you entered.

• After you find the appropriate model using your calculator, you can graph this equation on top of the scatter plot to verify that it is a reasonable model.

Graphing Equations Using a TI-83/84:



Step 3: Enter x-values into L1.





Lesson 5.9.2: Fitting a Function to Data


Graphing Equations Using a TI-Nspire:


Step 2: Name Column A “x” and Column B “y.”

Step 3: Enter x-values under Column A.


Step 5: Select Menu then 3: Data, and then 6: Quick Graph.


Step 7: Move the cursor to the x-axis and choose x.

Step 8: Move the cursor to the y-axis and choose y.

• Your graphing calculator can help you to find a quadratic regression model after you input the data.

Entering Lists Using a TI-83/84:




Step 4: At the QuadReg screen, enter the parameters for the function (Xlist: L

1, Ylist: L

2, Store RegEQ: Y

1). To enter Y

1, press [VARS] and arrow

over to the right to “Y-VARS.” Select 1: Function. Select 1: Y1.

Step 5: Press [ENTER] twice to see the quadratic regression equation.

Step 6: Press [ZOOM][9] to view the graph of the scatter plot and the regression equation.

Entering Lists Using a TI-Nspire:




Example 1

The students in Ms. Swan’s class surveyed people of all ages to find out how many people in each of the age groups below exercise on a regular basis. Use your calculator to make a scatter plot of the data in the table and to find a quadratic regression of this data. Use the “Group number” column in the table to represent the age group, the x-values. Graph the regression equation on top of your scatter plot.

Age range 11–20 21–30 31–40 41–50 51–60 61–70

Group number (x) 1 2 3 4 5 6

Number of people who exercise (y) 21 38 43 41 26 11

1. Make a scatter plot of the data.

On a TI-83/84:








On a TI-Nspire:


Step 2: Name Column A “group” and Column B “people.”





Step 7: Move the cursor to the x-axis and choose “group.”(continued)



Step 8: Move the cursor to the y-axis and choose “people.”

0 1 2 3 4 5 6 7

–5

5

10

15

20

25

30

35

40

45

Group number

Num

ber o

f peo

ple

who

exe

rcis

e

2. Find the quadratic regression model that best fits this data.

On a TI-83/84:





1, Ylist: L

2, Store RegEQ: Y

1). To enter Y

1, press [VARS]

and arrow over to the right to “Y-VARS.” Select 1: Function. Select 1: Y

1.



(continued)


On a TI-Nspire:



A quadratic regression model for this problem is y = –4.286x2 + 27.486x – 1.2.

0 1 2 3 4 5 6 7

–5

5

10

15

20

25

30

35

40

45

Group number

Num

ber o

f peo

ple

who

exe

rcis

e


Example 2

The number of calories recommended for a healthy diet varies depending on your age and level of activity. The table below shows the recommended number of calories required for moderately active females at various ages. Use your calculator to make a scatter plot of this data and to find a quadratic regression model for this situation. Then evaluate the strengths and weaknesses of the regression model.

Age (years) 2 5 10 20 50

Daily calories required 1,000 1,600 2,000 2,400 2,200


On a TI-83/84:








On a TI-Nspire:


Step 2: Name Column A “age” and Column B “calories.”





Step 7: Move the cursor to the x-axis and choose “age.”

Step 8: Move the cursor to the y-axis and choose “calories.”

(continued)


0 10 20 30 40 50

90010001100120013001400150016001700180019002000210022002300240025002600270028002900

Age (years)

Dai

ly c

alor

ies

requ

ired

2. Find the quadratic regression model that best fits this data.

Use a graphing calculator to determine the model.

On a TI-83/84:





1, Ylist: L

2, Store RegEQ: Y

1). To enter Y

1, press [VARS]


1.



(continued)


On a TI-Nspire:



A quadratic regression model for this problem is y = –1.764x2 + 113.233x + 938.460.

0 10 20 30 40 50 60 70 80 90

90010001100120013001400150016001700180019002000210022002300240025002600270028002900

Age (years)

Dai

ly c

alor

ies

requ

ired

3. Evaluate the strengths and weaknesses of the regression model.

One possible strength of the model is its potential use for extrapolating the required calories needed for ages not represented in the table. A possible weakness is that this quadratic model suggests that women older than about 67 require a negative number of calories, which is clearly not true.


Example 3

Doctors recommend that most people exercise for 30 minutes every day to stay healthy. To get the best results, a person’s heart rate while exercising should reach between 50% and 75% of his or her maximum heart rate, which is usually found by subtracting your age from 220. The peak rate should occur at around the 25th minute of exercise. Alice is 30 years old, and her maximum heart rate is 190 beats per minute (bpm). Assume that her resting rate is 60 bpm. The table below shows Alice’s heart rate as it is measured every 5 minutes for 30 minutes while she exercises. Interpret the model.

Time (minutes) 0 5 10 15 20 25 30

Heart rate (bpm) 60 93 113 134 142 152 148

Make a scatter plot of the data. Use a graphing calculator to find a quadratic regression model for the data. Use your model to extrapolate Alice’s heart rate after 35 minutes of exercise.


On a TI-83/84:








On a TI-Nspire:


Step 2: Name Column A “time” and Column B “rate.”




Step 6: Move the cursor to the x-axis and choose “time.”

Step 7: Move the cursor to the y-axis and choose “rate.” (continued)


0 10 20 30 40 50 60 70 80 90

102030405060708090

100110120130140150160170180190

Time (minutes)

Hea

rt ra

te (b

eats

per

min

ute)

2. Find a quadratic regression model using your graphing calculator.

On a TI-83/84:





1, Ylist: L

2, Store RegEQ: Y

1). To enter Y

1, press [VARS]


1.

(continued)




On a TI-Nspire:



A quadratic regression model for this problem is y = –0.1243x2 + 6.6643x + 60.7143.

0 10 20 30 40 50 60 70 80 90

102030405060708090

100110120130140150160170180190

Time (minutes)

Hea

rt ra

te (b

eats

per

min

ute)


3. Use your model to extrapolate Alice’s heart rate after 35 minutes of exercise.

Substitute 35 for x in the regression model.

y = –0.1243(35)2 + 6.6643(35) + 60.7143

y ≈ 141.70

After 35 minutes of exercise, we can expect Alice’s heart rate to be approximately 141.7 beats per minute.

4. Interpret the model.

Alice appears to be reducing her heart rate and, therefore, reducing her exercise intensity after a peak at approximately 27 minutes. If she continues the trend, her heart rate will be back to her resting heart rate at approximately 54 minutes. Heart rates below her resting heart rate can be ignored.


PRACTICE


Use a graphing calculator as necessary to solve each problem.

1. Find a quadratic regression model to represent the data in the table below.

x 1 2 3 4 5 6

y 5 11 21 35 53 75

Use the data and information below to solve problems 2–4.

The following data show the median household income in Georgia for the given years.

Year 2007 2008 2009

Median income ($) 49,136 50,861 47,590

2. Assume that the data is quadratic and find the equation.

3. Use the equation from problem 2 to extrapolate the median income in 2006 and 2010.

4. Use the equation from problem 2 to extrapolate the median income in 2012. Is this prediction reasonable? Why or why not?

Use the given information to solve problems 5 and 6.

A soccer ball is kicked into the air from an initial height of 5 feet. After 1 second, the ball is 117 feet in the air. After 3 seconds, it is about 245 feet in the air.

5. Find the equation that best describes the data.

6. What is the maximum height attained by the soccer ball? At what time will it hit the ground? continued

Practice 5.9.2: Fitting a Function to Data


PRACTICE


The table below shows the number of people in millions in the United States workforce over various years. Use the table to complete problems 7 and 8.

Year 1960 1970 1980 1990 2000

Workforce (in millions) 70 82 106 125 141

7. Which type of function (exponential, linear, or quadratic) would best describe the data? Why? Find the equation that describes the data.

8. Use your equation from problem 7 to predict the number of people in the workforce for 1987 and 2010. Which prediction do you believe is the most accurate? Why?

Use the given information to solve problems 9 and 10.

The position of a bird in flight is noted by an observer at the points (4, 31), (8, 53), and (10, 42), where the x-coordinate is the time in seconds and the y-coordinate is the height in feet.

9. Which type of function (exponential, linear, or quadratic) would best describe the data? Why?

10. Find the equation that describes the data for the bird’s position. At about what time after the observer began watching did the bird take off?

AK-1Answer Key

Answer KeyLesson 1: Interpreting Structure in Expressions

Practice 5.1.1: Identifying Terms, Factors, and Coefficients, pp. 9–101. terms:30x2,–18x,72;coefficients:30,–18;constant:72;factors:30andx2,–18andx2.

3. –x2+x+52orx2–x+524.

5. Theexpressionisnotquadraticbecauseitcannotbewrittenintheformax2+bx +c.6.

7. Expression:2(2+x)+(1/2)x2=(1/2)x2+2x+4;terms:(1/2)x2,2x,4;coefficients:1/2,2;constant:4.Theexpressionisquadraticbecauseitcanbewrittenintheformax2+bx +c.8.

9. Expression:s3;term:s3;coefficient:1;constant:none.Theexpressionisnotquadraticbecauseitcannotbewrittenintheformax2+bx +c.

Practice 5.1.2: Interpreting Complicated Expressions, pp. 16–171. –4<x<62.

3. 18x2–21x–15;a=18;b=–21;c=–154.

5. Theexpressionisquadraticbecauseitcanbewrittenintheformax2+bx +c.6.

7. Thepopulationincreasesfrom50,000to68,000.8.

9. Theareaincreasesbyafactorof9.

Lesson 2: Creating and Solving Quadratic Equations in One Variable

Practice 5.2.1: Taking the Square Root of Both Sides, p. 261. x=± 22.

3. x i= ±2 2 4.

5. x=–66.

7. Aquadraticequationintheformax2+b =chasonlyonerealsolutionwhenc=b.8.

9. r= 2 5 ≈4.47units

Practice 5.2.2: Factoring, p. 35

1. 6b(2b–7)2.

3. prime4.

5. a=0ora=26.

7. x = −5

2orx=1

8.

9. (x–9)feet

Practice 5.2.3: Completing the Square, p. 411. c=121

2.

3. c =81

4

4.

5. x = ±4 14 6.

7. x=3orx=–78.

9. afterabout2.5seconds;youcannothavenegativetime,soonlyonevalueofxisreasonable.

AK-2Answer Key

Practice 5.2.4: Applying the Quadratic Formula, p. 491. 13; two real, irrational solutions2.

3. x = –14.

5. xi

=±7 119

66.

7. after approximately 8.57 seconds8.

9. after approximately 1.58 seconds

Practice 5.2.5: Solving Quadratic Inequalities, p. 57

1. x ≤−1

2 or x ≥ 1

– 3 – 2 – 1 10 2 3

–1/2 12.

3. no real solutions4.

5. x < –8.18 or x > 0.18

– 15 – 10 – 5 0 5

0.18–8.186.

7. x ≤ –0.91 or x ≥ 1.91

– 4 – 3 – 2 – 1 10 2 3

1.91–0.918.

9. The balloon is never more than 6 feet above the ground.

Lesson 3: Creating Quadratic Equations in Two or More Variables

Practice 5.3.1: Creating and Graphing Equations Using Standard Form, pp. 81–821.

x

y

–4 –2 0 2 4

2

4

6

2.

AK-3Answer Key

3.

x

y

–4 –2 0 2 4

2

4

6

4.

5. y-intercept: (0, 7); vertex: (–5, –93); the vertex is a minimum because a > 0.6.

7. No. The y-intercept is (0, –5) because the constant term is –5. However, the graph shows a y-intercept of –10.8.

9. 40 feet

Practice 5.3.2: Creating and Graphing Equations Using the x-intercepts, pp. 90–921. 1/16, 7, x = 3.532.

3. y = x2 + 6x + 84.

5.

x

y

–8 –6 –4 –2 0 2 4 6 8

–8

–6

–4

–2

2

4

6

8

6.

7. y = 3x2 + 15x – 18 8.

9. 8 feet

AK-4Answer Key

Practice 5.3.3: Creating and Graphing Equations Using Vertex Form, p. 1001. Vertex: (–4, 1); it is a maximum because a < 0.2.

3. y = 0.96(x – 5.5)2 – 6 4.

5. f(x) = (x – 1)2 – 3 6.

7.

x

y

–8 –6 –4 –2 00

5

10

15

20Axis of symmetry: x = –4

V (–4, 2)

(–8, 18) (0, 18)

(–1, 11)(–7, 11)

8.

9. y = –2(x – 6)2 + 72

Practice 5.3.4: Rearranging Formulas, pp. 107–108

1. x y= ± −360 4 2

2.

3. xy

= − ±74

4.

5. V PR= ±6.

7. c a b= +2 2

8.

9. y k r x h= ± − −( )2 2

AK-5Answer Key


Practice 5.4.1: Solving Systems Graphically, pp. 119–1211. no real solutions2.

3. no real solutions

xy

–8 –6 –4 –2 0 2 4 6 8

–18

–16

–14

–12

–10

–8

–6

–4

–2

0

4.

5. (–3, –18)xy

–7 –6 –5 –4 –3 –2 –1 0 1 2

–55

–50

–45

–40

–35

–30

–25

–20

–15

–10

–5

0

6.

AK-6Answer Key

7. ( –15.1, –1201.9) and (4.9, –751.8)xy

–40 –30 –20 –10 0 10 20 30 40

–1400

–1200

–1000

–800

–600

–400

–200

0

8.

9. Yes, the stocks were both about $9.33 per share after 2.7 months, and then they were both about $26.75 per share after 9 months.

y

–6 –4 –2 0 2 4 6 8 10 12

8

6

4

2

0

10

12

14

16

18

20

22

24

26

–2

–4

AK-7Answer Key

Practice 5.4.2: Solving Systems Algebraically, pp. 129–1301. (0, –5)2.

3. xi

=− ±3 7

44.

5. (0, 0) and (4, 4)6.

7. (1, 2)8.

9. After about 1 second, the hawk and the soccer ball will be about 36 feet above the ground, and after about 3.5 seconds, the hawk and the soccer ball will be about 3 feet above the ground.


Practice 5.5.1: Interpreting Key Features of Quadratic Functions, pp. 147–1481. x > 1.5, x < 1.5; minimum = – 8.25; x = – 1.37, x = 4.37; neither2.

3. x < 1, x > 1; maximum = 16; x = –1, x = 3; neither4.

5. x = –16.3; x > –16.3, x < –16.36.

7. x = –0.5; x < –0.5, x > –0.58.

9. The height of the football is increasing for the first second and then it is decreasing.

Practice 5.5.2: Identifying the Domain of a Quadratic Function, pp. 154–1561. all real numbers or −∞< <∞x2.

3. all real numbers or −∞< <∞x 4.

5. all real numbers; −∞< <∞x 6.

7. all real numbers; −∞< <∞x 8.

9. about 3 seconds

Practice 5.5.3: Identifying the Average Rate of Change, pp. 163–1641. –162.

3. –3/24.

5. –266.

7. between x = –1 and x = 08.

9. between x = –1 and x = 0


Practice 5.6.1: Graphing Quadratic Functions, p. 1791. x-intercepts: (–7, 0) and (1, 0); y-intercept: (0, –7); vertex: (–3, –16); minimum: –16

40

35

30

25

20

15

10

5

5

10

15

25

30

35

40

20 15 10 5 50 10

(–3, –16)

(–6, –7) (0, –7)

(–7, 0) (1, 0)

y

x

2.

AK-8Answer Key

3. x-intercepts: (–3, 0) and (–1, 0); y-intercept: (0, 3); vertex: (–2, –1); minimum: –1

12

11

10

9

8

6

5

4

3

2

1

1

3

4

5

6

7

8 6 4 2 2

(–4, 3)

(–3, 0) (–1, 0)

(–2, –1)

(0, 3)

y

x0

4.

5. x-intercepts: (–7, 0) and (–3, 0); y-intercept: (0, 21); vertex: (–5, –4); minimum: –4

15 10 5 5

40

35

30

20

15

10

5

5

10

15

20

25

y

x

(–10, 21) (0, 21)

(–3, 0)(–7, 0)

(–5, –4)

0

6.

7. x-intercepts: (–4, 0) and (–2, 0); y-intercept: (0, –16); vertex: (–3, 2); maximum: 2

12 10 8 6 4 2 2 4

8

4

2

2

4

6

8

10

12

14

16

18

20

22

24

26

28

30

32

34

36

38

40

42

44

(–2, 0)(–4, 0)

(–3, 2)

y

x

(–6, –16) (0, –16)

0

8.

9. x-intercepts: (–5, 0) and (15, 0); y-intercept: (0, 750); vertex: (5, 1000); maximum: 1000

AK-9Answer Key

Practice 5.6.2: Writing Equivalent Forms of Quadratic Functions, pp. 190–1911. a. y = (x – 4)2 – 4

b. minimum: –4c. x = 4d. y = (x – 2)(x – 6)e. (2, 0) and (6, 0)

2.

3. a. y t= − −

+163

236

2

b. maximum: 36

c. t =3

2d. y = –16t(t – 3)e. (0, 0) and (3, 0)4.

5. h(t) = –2(t – 1)2 + 52; (1, 52); the cliff diver reaches a maximum height of 52 feet 1 second after starting the dive; 2 seconds.6.

7. y = –0.03(x – 20)2 + 12; (20, 12); the football reaches a maximum height of 12 feet after it has traveled 20 feet in the horizontal direction; 40 feet.

8.

9. R(x) = –2(x – 5)2 + 200; (5, 200); the revenue reaches its maximum value of $200 when the price is increased by $5; x = 5; $10.

Practice 5.6.3: Comparing Properties of Functions Given in Different Forms, pp. 202–203 1. f(x) = 0.05x + 52.

3. The price for 600 bolts would be the lowest at Nuts for You. The table provided does not go up as high as 600 bolts, but the prices go up less and less for every 100 bolts purchased. We can see from the table that 400 bolts would be $27, so the price for 600 bolts is likely less than that. The price for 600 bolts at Nuts and Bolts would be $57 and the price for 600 bolts at Loose Screws would be $35.4.

5. Bertha and Charlotte are tied.6.

7. Bertha and Charlotte. They had the biggest gain from 5 to 6 seconds. This suggests that they will continue to make the larger gains as time progresses and, therefore, will reach 40 feet first. 8.

9. a parabola with two x-intercepts and a > 0

Lesson 7: Building Functions

Practice 5.7.1: Building Functions from Context, pp. 216–2171. f(x) = 3x2 + 13x + 12 2.

3. x(x + 1) = 402; 73 and 744.

5. f(x)f x x( ) ( )= + +11

1225175 252

6.

7. 800 by 500 pixels or 400,000 pixels8.

9. f(x) f x x( ) ( )= − − +1

9060 402

Practice 5.7.2: Operating on Functions, pp. 224–2251. h(x) = x2 + 10x + 252.

3. m(x) = 7x2 + 30x – 254.

5. k(x) = 3x2 + 9x – 206.

7. A(x) = x(x + 4); R(x) = (x – 6)(x – 2); d(x) = 12x – 128.

9. f(x) = 200 – x; g(x) = 300 – x; A(x) = (200 – x)(300 – x)

AK-10Answer Key

Lesson 8: Transforming Functions

Practice 5.8.1: Replacing f(x) with f(x) + k and f(x + k), pp. 238–2391. y = (x – 3)2 2.

3. y = (x + 6)2 – 14.

5.

–5 –4 –3 –2 –1 0 1 2 3 4 5

–3

–2

–1

1

2

3

4

5

6

7

8

9

10

6.

7. y = –(x + 2)2 + 5. For the horizontal translation, k = 2. For the vertical translation, k = 5.8.

9. f(x) = –0.05x2 + x + 19; the paper wad will land about 32 feet away.

Practice 5.8.2: Replacing f(x) with k • f(x) and f(k • x), pp. 256–2581. 3 • f(x) = 3x2 – 9x – 12. Since the entire function is being multiplied by 3, the roots of the two functions

are the same (–1 and 4). However, the second parabola, 3 • f(x), is stretched vertically because each y-coordinate is multiplied by 3. As a result, the new parabola is narrower than the original parabola.

–5 –4 –3 –2 –1 0 1 2 3 4 5

–20

–18

–16

–14

–12

–10

–8

–6

–4

–2

2

4

6

8

10

3 • f(x)

f(x)

2.

AK-11Answer Key

3. g(x) = 2x2 – 8x + 64.

5. Graph of f(x) and g(x):

–5 –4 –3 –2 –1 0 1 2 3 4 5

–10–9–8–7–6–5–4–3–2–1

123456789

10

f(x)

g(x)

Key features for graphing g(x): Since g is a multiple of f, it will have the same x-intercepts as f. Factor f to find these intercepts: f(x) = x2 + 2x – 3, so f(x) = (x – 1)(x + 3). The x-intercepts for both graphs are –3 and 1. Since the value of k is –2, the graph will be stretched vertically by a factor of 2, and it will reflect over the x-axis because k < 0.6.

7. Graph of f(x) and g(x):

–5 –4 –3 –2 –1 0 1 2 3 4 5

–10–9–8–7–6–5–4–3–2–1

123456789

10

f(x)

g(x)

Key features for graphing g(x): Since g is a multiple of f, it will have the same x-intercepts as f. Factor f to find

these intercepts: f(x) = x2 – 4, so f(x) = (x – 2)(x + 2). The x-intercepts for both graphs are –2 and 2. Since the

value of k is −1

2, the graph will be condensed vertically by a factor of

1

2, and it will reflect over the x-axis

because k < 0.

AK-12Answer Key

8.

9. f(x) represents the profit, so earning half of the profit is modeled by 1

2 • f(x).


Practice 5.9.1: Solving Problems Given Functions Fitted to Data, pp. 279–2801. The data does appear parabolic, so a quadratic regression model makes sense. Graph:

–1 0 1 2 3 4 5 6 7 8 9 10–2

2468

1012141618202224262830323436384042444648505254

Distance from the spout in inches

Hei

ght a

bove

gro

und

in in

ches

2.

3. about 0.33 inch and 6.07 inches4.

5. The y-intercept is 36. The height of the spout is 36 inches.6.

7. about 0.67 second8.

9. y = 0.5(x + 1)(x – 3); the data falls and then rises. The second difference is common and is 1.

–9–10

–8–7–6–5–4–3–2–1

123456789

10

–10 –9 –8 –7 –6 –5 –4 –3 –2 –1 0 1 2 3 4 5 6 7 8 9 10

AK-13Answer Key

Practice 5.9.2: Fitting a Function to Data, pp. 293–2941. y = 2x2 + 32.

3. $42,242 and $42,1504.

5. h(t) = –16t2 + 128t + 56.

7. Linear; it is a near-constant rate of change. y ≈ 1.85x – 3558.28.

9. Quadratic; the bird rises and falls.

AG SRB U5 - 072656 - DR. D. Dambreville's Math...

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