Chapter 4 Resource · PDF fileChapter 4 iv Glencoe Algebra 2 Teacher’s Guide to Using...

Chapter 4 Resource Masters

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All rights reserved. The contents, or parts thereof, may be reproduced in print form for non-profit educational use with Glencoe Algebra 2, provided such reproductions bear copyright notice, but may not be reproduced in any form for any other purpose without the prior written consent of The McGraw-Hill Companies, Inc., including, but not limited to, network storage or transmission, or broadcast for distance learning.

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Printed in the United States of America.

1 2 3 4 5 6 7 8 9 DOH 16 15 14 13 12 11

CONSUMABLE WORKBOOKS Many of the worksheets contained in the Chapter Resource Masters booklets are available as consumable workbooks in both English and Spanish.

MHID ISBNStudy Guide and Intervention Workbook 0-07-660301-6 978-0-07-660301-5Homework Practice Workbook 0-07-660299-0 978-0-07-660299-5

Spanish VersionHomework Practice Workbook 0-07-660300-8 978-0-07-660300-8

Answers For Workbooks The answers for Chapter 4 of these workbooks can be found in the back of this Chapter Resource Masters booklet.

ConnectED All of the materials found in this booklet are included for viewing, printing, and editing at connected.mcgraw-hill.com.

Spanish Assessment Masters (MHID: 0-07-660298-2, ISBN: 978-0-07-660298-8) These masters contain a Spanish version of Chapter 4 Test Form 2A and Form 2C.

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Chapter 4 iii Glencoe Algebra 2

Teacher’s Guide to Using the Chapter 4 Resource Masters ......................................... iv

Chapter ResourcesStudent-Built Glossary ....................................... 1Anticipation Guide (English) .............................. 3Anticipation Guide (Spanish) ............................. 4

Lesson 4-1Graphing Quadratic FunctionsStudy Guide and Intervention ............................ 5Skills Practice ................................................... 7Practice ............................................................. 8Word Problem Practice ...................................... 9Enrichment ...................................................... 10

Lesson 4-2Solving Quadratic Equations by GraphingStudy Guide and Intervention ...........................11Skills Practice .................................................. 13Practice ............................................................ 14Word Problem Practice .................................... 15Enrichment ...................................................... 16

Lesson 4-3Solving Quadratic Equations by FactoringStudy Guide and Intervention .......................... 17Skills Practice .................................................. 19Practice ........................................................... 20Word Problem Practice .................................... 21Enrichment ...................................................... 22Graphing Calculator Activity ............................ 23

Lesson 4-4Complex NumbersStudy Guide and Intervention .......................... 24Skills Practice .................................................. 26Practice ........................................................... 27Word Problem Practice .................................... 28Enrichment ...................................................... 29

Lesson 4-5Completing the SquareStudy Guide and Intervention .......................... 30Skills Practice .................................................. 32Practice ........................................................... 33Word Problem Practice .................................... 34Enrichment ...................................................... 35

Lesson 4-6The Quadratic Formula and the DiscriminantStudy Guide and Intervention .......................... 36Skills Practice .................................................. 38Practice ........................................................... 39Word Problem Practice .................................... 40Enrichment ...................................................... 41Spreadsheet Activity ........................................ 42

Lesson 4-7Transformations of Quadratic GraphsStudy Guide and Intervention .......................... 43Skills Practice .................................................. 45Practice ........................................................... 46Word Problem Practice .................................... 47Enrichment ...................................................... 48

Lesson 4-8Quadratic InequalitiesStudy Guide and Intervention .......................... 49Skills Practice .................................................. 51Practice ........................................................... 52Word Problem Practice .................................... 53Enrichment ...................................................... 54Graphing Calculator Activity ............................ 55

AssessmentStudent Recording Sheet ................................ 57Rubric for Extended Response ....................... 58Chapter 4 Quizzes 1 and 2 ............................. 59Chapter 4 Quizzes 3 and 4 ............................. 60Chapter 4 Mid-Chapter Test ............................ 61Chapter 4 Vocabulary Test ............................... 62Chapter 4 Test, Form 1 .................................... 63Chapter 4 Test, Form 2A ................................. 65Chapter 4 Test, Form 2B ................................. 67Chapter 4 Test, Form 2C ................................. 69Chapter 4 Test, Form 2D ................................. 71Chapter 4 Test, Form 3 .................................... 73Chapter 4 Extended-Response Test ................ 75Standardized Test Practice .............................. 76

Answers ........................................... A1–A36

Contents

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Chapter 4 iv Glencoe Algebra 2

Teacher’s Guide to Using the Chapter 4 Resource Masters

The Chapter 4 Resource Masters includes the core materials needed for Chapter 4. These materials include worksheets, extensions, and assessment options. The answers for these pages appear at the back of this booklet.

All of the materials found in this booklet are included for viewing, printing, and editing at connectED.mcgraw-hill.com.

Chapter ResourcesStudent-Built Glossary (pages 1–2) These masters are a student study tool that presents up to twenty of the key vocabulary terms from the chapter. Students are to record definitions and/or examples for each term. You may suggest that students highlight or star the terms with which they are not familiar. Give this to students before beginning Lesson 4-1. Encourage them to add these pages to their mathematics study notebooks. Remind them to complete the appropriate words as they study each lesson.

Anticipation Guide (pages 3–4) This master, presented in both English and Spanish, is a survey used before beginning the chapter to pinpoint what students may or may not know about the concepts in the chapter. Students will revisit this survey after they complete the chapter to see if their perceptions have changed.

Lesson ResourcesStudy Guide and Intervention These masters provide vocabulary, key concepts, additional worked-out examples and Check Your Progress exercises to use as a reteaching activity. It can also be used in conjunction with the Student Edition as an instructional tool for students who have been absent.

Skills Practice This master focuses more on the computational nature of the lesson. Use as an additional practice option or as homework for second-day teaching of the lesson.

Practice This master closely follows the types of problems found in the Exercises section of the Student Edition and includes word problems. Use as an additional practice option or as homework for second-day teaching of the lesson.

Word Problem Practice This master includes additional practice in solving word problems that apply the concepts of the lesson. Use as an additional practice or as homework for second-day teaching of the lesson.

Enrichment These activities may extend the concepts of the lesson, offer an histori-cal or multicultural look at the concepts, or widen students’ perspectives on the mathematics they are learning. They are written for use with all levels of students.

Graphing Calculator, TI–Nspire, or Spreadsheet Activities These activities present ways in which technology can be used with the concepts in some lessons of this chapter. Use as an alternative approach to some concepts or as an integral part of your lesson presentation.

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Chapter 4 v Glencoe Algebra 2

Assessment OptionsThe assessment masters in the Chapter 4 Resource Masters offer a wide range of assessment tools for formative (monitoring) assessment and summative (final) assessment.

Student Recording Sheet This master corresponds with the standardized test practice at the end of the chapter.

Extended Response Rubric This master provides information for teachers and stu-dents on how to assess performance on open-ended questions.

Quizzes Four free-response quizzes offer assessment at appropriate intervals in the chapter.

Mid-Chapter Test This 1-page test provides an option to assess the first half of the chapter. It parallels the timing of the Mid-Chapter Quiz in the Student Edition and includes both multiple-choice and free-response questions.

Vocabulary Test This test is suitable for all students. It includes a list of vocabulary words and 10 questions to assess students’ knowledge of those words. This can also be used in conjunction with one of the leveled chapter tests.

Leveled Chapter Tests

• Form 1 contains multiple-choice questions and is intended for use with below grade level students.

• Forms 2A and 2B contain multiple-choice questions aimed at on grade level students. These tests are similar in format to offer comparable testing situations.

• Forms 2C and 2D contain free-response questions aimed at on grade level students. These tests are similar in format to offer comparable testing situations.

• Form 3 is a free-response test for use with above grade level students.

All of the above mentioned tests include a free-response Bonus question.

Extended-Response Test Performance assessment tasks are suitable for all students. Sample answers and a scoring rubric are included for evaluation.

Standardized Test Practice These three pages are cumulative in nature. It includes three parts: multiple-choice questions with bubble-in answer format, griddable questions with answer grids, and short-answer free-response questions.

Answers• The answers for the Anticipation Guide

and Lesson Resources are provided as reduced pages.

• Full-size answer keys are provided for the assessment masters.

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Chapter 4 1 Glencoe Algebra 2

This is an alphabetical list of the key vocabulary terms you will learn in Chapter 4. As you study the chapter, complete each term’s definition or description. Remember to add the page number where you found the term. Add these pages to your Algebra Study Notebook to review vocabulary at the end of the chapter.

Vocabulary Term

Found on Page

Defi nition/Description/Example

completing the square

complex conjugates

complex number

constant term

discriminant

(dihs·KRIH·muh·nuhnt)

imaginary unit

linear term

maximum value

minimum value

pure imaginary number

(continued on the next page)

Student-Built Glossary

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Student-Built Glossary

Vocabulary TermFound

on PageDefi nition/Description/Example

quadratic equation

(kwah·DRA·tihk)

Quadratic Formula

quadratic inequality

quadratic term

root

standard form

vertex

vertex form

zero


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Before you begin Chapter 4

• Read each statement.

• Decide whether you Agree (A) or Disagree (D) with the statement.

• Write A or D in the first column OR if you are not sure whether you agree or disagree, write NS (Not Sure).

After you complete Chapter 4

• Reread each statement and complete the last column by entering an A or a D.

• Did any of your opinions about the statements change from the first column?

• For those statements that you mark with a D, use a piece of paper to write an example of why you disagree.

STEP 1 A, D, or NS

StatementSTEP 2 A or D

1. All quadratic functions have a term with the variable to the second power.

2. If the graph of the quadratic function y = ax2 + c opens up then c < 0.

3. A quadratic equation whose graph does not intersect the x-axis has no real solution.

4. Since graphing shows the exact solutions to a quadratic equation, no other method is necessary for solving.

5. If (x - 3)(x + 4) = 0, then either x - 3 = 0 or x + 4 = 0.

6. An imaginary number contains i, which equals the square root of -1.

7. A method called completing the square can be used to rewrite a quadratic expression as a perfect square.

8. The quadratic formula can only be used for quadratic equations that cannot be solved by graphing or completing the square.

9. The discriminant of a quadratic equation can be used to determine the direction the graph will open.

10. The graph of y = 2x2 is a dilation of the graph of y = x2.

11. The graph of y = (x + 2)2 will be two units to the right of the graph of y = x2.

12. The graph of a quadratic inequality containing the symbol < will be a parabola opening downward.

Step 2

Step 1

Anticipation GuideQuadratic Functions and Relations


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Capítulo 4 4 Álgebra 2 de Glencoe

PASO 1 A, D o NS

EnunciadoPASO 2 A o D

1. Todas las funciones cuadráticas tienen un término con la variable elevada a la segunda potencia.

2. Si la gráfica de la función cuadrática y = ax2 + c se abre hacia arriba, entonces c > 0.

3. Una ecuación cuadrática cuya gráfica no interseca el eje x no tiene solución real.

4. Dado que graficar muestra las soluciones exactas de una ecuación cuadrática, no se necesita ningún otro método para resolverla.

5. Si (x - 3)(x + 4) = 0, entonces x - 3 = 0 ó x + 4 = 0.

6. Un número imaginario contiene i, la cual es igual a la raíz cuadrada de -1.

7. El método de completar el cuadrado se puede usar para volver a plantear una expresión cuadrática como un cuadrado perfecto.

8. La fórmula cuadrática se puede usar sólo para ecuaciones cuadráticas que no pueden resolverse mediante la completación del cuadrado o una gráfica.

9. Se puede usar el discriminante de una ecuación cuadrática para determinar la dirección en que se abrirá la gráfica.

10. La gráfica de y = 2x2 es una dilatación de la gráfica de y = x2.

11. La gráfica de y = (x + 2)2 estará dos unidades a la derecha de la gráfica de y = x2.

12. La gráfica de una desigualdad cuadrática con el símbolo < será una parábola que se abre hacia abajo.

Paso 2

Paso 1 Antes de comenzar el Capítulo 4

• Lee cada enunciado.

• Decide si estás de acuerdo (A) o en desacuerdo (D) con el enunciado.

• Escribe A o D en la primera columna O si no estás seguro(a) de la respuesta, escribe NS (No estoy seguro(a)).

Después de completar el Capítulo 4

• Vuelve a leer cada enunciado y completa la última columna con una A o una D.

• ¿Cambió cualquiera de tus opiniones sobre los enunciados de la primera columna?

• En una hoja de papel aparte, escribe un ejemplo de por qué estás en desacuerdo con los enunciados que marcaste con una D.

Ejercicios preparatoriosFunciones cuadráticas y las relaciones


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4-1 Study Guide and InterventionGraphing Quadratic Functions

Graph Quadratic Functions

Quadratic Function A function defi ned by an equation of the form f(x) = ax2 + bx + c, where a ≠ 0

Graph of a

Quadratic Function

A parabola with these characteristics: y-intercept: c; axis of symmetry: x = -b − 2a

;

x- coordinate of vertex: -b − 2a

Find the y-intercept, the equation of the axis of symmetry, and the x-coordinate of the vertex for the graph of f(x) = x2 - 3x + 5. Use this information to graph the function.a = 1, b = -3, and c = 5, so the y-intercept is 5. The equation of the axis of symmetry is

x = -(-3) −

2(1) or 3 −

2 . The x-coordinate of the vertex is 3 −

2 .

Next make a table of values for x near 3 − 2 .

x x2 - 3x + 5 f(x) (x, f(x))

0 02 - 3(0) + 5 5 (0, 5)

1 12 - 3(1) + 5 3 (1, 3)

3 − 2 ( 3 −

2 )

2

- 3 ( 3 − 2 ) + 5 11

− 4 ( 3 −

2 , 11

− 4 )

2 22 - 3(2) + 5 3 (2, 3)

3 32 - 3(3) + 5 5 (3, 5)

ExercisesComplete parts a–c for each quadratic function.a. Find the y-intercept, the equation of the axis of symmetry, and the x-coordinate

of the vertex.b. Make a table of values that includes the vertex.c. Use this information to graph the function.

1. f(x) = x2 + 6x + 8 2. f(x) = -x2 - 2x + 2 3. f(x) = 2x2 - 4x + 3

Example

xO

f (x)

xO

f (x)

xO

f (x)

xO

f (x)

6

4

2

-2 2 4


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4-1 Study Guide and Intervention (continued)

Graphing Quadratic Functions

Maximum and Minimum Values The y-coordinate of the vertex of a quadratic function is the maximum value or minimum value of the function.

Maximum or Minimum Value

of a Quadratic Function

The graph of f (x ) = ax 2 + bx + c, where a ≠ 0, opens up and has a minimum

when a > 0. The graph opens down and has a maximum when a < 0.

Determine whether each function has a maximum or minimum value, and find that value. Then state the domain and range of the function.

ExercisesDetermine whether each function has a maximum or minimum value, and find that value. Then state the domain and range of the function. 1. f(x) = 2x2 - x + 10 2. f(x) = x2 + 4x - 7 3. f(x) = 3x2 - 3x + 1

4. f(x) = x2 + 5x + 2 5. f(x) = 20 + 6x - x2 6. f(x) = 4x2 + x + 3

7. f(x) = -x2 - 4x + 10 8. f(x) = x2 - 10x + 5 9. f(x) = -6x2 + 12x + 21

Example

a. f(x) = 3x2 - 6x + 7

For this function, a = 3 and b = -6.Since a > 0, the graph opens up, and the function has a minimum value.The minimum value is the y-coordinate of the vertex. The x-coordinate of the

vertex is -b − 2a

= -(-6)

− 2(3)

= 1.

Evaluate the function at x = 1 to find the minimum value.f(1) = 3(1)2 - 6(1) + 7 = 4, so the minimum value of the function is 4. The domain is all real numbers. The range is all reals greater than or equal to the minimum value, that is {f(x) | f(x) ≥ 4}.

b. f(x) = 100 - 2x - x2

For this function, a = -1 and b = -2.Since a < 0, the graph opens down, and the function has a maximum value.The maximum value is the y-coordinate of the vertex. The x-coordinate of the vertex is -b −

2a = - -2 −

2(-1) = -1.

Evaluate the function at x = -1 to find the maximum value.f (-1) = 100 - 2(-1) - (-1)2 = 101, so the maximum value of the function is 101. The domain is all real numbers. The range is all reals less than or equal to the maximum value, that is {f(x) ⎪ f(x) ≤ 101}.


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4-1 Skills PracticeGraphing Quadratic Functions

Complete parts a–c for each quadratic function.a. Find the y-intercept, the equation of the axis of symmetry, and the x-coordinate


1. f(x) = -2x2 2. f(x) = x2 - 4x + 4 3. f(x) = x2 - 6x + 8

Determine whether each function has a maximum or a minimum value, and find that value. Then state the domain and range of the function.

4. f(x) = 6x2 5. f(x) = -8x2 6. f(x) = x2 + 2x

7. f(x) = -2x2 + 4x - 3 8. f(x) = 3x2 + 12x + 3 9. f(x) = 2x2 + 4x + 1

10. f(x) = 3x2 11. f(x) = x2 + 1 12. f(x) = -x2 + 6x - 15

13. f(x) = 2x2 - 11 14. f(x) = x2 - 10x + 5 15. f(x) = -2x2 + 8x + 7

xO

f (x)

xO

f (x)

xO

f (x)


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4-1 PracticeGraphing Quadratic Functions

Complete parts a–c for each quadratic function.a. Find the y-intercept, the equation of the axis of symmetry, and the x-coordinate


1. f(x) = x2 - 8x + 15 2. f(x) = -x2 - 4x + 12 3. f(x) = 2x2 - 2x + 1

xO

f (x)

xO

f (x)

xO

f (x)

Determine whether each function has a maximum or minimum value, and find that value. Then state the domain and range of the function.

4. f (x) = x2 + 2x - 8 5. f (x) = x2 - 6x + 14 6. v(x) = -x2 + 14x - 57

7. f (x) = 2x2 + 4x - 6 8. f (x) = -x2 + 4x - 1 9. f (x) = - 2 − 3 x2 + 8x - 24

10. GRAVITATION From 4 feet above a swimming pool, Susan throws a ball upward with a velocity of 32 feet per second. The height h(t) of the ball t seconds after Susan throws it is given by h(t) = -16t2 + 32t + 4. For t ≥ 0, find the maximum height reached by the ball and the time that this height is reached.

11. HEALTH CLUBS Last year, the SportsTime Athletic Club charged $20 to participate in an aerobics class. Seventy people attended the classes. The club wants to increase the class price this year. They expect to lose one customer for each $1 increase in the price.

a. What price should the club charge to maximize the income from the aerobics classes?

b. What is the maximum income the SportsTime Athletic Club can expect to make?


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4-1 Word Problem PracticeGraphing Quadratic Functions

1. TRAJECTORIES A cannonball is launched from a cannon on the wall of Fort Chambly, Quebec. If the path of the cannonball is traced on apiece of graph paper aligned so that the cannon is situated on the y-axis, the equation that describes the path is

y = - 1

−1600

x2+

1−2

x + 20,

where x is the horizontal distance from the cliff and y is the vertical distance above the ground in feet. How high above the ground is the cannon?

2. TICKETING The manager of a symphony computes that the symphony will earn -40P2

+ 1100P dollars per concert if they charge P dollars for tickets. What ticket price should the symphony charge in order to maximize its profits?

3. ARCHES An architect decides to use a parabolic arch for the main entrance of a science museum. In one of his plans, the top edge of the arch is described by the graph of y = -

1−4

x2+

5−2

x + 15. What are the coordinates of the vertex of this parabola?

4. FRAMING A frame company offers a line of square frames. If the side length of the frame is s, then the area of the opening in the frame is given by the function a(s) = s2

- 10s + 24. Graph a(s).

O

5. WALKING Canal Street and Walker Street are perpendicular to each other. Evita is driving south on Canal Street and is currently 5 miles north of the intersection with Walker Street. Jack is at the intersection of Canal and Walker Streets and heading east on Walker. Jack and Evita are both driving 30 miles per hour.

a. When Jack is x miles east of the intersection, where is Evita?

b. The distance between Jack and Evita is given by the formula √��x2

+ (5 - x) 2 .For what value of x are Jack and Evita at their closest? (Hint: Minimize the square of the distance.)

c. What is the distance of closest approach?


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4-1 Enrichment

Finding the x-intercepts of a Parabola

As you know, if f (x) = ax2 + bx + c is a quadratic function, the values of x

that make f (x) equal to zero are -b + √ �� b2 - 4ac −

2a and -b - √ �� b2 - 4ac

− 2a

.

The average of these two number values is - b − 2a

.

O

f(x)

x

– –, f( ( (( b––2a

b––2a

b––2ax = –

f(x) = ax2 + bx + c

The function f (x) has its maximum or minimum

value when x = - b − 2a

. The x-intercepts of the parabola,

when they exist, are √ �� b2 - 4ac

− 2a

units to the left and right of the axis of symmetry.

Find the vertex, axis of symmetry, and x-intercepts for f(x) = 5x2 + 10x - 7.

Use x = - b − 2a

.

x = - 10 − 2(5)

= -1 The x-coordinate of the vertex is -1.

Substitute x = -1 in f (x) = 5x2 + 10x - 7.

f (-1) = 5(-1)2 + 10(-1) - 7 = -12. The vertex is (-1,-12).

The axis of symmetry is x = - b − 2a

, or x = -1.

The x-coordinates of the x-intercepts are –1 ± √ �� b2 - 4ac

− 2a

= –1 ± √ �� 102 - 4 � 5 � (-7)

−− 2 � 5

= –1 ± √ �� 240

− 10

. The x–intercepts are (–1 – 2 − 5 √ �� 15 , 0) and (–1 + 2 −

5 √ �� 15 , 0) .

ExercisesFind the vertex, axis of symmetry, and x-intercepts for the graph of each function using x = - b −

2a .

1. f (x) = x2 - 4x - 8 2. g(x) = -4x2 - 8x + 3

3. y = -x2 + 8x + 3 4. f (x) = 2x2 + 6x + 5

5. A(x) = x2 + 12x + 36 6. k(x) = -2x2 + 2x - 6

Example


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Study Guide and InterventionSolving Quadratic Equations by Graphing

Solve Quadratic Equations

Quadratic Equation A quadratic equation has the form ax 2 + bx + c = 0, where a ≠ 0.

Roots of a Quadratic Equation solution(s) of the equation, or the zero(s) of the related quadratic function

The zeros of a quadratic function are the x-intercepts of its graph. Therefore, finding the x-intercepts is one way of solving the related quadratic equation.

Solve x2 + x - 6 = 0 by graphing.

Graph the related function f (x) = x2 + x - 6.

The x-coordinate of the vertex is -b − 2a

= - 1 − 2 , and the equation of the

axis of symmetry is x = - 1 − 2 .

Make a table of values using x-values around - 1 − 2 .

x -1 - 1 − 2 0 1 2

f (x) -6 -6 1 − 4 -6 -4 0

From the table and the graph, we can see that the zeros of the function are 2 and -3.

Exercises

Use the related graph of each equation to determine its solution.

1. x2 + 2x - 8 = 0 2. x2 - 4x - 5 = 0 3. x2 - 5x + 4 = 0

xO

f (x)

xO

f (x)

xO

f (x)

4. x2 - 10x + 21 = 0 5. x2 + 4x + 6 = 0 6. 4x2 + 4x + 1 = 0

xO

f (x)

xO

f (x)

xO

f (x)

xO

-6

-4

-2

-4 -2 2

f (x)Example


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Study Guide and Intervention (continued)

Solving Quadratic Equations by Graphing

Estimate Solutions Often, you may not be able to find exact solutions to quadratic equations by graphing. But you can use the graph to estimate solutions.

Solve x2 - 2x - 2 = 0 by graphing. If exact roots cannot be found, state the consecutive integers between which the roots are located.

The equation of the axis of symmetry of the related function is

xO-2

-2

-4

2

2 4

4f (x)

x = - -2 −

2(1) = 1, so the vertex has x-coordinate 1. Make a table of values.

x -1 0 1 2 3

f (x) 1 -2 -3 -2 1

The x-intercepts of the graph are between 2 and 3 and between 0 and -1. So one solution is between 2 and 3, and the other solution is between 0 and -1.

ExercisesSolve the equations. If exact roots cannot be found, state the consecutive integers between which the roots are located.

1. x2 - 4x + 2 = 0 2. x2 + 6x + 6 = 0 3. x2 + 4x + 2= 0

xO

f (x)

xO

f (x)

xO

f (x)

4. -x2 + 2x + 4 = 0 5. 2x2 - 12x + 17 = 0 6. - 1 −

2 x2 + x + 5 −

2 = 0

xO

f (x)

xO

f (x)

xO

f (x)

Example


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Skills PracticeSolving Quadratic Equations By Graphing

Use the related graph of each equation to determine its solutions.

1. x2 + 2x - 3 = 0 2. -x2 - 6x - 9 = 0 3. 3x2 + 4x + 3 = 0

x

f (x)

O

-2

-4

2-2-4 4

2

f(x) = x2 + 2x - 3

x

f (x)

O

-2

-4

-6

-8

-2-6 -4

f(x) = -x2 - 6x - 9

xO

f(x) = 3x2 + 4x + 3

12

8

4

–2–4–6

f (x)

Solve each equation. If exact roots cannot be found, state the consecutive integers between which the roots are located.

4. x2 - 6x + 5 = 0 5. -x2 + 2x - 4 = 0 6. x2 - 6x + 4 = 0

xO

f (x)

xO

f (x)

xO

f (x)

7. -x2 - 4x = 0 8. -x2 + 36 = 0

xO

f (x)

xO

f (x)


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PracticeSolving Quadratic Equations By Graphing

Use the related graph of each equation to determine its solutions.

1. -3x2 + 3 = 0 2. 3x2 + x + 3 = 0 3. x2 - 3x + 2 = 0

xO-2

-2

-4

-4 2

2

f (x)4

xO-2-4 2

2

4

6

8

4

f (x)

xO-2-4 2

2

4

6

8

4

f (x)

Solve each equation. If exact roots cannot be found, state the consecutive integers between which the roots are located. 4. -2x2 - 6x + 5 = 0 5. x2 + 10x + 24 = 0 6. 2x2 - x - 6 = 0

xO

f (x)

xO

f (x)

xO

f (x)

7. -x2 + x + 6 = 0 8. -x2 + 5x - 8 = 0

xO

f (x)

xO

f (x)

9. GRAVITY Use the formula h(t) = v0t - 16t2, where h(t) is the height of an object in feet, v0 is the object’s initial velocity in feet per second, and t is the time in seconds.

a. Marta throws a baseball with an initial upward velocity of 60 feet per second. Ignoring Marta’s height, how long after she releases the ball will it hit the ground?

b. A volcanic eruption blasts a boulder upward with an initial velocity of 240 feet per second. How long will it take the boulder to hit the ground if it lands at the same elevation from which it was ejected?


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Word Problem PracticeSolving Quadratic Equations by Graphing

1. TRAJECTORIES David threw a baseball into the air. The function of the height of the baseball in feet is h = 80t -16t2, where t represents the time in seconds after the ball was thrown. Use this graph of the function to determine how long it took for the ball to fall back to the ground.

h

t1 2 3 4 5-1

-40

40

80

2. BRIDGES In 1895, a brick arch railway bridge was built on North Avenue in Baltimore, Maryland. The arch is described by the equation h = 9 – 1

−50

x2, where h is the height in yards and x is the distance in yards from the center of the bridge. Graph this equation and describe, to the nearest yard, where the bridge touches the ground.

3. LOGIC Wilma is thinking of two numbers. The sum is 2 and the product is -24. Use a quadratic equation to find the two numbers.

4. RADIO TELESCOPES The cross-section of a large radio telescope is a parabola. The dish is set into the ground. The equation that describes the cross-section is d =

2−75

x2-

4−3

x - 32−3

, where d givesthe depth of the dish below ground and x is the distance from the control center, both in meters. If the dish does not extend above the ground level, what is the diameter of the dish? Solve by graphing.

x

d

5. BOATS The distance between two boats is

d = √��t2- 10t + 35 ,

where d is distance in meters and t is time in seconds.

a. Make a graph of d2 versus t.d

tO

b. Do the boats ever collide?

x

h


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Enrichment

Graphing Absolute Value Equations

You can solve absolute value equations in much the same way you solved quadratic equations. Graph the related absolute value function for each equation using a graphing calculator. Then use the ZERO feature in the CALC menu to find its real solutions, if any. Recall that solutions are points where the graph intersects the x-axis.

For each equation, make a sketch of the related graph and find the solutions rounded to the nearest hundredth.

1. ⎪x + 5⎥ = 0 2. ⎪4x - 3⎥ + 5 = 0 3. ⎪x - 7⎥ = 0

4. ⎪x + 3⎥ - 8 = 0 5. - ⎪x + 3⎥ + 6 = 0 6. ⎪x - 2⎥ - 3 = 0

7. ⎪3x + 4⎥ = 2 8. ⎪x + 12⎥ = 10 9. ⎪x⎥ - 3 = 0

10. Explain how solving absolute value equations algebraically and finding zeros of absolute value functions graphically are related.


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Study Guide and InterventionSolving Quadratic Equations by Factoring

Factored Form To write a quadratic equation with roots p and q, let (x - p)(x - q) = 0. Then multiply using FOIL.

Write a quadratic equation in standard form with the given roots.

a. 3, -5

(x - p)(x - q) = 0 Write the pattern.

(x - 3)[x - (-5)] = 0 Replace p with 3, q with -5.

(x - 3)(x + 5) = 0 Simplify.

x2 + 2x - 15 = 0 Use FOIL.

The equation x2 + 2x - 15 = 0 has roots 3 and -5.

b. - 7 −

8 , 1 −

3

(x - p)(x - q) = 0

[x - (- 7 − 8 ) ] (x - 1 −

3 ) = 0

(x + 7 − 8 ) (x - 1 −

3 ) = 0

(8x + 7) −

8 � (3x - 1)

− 3 = 0

24 � (8x + 7)(3x - 1) −−

24 = 24 � 0

24x2 + 13x - 7 = 0

The equation 24x2 + 13x - 7 = 0 has

roots - 7 − 8 and 1 −

3 .

ExercisesWrite a quadratic equation in standard form with the given root(s).

1. 3, -4 2. -8, -2 3. 1, 9

4. -5 5. 10, 7 6. -2, 15

7. - 1 − 3 , 5 8. 2, 2 −

3 9. -7, 3 −

4

10. 3, 2 − 5 11. - 4 −

9 , -1 12. 9, 1 −

6

13. 2 − 3 , - 2 −

3 14. 5 −

4 , - 1 −

2 15. 3 −

7 , 1 −

5

16. - 7 − 8 , 7 −

2 17. 1 −

2 , 3 −

4 18. 1 −

8 , 1 −

6

Example


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Solving Quadratic Equations by Factoring

Solve Equations by Factoring When you use factoring to solve a quadratic equation, you use the following property.

Zero Product Property For any real numbers a and b, if ab = 0, then either a = 0 or b =0, or both a and b = 0.

Solve each equation by factoring.

a. 3x2 = 15x

3x2 = 15x Original equation

3x2 - 15x = 0 Subtract 15x from both sides.

3x(x - 5) = 0 Factor the binomial.

3x = 0 or x - 5 = 0 Zero Product Property

x = 0 or x = 5 Solve each equation.

The solution set is {0, 5}.

b. 4x2 - 5x = 21 4x2 - 5x = 21 Original equation

4x2 - 5x - 21 = 0 Subtract 21 from both sides.

(4x + 7)(x - 3) = 0 Factor the trinomial.

4x + 7 = 0 or x - 3 = 0 Zero Product Property

x = - 7 − 4 or x = 3 Solve each equation.

The solution set is {- 7 − 4 , 3} .

ExercisesSolve each equation by factoring.

1. 6x2 - 2x = 0 2. x2 = 7x 3. 20x2 = -25x

4. 6x2 = 7x 5. 6x2 - 27x = 0 6. 12x2 - 8x = 0

7. x2 + x - 30 = 0 8. 2x2 - x - 3 = 0 9. x2 + 14x + 33 = 0

10. 4x2 + 27x - 7 = 0 11. 3x2 + 29x - 10 = 0 12. 6x2 - 5x - 4 = 0

13. 12x2 - 8x + 1 = 0 14. 5x2 + 28x - 12 = 0 15. 2x2 - 250x + 5000 = 0

16. 2x2 - 11x - 40 = 0 17. 2x2 + 21x - 11 = 0 18. 3x2 + 2x - 21 = 0

19. 8x2 - 14x + 3 = 0 20. 6x2 + 11x - 2 = 0 21. 5x2 + 17x - 12 = 0

22. 12x2 + 25x + 12 = 0 23. 12x2 + 18x + 6 = 0 24. 7x2 - 36x + 5 = 0

Example


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Skills PracticeSolving Quadratic Equations by Factoring

Write a quadratic equation in standard form with the given root(s).

1. 1, 4 2. 6, -9

3. -2, -5 4. 0, 7

5. - 1 − 3 , -3 6. - 1 −

2 , 3 −

4

Factor each polynomial.

7. m2 + 7m - 18 8. 2x2 - 3x - 5

9. 4z2 + 4z - 15 10. 4p2 + 4p - 24

11. 3y2 + 21y + 36 12. c2 - 100


13. x2 = 64 14. x2 - 100 = 0

15. x2 - 3x + 2 = 0 16. x2 - 4x + 3 = 0

17. x2 + 2x - 3 = 0 18. x2 - 3x - 10 = 0

19. x2 - 6x + 5 = 0 20. x2 - 9x = 0

21. x2 - 4x = 21 22. 2x2 + 5x - 3 = 0

23. 4x2 + 5x - 6 = 0 24. 3x2 - 13x - 10 = 0

25. NUMBER THEORY Find two consecutive integers whose product is 272.


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PracticeSolving Quadratic Equations by Factoring

Write a quadratic equation in standard form with the given root(s).

1. 7, 2 2. 0, 3 3. -5, 8

4. -7, -8 5. -6, -3 6. 3, -4

7. 1, 1 − 2 8. 1 −

3 , 2 9. 0, - 7 −

2


10. r3 + 3r2 - 54r 11. 8a2 + 2a - 6 12. c2 - 49

13. x3 + 8 14. 16r2 - 169 15. b4 - 81


16. x2 - 4x - 12 = 0 17. x2 - 16x + 64 = 0

18. x2 - 6x + 8 = 0 19. x2 + 3x + 2 = 0

20. x2 - 4x = 0 21. 7x2 = 4x

22. 10x2 = 9x 23. x2 = 2x + 99

24. x2 + 12x = -36 25. 5x2 - 35x + 60 = 0

26. 36x2 = 25 27. 2x2 - 8x - 90 = 0

28. NUMBER THEORY Find two consecutive even positive integers whose product is 624.

29. NUMBER THEORY Find two consecutive odd positive integers whose product is 323.

30. GEOMETRY The length of a rectangle is 2 feet more than its width. Find the dimensions of the rectangle if its area is 63 square feet.

31. PHOTOGRAPHY The length and width of a 6-inch by 8-inch photograph are reduced by the same amount to make a new photograph whose area is half that of the original. By how many inches will the dimensions of the photograph have to be reduced?


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Word Problem PracticeSolving Quadratic Equations by Factoring

1. FLASHLIGHTS When Dora shines her flashlight on the wall at a certain angle, the edge of the lit area is in the shape of a parabola. The equation of the parabola is y = 2x2

+ 2x - 60. Factor this quadratic equation.

2. SIGNS David was looking through an old algebra book and came across this equation.

x2 6x + 8 = 0

The sign in front of the 6 was blotted out. How does the missing sign depend on the signs of the roots?

3. ART The area in square inches of the drawing Maisons prés de la mer by Claude Monet is approximated by the equation y = x2

– 23x + 130. Factor the equation to find the two roots, which are equal to the approximate length and width of the drawing.

4. PROGRAMMING Ray is a computer programmer. He needs to find the quadratic function of this graph for an algorithm related to a game involving dice. Provide such a function.

y

xO

-2

-4

2

2

4

4 6 8 10 12

5. ANIMATION A computer graphics animator would like to make a realistic simulation of a tossed ball. The animator wants the ball to follow the parabolic trajectory represented by the quadratic equation f(x) = -0.2(x + 5) (x - 5).

a. What are the solutions of f (x) = 0?

b. Write f (x) in standard form.

c. If the animator changes the equation to f(x) = -0.2x2 + 20, what are the solutions of f(x) = 0?


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Enrichment

Using Patterns to Factor

Study the patterns below for factoring the sum and the difference of cubes.

a3 + b3 = (a + b)(a2 - ab + b2)a3 - b3 = (a - b)(a2 + ab + b2)

This pattern can be extended to other odd powers. Study these examples.

Factor a5 + b5.Extend the first pattern to obtain a5 + b5 = (a + b)(a4 - a3b + a2b2 - ab3 + b4).Check: (a + b)(a4 - a3b + a2b2 - ab3 + b4) = a5 - a4b + a3b2 - a2b3 + ab4 + a4b - a3b2 + a2b3 - ab4 + b5

= a5 + b5

Factor a5 - b5.

Extend the second pattern to obtain a5 - b5 = (a - b)(a4 + a3b + a2b2 + ab3 + b4).Check: (a - b) (a4 + a3b + a2 b2 + ab3 + b4) = a5 + a4b + a3b2 + a2b3 + ab4 - a4b - a3b2 - a2b3 - ab4 - b5

= a5 - b5

In general, if n is an odd integer, when you factor an + bn or an - bn, one factor will be either (a + b) or (a - b), depending on the sign of the original expression. The other factor will have the following properties:

• The first term will be an - 1 and the last term will be bn - 1.• The exponents of a will decrease by 1 as you go from left to right.• The exponents of b will increase by 1 as you go from left to right.• The degree of each term will be n - 1.• If the original expression was an + bn, the terms will alternately have + and - signs.• If the original expression was an - bn, the terms will all have + signs.

Use the patterns above to factor each expression.

1. a7 + b7

2. c9 - d9

3. f 11 + g11

To factor x10 - y10, change it to (x5 + y5)(x5 - y5) and factor each binomial. Use this approach to factor each expression.

4. x10 - y10

5. a14 - b14

Example 1

Example 2


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Graphing Calculator ActivityUsing Tables to Factor by Grouping

The TABLE feature of a graphing calculator can be used to help factor a polynomial of the form ax2 + bx + c. (The same problems can be solved with the Lists and Spreadsheet application on the TI-Nspire.)

Factor 10x2 - 43x + 28 by grouping.

Make a table of the negative factors of 10 � 28 or 280. Look for a pair of factors whose sum is -43.

Enter the equation y = 280 − x in Y1 to find the factors of 280. Then,

find the sum of the factors using y = 280 − x + x in Y2. Set up the table

to display the negative factors of 280 by setting �Tbl = to -1. Examine the results.

Keystrokes: Y= 280 ÷

ENTER VARS ENTER ENTER + ENTER 2nd [TBLSET] (–) 1 ENTER (–) 1 ENTER 2nd

[TABLE].

The last line of the table shows that -43x may be replaced with -8x + (-35x).

10x2 - 43x + 28 = 10x2 - 8x + (-35x) + 28 = 2x(5x - 4) + (-7)(5x - 4) = (5x - 4)(2x - 7)Thus, 10x2 - 43x + 28 = (5x - 4)(2x - 7).


1. y2 - 20y - 96 2. 4z2 - 33z + 35 3. 4y2 + y -18 4. 6a2 + 2a - 15

5. 6m2 + 17m + 12 6. 24z2 - 46z + 15 7. 36y2 + 84y + 49 8. 4b2 + 36b - 403

Factor 12x2 - 7x - 12.

Look at the factors of 12(-12) or -144 for a pair with a sum of -7. Enter an equation to determine the factors in Y1 and an equation to find the sum of factors in Y2. Examine the table to find a sum of -7.Keystrokes: Y= (–) 144

÷

ENTER VARS ENTER ENTER + ENTER 2nd [TBLSET] 1 ENTER 1 ENTER 2nd

[TABLE].12x2 - 7x - 12 = 12x2 + 9x + (-16x) - 12 = 3x(4x + 3) - 4(4x + 3) = (4x + 3)(3x - 4) Thus, 12x2 - 7x - 12 = (4x + 3)(3x - 4).

Exercises

Example 1

Example 2


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Study Guide and InterventionComplex Numbers

Solve x2 + 5 = 0.

x2 + 5 = 0 Original equation.

x2 = -5 Subtract 5 from each side.

x = ± √ � 5 i Square Root Property.

ExercisesSimplify.

1. √ �� -72 2. √ �� -24

3. √ �� -84 4. (2 + i) (2 - i)

Solve each equation.

5. 5x2 + 45 = 0 6. 4x2 + 24 = 0

7. -9x2 = 9 8. 7x2 + 84 = 0

Pure Imaginary Numbers A square root of a number n is a number whose square

is n. For nonnegative real numbers a and b, √ �� ab = √ � a � √ � b and √ � a − b =

√ � a −

√ � b , b ≠ 0.

• The imaginary unit i is defined to have the property that i2 = -1.

• Simplified square root expressions do not have radicals in the denominator, and any number remaining under the square root has no perfect square factor other than 1.

a. Simplify √ �� -48 . √ �� -48 = √ �� 16 � (-3) = √ �� 16 � √ � 3 � √ �� -1 = 4i √ � 3

b. Simplify √ �� -63 . √ �� -63 = √ �� -1 � 7 � 9 = √ �� -1 � √ � 7 . √ � 9 = 3i √ � 7

a. Simplify -3i � 4i.-3i � 4i = -12i2

= -12(-1) = 12

b. Simplify √ �� -3 � √ �� -15 . √ �� -3 � √ �� -15 = i √ � 3 � i √ �� 15 = i2 √ �� 45 = -1 � √ � 9 � √ � 5 = -3 √ � 5

Example 1 Example 2

Example 3


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Complex Numbers

Operations with Complex Numbers

Complex Number

A complex number is any number that can be written in the form a + bi, where a and b are

real numbers and i is the imaginary unit (i 2 = -1). a is called the real part, and b is called

the imaginary part.

Addition and

Subtraction of

Complex Numbers

Combine like terms.

(a + bi ) + (c + di ) = (a + c) + (b + d )i

(a + bi ) - (c + di ) = (a - c) + (b - d )i

Multiplication of

Complex NumbersUse the defi nition of i 2 and the FOIL method:

(a + bi )(c + di ) = (ac - bd ) + (ad + bc)i

Complex Conjugatea + bi and a - bi are complex conjugates. The product of complex conjugates is always a

real number.

To divide by a complex number, first multiply the dividend and divisor by the complex conjugate of the divisor.

ExercisesSimplify.

1. (-4 + 2i) + (6 - 3i) 2. (5 - i) - (3 - 2i) 3. (6 - 3i) + (4 - 2i)

4. (-11 + 4i) - (1 - 5i) 5. (8 + 4i) + (8 - 4i) 6. (5 + 2i) - (-6 - 3i)

7. (2 + i)(3 - i) 8. (5 - 2i)(4 - i) 9. (4 - 2i)(1 - 2i)

10. 5 − 3 + i

11. 7 - 13i − 2i

12. 6 - 5i − 3i

Simplify (6 + i) + (4 - 5i).

(6 + i) + (4 - 5i) = (6 + 4) + (1 - 5)i = 10 - 4i

Simplify (2 - 5i) � (-4 + 2i).

(2 - 5i) � (-4 + 2i) = 2(-4) + 2(2i) + (-5i)(-4) + (-5i)(2i) = -8 + 4i + 20i - 10i2

= -8 + 24i - 10(-1) = 2 + 24i

Simplify (8 + 3i) - (6 - 2i).(8 + 3i) - (6 - 2i) = (8 - 6) + [3 - (-2)]i = 2 + 5i

Simplify 3 - i − 2 + 3i

.

3 - i − 2 + 3i

= 3 - i − 2 + 3i

� 2 - 3i − 2 - 3i

= 6 - 9i - 2i + 3 i 2 −− 4 - 9 i 2

= 3 - 11i − 13

= 3 − 13

- 11 − 13

i

Example 1 Example 2

Example 3 Example 4


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Skills PracticeComplex Numbers

Simplify.

1. √ �� 99 2. √ �� 27 −

49

3. √ �� 52x3y5 4. √ �� -108x7

5. √ �� -81x6 6. √ �� -23 � √ �� -46

7. (3i)(-2i)(5i) 8. i11

9. i65 10. (7 - 8i) + (-12 - 4i)

11. (-3 + 5i) + (18 - 7i) 12. (10 - 4i) - (7 + 3i)

13. (7 - 6i)(2 - 3i) 14. (3 + 4i)(3 - 4i)

15. 8 - 6i − 3i

16. 3i − 4 + 2i


17. 3x2 + 3 = 0 18. 5x2 + 125 = 0

19. 4x2 + 20 = 0 20. -x2 - 16 = 0

21. x2 + 18 = 0 22. 8x2 + 96 = 0

Find the values of � and m that make each equation true.

23. 20 - 12i = 5� + (4m)i 24. � - 16i = 3 - (2m)i

25. (4 + �) + (2m)i = 9 + 14i 26. (3 - m) + (7� - 14)i = 1 + 7i


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PracticeComplex Numbers

Simplify.

1. √ �� -36 2. √ �� -8 � √ �� -32 3. √ �� -15 � √ �� -25

4. (-3i) (4i)(-5i) 5. (7i)2(6i) 6. i42

7. i55 8. i89 9. (5 - 2i) + (-13 - 8i)

10. (7 - 6i) + (9 + 11i) 11. (-12 + 48i) + (15 + 21i) 12. (10 + 15i) - (48 - 30i)

13. (28 - 4i) - (10 - 30i) 14. (6 - 4i) (6 + 4i) 15. (8 - 11i) (8 - 11i)

16. (4 + 3i) (2 - 5i) 17. (7 + 2i) (9 - 6i) 18. 6 + 5i − -2i

19. 2 − 7 - 8i

20. 3 - i − 2 - i

21. 2 - 4i − 1 + 3i


22. 5n2 + 35 = 0 23. 2m2 + 10 = 0

24. 4m2 + 76 = 0 25. -2m2 - 6 = 0

26. -5m2 - 65 = 0 27. 3 − 4 x2 + 12 = 0

Find the values of ℓ and m that make each equation true.

28. 15 - 28i = 3ℓ + (4m)i 29. (6 - ℓ) + (3m)i = -12 + 27i

30. (3ℓ + 4) + (3 - m)i = 16 - 3i 31. (7 + m) + (4ℓ - 10)i = 3 - 6i

32. ELECTRICITY The impedance in one part of a series circuit is 1 + 3j ohms and the impedance in another part of the circuit is 7 - 5j ohms. Add these complex numbers to find the total impedance in the circuit.

33. ELECTRICITY Using the formula E = IZ, find the voltage E in a circuit when the current I is 3 - j amps and the impedance Z is 3 + 2j ohms.


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Word Problem PracticeComplex Numbers

1. SIGN ERRORS Jennifer and Jessica come up with different answers to the same problem. They had to multiply (4 + i)(4 - i) and give their answer as a complex number. Jennifer claims that the answer is 15 and Jessica claims that the answer is 17. Who is correct? Explain.

2. COMPLEX CONJUGATES You have seen that the product of complex conjugates is always a real number. Show that the sum of complex conjugates is also always a real number.

3. PYTHAGOREAN TRIPLES If three integers a, b, and c satisfy a2

+ b2= c2,

then they are called a Pythagorean triple. Suppose that a, b, and c are a Pythagorean triple. Show that the real and imaginary parts of (a + bi)2, together with the number c2, form another Pythagorean triple.

4. ROTATIONS Complex numbers can be used to perform rotations in the plane. For example, if (x, y) are the coordinates of a point in the plane, then the real and imaginary parts of i(x + yi) are the horizontal and vertical coordinates of the 90° counterclockwise rotation of (x, y) about the origin. What are the real and imaginary parts of i(x + yi)?

5. ELECTRICAL ENGINEERING Alternating current (AC) in an electrical circuit can be described by complex numbers. In any electrical circuit, Z, the impedance in the circuit, is related to the voltage V and the current I by the formula Z = V −

I . The standard electrical

voltage in Europe is 220 volts, so in these problems use V = 220.

a. Find the impedance in a standard European circuit if the current is 22 – 11i amps.

b. Find the current in a standard European circuit if the impedance is 10 – 5i watts.

c. Find the impedance in a standard European circuit if the current is 20i amps.


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Enrichment

Conjugates and Absolute Value

When studying complex numbers, it is often convenient to represent a complex number by a single variable. For example, we might let z = x + yi. We denote the conjugate of z by − z . Thus, − z = x - yi.

We can define the absolute value of a complex number as follows.

⎪z⎥ = ⎪x + yi⎥ = √ �� x2 + y2

There are many important relationships involving conjugates and absolute values of complex numbers.

Show ⎪z⎥ 2 = z − z for any complex number z.

Let z = x + yi. Then, zz − z = (x + yi)(x - yi) = x2 + y2

= √ �� (x2 + y2)2

= ⎪z⎥ 2

Show − z − ⎪z⎥ 2

is the multiplicative inverse for any nonzero

complex number z.

We know ⎪z⎥ 2 = z − z . If z ≠ 0, then we have z ( − z − ⎪z⎥ 2

) = 1.

Thus, − z − ⎪z⎥ 2

is the multiplicative inverse of z.

ExercisesFor each of the following complex numbers, find the absolute value and multiplicative inverse.

1. 2i 2. -4 - 3i 3. 12 - 5i

4. 5 - 12i 5. 1 + i 6. √ � 3 - i

7. √ � 3

− 3 +

√ � 3 −

3 i 8.

√ � 2 −

2 -

√ � 2 −

2 i 9. 1 −

2 -

√ � 3 −

2 i

Example 1

Example 2


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Study Guide and InterventionCompleting the Square

Square Root Property Use the Square Root Property to solve a quadratic equation that is in the form “perfect square trinomial = constant.”

Solve each equation by using the Square Root Property. Round to the nearest hundredth if necessary.

a. x2 - 8x + 16 = 25 x2 - 8x + 16 = 25 (x - 4)2 = 25 x - 4 = √ �� 25 or x - 4 = - √ �� 25 x = 5 + 4 = 9 or x = -5 + 4 = -1

The solution set is {9, -1}.

b. 4x2 - 20x + 25 = 32 4x2 - 20x + 25 = 32 (2x - 5)2 = 322x - 5 = √ �� 32 or 2x - 5 = - √ �� 32 2x - 5 = 4 √ � 2 or 2x - 5 = -4 √ � 2

x = 5 ± 4 √ � 2 −

2

The solution set is {

5 ± 4 √ � 2 −

2 } .

ExercisesSolve each equation by using the Square Root Property. Round to the nearest hundredth if necessary.

1. x2 - 18x + 81 = 49 2. x2 + 20x + 100 = 64 3. 4x2 + 4x + 1 = 16

4. 36x2 + 12x + 1 = 18 5. 9x2 - 12x + 4 = 4 6. 25x2 + 40x + 16 = 28

7. 4x2 - 28x + 49 = 64 8. 16x2 + 24x + 9 = 81 9. 100x2 - 60x + 9 = 121

10. 25x2 + 20x + 4 = 75 11. 36x2 + 48x + 16 = 12 12. 25x2 - 30x + 9 = 96

Example


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Completing the Square

Complete the Square To complete the square for a quadratic expression of the form x2 + bx, follow these steps.

1. Find b − 2 . 2. Square b −

2 . 3. Add ( b −

2 )

2 to x2 + bx.

Find the value of c that makes x2 + 22x + c a perfect square trinomial. Then write the trinomial as the square of a binomial.

Step 1 b = 22; b − 2 = 11

Step 2 112 = 121Step 3 c = 121

The trinomial is x2 + 22x + 121, which can be written as (x + 11)2.

Solve 2x2 - 8x - 24 = 0 by completing the square.

2x2 - 8x - 24 = 0 Original equation

2x2 - 8x - 24 − 2 = 0 −

2 Divide each side by 2.

x2 - 4x - 12 = 0 x2 - 4x - 12 is not a perfect square.

x2 - 4x = 12 Add 12 to each side.

x2 - 4x + 4 = 12 + 4 Since ( 4 − 2 )

2

= 4, add 4 to each side.

(x - 2)2 = 16 Factor the square.

x - 2 = ±4 Square Root Property

x = 6 or x = - 2 Solve each equation.

The solution set is {6, -2}.

Exercises

Find the value of c that makes each trinomial a perfect square. Then write the trinomial as a perfect square. 1. x2 - 10x + c 2. x2 + 60x + c 3. x2 - 3x + c

4. x2 + 3.2x + c 5. x2 + 1 − 2 x + c 6. x2 - 2.5x + c

Solve each equation by completing the square. 7. y2 - 4y - 5 = 0 8. x2 - 8x - 65 = 0 9. w2 - 10w + 21 = 0

10. 2x2 - 3x + 1 = 0 11. 2x2 - 13x - 7 = 0 12. 25x2 + 40x - 9 = 0

13. x2 + 4x + 1 = 0 14. y2 + 12y + 4 = 0 15. t2 + 3t - 8 = 0

Example 1 Example 2


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Skills PracticeCompleting the Square


1. x2 - 8x + 16 = 1 2. x2 + 4x + 4 = 1

3. x2 + 12x + 36 = 25 4. 4x2 - 4x + 1 = 9

5. x2 + 4x + 4 = 2 6. x2 - 2x + 1 = 5

7. x2 - 6x + 9 = 7 8. x2 + 16x + 64 = 15

Find the value of c that makes each trinomial a perfect square. Then write the trinomial as a perfect square.

9. x2 + 10x + c 10. x2 - 14x + c

11. x2 + 24x + c 12. x2 + 5x + c

13. x2 - 9x + c 14. x2 - x + c

Solve each equation by completing the square.

15. x2 - 13x + 36 = 0 16. x2 + 3x = 0

17. x2 + x - 6 = 0 18. x2 - 4x - 13 = 0

19. 2x2 + 7x - 4 = 0 20. 3x2 + 2x - 1 = 0

21. x2 + 3x - 6 = 0 22. x2 - x - 3 = 0

23. x2 = -11 24. x2 - 2x + 4 = 0


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PracticeCompleting the Square


1. x2 + 8x + 16 = 1 2. x2 + 6x + 9 = 1 3. x2 + 10x + 25 = 16

4. x2 - 14x + 49 = 9 5. 4x2 + 12x + 9 = 4 6. x2 - 8x + 16 = 8

7. x2 - 6x + 9 = 5 8. x2 - 2x + 1 = 2 9. 9x2 - 6x + 1 = 2

Find the value of c that makes each trinomial a perfect square. Then write the trinomial as a perfect square.

10. x2 + 12x + c 11. x2 - 20x + c 12. x2 + 11x + c

13. x2 + 0.8x + c 14. x2 - 2.2x + c 15. x2 - 0.36x + c

16. x2 + 5 − 6 x + c 17. x2 - 1 −

4 x + c 18. x2 - 5 −

3 x + c


19. x2 + 6x + 8 = 0 20. 3x2 + x - 2 = 0 21. 3x2 - 5x + 2 = 0

22. x2 + 18 = 9x 23. x2 - 14x + 19 = 0 24. x2 + 16x - 7 = 0

25. 2x2 + 8x - 3 = 0 26. x2 + x - 5 = 0 27. 2x2 - 10x + 5 = 0

28. x2 + 3x + 6 = 0 29. 2x2 + 5x + 6 = 0 30. 7x2 + 6x + 2 = 0

31. GEOMETRY When the dimensions of a cube are reduced by 4 inches on each side, the surface area of the new cube is 864 square inches. What were the dimensions of the original cube?

32. INVESTMENTS The amount of money A in an account in which P dollars are invested for 2 years is given by the formula A = P(1 + r)2, where r is the interest rate compounded annually. If an investment of $800 in the account grows to $882 in two years, at what interest rate was it invested?


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Word Problem PracticeCompleting the Square

1. COMPLETING THE SQUARESamantha needs to solve the equation

x2- 12x = 40.

What must she do to each side of the equation to complete the square?

2. ART The area in square inches of the drawing Foliage by Paul Cézanne is approximated by the equation y = x2

– 40x + 396. Complete the square and find the two roots, which are equal to the approximate length and width of the drawing.

3. COMPOUND INTEREST Nikki invested $1000 in a savings account with interest compounded annually. After two years the balance in the account is $1210. Use the compound interest formula A = P(1 + r)t to find the annual interest rate.

4. REACTION TIME Lauren was eating lunch when she saw her friend Jason approach. The room was crowded and Jason had to lift his tray to avoid obstacles. Suddenly, a glass on Jason’s lunch tray tipped and fell off the tray. Lauren lunged forward and managed to catch the glass just before it hit the ground. The height h, in feet, of the glass t seconds after it was dropped is given by h = -16t2

+ 4.5. Lauren caught the glass when it was six inches off the ground. How long was the glass in the air before Lauren caught it?

5. PARABOLAS A parabola is modeled by y = x2

- 10x + 28. Jane’s homework problem requires that she find the vertex of the parabola. She uses the completing square method to express the function in the form y = (x - h)2

+ k, where (h, k) is the vertex of the parabola. Write the function in the form used by Jane.

6. AUDITORIUM SEATING The seats in an auditorium are arranged in a square grid pattern. There are 45 rows and 45 columns of chairs. For a special concert, organizers decide to increase seating by adding n rows and n columns to make a square pattern of seating 45 + n seats on a side.

a. How many seats are there after the expansion?

b. What is n if organizers wish to add 1000 seats?

c. If organizers do add 1000 seats, what is the seating capacity of the auditorium?


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Enrichment

The Golden Quadratic Equations

A golden rectangle has the property that its length

a

a

a

b

b

a

can be written as a + b, where a is the width of the

rectangle and a + b − a = a − b . Any golden rectangle can be

divided into a square and a smaller golden rectangle, as shown.

The proportion used to define golden rectangles can be used to derive two quadratic equations. These are sometimes called golden quadratic equations.

Solve each problem.

1. In the proportion for the golden rectangle, let a equal 1. Write the resulting quadratic equation and solve for b.

2. In the proportion, let b equal 1. Write the resulting quadratic equation and solve for a.

3. Describe the difference between the two golden quadratic equations you found in exercises 1 and 2.

4. Show that the positive solutions of the two equations in exercises 1 and 2 are reciprocals.

5. Use the Pythagorean Theorem to find a radical expression for the diagonal of a golden rectangle when a = 1.

6. Find a radical expression for the diagonal of a golden rectangle when b = 1.


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Study Guide and InterventionThe Quadratic Formula and the Discriminant

Quadratic Formula The Quadratic Formula can be used to solve any quadratic equation once it is written in the form ax2 + bx + c = 0.

Quadratic Formula The solutions of ax 2 + bx + c = 0, with a ≠ 0, are given by x = -b ± √ �� b2 - 4ac

− 2a

.

Solve x2 - 5x = 14 by using the Quadratic Formula.

Rewrite the equation as x2 - 5x - 14 = 0.

x = -b ± √ �� b2 - 4ac −

2a Quadratic Formula

= -(-5) ± √ �� (-5)2 - 4(1)(-14)

−− 2(1)

Replace a with 1, b with -5, and c with -14.

= 5 ± √ �� 81 −

2 Simplify.

= 5 ± 9 − 2

= 7 or -2

The solutions are -2 and 7.

ExercisesSolve each equation by using the Quadratic Formula.

1. x2 + 2x - 35 = 0 2. x2 + 10x + 24 = 0 3. x2 - 11x + 24 = 0

4. 4x2 + 19x - 5 = 0 5. 14x2 + 9x + 1 = 0 6. 2x2 - x - 15 = 0

7. 3x2 + 5x = 2 8. 2y2 + y - 15 = 0 9. 3x2 - 16x + 16 = 0

10. 8x2 + 6x - 9 = 0 11. r2 - 3r − 5 + 2 −

25 = 0 12. x2 - 10x - 50 = 0

13. x2 + 6x - 23 = 0 14. 4x2 - 12x - 63 = 0 15. x2 - 6x + 21 = 0

Example


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Roots and the Discriminant

DiscriminantThe expression under the radical sign, b2 - 4ac, in the Quadratic Formula is called

the discriminant.

Discriminant Type and Number of Roots

b2 - 4ac > 0 and a perfect square 2 rational roots

b2 - 4ac > 0, but not a perfect square 2 irrational roots

b2 - 4ac = 0 1 rational root

b2 - 4ac < 0 2 complex roots

Find the value of the discriminant for each equation. Then describe the number and type of roots for the equation.

a. 2x2 + 5x + 3The discriminant is b2 - 4ac = 52 - 4(2) (3) or 1.The discriminant is a perfect square, so the equation has 2 rational roots.

b. 3x2 - 2x + 5The discriminant is b2 - 4ac = (-2)2 - 4(3) (5) or -56.The discriminant is negative, so the equation has 2 complex roots.

ExercisesComplete parts a-c for each quadratic equation.a. Find the value of the discriminant.b. Describe the number and type of roots.c. Find the exact solutions by using the Quadratic Formula.

1. p2 + 12p = -4 2. 9x2 - 6x + 1 = 0 3. 2x2 - 7x - 4 = 0

4. x2 + 4x - 4 = 0 5. 5x2 - 36x + 7 = 0 6. 4x2 - 4x + 11 = 0

7. x2 - 7x + 6 = 0 8. m2 - 8m = -14 9. 25x2 - 40x = -16

10. 4x2 + 20x + 29 = 0 11. 6x2 + 26x + 8 = 0 12. 4x2 - 4x - 11 = 0

Example


The Quadratic Formula and the Discriminant


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Skills PracticeThe Quadratic Formula and the Discriminant

Complete parts a-c for each quadratic equation.a. Find the value of the discriminant.b. Describe the number and type of roots.c. Find the exact solutions by using the Quadratic Formula.

1. x2 - 8x + 16 = 0 2. x2 - 11x - 26 = 0

3. 3x2 - 2x = 0 4. 20x2 + 7x - 3 = 0

5. 5x2 - 6 = 0 6. x2 - 6 = 0

7. x2 + 8x + 13 = 0 8. 5x2 - x - 1 = 0

9. x2 - 2x - 17 = 0 10. x2 + 49 = 0

11. x2 - x + 1 = 0 12. 2x2 - 3x = -2

Solve each equation by using the Quadratic Formula.

13. x2 = 64 14. x2 - 30 = 0

15. x2 - x = 30 16. 16x2 - 24x - 27 = 0

17. x2 - 4x - 11 = 0 18. x2 - 8x - 17 = 0

19. x2 + 25 = 0 20. 3x2 + 36 = 0

21. 2x2 + 10x + 11 = 0 22. 2x2 - 7x + 4 = 0

23. 8x2 + 1 = 4x 24. 2x2 + 2x + 3 = 0

25. PARACHUTING Ignoring wind resistance, the distance d(t) in feet that a parachutist falls in t seconds can be estimated using the formula d(t) = 16t2. If a parachutist jumps from an airplane and falls for 1100 feet before opening her parachute, how many seconds pass before she opens the parachute?


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PracticeThe Quadratic Formula and the Discriminant

Solve each equation by using the Quadratic Formula.

1. 7x2 - 5x = 0 2. 4x2 - 9 = 0

3. 3x2 + 8x = 3 4. x2 - 21 = 4x

5. 3x2 - 13x + 4 = 0 6. 15x2 + 22x = -8

7. x2 - 6x + 3 = 0 8. x2 - 14x + 53 = 0

9. 3x2 = -54 10. 25x2 - 20x - 6 = 0

11. 4x2 - 4x + 17 = 0 12. 8x - 1 = 4x2

13. x2 = 4x - 15 14. 4x2 - 12x + 7 = 0

Complete parts a-c for each quadratic equation.a. Find the value of the discriminant.b. Describe the number and type of roots.c. Find the exact solutions by using the Quadratic Formula.

15. x2 - 16x + 64 = 0 16. x2 = 3x 17. 9x2 - 24x + 16 = 0

18. x2 - 3x = 40 19. 3x2 + 9x - 2 = 0 20. 2x2 + 7x = 0

21. 5x2 - 2x + 4 = 0 22. 12x2 - x - 6 = 0 23. 7x2 + 6x + 2 = 0

24. 12x2 + 2x - 4 = 0 25. 6x2 - 2x - 1 = 0 26. x2 + 3x + 6 = 0

27. 4x2 - 3x2 - 6 = 0 28. 16x2 - 8x + 1 = 0 29. 2x2 - 5x - 6 = 0

30. GRAVITATION The height h(t) in feet of an object t seconds after it is propelled straight up from the ground with an initial velocity of 60 feet per second is modeled by the equation h(t) = -16t2 + 60t. At what times will the object be at a height of 56 feet?

31. STOPPING DISTANCE The formula d = 0.05s2 + 1.1s estimates the minimum stopping distance d in feet for a car traveling s miles per hour. If a car stops in 200 feet, what is the fastest it could have been traveling when the driver applied the brakes?


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Word Problem PracticeThe Quadratic Formula and the Discriminant

1. PARABOLAS The graph of a quadratic equation of the form y = ax2

+ bx + c is shown below.

y

xO

5

-5

Is the discriminant b2- 4ac positive,

negative, or zero? Explain.

2. TANGENT Kathleen is trying to find bso that the x-axis is tangent to the parabola y = x2

+ bx + 4. She finds one value that works, b = 4. Is this the only value that works? Explain.

3. SPORTS In 1990, American Randy Barnes set the world record for the shot put. His throw can be described by the equation y = –16x2

+ 368x. Use the Quadratic Formula to find how far his throw was to the nearest foot.

4. EXAMPLES Give an example of a quadratic function f (x) that has the following properties.

I. The discriminant of f is zero.

II. There is no real solution of the equation f(x) = 10.

Sketch the graph of x = f(x).

yx

O

5. TANGENTS The graph of y = x2 is a parabola that passes through the point at (1, 1). The line y = mx - m + 1, where m is a constant, also passes through the point at (1, 1).

a. To find the points of intersection between the line y = mx - m + 1 and the parabola y = x2, set x2 = mx - m + 1 and then solve for x. Rearranging terms, this equation becomes x2 - mx + m - 1 = 0. What is the discriminant of this equation?

b. For what value of m is there only one point of intersection? Explain the meaning of this in terms of the corresponding line and the parabola.


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Enrichment

Sum and Product of Roots

Sometimes you may know the roots of a quadratic equation without knowing the equation itself. Using your knowledge of factoring to solve an equation, you can work backward to find the quadratic equation. The rule for finding the sum and product of roots is as follows:

Sum and Product of RootsIf the roots of ax2 + bx + c = 0, with a ≠ 0, are s

1 and

s2, then s

1 + s

2 = - a −

b and s

1 � s

2 = c − a .

Write a quadratic equation that has the roots 3 and -8.

The roots are x = 3 and x = -8.

x

y

O

(–5–2, –30–1

4)

10

–10

–20

–30

2 4–2–4–6–8

3 + (-8) = -5 Add the roots.

3(-8) = -24 Multiply the roots. Equation: x2 + 5x - 24 = 0

ExercisesWrite a quadratic equation that has the given roots.

1. 6, -9 2. 5, -1 3. 6, 6

4. 4 ± √ � 3 5. - 2 − 5 , 2 −

7 6. -2 ± 3 √ � 5

− 7

Find k such that the number given is a root of the equation.

7. 7; 2x2 + kx - 21 = 0 8. -2; x2 - 13x + k = 0

Example


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Spreadsheet ActivityApproximating the Real Zeros of Polynomials

You have learned the Location Principle, which can be used to approximate the real zeros of a polynomial.

A12345

C

xf(x)

B

5

G

0–2

D

–523

J

523

E

–3.33333339.1111111

H

1.66666670.7777778

I

3.33333339.1111111

F

–1.66666670.7777778

–5

Sheet 1 Sheet 2 Sheet 3

In the spreadsheet above, the positive real zero of ƒ(x) = x2 - 2 can be approximated in the following way. Set the spreadsheet preference to manual calculation. The values in A2 and B2 are the endpoints of a range of values. The values in D2 through J2 are values equally in the interval from A2 to B2. The formulas for these values are A2, A2 + (B2 - A2)/6, A2 + 2*(B2 - A2)/6, A2 + 3*(B2 - A2)/6, A2 + 4*(B2 - A2)/6, A2 + 5*(B2 - A2)/6, and B2, respectively.

Row 3 gives the function values at these points. The function ƒ(x) = x2 - 2 is entered into the spreadsheet in Cell D3 as D2^2 - 2. This function is then copied to the remaining cells in the row.

You can use this spreadsheet to study the function values at the points in cells D2 through J2. The value in cell F3 is positive and the value in cell G3 is negative, so there must be a zero between -1.6667 and 0. Enter these values in cells A2 and B2, respectively, and recalculate the spreadsheet. (You will have to recalculate a number of times.) The result is a new table from which you can see that there is a zero between 1.41414 and 1.414306. Because these values agree to three decimal places, the zero is about 1.414. This can be verified by using algebra.By solving x2 - 2 = 0, we obtain x = ± √ � 2 . The positive root is x = ± √ � 2 = 1.414213. . . , which verifies the result.

Exercises 1. Use a spreadsheet like the one above to approximate the zero of ƒ(x) = 3x - 2 to three

decimal places. Then verify your answer by using algebra to find the exact value of the root.

2. Use a spreadsheet like the one above to approximate the real zeros of f(x) = x2 + 2x + 0.5. Round your answer to four decimal places. Then, verify your answer by using the quadratic formula.

3. Use a spreadsheet like the one above to approximate the real zero of ƒ(x) = x3 - 3 −

2 x2 - 6x - 2 between -0.4 and -0.3.


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Study Guide and InterventionTransformations of Quadratic Graphs

Write Quadratic Equations in Vertex Form A quadratic function is easier to graph when it is in vertex form. You can write a quadratic function of the form y = ax2 + bx + c in vertex from by completing the square.

Write y = 2x2 - 12x + 25 in vertex form. Then graph the function.

y = 2x2 - 12x + 25

x

y

O 2

2

4

6

8

4 6

y = 2(x2 - 6x) + 25y = 2(x2 - 6x + 9) + 25 - 18y = 2(x - 3)2 + 7

The vertex form of the equation is y = 2(x - 3)2 + 7.

Exercises

Write each equation in vertex form. Then graph the function.

1. y = x2 - 10x + 32 2. y = x2 + 6x 3. y = x2 - 8x + 6

x

y

O

x

y

O

x

y

O

4. y = -4x2 + 16x - 11 5. y = 3x2 - 12x + 5 6. y = 5x2 - 10x + 9

x

y

O

x

y

O

x

y

O

Example


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Transformations of Quadratic Graphs

Transformations of Quadratic Graphs Parabolas can be transformed by changing the values of the constants a, h, and k in the vertex form of a quadratic equation: y = a(x – h ) 2 + k.

• The sign of a determines whether the graph opens upward (a > 0) or downward (a < 0).

• The absolute value of a also causes a dilation (enlargement or reduction) of the parabola. The parabola becomes narrower if ⎪a⎥ >1 and wider if ⎪a⎥ < 1.

• The value of h translates the parabola horizontally. Positive values of h slide the graph to the right and negative values slide the graph to the left.

• The value of k translates the graph vertically. Positive values of k slide the graph upward and negative values slide the graph downward.

Graph y = (x + 7)2 + 3.

• Rewrite the equation as y = [x – (–7)]2 + 3.

• Because h = –7 and k = 3, the vertex is at (–7, 3). The axis of symmetry is x = –7. Because a = 1, we know that the graph opens up, and the graph is the same width as the graph of y = x2.

• Translate the graph of y = x2 seven units to the left and three units up.

Exercises

Graph each function.

1. y = –2x2 + 2 2. y = – 3(x – 1)2 3. y = 2(x + 2)2 + 3

Example

x

15

5

-15

-15

-55 15

y

-5

x

y

x

y

x

y


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Skills PracticeTransformations of Quadratic Graphs

Write each quadratic function in vertex form. Then identify the vertex, axis of symmetry, and direction of opening.

1. y = (x - 2)2 2. y = -x2 + 4 3. y = x2 - 6

4. y = -3(x + 5)2 5. y = -5x2 + 9 6. y = (x - 2)2 - 18

7. y = x2 - 2x - 5 8. y = x2 + 6x + 2 9. y = -3x2 + 24x


10. y = (x - 3)2 - 1 11. y = (x + 1)2 + 2 12. y = -(x - 4)2 - 4

x

y

O

x

y

O

x

y

O

13. y = - 1 − 2 (x + 2)2 14. y = -3x2 + 4 15. y = x2 + 6x + 4

x

y

O

x

y

O

x

y

O


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PracticeTransformations of Quadratic Graphs

Write each equation in vertex form. Then identify the vertex, axis of symmetry, and direction of opening. 1. y = -6x2

- 24x - 25 2. y = 2x2 + 2 3. y = -4x2 + 8x

4. y = x2 + 10x + 20 5. y = 2x2 + 12x + 18 6. y = 3x2 - 6x + 5

7. y = -2x2 - 16x - 32 8. y = -3x2 + 18x - 21 9. y = 2x2 + 16x + 29


10. y = (x + 3)2 - 1 11. y = -x2 + 6x - 5 12. y = 2x2 - 2x + 1

x

y

O

x

y

O

x

y

O

13. Write an equation for a parabola with vertex at (1, 3) that passes through (-2, -15).

14. Write an equation for a parabola with vertex at (-3, 0) that passes through (3, 18).

15. BASEBALL The height h of a baseball t seconds after being hit is given by h(t) = -16t2 + 80t + 3. What is the maximum height that the baseball reaches, and when does this occur?

16. SCULPTURE A modern sculpture in a park contains a parabolic arc that starts at the ground and reaches a maximum height of 10 feet after a horizontal distance of 4 feet. Write a quadratic function in vertex form that describes the shape of the outside of the arc, where y is the height of a point on the arc and x is its horizontal distance from the left-hand starting point of the arc. 10 ft

4 ft


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Word Problem PracticeTransformations of Quadratic Graphs

1. ARCHES A parabolic arch is used as a bridge support. The graph of the arch is shown below.

y

xO 5

5

-5

If the equation that corresponds to this graph is written in the form y + a(x - h)2

+ k, what are h and k?

2. TRANSLATIONS For a computer animation, Barbara uses the quadratic function f(x) = -42(x - 20)2 + 16800 to help her simulate an object tossed on another planet. For one skit, she had to use the function f (x + 5) - 8000 instead of f (x). Where is the vertex of the graph of y = f (x + 5) - 8000?

3. BRIDGES The shape formed by the main cables of the Golden Gate Bridge approximately follows the equation y = 0.0002x2 - 0.23x + 227. Graph the parabola formed by one of the cables.

4. WATER JETS The graph shows the path of a jet of water.

y

xO

5

The equation corresponding to this graph is y = a(x - h) 2

+ k. What are a, h, and k?

5. PROFIT A theater operator predicts that the theater can make -4x2 + 160x dollars per show if tickets are priced at x dollars.

a. Rewrite the equation y = -4x2 + 160x in the form y = a(x - h) 2 + k.

b. What is the vertex of the parabola and what is its axis of symmetry?

c. Graph the parabola.

y

xO

Width

Heig

ht

x

y


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Enrichment

A Shortcut to Complex RootsWhen graphing a quadratic function, the real roots are shown in the graph. You have learned that quadratic functions can also have imaginary roots that cannot be seen on the graph of the function. However, there is a way to graphically represent the complex roots of a quadratic function.

Find the complex roots of the quadratic function y = x2 - 4x + 5.

Step 1 Graph the function.

y

xO

6

5

Step 2 Reflect the graph over the horizontal line containing the vertex. In this example, the vertex is (2, 1).

y

xO

Step 3 The real part of the complex root is the point halfway between the x-intercepts of the reflected graph and the imaginary part of the complex roots are + and - half the distance between the x-intercepts of the reflected graph. So, in this example, the complex roots are 2 + 1i and 2 - 1i.

ExercisesUsing this method, find the complex roots of the following quadratic functions.

1. y = x2 + 2x + 5 2. y = x2 + 4x + 8

3. y = x2 + 6x + 13 4. y = x2 + 2x + 17

Example


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Study Guide and InterventionQuadratic Inequalities

Graph Quadratic Inequalities To graph a quadratic inequality in two variables, use the following steps:

1. Graph the related quadratic equation, y = ax2 + bx + c. Use a dashed line for < or >; use a solid line for ≤ or ≥.

2. Test a point inside the parabola. If it satisfies the inequality, shade the region inside the parabola; otherwise, shade the region outside the parabola.

Graph the inequality y > x2 + 6x + 7.

First graph the equation y = x2 + 6x + 7. By completing the

x

y

O 2

-2

2

4

-4 -2

square, you get the vertex form of the equation y = (x + 3)2 - 2, so the vertex is (-3, -2). Make a table of values around x = -3, and graph. Since the inequality includes >, use a dashed line.Test the point (-3, 0), which is inside the parabola. Since (-3)2 + 6(-3) + 7 = -2, and 0 > -2, (-3, 0) satisfies the inequality. Therefore, shade the region inside the parabola.

ExercisesGraph each inequality.

1. y > x2 - 8x + 17 2. y ≤ x2 + 6x + 4 3. y ≥ x2 + 2x + 2

x

y

O

x

y

O

x

y

O

4. y < -x2 + 4x - 6 5. y ≥ 2x2 + 4x 6. y > -2x2 - 4x + 2

x

y

O

x

y

O

x

y

O

Example


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Quadratic Inequalities

Solve Quadratic Inequalities Quadratic inequalities in one variable can be solved graphically or algebraically.

Graphical Method

To solve ax 2 + bx + c < 0:

First graph y = ax 2 + bx + c. The solution consists of the x-values

for which the graph is below the x-axis.

To solve ax 2 + bx + c > 0:

First graph y = ax 2 + bx + c. The solution consists of the x-values

for which the graph is above the x-axis.

Algebraic Method

Find the roots of the related quadratic equation by factoring,

completing the square; or using the Quadratic Formula.

2 roots divide the number line into 3 intervals.

Test a value in each interval to see which intervals are solutions.

If the inequality involves ≤ or ≥, the roots of the related equation are included in the solution set.

Solve the inequality x2 - x - 6 ≤ 0.

First find the roots of the related equation x2 - x - 6 = 0. The

x

y

O

equation factors as (x - 3)(x + 2) = 0, so the roots are 3 and -2. The graph opens up with x-intercepts 3 and -2, so it must be on or below the x-axis for -2 ≤ x ≤ 3. Therefore the solution set is {x|-2 ≤ x ≤ 3}.

ExercisesSolve each inequality.

1. x2 + 2x < 0 2. x2 - 16 < 0 3. 0 < 6x - x2 - 5

4. c2 ≤ 4 5. 2m2 - m < 1 6. y2 < -8

7. x2 - 4x - 12 < 0 8. x2 + 9x + 14 > 0 9. -x2 + 7x - 10 ≥ 0

10. 2x2 + 5x- 3 ≤ 0 11. 4x2 - 23x + 15 > 0 12. -6x2 - 11x + 2 < 0

13. 2x2 - 11x + 12 ≥ 0 14. x2 - 4x + 5 < 0 15. 3x2 - 16x + 5 < 0

Example


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Skills PracticeQuadratic Inequalities

Graph each inequality.

1. y ≥ x2 - 4x + 4 2. y ≤ x2 - 4 3. y > x2 + 2x - 5

x

y

O

x

y

O

x

y

O

Solve each inequality by graphing.

4. x2 - 6x + 9 ≤ 0 5. -x2 - 4x + 32 ≥ 0 6. x2 + x - 10 > 10

xO

y

x

y

O

Solve each inequality algebraically.

7. x2 - 3x - 10 < 0 8. x2 + 2x - 35 ≥ 0

9. x2 - 18x + 81 ≤ 0 10. x2 ≤ 36

11. x2 - 7x > 0 12. x2 + 7x + 6 < 0

13. x2 + x - 12 > 0 14. x2 + 9x + 18 ≤ 0

15. x2 - 10x + 25 ≥ 0 16. -x2 - 2x + 15 ≥ 0

17. x2 + 3x > 0 18. 2x2 + 2x > 4

19. -x2 - 64 ≤ -16x 20. 9x2 + 12x + 9 < 0

x

y

O


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PracticeQuadratic Inequalities

Graph each inequality.

1. y ≤ x2 + 4 2. y > x2 + 6x + 6 3. y < 2x2 - 4x - 2

x

y

O

x

y

O

x

y

O

Solve each inequality.

4. x2 + 2x + 1 > 0 5. x2 - 3x + 2 ≤ 0 6. x2 + 10x + 7 ≥ 0

7. x2 - x - 20 > 0 8. x2 - 10x + 16 < 0 9. x2 + 4x + 5 ≤ 0

10. x2 + 14x + 49 ≥ 0 11. x2 - 5x > 14 12. -x2 - 15 ≤ 8x

13. -x2 + 5x - 7 ≤ 0 14. 9x2 + 36x + 36 ≤ 0 15. 9x ≤ 12x2

16. 4x2 + 4x + 1 > 0 17. 5x2 + 10 ≥ 27x 18. 9x2 + 31x + 12 ≤ 0

19. FENCING Vanessa has 180 feet of fencing that she intends to use to build a rectangular play area for her dog. She wants the play area to enclose at least 1800 square feet. What are the possible widths of the play area?

20. BUSINESS A bicycle maker sold 300 bicycles last year at a profit of $300 each. The maker wants to increase the profit margin this year, but predicts that each $20 increase in profit will reduce the number of bicycles sold by 10. How many $20 increases in profit can the maker add in and expect to make a total profit of at least $100,000?


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Word Problem PracticeQuadratic Inequalities

1. HUTS The space inside a hut is shaded in the graph. The parabola is described

by the equation y = - 4−5

(x - 1)2+ 4.

y

xO

Write an inequality that describes the shaded region.

2. DISCRIMINANTS Consider the equation ax2 + bx + c = 0. Assume that the discriminant is zero and that a is positive. What are the solutions of the inequality ax2 + bx + c ≤ 0?

3. KIOSKS Caleb is designing a kiosk by wrapping a piece of sheet metal with dimensions x + 5 inches by 4x + 8 inches into a cylindrical shape. Ignoring cost, Caleb would like a kiosk that has a surface area of at least 4480 square inches. What values of x satisfy this condition?

4. DAMS The Hoover Dam is a concrete arch dam designed to hold the water of Lake Mead. At its center, the dam’s height is approximately 725 feet, and the dam varies from 45 to 660 feet thick. The dark line on this sketch of the cross-section of the dam is a parabola.

y

xThickness

Heig

ht

Cross-Section ofHoover Dam

Dam

Lake Mead

a. Write an equation for the Hoover Dam parabola. Let the height be the y-value of the parabola and the thickness be the x-value of the parabola. (Hint: the equation will be in the form: y = k(x – maximum thickness) + maximum height.)

b. Using your equation, graph the parabola of the Hoover Dam for 45 ≤ x ≤ 660.

Thickness

Heig

ht

x

y

c. Estimate to the nearest foot the thickness of the dam when the height is 200 feet.


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Enrichment

Graphing Absolute Value Inequalities

You can solve absolute value inequalities by graphing in much the same manner you graphed quadratic inequalities. Graph the related absolute function for each inequality by using a graphing calculator. For > and ≥, identify the x-values, if any, for which the graph lies below the x-axis. For < and ≤, identify the x values, if any, for which the graph lies above the x-axis.

For each inequality, make a sketch of the related graph and find the solutions rounded to the nearest hundredth.

1. ⎪x - 3⎥ > 0 2. ⎪x⎥ - 6 < 0 3. - ⎪x + 4⎥ + 8 < 0

4. 2 ⎪x + 6⎥ - 2 ≥ 0 5. ⎪3x - 3⎥ ≥ 0 6. ⎪x - 7⎥ < 5

7. ⎪7x - 1⎥ > 13 8. ⎪x - 3.6⎥ ≤ 4.2 9. ⎪2x + 5⎥ ≤ 7


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Graphing Calculator ActivityQuadratic Inequalities and the Test Menu

The inequality symbols, called relational operators, in the TEST menu can be used to display the solution of a quadratic inequality. Another method that can be used to find the solution set of a quadratic inequality is to graph each side of an inequality separately. Examine the graphs and use the intersect function to determine the range of values for which the inequality is true.

ExercisesSolve each inequality.

1. -x2 - 10x - 21 < 0 2. x2 - 9 < 0 3. x2 + 10x + 25 ≤ 0

4. x2 + 3x ≤ 28 5. 2x2 + x ≥ 3 6. 4x2 + 12x + 9 > 0

7. 23 > -x2 + 10x 8. x2 - 4x - 13 ≤ 0 9. (x + 1)(x -3) > 0

Solve x2 + x ≥ 6.

Place the calculator in Dot mode. Enter the inequality into Y1. Then trace the graph and describe the solution as an inequality.

Keystrokes: Y= x2 + 2nd [TEST] 4 6 ZOOM 4.

Use TRACE to determine the endpoints of the segments. Theses values are used to express the solution of the inequality, { x | x ≥ - 3 or x ≥ 2 }.

Solve 2x2 + 4x - 5 ≤ 3.

Place the left side of the inequality in Y1 and the right side in Y2. Determine the points of intersection. Use the intersection points to express the solution set of the inequality. Be sure to set the calculator to Connected mode.

Keystrokes: Y= 2 x2 + 4 — 5 ENTER 3 ENTER ZOOM 6.

Press 2nd [CALC] 5 and use the key to move the cursor to the left of the first intersection point. Press ENTER . Then move

the cursor to the right of the intersection point and press ENTER

ENTER . One of the values used in the solution set is displayed. Repeat the procedure on the other intersection point.

The solution is { x | -3.24 ≤ x ≤ 1.24}.

Example 1

Example 2

[-4.7, 4.7] scl:1 by [-3.1, 3.1] scl:1

[-10, 10] scl:1 by [-10, 10] scl:1

[-10, 10] scl:1 by [-10, 10] scl:1


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4 Student Recording Sheet

Read each question. Then fill in the correct answer.

1. A B C D

2. F G H J

3. A B C D

4. F G H J

5. A B C D

6. F G H J

7. A B C D

8. F G H J

Multiple Choice

Short Response/Gridded Response

Record your answer in the blank.

For gridded response, also enter your answer in the grid by writing each number or symbol in a box. Then fill in the corresponding circle for that number or symbol.

9. ———————— (grid in)

10. ———————— (grid in)

11. a. ———————

b. ———————

c. ———————

12. ————————

13. ————————

9. 10.

Extended Response

Record your answers for Questions 14–16 on the back of this paper.

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General Scoring Guidelines

• If a student gives only a correct numerical answer to a problem but does not show how he or she arrived at the answer, the student will be awarded only 1 credit. All extended response questions require the student to show work.

• A fully correct answer for a multiple-part question requires correct responses for all parts of the question. For example, if a question has three parts, the correct response to one or two parts of the question that required work to be shown is not considered a fully correct response.

• Students who use trial and error to solve a problem must show their method. Merely showing that the answer checks or is correct is not considered a complete response for full credit.

Exercises 14–16 Rubric

Score Specifi c Criteria

4 For Exercise 14, the three possible results of the discriminant (positive, negative, or zero) are explained. Positive has two real roots, zero has one real root, and negative has no real roots.For Exercise 15, a system of equation is written in part a. One equation should be based on the cost of the printers; the second equation will be based on the number of each that exist. In part b, the system of equations is set up in a matrix and used in part c to find the inverse to solve the system of equations.For Exercise 16, the quadratic equation is graphed in part a. The vertex of this graph will be used in parts b and c to solve for maximum height and time.

3 A generally correct solution, but may contain minor flaws in reasoning or computation.

2 A partially correct interpretation and/or solution to the problem.

1 A correct solution with no evidence or explanation.

0 An incorrect solution indicating no mathematical understanding of the concept or task, or no solution given.

Rubric for Scoring Extended Response


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For Questions 1 and 2, consider f(x) = x2 + 2x - 3. 1. Find the y-intercept, the equation of the axis of symmetry,

and the x-coordinate of the vertex. 1.

2. Graph the function, labeling the y-intercept, vertex, and 2. axis of symmetry.

3. MULTIPLE CHOICE Determine the maximum or minimum value of f (x) = 2x2 - 8x + 9.

A min. 1 B max. 1 C min. 2 D max. 2 3.

Solve each equation. If exact roots cannot be found, state the consecutive integers between which the roots are located.

4. x2 - 2x = 3 5. x2 + 4x - 7 = 0 4.

5.

For Questions 1 and 2, solve each equation by factoring.

1. 3x2 = 10 - 13x 2. x2 + 4x = 45

3. MULTIPLE CHOICE Solve 5x2 + 100 = 0.

A ±2 √ � 5 B ±10 √ � 5 C ±2i √ � 5 D ±10i √ � 5

Write a quadratic equation with the given roots. Write the equation in the form ax2 + bx + c = 0, where a, b, and c are integers.

4. -6 and 2 5. 2 − 3 and -4

Simplify.

6. √ �� -80 7. √ �� -6 � √ �� -12

8. (6 - 9i) - (17 - 12i) 9. (7 - 3i)(8 + 4i)

10. 2 + i − 3 - i

xO

f(x)

1.

2.

3.

4.

5.

6.

7.

8.

9.

10.

4

Chapter 4 Quiz 1(Lessons 4-1 and 4-2)



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4

4

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Solve each equation by using the Square Root Property.

1. x2 + 8x + 16 = 36 2. x2 - 2x + 1 = 45 1.

2.

3. MULTIPLE CHOICE What is the solution of x2 - 10x = 11?

A {-11, 1} C -5 ± √ �� 14

B {-1, 11} D 5 ± √ �� 14 3.

4. Solve x2 - 4x = 1 by using the Quadratic Formula. Find exact solutions. 4.

5. Find the value of the discriminant for 3x2 = 6x - 11. Then describe the number and type of roots for the equation. 5.

1. Graph y = -(x - 2)2 - 1. Show and label the vertex and 1. axis of symmetry.

2. Write an equation for the parabola whose vertex is at (-5, 0) and passes through (0, 50). 2.

3. Graph y ≤ - 1 − 3 (x + 2)2 + 3. 3.

4. Use the graph of its related function to write the solutions of -x2 + 6x - 5 > 0. 4.

5. MULTIPLE CHOICE What is the solution of 4x2 + 1 ≥ 4x?

A all reals B empty set C {x ⎪x ≥ 1 − 2 } D {x ⎪x ≤ 1 −

2 } 5.

xO

y


y

xO



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4

Part I Write the letter for the correct answer in the blank at the right of each question.

1. Which function is graphed?

A f (x) = x2- 2x - 3

B f (x) = x2+ 2x - 3

C f (x) = x2+ x - 3

D f (x) = (x - 3)2 1.

2. By the Zero Product Property, if (2x - 1)(x - 5) = 0, then .

F x = 1 or x = 5 H x =1−2

or x = 5

G x = - 1−2

or x = -5 J x = -1 or x = -5 2.

3. Write a quadratic equation with 7 and 2−5

as its roots. Write the equation in the form ax2

+ bx + c = 0, where a, b, and c are integers.

A 5x2- 37x + 14 = 0 C 5x2

+ 37x + 14 = 0 B 2x2

+ 9x - 35 = 0 D 2x2- 9x - 35 = 0 3.

4. The current in one part of a series circuit is 3 - 2j amps. The current in another part of the circuit is 2 + 4j amps. Find the total amps in the circuit.

F 5 + 2j H 1 + 2j G 6 - 8j J 7j 4.

5. Solve x2+ 6x = -6. If exact roots cannot be found, state the

consecutive integers between which the roots are located.

A -2, -3 C between -4 and -3; between -2 and -1 B -3 D between -5 and -4; between -2 and -1 5.

Part II

6. Solve x2- 4x + 3 = 0 by graphing. 6.

7. Determine whether f (x) = 1−2

x2- x - 9 7.

has a maximum or a minimum value and find that value.

For Questions 8 and 9, solve each equation by factoring.

8. x2- 7x = 18 9. 4x2

= x 9.

10. Simplify 5i−3 - 5i

. 10.

y

xO

Chapter 4 Mid-Chapter Test(Lessons 4-1 through 4-4)

8.

O

f (x)4

2

2 4

–2

–4

-4 -2 x


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Write whether each sentence is true or false. If false, replace the underlined word or words to make a true sentence.

1. Two complex numbers of the form a + bi and a - bi are called imaginary units. 1.

2. In f(x) = 3x2 - 2x + 5, the linear term is 5. 2.

3. 2x2 + 3x - 4 ≤ 0 is an example of a quadratic equation. 3.

4. The solutions of a quadratic equation are called its zeros. 4.

5. The quadratic equation y = 2(x + 3)2 - 1 is written in vertex form. 5.

6. If a parabola opens upward, the y-coordinate of the vertex is the maximum value. 6.

7. In f(x) = -x2 + 2x - 1, the constant term is -x2. 7.

8. Pure imaginary numbers are square roots of negative real numbers. 8.

9. The highest or lowest point on a parabola is called the vertex. 9.

10. In the Quadratic Formula, the expression b2 - 4ac is called the quadratic term. 10.

Define each term in your own words.

11. discriminant 11.

12. standard form 12.

completing the square

complex conjugates

complex number

constant term

discriminant

imaginary unit

linear term

maximum value

minimum value

pure imaginary number

quadratic equation

Quadratic Formula

quadratic inequality

quadratic term

root

standard form

vertex

vertex form

zero

Chapter 4 Vocabulary Test


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Write the letter for the correct answer in the blank at the right of each question.

1. Find the y-intercept for f (x) = -(x + 1)2. A 1 B -1 C x D 0 1.

2. What is the equation of the axis of symmetry of y = -3(x + 6)2 + 12? F x = 2 G x = -6 H x = 6 J x = -18 2.

3. The graph of f (x) = -2x2 + x opens and has a value. A down; maximum C up; maximum B down; minimum D up; minimum 3.

4. The related graph of a quadratic equation is shown at the right. Use the graph to determine the solutions of the equation.

F -2, 3 H -3, 2 G 0, -6 J 0, 2 4.

5. The quadratic function f (x) = x2 has . A no zeros C exactly two zeros B exactly one zero D more than two zeros 5.

6. Solve x2 - 3x - 10 = 0 by factoring. F {-5, 2} G (-2, -5) H {-2, 5} J {-10, 1} 6.

7. Which quadratic equation has roots -2 and 3? A x2 + x + 6 = 0 C x2 - 6x + 1 = 0 B x2 - x - 6 = 0 D x2 + x - 6 = 0 7.

8. Simplify (5 + 2i)(1 + 3i). F 5 + 6i G -1 H -1 + 17i J 11 + 17i 8.

9. ELECTRICITY The total impedance of a series circuit is the sum of the impedances of all parts of the circuit. A technician determined that the impedance of the first part of a particular circuit was 2 + 5j ohms. The impedance of the remaining part of the circuit was 3 - 2j ohms. What was the total impedance of the circuit?

A 5 + 3j ohms C -1 + 7j ohms B 5 + 7j ohms D 16 + 11j ohms 9.

10. To solve x2 + 8x + 16 = 25 by using the Square Root Property, you would first rewrite the equation as .

F (x + 4)2 = 25 H x2 + 8x - 9 = 0 G (x + 4)2 = 5 J x2 + 8x = 9 10.

xO

f (x)2

2 4

–2

–4

–6

-4 -2

Chapter 4 Test, Form 1


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11. Find the value of c that makes x2 + 10x + c a perfect square. A 100 B 25 C 10 D 50 11.

12. The quadratic equation x2 + 6x = 1 is to be solved by completing the square. Which equation would be the first step in that solution?

F x2 + 6x - 1 = 0 H x2 + 6x + 36 = 1 + 36 G x(x + 6) = 1 J x2 + 6x + 9 = 1 + 9 12.

13. Find the exact solutions to x2 - 3x + 1 = 0 by using the Quadratic Formula.

A -3 ± √ � 5 −

2 B 3 ± √ �� 13

− 2 C -3 ± √ �� 13

− 2 D 3 ± √ � 5

− 2 13.

For Questions 14 and 15, use the value of the discriminant to determine the number and type of roots for each equation.

14. x2 - 3x + 7 = 0 F 2 complex roots H 2 real, irrational roots G 2 real, rational roots J 1 real, rational root 14.

15. x2 = 4x - 4 A 2 real, rational roots C 1 real, rational root B 2 real, irrational roots D no real roots

16. What is the vertex of y = 2(x - 3)2 + 6? F (-3, -6) G (3, -6) H (-3, 6) J (3, 6)

17. What is the equation of the axis of symmetry of y = -3(x + 6)2 + 1? A x = 2 B x = -6 C x = -3 D x = 6 17.

18. Which quadratic function has its vertex at (2, 3) and passes through (1, 0)? F y = 2(x - 2)2 + 3 H y = -3(x + 2)2 + 3

G y = -3(x - 2)2 + 3 J y = 2(x - 2)2 - 3 18.

19. Which quadratic inequality is graphed at the right?

A y ≥ (x + 1)2 + 4 B y ≤ -(x + 1)2 + 4 C y ≤ -(x - 1)2 + 4

D y ≤ -(x - 1)2 - 4 19.

20. Solve (x - 4)(x + 2) ≤ 0. F {x ⎪x ≤ -2 or x ≥ 4} H {x ⎪-4 ≤ x ≤ 2} G {x ⎪-2 ≤ x ≤ 4} J {x ⎪x = -2 or x = 4} 20.

Bonus Find the x-intercepts and the y-intercept of the graph of y = 2(x - 4)2 - 18. B:

y

xO

4

2

2 4

–2

–4

-4 -2

Chapter 4 Test, Form 1 (continued)

15.

16.


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1. Identify the y-intercept and the axis of symmetry for the graph of f (x) = 10x2 + 40x + 42.

A 42; x = 4 B 0; x = -4 C 42; x = -2 D -42; x = 2 1.

2. Identify the quadratic function graphed at the right. F f (x) = -x2 - 2x G f (x) = -x2 + 2x H f (x) = x2 - 2x

J f (x) = -(x + 2)2 2.

3. Determine whether f (x) = 4x2 - 16x + 6 has a maximum or a minimum value and find that value.

A minimum; -10 B minimum; 2 C maximum; -10 D maximum; 2 3.

4. Solve -x2 = 4x by graphing. If exact roots cannot be found, state the consecutive integers between which the roots are located.

F 4, 0 H -4, 0 G between -4 and 4 J -2, 4 4.

5. Solve x2 - 3x = 18 by factoring.

A {6} B {-6, 3} C {-9, 2} D {-3, 6} 5.

6. Which quadratic equation has roots -2 and 1 − 5 ?

F x2 + 4x + 4 = 0 H 5x2 - 9x - 2 = 0 G 5x2 + 9x - 2 = 0 J 5x2 - 11x + 2 = 0 6.

7. Simplify (4 - 12i) - (-8 + 4i). A 12 - 8 B 28 C 12 - 16i D 12 + 16i 7.

8. Simplify 4 - 2i − 7 + 3i

.

F 11 − 29

- 13 − 29

i G 11 − 29

- 14 − 29

i H 13 − 29

- 17 − 29

i J 11 − 29

- 13 − 29

i 8.

9. To solve 9x2 - 12x + 4 = 49 by using the Square Root Property, you would first rewrite the equation as .

A 9x2 - 12x - 45 = 0 C (3x - 2)2 = 7 B (3x - 2)2 = ±49 D (3x - 2)2 = 49 9.

10. Find the value of c that makes x2 - 9x + c a perfect square.

F 81 − 4 G 9 −

2 H - 81 −

4 J 81 10.

xO

f(x)2

2 4

–2

–4

-4 -2

Chapter 4 Test, Form 2A


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11. The quadratic equation x2 - 8x = -20 is to be solved by completing the square. Which equation would be a step in that solution? A (x - 4)2 = 4 C x2 - 8x + 20 = 0

B x - 4 = ± 2i D x2 - 8x + 16 = -20 11.

12. Find the exact solutions to 3x2 = 5x - 1 by using the Quadratic Formula.

F -5 ± √ �� 13 −

6 G 5 ± √ �� 13

− 6 H 5 ± √ �� 37

− 6 J 5 ±

√ �� 13 −

6


13. 2x2 - 7x + 9 = 0 A 2 real, rational C 2 complex B 2 real, irrational D 1 real, rational 13.

14. x2 + 20 = 12x - 16 F 1 real, irrational H 2 real, rational G no real J 1 real, rational 14.

15. Identify the vertex, axis of symmetry, and direction of opening for y = 1 −

2 (x - 8)2 + 2.

A (-8, 2); x = -8; up C (8, -2); x = 8; up B (-8, -2); x = -8; down D (8, 2); x = 8; up 15.

16. Which quadratic function has its vertex at (-2, 7) and opens down? F y = -3(x + 2)2 + 7 H y = (x - 2)2 + 7 G y = -12(x + 2)2 - 7 J y = -2(x - 2)2 + 7 16.

17. Write y = x2 + 4x - 1 in vertex form. A y = (x - 2)2 + 5 C y = (x + 2)2 - 1 B y = (x + 2)2 - 5 D y = (x + 2)2 + 3 17.

18. Write an equation for the parabola whose vertex is at (-8, 4) and passes through (-6, -2).

F y = - 3 − 2 (x + 8)2 + 4 H y = - 1 −

4 (x + 8)2 + 4

G y = 3 − 2 (x + 6)2 - 2 J y = - 3 −

2 (x - 8)2 + 4 18.

19. Which quadratic inequality is graphed at the right?A y ≥ (x - 2)(x + 3) C y > (x + 2)(x - 3)B y > (x - 2)(x + 3) D y < (x + 2)(x - 3) 19.

20. Solve x2 ≥ 2x + 24. F {x ⎪-4 ≤ x ≤ 6} H {x ⎪-6 ≤ x ≤ 4} G {x ⎪x ≤ -6 or x ≥ 4} J {x ⎪x ≤ -4 or x ≥ 6} 20.

Bonus Write a quadratic equation with roots ± i √ � 3 −

4 . B:

yxO 2 4

–2

–4

–6

-4 -2

Chapter 4 Test, Form 2A (continued)

12.


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1. Identify the y-intercept and the axis of symmetry for the graph of f (x) = -3x2 + 6x + 12.

A 2; x = -12 B 12; x = 1 C -2; x = 0 D -12; x = -1 1.

2. Identify the quadratic function graphed at the right. F f (x) = x2 - 4xG f (x) = -x2 + 4x H f (x) = -x2 - 4xJ f (x) = -(x + 4)2 2.

3. Determine whether f (x) = -5x2 - 10x + 6 has a maximum or a minimum value and find that value.

A minimum; -1 B maximum; 11 C maximum; -1 D minimum; 11 3.

4. Solve x2 = 4x by graphing. If exact roots cannot be found, state the consecutive integers between which the roots are located.

F -4, 0 H between -4 and 4 G 2, -4 J 0, 4 4.

5. Solve x2 - 3x = 28 by factoring. A {-4, 7} B {-14, 2} C {-7, 4} D {-2, 14} 5.

6. Which quadratic equation has roots 7 and - 2 − 3 ?

F 2x2 - 11x - 21 = 0 H 3x2 - 19x - 14 = 0 G 3x2 + 23x + 14 = 0 J 2x2 + 11x - 21 = 0 6.

7. Simplify (15 - 13i) - (-1 + 17i). A 16 - 30i B 16 + 4i C 16 + 30i D 46 7.

8. Simplify 1 + 2i − 2 - 3i

.

F 8 − 7 + 1 −

7 i G 8 −

7 + i H -4 + 7i J - 4 −

13 + 7 −

13 i 8.

9. To solve 4x2 - 28x + 49 = 25 by using the Square Root Property, you would first rewrite the equation as .

A (2x - 7)2 = 25 C (2x - 7)2 = ±5 B (2x - 7)2 = 5 D 4x2 - 28x + 24 = 0 9.

10. Find the value of c that makes x2 + 5x + c a perfect square trinomial.

F 25 − 16

G 5 − 4 H 25 −

4 J 25 −

4

xO

f(x)4

2

2-4 -2

Chapter 4 Test, Form 2B

10.


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11. The quadratic equation x2 - 18x = -106 is to be solved by completing the square. Which equation would be a step in that solution? A (x - 9)2 = 25 C x - 9 = ±5i B x2 - 18x + 106 = 0 D x2 - 18x + 81 = -106 11.

12. Find the exact solutions to 2x2 = 5x - 1 by using the Quadratic Formula.

F -5 ± √ �� 17 −

4 G 5 ± √ �� 17

− 4 H 5 ± √ �� 33

− 4 J 5 ± √ �� 17

− 2


13. 3x2 - x - 12 = 0 A 2 complex roots C 2 real, rational roots B 1 real, rational root D 2 real, irrational roots 13.

14. x2 + 10 = 3x - 3 F 2 complex roots H 2 real, irrational roots G 1 real, rational root J 2 real, rational roots 14.

15. Identify the vertex, axis of symmetry, and direction of opening for y = -8(x + 2)2.

A (-8, -2); x = -8 up C (2, 0); x = 2; down B (-2, 0); x = -2; down D (-2, -8); x = -2; down 15.

16. Which quadratic function has its vertex at (-3, 5) and opens down? F y = (x - 3)2 + 5 H y = (x + 3)2 - 5 G y = -(x + 3)2 + 5 J y = -(x - 3)2 + 5 16.

17. Write y = x2 - 18x + 52 in vertex form. A y = (x - 9)2 + 113 C y = (x - 9)2 + 52 B y = (x + 9)2 - 29 D y = (x - 9)2 - 29 17.

18. Write an equation for the parabola whose vertex is at (-5, 7) and passes through (-3, -1).

F y = - 1 − 11

(x + 5)2 + 7 H y = -2(x + 5)2 + 7

G y = - 1 − 2 (x + 5)2 + 7 J y = - 1 −

2 (x - 5)2 + 7 18.

19. Which quadratic inequality is graphed at the right? A y ≤ (x - 3)(x + 1) C y ≥ (x + 3)(x - 1) B y > (x - 3)(x + 1) D y > (x + 3)(x - 1) 19.

20. Solve 2x + 3 ≥ x2. F {x ⎪-1 ≤ x ≤ 3} H {x ⎪-3 ≤ x ≤ 1} G {x ⎪x ≤ -1 or x ≥ 3} J {x ⎪x ≤ -3 or x ≥ 1} 20.

Bonus Write a quadratic equation with roots ± i √ � 2 −

3 . B:

y

xO 2 4

–2

–4

-4 -2

Chapter 4 Test, Form 2B (continued)

12.


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1. Graph f (x) = -5x2 + 10x, labeling the y-intercept, vertex, 1. and axis of symmetry.

2. Determine whether f (x) = -3x2 + 6x + 1 has a maximum or a minimum value and find that value. 2.

3. Solve x2 = 6x - 8 by graphing. If exact roots cannot befound, state the consecutive integers between which the roots are located. 3.

4. Solve 5x2 + 13x = 6 by factoring. 4.

5. GEOMETRY The length of a rectangle is 7 inches longer than its width. If the area of the rectangle is 144 square inches, what are its dimensions? 5.

6. ELECTRICITY The total impedance of a series circuit is the sum of the impedances of all parts of the circuit. Suppose that the first part of a circuit has an impedance of 6 - 5j ohms and that the total impedance of the circuit was 12 + 7j ohms. What is the impedance of the remainder of the circuit? 6.

7. ELECTRICITY In an AC circuit, the voltage E (in volts), current I (in amps), and impedance Z (in ohms) are related by the formula E = I � Z. Find the current in a circuit with voltage 10 - 3j volts and impedance 4 + j ohms. 7.

8. Write a quadratic equation with -6 and 3 − 4 as its roots.

Write the equation in the form ax2 + bx + c = 0, where a, b, and c are integers. 8.

9. Solve x2 + 6x + 9 = 25 by using the Square Root Property. 9.

xO

f (x)

xO

y

Chapter 4 Test, Form 2C


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For Questions 10 and 11, solve each equation by completing the square.

10. x2 + 4x - 9 = 0 10.

11. 2x2 + 3x - 2 = 0 11.

12. Find the exact solutions to 5x2 = 3x - 2 by using the Quadratic Formula. 12.

For Questions 13 and 14, find the value of the discriminant for each quadratic equation. Then describe the number and type of roots for the equation.

13. 9x2 - 12x + 4 = 0 13.

14. 4x2 + 1 = 9x - 2 14.

15. Identify the vertex, axis of symmetry, and direction of

opening for y = - 2 − 3 (x + 5)2 - 7. 15.

16. Write an equation for the parabola with vertex at (2, -1) and y-intercept 5. 16.

17. Write y = x2 - 6x + 8 in vertex form. 17.

18. PHYSICS The height h (in feet) of a certain rocket t seconds after it leaves the ground is modeled by h(t) = -16t2 + 48t + 15. Write the function in vertex form and find the maximum height reached by the rocket. 18.

19. Graph y < x2 + 6x + 9. 19.

20. Solve 2x2 - 5x - 3 ≥ 0 algebraically. 20.

Bonus Write a quadratic equation with roots ± √ � 7

− 3 . Write the

equation in the form ax2 + bx + c = 0, where a, b, and c are integers. B:

y

xO

Chapter 4 Test, Form 2C (continued)


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1. Graph f (x) = x2 - 4x + 3, labeling the y-intercept, vertex, 1. and axis of symmetry.

2. Determine whether f (x) = 5x2 - 20x + 3 has a maximum or a minimum value and find that value. 2.

3. Solve x2 + 2x - 3 = 0 by graphing. If exact roots cannot be found, state the consecutive integers between which the roots are located. 3.

4. Solve 3x2 - x = 4 by factoring. 4.

5. GEOMETRY The length of a rectangle is 10 inches longer than its width. If the area of the rectangle is 144 square inches, what are its dimensions? 5.

6. ELECTRICITY The total impedance of a series circuit is the sum of the impedances of all parts of the circuit. Suppose that the first part of a circuit has an impedance of 7 + 4j ohms and that the total impedance of the circuit was 16 - 2j ohms. What is the impedance of the remainder of the circuit? 6.

7. ELECTRICITY In an AC circuit, the voltage E (in volts), current I (in amps), and impedance Z (in ohms) are related by the formula E = I � Z. Find the impedance in a circuit with voltage 12 + 2j volts and current 3 + 5j amps. 7.

8. Write a quadratic equation with -4 and 3 − 2 as its roots.


y

xO

xO

f(x)

Chapter 4 Test, Form 2D


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9. Solve 9x2 + 12x + 4 = 6 by using the Square Root Property. 9.


10. x2 - 8x + 14 = 0 10.

11. 3x2 + x - 2 = 0 11.

12. Find the exact solutions to 2x2 = 9x - 5 by using the Quadratic Formula. 12.

For Questions 13 and 14, find the value of the discriminant for each quadratic equation. Then describe the number and type of roots for the equation.

13. 25x2 - 20x + 4 = 0 13.

14. 2x2 + 10x + 9 = 2x 14.

15. Identify the vertex, axis of symmetry, and direction of opening for y = -(x - 6)2 - 5. 15.

16. Write an equation for the parabola with vertex at (-4, 2) and y-intercept -2. 16.

17. Write y = x2 + 4x + 8 in vertex form. 17.

18. PHYSICS The height h (in feet) of a certain rocket t seconds after it leaves the ground is modeled by h(t) = -16t2 + 64t + 12. Write the function in vertex form and find the maximum height reached by the rocket. 18.

19. Graph y ≥ x2 - 4x + 4. 19.

20. Solve 2x2 - 7x - 15 < 0 algebraically. 20.

Bonus Write a quadratic equation with roots ± √ � 5

− 4 .

Write the equation in the form ax2 + bx + c = 0, where a, b, and c are integers. B:

y

xO

Chapter 4 Test, Form 2D (continued)


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1. Graph f (x) = 3 + 3x2 + 2x, labeling the y-intercept, vertex, 1. and axis of symmetry.

2. Determine whether f (x) = 1 - 3 − 5 x + 3 −

4 x2

has a maximum or a minimum value and find that value.

3. BUSINESS Khalid charges $10 for a one-year subscription to his on-line newsletter. Khalid currently has 600 subscribers and he estimates that for each $1 decrease in the subscription price, he would gain 100 new subscribers. What subscription price will maximize Khalid’s 2. income? If he charges this price, how much income should Khalid expect? 3.

For Questions 4 and 5, solve each equation by graphing. If exact roots cannot be found, state the consecutive integers between which the roots are located.

4. 0.5x2 + 9 = 4.5x 4.

5. 2 − 3 x + 3 = 1 −

3 x2 5.

6. Solve 18x2 + 15 = 39x by factoring. 6.

7. Write a quadratic equation with - 2 − 3 and 1.75 as its roots.


8. Simplify (5 - i) + (2 - 4i) - (3 + i). 8.

9. Simplify 2 - i √ � 5 −

2 + i √ � 5 . 9.

y

xO

y

xO

xO

f (x)

Chapter 4 Test, Form 3


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10. Solve 4x2 - 2x + 0.25 = 1.44 by using the Square Root Property. 10.

For Questions 11 and 12, solve each equation by completing the square.

11. 2x2 - 5 − 2 x + 2 = 0 11.

12. x2 + 2.5x - 3 = 0.5 12.

13. Find the exact solutions to 1 − 4 x2 - 3x + 1 = 0 by using the

Quadratic Formula. 13.

14. Find the value of the discriminant for 3x(0.2x - 0.4) + 1 = 0.9. Then describe the number and type of roots for the equation. 14.

15. Find all values of k such that x2 + kx + 1 = 0 has two complex roots. 15.

16. Write an equation of the parabola with equation

y = - 3 − 5 (x - 1 −

2 )

2

- 5 − 2 , translated 4 units left and 2 units up.

Then identify the vertex, axis of symmetry, and direction of opening of your function. 16.

17. PHYSICS The height h (in feet) of a certain aircraft t seconds after it leaves the ground is modeled by h(t) = -9.1t2 + 591.5t + 20,388.125. Write the function in vertex form and find the maximum height reached by the aircraft. 17.

18. Write an equation for the parabola that has the same

vertex as y = 1 − 3 x2 + 6x + 83 −

2 and x-intercept 1. 18.

19. Graph y < -(x2 + 2x) + 5.25. 19.

20. Solve (x + 7 − 2 ) (x - 1) 2 ≤ 0. 20.

Bonus Write a quadratic equation with roots -3 ± 2i √ � 5 −

4 .

Write the equation in the form ax2 + bx + c = 0, where a, b, and c are integers. B:

y

xO

Chapter 4 Test, Form 3 (continued)


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Demonstrate your knowledge by giving a clear, concise solution to each problem. Be sure to include all relevant drawings and justify your answers. You may show your solution in more than one way or investigate beyond the requirements of the problem.

1. Mr. Moseley asked the students in his Algebra class to work in groups to solve (x - 3)2 = 25, stating that each student in the first group to solve the equation correctly would earn five bonus points on the next quiz. Mi-Ling’s group solved the equation using the Square Root Property. Emilia’s group used the Quadratic Formula to find the solutions. In which group would you prefer to be? Explain your reasoning.

2. The next day, Mr. Moseley had his students work in pairs to review for their chapter exam. He asked each student to write a practice problem for his or her partner. Len wrote the following problem for his partner, Jocelyn: Write an equation for the parabola whose vertex is (-3, -4), that passes through (-1, 0), and opens down.

a. Jocelyn had trouble solving Len’s problem. Explain why.b. How would you change Len’s problem?c. Make the change you suggested in part b and complete the

problem.

3. a. Write a quadratic function in vertex form whose maximum value is 8.

b. Write a quadratic function that transforms the graph of your function from part a so that it is shifted horizontally. Explain the change you made and describe the transformation that results from this change.

4. When asked to write f (x) = 2x2 + 12x - 5 in vertex form, Joseph wrote:

f (x) = 2x2 + 12x - 5 Step 1 f (x) = 2(x2 + 6x) - 5 Step 2 f (x) = 2(x2 + 6x + 9) - 5 + 9 Step 3 f (x) = 2(x + 3)2 + 4 Is Joseph’s answer correct? Explain your reasoning.

5. The graph of y = x2 + 4x + 4 is shown. Susan used this graph to solve three quadratic inequalities. Her three solutions are given below. Replace each ● with an inequality symbol (<, >, ≤, ≥) so that each solution is correct. Explain your reasoning for each.

a. The solution of x2 + 4x + 4 ● 0 is {x ⎪x < -2 or x > -2}.

b. The solution of x2 + 4x + 4 ● 0 is ∅.c. The solution of x2 + 4x + 4 ● 0 is all real numbers.

y

xO

4

2

2 4

–2

–4

-4 -2

Chapter 4 Extended-Response Test


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Part 1: Multiple Choice

Instructions: Fill in the appropriate circle for the best answer.

1. If a − b = 3 −

2 ,then 8a equals which of the following?

A 16b B 12b C 3b − 2 D 8 −

3 b 1. A B C D

2. 20% of 3 yards is how many fifths of 9 feet?

F 1 G 6 H 10 J 15 2. F G H J

3. If u > v and t > 0, which of the following are true? I. ut > vt II. u + t > v + t III. u - t > v - t A I only C I and II only B III only D I, II, and III 3. A B C D

4. Which of the following is the greatest?

F 2 − 3 G 7 −

9 H 10 −

15 J 8 −

11 4. F G H J

5. If 2a + 3b represents the perimeter of a rectangle and a - 2b represents its width, the length is .

A 7b B b C 7b − 2 D 14b 5. A B C D

6. In the figure, what is the area of the shaded region?

F 30 units2 H 36 units2

G 54 units2 J 27 units2 6. F G H J

7. Mr. Salazár rented a car for d days. The rental agency charged x dollars per day plus c cents per mile for the model he selected. When Mr. Salazár returned the car, he paid a total of T dollars. In terms of d, x, c, and T, how many miles did he drive?

A T - (xd + c) B T - xd − c C T − xd + c

D T - xd − c 7. A B C D

8. If P(3, 2) and Q(7, 10) are the endpoints of the diameter of a circle, what is the area of the circle in square units?

F 2 √ � 5 π G 80π H 4 √ � 5 π J 20π 8. F G H J

9. If (x - y)2 = 100 and xy = 20, what is the value of x2 + y2?

A 120 B 140 C 80 D 60 9. A B C D

10. The tenth term in the sequence 7, 12, 19, 28, . . . is . F 124 G 103 H 57 J 147 10. F G H J

6

3

15

Standardized Test Practice(Chapters 1–4)


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Part 2: Gridded Response

Instructions: Enter your answer by writing each digit of the answer in a column box

and then shading in the appropriate circle that corresponds to that entry.

15. The bar graph shows the distribution of votes 15. among the candidates for senior class president. If 220 seniors voted in all, how many students voted for either Theo or Pam?

16. Find the median of x, 2x + 1, x − 2 - 13, 45, and 16.

x + 22 if the mean of this set of numbers is 83.

JoeyAnaPamTheo

20

30

10

0

40

Perc

ent

of

vote

s re

ceiv

ed 50

Candidates

11. If t2 + 6t = -9, what is the value of (t - 1 − 2 )

2?

A -3 B 12 1 − 4 C 6 1 −

4 D -12 1 −

4 11. A B C D

12. All four walls of a rectangular room that is 14 feet wide, 20 feet long, and 8 feet high, are to be painted. What is the minimum cost of paint if one gallon covers at most 130 square feet and the paint costs $22 per gallon?

F $92 G $102 H $110 J $190 12. F G H J

13. If i 2 = -1, then what is the value of i32? A -1 B 1 C -i D i 13. A B C D

14. Which of the following is the sum of both solutions of the equation x2 + x - 42 = 0?

F 13 G -1 H -13 J 1 14. F G H J

Standardized Test Practice (continued)

9

8

7

6

5

4

3

2

1

0

9

8

7

6

5

4

3

2

1

0

9

8

7

6

5

4

3

2

1

0

9

8

7

6

5

4

3

2

1

0

9

8

7

6

5

4

3

2

1

0

9

8

7

6

5

4

3

2

1

0

9

8

7

6

5

4

3

2

1

0

9

8

7

6

5

4

3

2

1

0

9

8

7

6

5

4

3

2

1

0

9

8

7

6

5

4

3

2

1

0


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4

17. Find the value of 12 + 36 ÷ 4 - (5 - 7)2. 17.

18. Find the slope of the line that is parallel to the line with equation 3x + 4y = 10. 18.

19. Describe the system 2x - 3y = 21 and y - 5 = 2 − 3 x as

consistent and independent, consistent and dependent, or inconsistent. 19.

20. Find the coordinates of the vertices of the figure formed by the system of inequalities. x ≥ -2 x + y ≤ 6y ≥ -2 x + y ≥ -2 20.

21. Find the value of ⎪ 5 12 -6 4

⎥ . 21.

22. Solve [ 4 -1 -2 3

] � [a b

] = [ 11 -13

] by using inverse matrices. 22.

23. Solve 2x2 + 40 = 0. 23.

24. PHYSICS An object is thrown straight up from the top of a 100-foot platform at a velocity of 48 feet per second. The height h(t) of the object t seconds after being thrown is given by h(t) = -16t2 + 48t + 100. Find the maximum height reached by the object and the time it takes to achieve this height. 24.

25. Solve x2 = 2x + 3 by graphing. 25.

26. Solve 4x2 - 4x = 24 by factoring. 26.

27. Find the value of the discriminant for 7x2 + 5x + 1 = 0. Then describe the number and type of roots for the equation. 27.

28. Use y = x2 - 7x + 5 for parts a – c.a. Write the equation in vertex form. 28a.

b. Identify the vertex. 28b.

c. Identify the axis of symmetry. 28c.

y

xO

Standardized Test Practice (continued)

Part 3: Short Response

Instructions: Write your answer in the space provided.


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Chapter 4 A1 Glencoe Algebra 2

Copyright © Glencoe/McGraw-Hill, a division of The McGraw-Hill Companies, Inc.

Lesson 4-1

NA

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Cha

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) + 5

5(0

, 5)

112 -

3(1

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3 −

2 ( 3 −

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3 −

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5 11

−

4 ( 3 −

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222 -

3(2

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332 -

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f(x)

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x +

3

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xO12

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4-

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, x =

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2, x

= -

1, -

1 3

, x =

1, 1

x-

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x)-

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Chapter Resources

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rite

A o

r D

in t

he fi

rst

colu

mn

OR

if y

ou a

re n

ot s

ure

whe

ther

you

agr

ee o

r di

sagr

ee,

wri

te N

S (N

ot S

ure)

.

A

fter

you

com

plet

e C

hapt

er 4

• R

erea

d ea

ch s

tate

men

t an

d co

mpl

ete

the

last

col

umn

by e

nter

ing

an A

or

a D

.

• D

id a

ny o

f you

r op

inio

ns a

bout

the

sta

tem

ents

cha

nge

from

the

firs

t co

lum

n?

• Fo

r th

ose

stat

emen

ts t

hat

you

mar

k w

ith

a D

, use

a p

iece

of p

aper

to

wri

te a

n ex

ampl

e of

why

you

dis

agre

e.

STE

P 1

A

, D, o

r N

SS

tate

men

tST

EP

2

A o

r D

1.

All

quad

rati

c fu

ncti

ons

have

a t

erm

wit

h th

e va

riab

le t

o th

e se

cond

pow

er.

2.

If t

he g

raph

of t

he q

uadr

atic

func

tion

y =

ax2 +

c o

pens

up

then

c <

0.

3.

A q

uadr

atic

equ

atio

n w

hose

gra

ph d

oes

not

inte

rsec

t th

e x-

axis

has

no

real

sol

utio

n.

4.

Sinc

e gr

aphi

ng s

how

s th

e ex

act

solu

tion

s to

a q

uadr

atic

eq

uati

on, n

o ot

her

met

hod

is n

eces

sary

for

solv

ing.

5.

If (x

- 3

)(x +

4) =

0, t

hen

eith

er x

- 3

= 0

or

x +

4 =

0.

6.

An

imag

inar

y nu

mbe

r co

ntai

ns i,

whi

ch e

qual

s th

e sq

uare

roo

t of

-1.

7.

A m

etho

d ca

lled

com

plet

ing

the

squa

re c

an b

e us

ed t

o re

wri

te a

qua

drat

ic e

xpre

ssio

n as

a p

erfe

ct s

quar

e.

8.

The

quad

rati

c fo

rmul

a ca

n on

ly b

e us

ed fo

r qu

adra

tic

equa

tion

s th

at c

anno

t be

sol

ved

by g

raph

ing

or

com

plet

ing

the

squa

re.

9.

The

disc

rim

inan

t of

a q

uadr

atic

equ

atio

n ca

n be

use

d to

de

term

ine

the

dire

ctio

n th

e gr

aph

will

ope

n.

10.

The

grap

h of

y =

2x2 i

s a

dila

tion

of t

he g

raph

of y

= x

2 .

11.

The

grap

h of

y =

(x +

2)2 w

ill b

e tw

o un

its

to t

he r

ight

of

the

grap

h of

y =

x2 .

12.

The

grap

h of

a q

uadr

atic

ineq

ualit

y co

ntai

ning

the

sy

mbo

l < w

ill b

e a

para

bola

ope

ning

dow

nwar

d.

Step

2

Step

1

Antic

ipat

ion

Guid

eQ

uad

rati

c F

un

cti

on

s an

d R

ela

tio

ns

A D A A A A AD D D D D

001_

010_

ALG

2_A

_CR

M_C

04_C

R_6

6078

5.in

dd

312

/20/

10

9:05

PM

Answers (Anticipation Guide and Lesson 4-1)

A01_A13_ALG2_A_CRM_C04_AN_660785.indd A1A01_A13_ALG2_A_CRM_C04_AN_660785.indd A1 12/20/10 10:57 PM12/20/10 10:57 PM

Co

pyrig

ht ©

Gle

nco

e/M

cG

raw

-Hill, a

div

isio

n o

f Th

e M

cG

raw

-Hill C

om

pa

nie

s, In

c.

PDF Pass



NA

ME

DAT

E

P

ER

IOD

Cha

pter

4

6 G

lenc

oe A

lgeb

ra 2

4-1

Stud

y Gu

ide

and

Inte

rven

tion

(con

tinu

ed)

Gra

ph

ing

Qu

ad

rati

c F

un

cti

on

s

Max

imum

and

Min

imum

Val

ues

The

y-co

ordi

nate

of t

he v

erte

x of

a q

uadr

atic

fu

ncti

on is

the

max

imum

val

ue o

r m

inim

um v

alue

of t

he fu

ncti

on.

Max

imum

or

Min

imum

Val

ue

of a

Qua

drat

ic F

unct

ion

The

grap

h of

f(x

) = a

x2 + b

x +

c, w

here

a ≠

0, o

pens

up

and

has

a m

inim

um

whe

n a

> 0

. The

gra

ph o

pens

dow

n an

d ha

s a

max

imum

whe

n a

< 0

.

D

eter

min

e w

heth

er e

ach

func

tion

has

a m

axim

um o

r m

inim

um

valu

e, a

nd f

ind

that

val

ue. T

hen

stat

e th

e do

mai

n an

d ra

nge

of t

he f

unct

ion.

Exer

cise

sD

eter

min

e w

heth

er e

ach

func

tion

has

a m

axim

um o

r m

inim

um v

alue

, and

fin

d th

at v

alue

. The

n st

ate

the

dom

ain

and

rang

e of

the

fun

ctio

n. 1

. f(x

) = 2

x2 - x

+ 1

0 2.

f(x)

= x

2 + 4

x -

7

3. f(

x) =

3x2 -

3x

+ 1

4. f

(x) =

x2 +

5x

+ 2

5.

f(x)

= 2

0 +

6x

- x

2 6.

f(x)

= 4

x2 + x

+ 3

7. f

(x) =

-x2 -

4x

+ 1

0 8.

f(x)

= x

2 - 1

0x +

5

9. f(

x) =

-6x

2 + 1

2x +

21

Exam

ple

a. f

(x)

= 3

x2 - 6

x +

7

For

this

func

tion

, a =

3 a

nd b

= -

6.Si

nce

a >

0, t

he g

raph

ope

ns u

p, a

nd t

he

func

tion

has

a m

inim

um v

alue

.Th

e m

inim

um v

alue

is t

he y

-coo

rdin

ate

of

the

vert

ex. T

he x

-coo

rdin

ate

of t

he

vert

ex is

-b

−

2a =

-

(-

6)

−

2(3)

=

1.

Eva

luat

e th

e fu

ncti

on a

t x

= 1

to

find

the

min

imum

val

ue.

f(1)

= 3

(1)2 -

6(1

) + 7

= 4

, so

the

min

imum

val

ue o

f the

func

tion

is 4

. The

do

mai

n is

all

real

num

bers

. The

ran

ge is

al

l rea

ls g

reat

er t

han

or e

qual

to

the

min

imum

val

ue, t

hat

is {f

(x) |

f(x)

≥ 4

}.

b. f

(x)

= 1

00 -

2x

- x

2

For

this

func

tion

, a =

-1

and

b =

-2.

Sinc

e a

< 0

, the

gra

ph o

pens

dow

n, a

nd

the

func

tion

has

a m

axim

um v

alue

.Th

e m

axim

um v

alue

is t

he y

-coo

rdin

ate

of

the

vert

ex. T

he x

-coo

rdin

ate

of t

he v

erte

x is

-b

−

2a =

-

-2

−

2(-

1) =

-1.

Eva

luat

e th

e fu

ncti

on a

t x

= -

1 to

find

th

e m

axim

um v

alue

.f(

-1)

= 1

00 -

2(-

1) -

(-1)

2 = 1

01, s

o th

e m

axim

um v

alue

of t

he fu

ncti

on is

101

. The

do

mai

n is

all

real

num

bers

. The

ran

ge is

al

l rea

ls le

ss t

han

or e

qual

to

the

max

imum

val

ue, t

hat

is {f

(x) ⎪

f(x)

≤ 1

01}.

m

in.,

9 7 −

8 ; al

l rea

ls;

min

., -

11; a

ll re

als;

m

in.,

1 −

4 ; al

l rea

ls;

{ f

(x)⎥

f(x)

≥ 9

7 −

8 }

{f(x

)⎥ f(

x) ≥

-11

}

{f(

x)⎥ f(

x) ≥

1 −

4 }

m

in.,

- 17

−

4 ;

all

real

s;

max

., 29

; all

real

s;

min

., 2

15

−

16 ;

all r

eals

;

{

f(x)

⎥ f(

x) ≥

- 17

−

4 }

{

f(x)

⎥ f(

x) ≤

29 }

{

f(x)

⎥ f(

x) ≥

2 15

−

16

}

m

ax.,

14; a

ll re

als;

m

in.,

-20

; all

real

s;

max

., 27

; all

real

s;

{f(

x)⎥

f(x

) ≤ 1

4 }

{f(

x)⎥

f(x

) ≥ -

20}

{f(

x)⎥

f(x

) ≤ 2

7 }

001_

010_

ALG

2_A

_CR

M_C

04_C

R_6

6078

5.in

dd

612

/20/

10

9:05

PM


Lesson 4-1

NA

ME

DAT

E

P

ER

IOD

Cha

pter

4

7 G

lenc

oe A

lgeb

ra 2

4-1

Skill

s Pr

actic

eG

rap

hin

g Q

uad

rati

c F

un

cti

on

s

Com

plet

e pa

rts

a–c

for

each

qua

drat

ic f

unct

ion.

a. F

ind

the

y-in

terc

ept,

the

equa

tion

of

the

axis

of

sym

met

ry, a

nd t

he x

-coo

rdin

ate

of t

he v

erte

x.b.

Mak

e a

tabl

e of

val

ues

that

incl

udes

the

ver

tex.

c. U

se t

his

info

rmat

ion

to g

raph

the

fun

ctio

n.

1. f

(x) =

-2x

2 2.

f(x)

= x

2 - 4

x +

4

3. f(

x) =

x2 -

6x

+ 8

Det

erm

ine

whe

ther

eac

h fu

ncti

on h

as a

max

imum

or

a m

inim

um v

alue

, and

fi

nd t

hat

valu

e. T

hen

stat

e th

e do

mai

n an

d ra

nge

of t

he f

unct

ion.

4. f

(x) =

6x2

5. f(

x) =

-8x

2 6.

f(x)

= x

2 + 2

x

7. f

(x) =

-2x

2 + 4

x -

3

8. f(

x) =

3x2 +

12x

+ 3

9.

f(x)

= 2

x2 + 4

x +

1

10. f

(x) =

3x2

11. f

(x) =

x2 +

1

12. f

(x) =

-x2 +

6x

- 1

5

13. f

(x) =

2x2 -

11

14. f

(x) =

x2 -

10x

+ 5

15

. f(x

) = -

2x2 +

8x

+ 7

xO

f(x)

( 0, 0

)

-2

-4

-6

-8

2-

2-

44

2

xO16 12 8 4

2–

24

6

f(x) ( 2

, 0)

xO

( 3, –

1)

f(x)

8 6 4 2

2

–2

46

0

; x =

0; 0

4

; x =

2; 2

8

; x =

3; 3

x-

2-

10

12

f(x)

-8

-2

0-

2-

2x

-2

02

46

f(x)

164

04

16x

02

34

6f(

x)8

0-

10

8

m

ax.;

-1;

m

in.;

-9;

m

in.;

-1;

D

= {

all r

eal n

umbe

rs};

D

= {

all r

eal n

umbe

rs};

D

= {

all r

eal n

umbe

rs};

R

= {

f(x) ⎥

f(x

) ≤ -

1}

R =

{f(

x) ⎥

f(x

) ≥ -

9}

R =

{f(

x) ⎥

f(x

) ≥ -

1}

m

in.;

0;

max

.; 0;

m

in.;

-1;

D =

{al

l rea

l num

bers

};

D =

{al

l rea

l num

bers

};

D =

{al

l rea

l num

bers

};

R =

{f(

x) ⎥

f(x

) ≥ 0

} R

= {

f(x) ⎥

f(x

) ≤ 0

} R

= {

f(x) ⎥

f(x

) ≥ -

1}

m

in.;

0;

min

.; 1;

m

ax.;

-6;

D =

{al

l rea

l num

bers

};

D =

{al

l rea

l num

bers

};

D =

{al

l rea

l num

bers

};

R =

{f(

x) ⎥

f(x

) ≥ 0

} R

= {

f(x) ⎥

f(x

) ≥ 1

} R

= {

f(x) ⎥

f(x

) ≤ -

6}

m

in.;

-11

; m

in.;

-20

; m

ax.;

15;

D

= {

all r

eal n

umbe

rs};

D

= {

all r

eal n

umbe

rs};

D

= {

all r

eal n

umbe

rs};

R

= {

f(x) ⎥

f(x

) ≥ -

11}

R =

{f(

x) ⎥

f(x

) ≥ -

20}

R =

{f(

x) ⎥

f(x

) ≤ 1

5}

001_

010_

ALG

2_A

_CR

M_C

04_C

R_6

6078

5.in

dd

712

/20/

10

9:06

PM

Answers (Lesson 4-1)


Copyri

ght

© G

lencoe/M

cG

raw

-Hill

, a

div

isio

n o

f T

he

McG

raw

-Hill

Co

mp

an

ies,

Inc.

PDF Pass



NA

ME

DAT

E

P

ER

IOD

Cha

pter

4

8 G

lenc

oe A

lgeb

ra 2

4-1

Prac

tice

Gra

ph

ing

Qu

ad

rati

c F

un

cti

on

sC

ompl

ete

part

s a–

c fo

r ea

ch q

uadr

atic

fun

ctio

n.a.

Fin

d th

e y-

inte

rcep

t, th

e eq

uati

on o

f th

e ax

is o

f sy

mm

etry

, and

the

x-c

oord

inat

e of

the

ver

tex.

b. M

ake

a ta

ble

of v

alue

s th

at in

clud

es t

he v

erte

x.c.

Use

thi

s in

form

atio

n to

gra

ph t

he f

unct

ion.

1. f

(x) =

x2 -

8x

+ 1

5 2.

f(x)

= -

x2 - 4

x +

12

3. f(

x) =

2x2 -

2x

+ 1

xO16 12 8 4

28

f(x)

( 4, -

1)64

16 12 8 4

xO

-2

-4

-6

( -2,

16)

f(x)

2

xO

f(x)

( 0.5

, 0.5

)2

2468

-2

-4

4

Det

erm

ine

whe

ther

eac

h fu

ncti

on h

as a

max

imum

or

min

imum

val

ue, a

nd f

ind

that

val

ue. T

hen

stat

e th

e do

mai

n an

d ra

nge

of t

he f

unct

ion.

4. f

(x) =

x2 +

2x

- 8

5.

f(x

) = x

2 - 6

x +

14

6. v

(x) =

-x2 +

14x

- 5

7

7. f

(x) =

2x2 +

4x

- 6

8.

f(x

) = -

x2 + 4

x -

1

9. f

(x) =

- 2 −

3 x2 + 8

x -

24

10. G

RAV

ITAT

ION

Fro

m 4

feet

abo

ve a

sw

imm

ing

pool

, Sus

an t

hrow

s a

ball

upw

ard

wit

h a

velo

city

of 3

2 fe

et p

er s

econ

d. T

he h

eigh

t h(

t) o

f the

bal

l t s

econ

ds a

fter

Sus

an t

hrow

s it

is

giv

en b

y h(

t) =

-16

t2 + 3

2t +

4. F

or t

≥ 0

, fin

d th

e m

axim

um h

eigh

t re

ache

d by

the

ba

ll an

d th

e ti

me

that

thi

s he

ight

is r

each

ed.

11. H

EALT

H C

LUBS

Las

t ye

ar, t

he S

port

sTim

e A

thle

tic

Clu

b ch

arge

d $2

0 to

par

tici

pate

in

an a

erob

ics

clas

s. S

even

ty p

eopl

e at

tend

ed t

he c

lass

es. T

he c

lub

wan

ts t

o in

crea

se t

he

clas

s pr

ice

this

yea

r. Th

ey e

xpec

t to

lose

one

cus

tom

er fo

r ea

ch $

1 in

crea

se in

the

pri

ce.

a. W

hat

pric

e sh

ould

the

clu

b ch

arge

to

max

imiz

e th

e in

com

e fr

om t

he a

erob

ics

clas

ses?

b. W

hat

is t

he m

axim

um in

com

e th

e Sp

orts

Tim

e A

thle

tic

Clu

b ca

n ex

pect

to

mak

e?

1

5; x

= 4

; 4

12;

x =

-2;

-2

1; x

= 0

.5; 0

.5

m

in.;

-9;

all

real

s;

min

.; 5;

all

real

s;

max

.; -

8; a

ll re

als;

{f

(x) ⎥

f(x

) ≥ -

9}

{f(

x) ⎥

f(x

) ≥ 5

} {

f(x) ⎥

f(x

) ≤ -

8}

m

in.;

-8;

all

real

s;

max

.; 3;

all

real

s;

max

.; 0;

all

real

s;

{f(x

) ⎥ f(x

) ≥ -

8}

{f(

x) ⎥

f(x

) ≤ 3

} {

f(x) ⎥

f(x

) ≤ 0

}

20 ft

; 1 s

$202

5

$45

x0

24

68

f(x)

153

-1

315

x-

6-

4-

20

2f(

x)0

1216

120

x-

10

0.5

12

f(x)

51

0.5

15

001_

010_

ALG

2_A

_CR

M_C

04_C

R_6

6078

5.in

dd

812

/20/

10

9:06

PM


Lesson 4-1

NA

ME

DAT

E

P

ER

IOD

Cha

pter

4

9 G

lenc

oe A

lgeb

ra 2

4-1

Wor

d Pr

oble

m P

ract

ice

Gra

ph

ing

Qu

ad

rati

c F

un

cti

on

s

1. T

RAJE

CTO

RIES

A c

anno

nbal

l is

laun

ched

from

a c

anno

n on

the

wal

l of

Fort

Cha

mbl

y, Q

uebe

c. If

the

pat

h of

the

ca

nnon

ball

is

trac

ed o

n a

piec

e of

gra

ph

pape

r al

igne

d so

th

at t

he c

anno

n is

sit

uate

d on

th

e y-

axis

, the

eq

uati

on t

hat

desc

ribe

s th

e pa

th is

y

=-

1−

1600

x2+

1−

2x +

20,

whe

re x

is t

he h

oriz

onta

l dis

tanc

e fr

om

the

cliff

and

y is

the

ver

tica

l dis

tanc

e ab

ove

the

grou

nd in

feet

. How

hig

h ab

ove

the

grou

nd is

the

can

non?

2. T

ICK

ETIN

G T

he m

anag

er o

f a

sym

phon

y co

mpu

tes

that

the

sym

phon

y w

ill e

arn

-40

P2+

110

0P d

olla

rs p

er

conc

ert

if th

ey c

harg

e P

dolla

rs fo

r ti

cket

s. W

hat

tick

et p

rice

sho

uld

the

sym

phon

y ch

arge

in o

rder

to

max

imiz

e it

s pr

ofit

s?

3. A

RCH

ES A

n ar

chit

ect

deci

des

to u

se a

pa

rabo

lic a

rch

for

the

mai

n en

tran

ce o

f a

scie

nce

mus

eum

. In

one

of h

is p

lans

, the

to

p ed

ge o

f the

arc

h is

des

crib

ed b

y th

e gr

aph

of y

=-

1−

4x2

+

5−

2x

+ 1

5. W

hat

are

the

coor

dina

tes

of t

he v

erte

x of

thi

s pa

rabo

la?

4. F

RAM

ING

A fr

ame

com

pany

offe

rs a

lin

e of

squ

are

fram

es. I

f the

sid

e le

ngth

of

the

fram

e is

s, t

hen

the

area

of t

he

open

ing

in t

he fr

ame

is g

iven

by

the

func

tion

a(s

) =s2

- 1

0s+

24.

G

raph

a(s

).

a

sO5

5

5. W

ALK

ING

Can

al S

tree

t an

d W

alke

r St

reet

are

per

pend

icul

ar t

o ea

ch o

ther

. E

vita

is d

rivi

ng s

outh

on

Can

al S

tree

t an

d is

cur

rent

ly 5

mile

s no

rth

of t

he

inte

rsec

tion

wit

h W

alke

r St

reet

. Jac

k is

at

the

inte

rsec

tion

of C

anal

and

Wal

ker

Stre

ets

and

head

ing

east

on

Wal

ker.

Jack

and

Evi

ta a

re b

oth

driv

ing

30 m

iles

per

hour

.

a. W

hen

Jack

is x

mile

s ea

st o

f the

in

ters

ecti

on, w

here

is E

vita

?

b. T

he d

ista

nce

betw

een

Jack

and

Evi

ta is

gi

ven

by t

he fo

rmul

a √

��

��

�

x2+

(5 -

x) 2 .

For

wha

t va

lue

of x

are

Jac

k an

d E

vita

at

thei

r cl

oses

t?

(Hin

t: M

inim

ize

the

squa

re o

f the

di

stan

ce.)

c. W

hat

is t

he d

ista

nce

of c

lose

st

appr

oach

?

20 ft

$13.

75

(5, 2

1.25

)

5

- x

mi n

orth

of t

he

inte

rsec

tion

5 √

�

2 −

2

mi

x =

2.5

001_

010_

ALG

2_A

_CR

M_C

04_C

R_6

6078

5.in

dd

912

/20/

10

9:06

PM



Co

pyrig

ht ©

Gle

nco

e/M

cG

raw

-Hill, a

div

isio

n o

f Th

e M

cG

raw

-Hill C

om

pa

nie

s, In

c.

PDF Pass



NA

ME

DAT

E

P

ER

IOD

Cha

pter

4

10

Gle

ncoe

Alg

ebra

2

4-1

Enri

chm

ent

Fin

din

g t

he x

-inte

rcep

ts o

f a P

ara

bo

la

As

you

know

, if f

(x) =

ax2 +

bx

+ c

is a

qua

drat

ic fu

ncti

on, t

he v

alue

s of

x

that

mak

e f(

x) e

qual

to

zero

are

-

b +

√

��

��

b2 -

4ac

−

2a

and

-

b -

√

��

��

b2 -

4ac

−

2a

.

The

aver

age

of t

hese

tw

o nu

mbe

r va

lues

is -

b −

2a .

Of(x)

x

––

,f(

(

((

b –– 2a b –– 2a

b –– 2ax

= –

f(x) =

ax2 +

bx

+ c

The

func

tion

f(x

) has

its

max

imum

or

min

imum

valu

e w

hen

x =

- b −

2a . T

he x

-inte

rcep

ts o

f the

par

abol

a,

whe

n th

ey e

xist

, are

√

��

��

b2 -

4ac

−

2a

uni

ts t

o th

e le

ft a

nd

righ

t of

the

axi

s of

sym

met

ry.

F

ind

the

vert

ex, a

xis

of s

ymm

etry

, and

x-i

nter

cept

s fo

r f(

x) =

5x2 +

10x

- 7

.

Use

x =

- b −

2a .

x =

- 10

−

2(5)

= -

1

The

x-c

oord

inat

e of

the

ver

tex

is -

1.

Subs

titu

te x

= -

1 in

f(x

) = 5

x2 + 1

0x -

7.

f(-

1) =

5(-

1)2 +

10(

-1)

- 7

= -

12. T

he v

erte

x is

(-1,

-12

).

The

axis

of s

ymm

etry

is x

= -

b −

2a , o

r x

= -

1.

The

x-co

ordi

nate

s of

the

x-in

terc

epts

are

–1

± √

��

��

b2 -

4ac

−

2a

= –

1 ±

√

��

��

��

�

102 -

4 �

5 � (-

7)

−

−

2 � 5

= –

1 ±

√

��

24

0 −

10

. The

x–i

nter

cept

s ar

e (–

1 –

2 −

5 √

��

15

, 0)

and

(–1

+ 2

− 5 √

��

15

, 0)

.

Exer

cise

sF

ind

the

vert

ex, a

xis

of s

ymm

etry

, and

x-i

nter

cept

s fo

r th

e gr

aph

of e

ach

func

tion

us

ing

x =

- b −

2a .

1. f

(x) =

x2 -

4x

- 8

2. g

(x)

= -

4x2 -

8x

+ 3

3. y

= -

x2 + 8

x +

34.

f(x

) = 2

x2 + 6

x +

5

5. A

(x) =

x2 +

12x

+ 3

66.

k(x

) = -

2x2 +

2x

- 6

Exam

ple

(4, 1

9); x

= 4

;4

± √

��

19

(2, -

12);

x =

2;

2 ±

2 √

� 3

(-1,

7);

x =

-1;

-1

± 2

√ �

7

(-6,

0);

x =

-6;

-6

( 1 −

2 , -5

1 −

2 ) ; x

= 1 −

2 ;

no x

-inte

rcep

ts

(- 3 −

2 , 1 −

2 ) ; x

= -

3 −

2 ;

no x

-inte

rcep

ts

001_

010_

ALG

2_A

_CR

M_C

04_C

R_6

6078

5.in

dd

1012

/20/

10

9:06

PM


NA

ME

DAT

E

P

ER

IOD

Lesson 4-2

4-2

Cha

pter

4

11

Gle

ncoe

Alg

ebra

2

Stud

y Gu

ide

and

Inte

rven

tion

So

lvin

g Q

uad

rati

c E

qu

ati

on

s b

y G

rap

hin

g

Solv

e Q

uadr

atic

Equ

atio

ns

Qua

drat

ic E

quat

ion

A q

uadr

atic

equ

atio

n ha

s th

e fo

rm a

x2 + b

x +

c =

0, w

here

a ≠

0.

Roo

ts o

f a Q

uadr

atic

Equ

atio

nso

lutio

n(s)

of t

he e

quat

ion,

or t

he z

ero(

s) o

f the

rela

ted

quad

ratic

func

tion

The

zero

s of

a q

uadr

atic

func

tion

are

the

x-in

terc

epts

of i

ts g

raph

. The

refo

re, f

indi

ng t

he

x-in

terc

epts

is o

ne w

ay o

f sol

ving

the

rel

ated

qua

drat

ic e

quat

ion.

So

lve

x2 + x

- 6

= 0

by

grap

hing

.

Gra

ph t

he r

elat

ed fu

ncti

on f

(x) =

x2 +

x -

6.

The

x-co

ordi

nate

of t

he v

erte

x is

-b

−

2a =

- 1 −

2 , and

the

equ

atio

n of

the

axis

of s

ymm

etry

is x

= -

1 −

2 .

Mak

e a

tabl

e of

val

ues

usin

g x-

valu

es a

roun

d -

1 −

2 .

x-

1-

1 −

2 0

12

f(x)

-6

-6

1 −

4 -

6-

40

Fro

m t

he t

able

and

the

gra

ph, w

e ca

n se

e th

at t

he z

eros

of t

he fu

ncti

on a

re 2

and

-3.

Exer

cise

s

Use

the

rel

ated

gra

ph o

f ea

ch e

quat

ion

to d

eter

min

e it

s so

luti

on.

1. x

2 + 2

x -

8 =

0

2. x

2 - 4

x -

5 =

0

3. x

2 - 5

x +

4 =

0

xO

f(x)

-2

-2

-4

-6

-8

-4

2

xO

-2

-4

-6

-8

-2

24

f(x)

xO

f(x)

-2

-2

-4

2

246

4

4. x

2 - 1

0x +

21

= 0

5.

x2 +

4x

+ 6

= 0

6.

4x2 +

4x

+ 1

= 0

xO

-2

-4

2

2

46

f(x)

xO

-2

-4

2

246

f(x)

xO

-2

-4

2

2468

4

f(x)

xO

-6

-4

-2

-4

-2

2

f(x)

Exam

ple

3

, 7

no

real

sol

utio

ns

- 1 −

2

2, -

45,

-1

1, 4

011_

029_

ALG

2_A

_CR

M_C

04_C

R_6

6078

5.in

dd

1112

/20/

10

9:06

PM

Answers (Lesson 4-1 and Lesson 4-2)


Copyri

ght

© G

lencoe/M

cG

raw

-Hill

, a

div

isio

n o

f T

he

McG

raw

-Hill

Co

mp

an

ies,

Inc.

PDF Pass



NA

ME

DAT

E

P

ER

IOD

4-2

Cha

pter

4

12

Gle

ncoe

Alg

ebra

2

Stud

y Gu

ide

and

Inte

rven

tion

(con

tinu

ed)

So

lvin

g Q

uad

rati

c E

qu

ati

on

s b

y G

rap

hin

g

Esti

mat

e So

luti

ons

Oft

en, y

ou m

ay n

ot b

e ab

le t

o fin

d ex

act

solu

tion

s to

qua

drat

ic

equa

tion

s by

gra

phin

g. B

ut y

ou c

an u

se t

he g

raph

to

esti

mat

e so

luti

ons.

So

lve

x2 - 2

x -

2 =

0 b

y gr

aphi

ng. I

f ex

act

root

s ca

nnot

be

foun

d,

stat

e th

e co

nsec

utiv

e in

tege

rs b

etw

een

whi

ch t

he r

oots

are

loca

ted.

The

equa

tion

of t

he a

xis

of s

ymm

etry

of t

he r

elat

ed fu

ncti

on is

xO

-2

-2

-42

24

4f(x

)

x =

- -

2 −

2(1)

= 1

, so

the

vert

ex h

as x

-coo

rdin

ate

1. M

ake

a ta

ble

of v

alue

s.

x-

10

12

3

f(x)

1-

2-

3-

21

The

x-in

terc

epts

of t

he g

raph

are

bet

wee

n 2

and

3 an

d be

twee

n 0

and

-1.

So

one

solu

tion

is b

etw

een

2 an

d 3,

and

the

oth

er s

olut

ion

is

betw

een

0 an

d -

1.

Exer

cise

sSo

lve

the

equa

tion

s. I

f ex

act

root

s ca

nnot

be

foun

d, s

tate

the

con

secu

tive

inte

gers

be

twee

n w

hich

the

roo

ts a

re lo

cate

d.

1. x

2 - 4

x +

2 =

0

2. x

2 + 6

x +

6 =

0

3. x

2 + 4

x +

2=

0

xO

-2

-2

-4

-4

2

24

4

f(x)

xO

-2

-2

-4

-4

2

2

f(x)

xO

-2

-2

-4

-4

2

24f(x

)

4. -

x2 + 2

x +

4 =

0

5. 2

x2 - 1

2x +

17

= 0

6.

- 1 −

2 x2 + x

+ 5 −

2 = 0

xO

-2

2

246

-2

4

f(x)

xO

2

246

46

f(x)

xO

-2

-2

-4

2

24

4

f(x)

Exam

ple

b

etw

een

0 an

d 1;

b

etw

een

-2

and

-1;

b

etw

een

-1

and

0;be

twee

n 3

and

4 b

etw

een

-5

and

-4

bet

wee

n -

4 an

d -

3

b

etw

een

3 an

d 4;

b

etw

een

2 an

d 3;

b

etw

een

-2

and

-1;

betw

een

-2

and

-1

bet

wee

n 3

and

4 b

etw

een

3 an

d 4

011_

029_

ALG

2_A

_CR

M_C

04_C

R_6

6078

5.in

dd

1212

/20/

10

9:06

PM


NA

ME

DAT

E

P

ER

IOD

Lesson 4-2

4-2

Cha

pter

4

13

Gle

ncoe

Alg

ebra

2

Skill

s Pr

actic

eS

olv

ing

Qu

ad

rati

c E

qu

ati

on

s B

y G

rap

hin

g

Use

the

rel

ated

gra

ph o

f ea

ch e

quat

ion

to d

eter

min

e it

s so

luti

ons.

1. x

2 + 2

x -

3 =

0

2. -

x2 - 6

x -

9 =

0

3. 3

x2 + 4

x +

3 =

0

x

f(x)

O

-2

-4

2-

2-

44

2

f(x) =

x2 +

2x

- 3

x

f(x)

O

-2

-4

-6

-8

-2

-6

-4

f(x) =

-x2 -

6x

- 9

xO

f(x) =

3x2 +

4x +

3

12 8 4

–2

–4

–6

f(x)

Solv

e ea

ch e

quat

ion.

If

exac

t ro

ots

cann

ot b

e fo

und,

sta

te t

he c

onse

cuti

ve in

tege

rs

betw

een

whi

ch t

he r

oots

are

loca

ted.

4. x

2 - 6

x +

5 =

0

5. -

x2 + 2

x -

4 =

0

6. x

2 - 6

x +

4 =

0

xO

f(x)

4 2

2

–2

–4

46

f(x) =

x2 -

6x

+ 5

xO

f(x)

2

–2

–2

–4

–6

–8

46

f(x) =

-x2 +

2x

- 4

xO

f(x)

4 2

2

–2

–4

46

f(x) =

x2 -

6x

+ 4

7. -

x2 - 4

x =

0

8. -

x2 + 3

6 =

0

xO

f(x)

4 2

2

–2

–4

-4

-2

-6f(x

) = -

x2 - 4

x

xO

12–1

2

36 24 12

f(x)

–66f(x

) = -

x2 + 3

6

0

, -4

-

6, 6

1

, 5

no

real

sol

utio

ns

bet

wee

n 0

and

1;

betw

een

5 an

d 6

-

3, 1

-

3 n

o re

al s

olut

ions

011_

029_

ALG

2_A

_CR

M_C

04_C

R_6

6078

5.in

dd

1312

/20/

10

9:06

PM



Co

pyrig

ht ©

Gle

nco

e/M

cG

raw

-Hill, a

div

isio

n o

f Th

e M

cG

raw

-Hill C

om

pa

nie

s, In

c.

PDF Pass



NA

ME

DAT

E

P

ER

IOD

4-2

Cha

pter

4

14

Gle

ncoe

Alg

ebra

2

Prac

tice

So

lvin

g Q

uad

rati

c E

qu

ati

on

s B

y G

rap

hin

g

3, -

2no

suc

h re

al

num

bers

exi

st

Use

the

rel

ated

gra

ph o

f ea

ch e

quat

ion

to d

eter

min

e it

s so

luti

ons.

1. -

3x2 +

3 =

0

2. 3

x2 + x

+ 3

= 0

3.

x2 -

3x

+ 2

= 0

xO

-2

-2

-4

-4

2

2

f(x)

4

xO

-2

-4

2

2468

4

f(x)

xO

-2

-4

2

2468

4

f(x)

Solv

e ea

ch e

quat

ion.

If

exac

t ro

ots

cann

ot b

e fo

und,

sta

te t

he c

onse

cuti

ve in

tege

rs

betw

een

whi

ch t

he r

oots

are

loca

ted.

4. -

2x2 -

6x

+ 5

= 0

5.

x2 +

10x

+ 2

4 =

0

6. 2

x2 - x

- 6

= 0

xO

-2

-6

f(x)

12 8 4

-4

xO

f(x)

246

-2

-4

-6

xO

f(x)

-2

-2

-4

-6

24

6

7. -

x2 + x

+ 6

= 0

8.

-x2 +

5x

- 8

= 0

x

O

f(x)

-2

2

246

46

xO

f(x)

-2

2

-2

-4

-6

-8

46

9. G

RAV

ITY

Use

the

form

ula

h(t)

= v

0t -

16t

2 , w

here

h(t

) is

the

heig

ht o

f an

obje

ct in

feet

, v 0 i

s th

e ob

ject

’s in

itia

l vel

ocit

y in

feet

per

sec

ond,

and

t is

the

tim

e in

sec

onds

.

a. M

arta

thr

ows

a ba

seba

ll w

ith

an in

itia

l upw

ard

velo

city

of 6

0 fe

et p

er s

econ

d.

Igno

ring

Mar

ta’s

heig

ht, h

ow lo

ng a

fter

she

rel

ease

s th

e ba

ll w

ill it

hit

the

gro

und?

b. A

vol

cani

c er

upti

on b

last

s a

boul

der

upw

ard

wit

h an

init

ial v

eloc

ity

of

240

feet

per

sec

ond.

How

long

will

it t

ake

the

boul

der

to h

it t

he g

roun

d if

it la

nds

at

the

sam

e el

evat

ion

from

whi

ch it

was

eje

cted

?

-

1, 1

n

o re

al s

olut

ions

1

, 2

b

etw

een

0 an

d 1;

-

6, -

4 -

1.5,

2be

twee

n -

4 an

d -

3

15 s

3.75

s

011_

029_

ALG

2_A

_CR

M_C

04_C

R_6

6078

5.in

dd

1412

/20/

10

9:06

PM


NA

ME

DAT

E

P

ER

IOD

Lesson 4-2

4-2

Cha

pter

4

15

Gle

ncoe

Alg

ebra

2

Wor

d Pr

oble

m P

ract

ice

So

lvin

g Q

uad

rati

c E

qu

ati

on

s b

y G

rap

hin

g

1. T

RAJE

CTO

RIES

Dav

id t

hrew

a b

aseb

all

into

the

air.

The

func

tion

of t

he h

eigh

t of

the

bas

ebal

l in

feet

is h

= 8

0t-

16t2 ,

whe

re t

repr

esen

ts t

he t

ime

in s

econ

ds

afte

r th

e ba

ll w

as t

hrow

n. U

se t

his

grap

h of

the

func

tion

to

dete

rmin

e ho

w

long

it t

ook

for

the

ball

to fa

ll ba

ck t

o th

e gr

ound

. h

t1

23

45

-1

-40

4080

2. B

RID

GES

In

1895

, a b

rick

arc

h ra

ilway

br

idge

was

bui

lt o

n N

orth

Ave

nue

in

Bal

tim

ore,

Mar

ylan

d. T

he a

rch

is

desc

ribe

d by

the

equ

atio

n h

= 9

–1−

50x2 ,

whe

re h

is t

he h

eigh

t in

yar

ds a

nd x

is

the

dist

ance

in y

ards

from

the

cen

ter

of

the

brid

ge. G

raph

thi

s eq

uati

on a

nd

desc

ribe

, to

the

near

est

yard

, whe

re t

he

brid

ge t

ouch

es t

he g

roun

d.

3. L

OG

IC W

ilma

is t

hink

ing

of t

wo

num

bers

. The

sum

is 2

and

the

pro

duct

is

-24

. Use

a q

uadr

atic

equ

atio

n to

find

th

e tw

o nu

mbe

rs.

4. R

AD

IO T

ELES

COPE

S Th

e cr

oss-

sect

ion

of a

larg

e ra

dio

tele

scop

e is

a p

arab

ola.

Th

e di

sh is

set

into

the

gro

und.

The

eq

uati

on t

hat

desc

ribe

s th

e cr

oss-

sect

ion

is d

=

2−

75x2

-

4−

3x

-

32−

3 , w

here

d g

ives

the

dept

h of

the

dis

h be

low

gro

und

and

x is

the

dis

tanc

e fr

om t

he c

ontr

ol c

ente

r, bo

th in

met

ers.

If t

he d

ish

does

not

ex

tend

abo

ve t

he g

roun

d le

vel,

wha

t is

th

e di

amet

er o

f the

dis

h? S

olve

by

grap

hing

.

x40

6080

20-

20

-2020406080

d

5. B

OAT

S Th

e di

stan

ce b

etw

een

two

boat

s is

d

=√

��

��

��

t2-

10t

+ 3

5 ,w

here

d is

dis

tanc

e in

met

ers

and

t is

tim

e in

sec

onds

.

a. M

ake

a gr

aph

of d

2 ver

sus

t.d

tO

1015

2025

5

102030

b. D

o th

e bo

ats

ever

col

lide?

N

o

5

sec

onds

6 an

d -

4

21 y

ards

from

the

cent

er o

f the

br

idge

on

eith

er s

ide

at (-

21, 0

) an

d (2

1, 0

)

x

30 10

-20

-40

-30

-10

2040

h

6

4 m

011_

029_

ALG

2_A

_CR

M_C

04_C

R_6

6078

5.in

dd

1512

/20/

10

9:06

PM



Copyri

ght

© G

lencoe/M

cG

raw

-Hill

, a

div

isio

n o

f T

he

McG

raw

-Hill

Co

mp

an

ies,

Inc.

PDF Pass



NA

ME

DAT

E

P

ER

IOD

4-2

Cha

pter

4

16

Gle

ncoe

Alg

ebra

2

Enri

chm

ent

Gra

ph

ing

Ab

solu

te V

alu

e E

qu

ati

on

s

You

can

solv

e ab

solu

te v

alue

equ

atio

ns in

muc

h th

e sa

me

way

you

sol

ved

quad

rati

c eq

uati

ons.

Gra

ph t

he r

elat

ed a

bsol

ute

valu

e fu

ncti

on fo

r ea

ch

equa

tion

usi

ng a

gra

phin

g ca

lcul

ator

. The

n us

e th

e ZE

RO fe

atur

e in

the

CA

LC m

enu

to fi

nd it

s re

al s

olut

ions

, if a

ny. R

ecal

l tha

t so

luti

ons

are

poin

ts

whe

re t

he g

raph

inte

rsec

ts t

he x

-axi

s.

For

each

equ

atio

n, m

ake

a sk

etch

of

the

rela

ted

grap

h an

d fi

nd t

he

solu

tion

s ro

unde

d to

the

nea

rest

hun

dred

th.

1. ⎪

x +

5⎥

= 0

2.

⎪ 4x

- 3

⎥

+ 5

= 0

3.

⎪ x -

7⎥

= 0

4. ⎪

x +

3⎥

- 8

= 0

5.

- ⎪ x

+ 3

⎥

+ 6

= 0

6.

⎪ x -

2⎥

- 3

= 0

7. ⎪

3x +

4⎥

= 2

8.

⎪ x +

12⎥

= 1

0 9.

⎪ x⎥

- 3

= 0

10. E

xpla

in h

ow s

olvi

ng a

bsol

ute

valu

e eq

uati

ons

alge

brai

cally

and

find

ing

zero

s of

abs

olut

e va

lue

func

tion

s gr

aphi

cally

are

rel

ated

.

-

5 N

o so

lutio

ns

7

Sam

ple

answ

er: v

alue

s of

x w

hen

solv

ing

alge

brai

cally

are

the

x-in

terc

epts

(or

zero

s) o

f the

func

tion

whe

n gr

aphe

d.

-

11, 5

-

9, 3

-

1, 5

-

2, -

2 −

3 -

22, -

2 -

3, 3

011_

029_

ALG

2_A

_CR

M_C

04_C

R_6

6078

5.in

dd

1612

/20/

10

9:06

PM


NA

ME

DAT

E

P

ER

IOD

Lesson 4-3

4-3

Cha

pter

4

17

Gle

ncoe

Alg

ebra

2

Stud

y Gu

ide

and

Inte

rven

tion

So

lvin

g Q

uad

rati

c E

qu

ati

on

s b

y Fa

cto

rin

g

Fact

ored

For

m T

o w

rite

a q

uadr

atic

equ

atio

n w

ith

root

s p

and

q, le

t (x

- p

)(x -

q) =

0.

Then

mul

tipl

y us

ing

FOIL

.

W

rite

a q

uadr

atic

equ

atio

n in

sta

ndar

d fo

rm w

ith

the

give

n ro

ots.

a. 3

, -5

(x

- p

)(x -

q) =

0

Writ

e th

e pa

ttern

.

(x

- 3

)[x -

(-5)

] = 0

R

epla

ce p

with

3, q

with

-5.

(x

- 3

)(x +

5) =

0

Sim

plify

.

x2 +

2x

- 1

5 =

0

Use

FO

IL.

The

equa

tion

x2 +

2x

- 1

5 =

0 h

as r

oots

3

and

-5.

b. -

7 −

8 , 1 −

3

(x

- p

)(x -

q) =

0

[x -

(-

7 −

8 ) ] (

x -

1 −

3 ) =

0

(

x +

7 −

8 ) (x

- 1 −

3 ) =

0

(8

x +

7)

−

8 �

(3x

- 1

) −

3 =

0

24

� (8

x +

7)(3

x -

1)

−

−

24

=

24

� 0

24

x2 + 1

3x -

7 =

0

The

equa

tion

24x

2 + 1

3x -

7 =

0 h

as

root

s -

7 −

8 and

1 −

3 .

Exer

cise

sW

rite

a q

uadr

atic

equ

atio

n in

sta

ndar

d fo

rm w

ith

the

give

n ro

ot(s

).

1. 3

, -4

2. -

8, -

2 3.

1, 9

4. -

5 5.

10,

7

6. -

2, 1

5

7. -

1 −

3 , 5

8. 2

, 2 −

3 9.

-7,

3 −

4

10. 3

, 2 −

5 11

. - 4 −

9 , -1

12. 9

, 1 −

6

13. 2

− 3 , -

2 −

3 14

. 5 −

4 , - 1 −

2 15

. 3 −

7 , 1 −

5

16. -

7 −

8 , 7 −

2 17

. 1 −

2 , 3 −

4 18

. 1 −

8 , 1 −

6

Exam

ple

1

6x2 -

42x

- 4

9 =

0

8x

2 - 1

0x +

3 =

0

48x

2 - 1

4x +

1 =

0

9

x2 -

4 =

0

8x

2 - 6

x -

5 =

0

35x

2 - 2

2x +

3 =

0

x

2 + x

- 1

2 =

0

x2 +

10x

+ 1

6 =

0

x2 -

10x

+ 9

= 0

x

2 + 1

0x +

25

= 0

x

2 - 1

7x +

70

= 0

x

2 - 1

3x -

30

= 0

5

x2 -

17x

+ 6

= 0

9

x2 +

13x

+ 4

= 0

6

x2 -

55x

+ 9

= 0

3

x2 -

14x

- 5

= 0

3

x2 -

8x

+ 4

= 0

4

x2 +

25x

- 2

1 =

0

011_

029_

ALG

2_A

_CR

M_C

04_C

R_6

6078

5.in

dd

1712

/20/

10

9:06

PM



Co

pyrig

ht ©

Gle

nco

e/M

cG

raw

-Hill, a

div

isio

n o

f Th

e M

cG

raw

-Hill C

om

pa

nie

s, In

c.

PDF Pass



NA

ME

DAT

E

P

ER

IOD

4-3

Cha

pter

4

18

Gle

ncoe

Alg

ebra

2

Stud

y Gu

ide

and

Inte

rven

tion

(con

tinu

ed)

So

lvin

g Q

uad

rati

c E

qu

ati

on

s b

y Fa

cto

rin

g

Solv

e Eq

uati

ons

by F

acto

ring

Whe

n yo

u us

e fa

ctor

ing

to s

olve

a q

uadr

atic

equ

atio

n,

you

use

the

follo

win

g pr

oper

ty.

Zero

Pro

duct

Pro

pert

yFo

r any

real

num

bers

a a

nd b

, if a

b =

0, t

hen

eith

er a

= 0

or b

=0,

or b

oth

a an

d b

= 0

.

So

lve

each

equ

atio

n by

fac

tori

ng.

a. 3

x2 = 1

5x

3x2 =

15x

O

rigin

al e

quat

ion

3x

2 - 1

5x =

0

Sub

tract

15x

from

bot

h si

des.

3x

(x -

5) =

0

Fact

or th

e bi

nom

ial.

3x =

0 o

r x

- 5

= 0

Ze

ro P

rodu

ct P

rope

rty

x

= 0

or

x

= 5

S

olve

eac

h eq

uatio

n.

Th

e so

luti

on s

et is

{0, 5

}.

b. 4

x2 - 5

x =

21

4x2 -

5x

= 2

1 O

rigin

al e

quat

ion

4x

2 - 5

x -

21

= 0

S

ubtra

ct 2

1 fro

m b

oth

side

s.

(4x

+ 7

)(x -

3) =

0

Fact

or th

e tri

nom

ial.

4x +

7 =

0

or x

- 3

= 0

Ze

ro P

rodu

ct P

rope

rty

x

= -

7 −

4 or

x =

3

Sol

ve e

ach

equa

tion.

The

solu

tion

set

is {

- 7 −

4 , 3} .

Exer

cise

sSo

lve

each

equ

atio

n by

fac

tori

ng.

1. 6

x2 - 2

x =

0

2. x

2 = 7

x 3.

20x

2 = -

25x

4. 6

x2 = 7

x 5.

6x2 -

27x

= 0

6.

12x

2 - 8

x =

0

7. x

2 + x

- 3

0 =

0

8. 2

x2 - x

- 3

= 0

9.

x2 +

14x

+ 3

3 =

0

10. 4

x2 + 2

7x -

7 =

0

11. 3

x2 + 2

9x -

10

= 0

12

. 6x2 -

5x

- 4

= 0

13. 1

2x2 -

8x

+ 1

= 0

14

. 5x2 +

28x

- 1

2 =

0

15.

2x2 -

250

x +

500

0 =

0

16. 2

x2 - 1

1x -

40

= 0

17

. 2x2 +

21x

- 1

1 =

0

18. 3

x2 + 2

x -

21

= 0

19. 8

x2 - 1

4x +

3 =

0

20. 6

x2 + 1

1x -

2 =

0

21. 5

x2 + 1

7x -

12

= 0

22. 1

2x2 +

25x

+ 1

2 =

0

23. 1

2x2 +

18x

+ 6

= 0

24

. 7x2 -

36x

+ 5

= 0

Exam

ple

{

5, -

6}

{ 3 −

2 , -

1 }

{-

11, -

3}

{

1 −

6 , 1 −

2 }

{ 2 −

5 , -6 }

{

100,

25}

{

- 4 −

3 , - 3 −

4 }

{-

1 −

2 , -1 }

{

1 −

7 , 5}

{

0, 7 −

6 }

{0,

9 −

2 }

{0,

2 −

3 }

{

0, 1 −

3 }

{0,

7}

{0,

- 5 −

4 }

{

8, -

5 −

2 }

{-

11, 1

− 2 }

{

7 −

3 , -

3 }

{

3 −

2 , 1 −

4 }

{-

2, 1 −

6 }

{ 3 −

5 , -

4 }

{

1 −

4 , -

7 }

{-

10, 1

− 3 }

{

- 1 −

2 , 4 −

3 }

011_

029_

ALG

2_A

_CR

M_C

04_C

R_6

6078

5.in

dd

1812

/20/

10

9:06

PM


NA

ME

DAT

E

P

ER

IOD

Lesson 4-3

4-3

Cha

pter

4

19

Gle

ncoe

Alg

ebra

2

Skill

s Pr

actic

eS

olv

ing

Qu

ad

rati

c E

qu

ati

on

s b

y Fa

cto

rin

g

Wri

te a

qua

drat

ic e

quat

ion

in s

tand

ard

form

wit

h th

e gi

ven

root

(s).

1. 1

, 4

2. 6

, -9

3. -

2, -

5 4.

0, 7

5. -

1 −

3 , -3

6. -

1 −

2 , 3 −

4

Fact

or e

ach

poly

nom

ial.

7. m

2 + 7

m -

18

8.

2x2 -

3x

- 5

9. 4

z2 + 4

z -

15

10

. 4p2 +

4p

- 2

4

11. 3

y2 + 2

1y +

36

12

. c2 -

100

Solv

e ea

ch e

quat

ion

by f

acto

ring

.

13. x

2 = 6

4

14. x

2 - 1

00 =

0

15. x

2 - 3

x +

2 =

0

16. x

2 - 4

x +

3 =

0

17. x

2 + 2

x -

3 =

0

18. x

2 - 3

x -

10

= 0

19. x

2 - 6

x +

5 =

0

20. x

2 - 9

x =

0

21. x

2 - 4

x =

21

22

. 2x2 +

5x

- 3

= 0

23. 4

x2 + 5

x -

6 =

0

24. 3

x2 - 1

3x -

10

= 0

25. N

UM

BER

THEO

RY F

ind

two

cons

ecut

ive

inte

gers

who

se p

rodu

ct is

272

.

x2 -

5x

+ 4

= 0

x2 +

3x

- 5

4 =

0

x2 +

7x

+ 1

0 =

0x

2 - 7

x =

0

3x2 +

10x

+ 3

= 0

8x2 -

2x

- 3

= 0

(

m -

2)(

m +

9)

(2x

- 5

)(x

+ 1

)

(

2z +

5)(

2z -

3)

4(p

- 2

)(p

+ 3

)

3

(y +

4)(

y +

3)

(c +

10)

(c -

10)

{-8,

8} {1

, 2}

{1, 3

}

{5, -

2}

{1, 5

}{0

, 9}

{-3,

7}

{

1 −

2 , -3 }

{

3 −

4 , -2 }

{

- 2 −

3 , 5}

16, 1

7 or

-16

, -17

{10,

-10

}

{1, -

3}

011_

029_

ALG

2_A

_CR

M_C

04_C

R_6

6078

5.in

dd

1912

/20/

10

9:06

PM



Copyri

ght

© G

lencoe/M

cG

raw

-Hill

, a

div

isio

n o

f T

he

McG

raw

-Hill

Co

mp

an

ies,

Inc.

PDF Pass



NA

ME

DAT

E

P

ER

IOD

4-3

Cha

pter

4

20

Gle

ncoe

Alg

ebra

2

Prac

tice

So

lvin

g Q

uad

rati

c E

qu

ati

on

s b

y Fa

cto

rin

g

Wri

te a

qua

drat

ic e

quat

ion

in s

tand

ard

form

wit

h th

e gi

ven

root

(s).

1. 7

, 2

2. 0

, 3

3. -

5, 8

4. -

7, -

8 5.

-6,

-3

6. 3

, -4

7. 1

, 1 −

2 8.

1 −

3 , 2

9. 0

, - 7 −

2

Fact

or e

ach

poly

nom

ial.

10. r

3 + 3

r2 - 5

4r

11. 8

a2 + 2

a -

6

12. c

2 - 4

9

13. x

3 + 8

14

. 16r

2 - 1

69

15. b

4 - 8

1

Solv

e ea

ch e

quat

ion

by f

acto

ring

.

16. x

2 - 4

x -

12

= 0

17

. x2 -

16x

+ 6

4 =

0

18. x

2 - 6

x +

8 =

0

19. x

2 + 3

x +

2 =

0

20. x

2 - 4

x =

0

21. 7

x2 = 4

x

22. 1

0x2 =

9x

23. x

2 = 2

x +

99

24. x

2 + 1

2x =

-36

25

. 5x2 -

35x

+ 6

0 =

0

26. 3

6x2 =

25

27. 2

x2 - 8

x -

90

= 0

28. N

UM

BER

THEO

RY F

ind

two

cons

ecut

ive

even

pos

itiv

e in

tege

rs w

hose

pro

duct

is 6

24.

29. N

UM

BER

THEO

RY F

ind

two

cons

ecut

ive

odd

posi

tive

inte

gers

who

se p

rodu

ct is

323

.

30. G

EOM

ETRY

The

leng

th o

f a r

ecta

ngle

is 2

feet

mor

e th

an it

s w

idth

. Fin

d th

e di

men

sion

s of

the

rec

tang

le if

its

area

is 6

3 sq

uare

feet

.

31. P

HO

TOG

RAPH

Y T

he le

ngth

and

wid

th o

f a 6

-inch

by

8-in

ch p

hoto

grap

h ar

e re

duce

d by

th

e sa

me

amou

nt t

o m

ake

a ne

w p

hoto

grap

h w

hose

are

a is

hal

f tha

t of

the

ori

gina

l. B

y ho

w m

any

inch

es w

ill t

he d

imen

sion

s of

the

pho

togr

aph

have

to

be r

educ

ed?

x

2 - 9

x +

14

= 0

x

2 - 3

x =

0

x2 -

3x

- 4

0 =

0

x

2 + 1

5x +

56

= 0

x

2 + 9

x +

18

= 0

x

2 + x

- 1

2 =

0

r

(r +

9)(

r -

6)

2

(4a -

3)(

a +

1)

(c

- 7

)(c

+ 7

)

(

x +

2)(

x2 -

2x

+ 4

) (

4r +

13)

(4r

- 1

3)

(b

2 + 9

)(b

+ 3

)(b

- 3

)

{8}

{-2,

-1}

{

0, 4 −

7 }

{-9,

11}

{-6}

{3, 4

}

{9, -

5}

7 ft

by 9

ft

2

x2 -

3x

+ 1

= 0

3

x2 -

7x

+ 2

= 0

2

x2 +

7x

= 0 2

in.

{6, -

2}

{2, 4

}

{0, 4

}

{

0, 9 −

10

}

{

5 −

6 , - 5 −

6 }

24, 2

6

17, 1

9

011_

029_

ALG

2_A

_CR

M_C

04_C

R_6

6078

5.in

dd

2012

/20/

10

9:06

PM


NA

ME

DAT

E

P

ER

IOD

Lesson 4-3

4-3

Cha

pter

4

21

Gle

ncoe

Alg

ebra

2

Wor

d Pr

oble

m P

ract

ice

So

lvin

g Q

uad

rati

c E

qu

ati

on

s b

y Fa

cto

rin

g

1. F

LASH

LIG

HTS

Whe

n D

ora

shin

es h

er

flash

light

on

the

wal

l at

a ce

rtai

n an

gle,

th

e ed

ge o

f the

lit

area

is in

the

sha

pe o

f a

para

bola

. The

equ

atio

n of

the

par

abol

a is

y=

2x2

+ 2

x-

60.

Fac

tor

this

qu

adra

tic

equa

tion

.

2. S

IGN

S D

avid

was

look

ing

thro

ugh

an

old

alge

bra

book

and

cam

e ac

ross

thi

s eq

uati

on.

x

26x

+ 8

= 0

The

sign

in fr

ont

of t

he 6

was

blo

tted

ou

t. H

ow d

oes

the

mis

sing

sig

n de

pend

on

the

sig

ns o

f the

roo

ts?

3. A

RT T

he a

rea

in s

quar

e in

ches

of t

he

draw

ing

Mai

sons

pré

s de

la m

er b

y C

laud

e M

onet

is a

ppro

xim

ated

by

the

equa

tion

y=

x2– 2

3x+

130

. Fa

ctor

the

eq

uati

on t

o fin

d th

e tw

o ro

ots,

whi

ch a

re

equa

l to

the

appr

oxim

ate

leng

th a

nd

wid

th o

f the

dra

win

g.

4. P

ROG

RAM

MIN

G R

ay is

a c

ompu

ter

prog

ram

mer

. He

need

s to

find

the

qu

adra

tic

func

tion

of t

his

grap

h fo

r an

al

gori

thm

rel

ated

to

a ga

me

invo

lvin

g di

ce. P

rovi

de s

uch

a fu

ncti

on.

y

xO

-2

-4

2

24

46

810

12

5. A

NIM

ATIO

N A

com

pute

r gr

aphi

cs

anim

ator

wou

ld li

ke t

o m

ake

a re

alis

tic

sim

ulat

ion

of a

tos

sed

ball.

The

ani

mat

or

wan

ts t

he b

all t

o fo

llow

the

par

abol

ic

traj

ecto

ry r

epre

sent

ed b

y th

e qu

adra

tic

equa

tion

f(x

) = -

0.2(

x +

5) (

x -

5).

a. W

hat

are

the

solu

tion

s of

f(x

) = 0

?

b. W

rite

f(x

) in

stan

dard

form

.

c. I

f the

ani

mat

or c

hang

es t

he e

quat

ion

to f

(x) =

-0.

2x2 +

20,

wha

t ar

e th

e so

luti

ons

of f(

x) =

0?

2(x

- 5

)(x

+ 6

)

The

mis

sing

sig

n is

the

oppo

site

of

the

sign

of t

he tw

o ro

ots,

be

caus

e th

eir

prod

uct i

s a

posi

tive

num

ber,

8.

10 in

ches

by

13 in

ches

f

(x) =

x2 -

18x

+ 7

7

x

= -

5 or

x =

5

f

(x) =

-0.

2x2 +

5

x

= -

10 o

r x

= 1

0

011_

029_

ALG

2_A

_CR

M_C

04_C

R_6

6078

5.in

dd

2112

/20/

10

9:06

PM



Co

pyrig

ht ©

Gle

nco

e/M

cG

raw

-Hill, a

div

isio

n o

f Th

e M

cG

raw

-Hill C

om

pan

ies, In

c.

PDF Pass



NA

ME

DAT

E

P

ER

IOD

4-3

Ch

ap

ter

4

22

Gle

ncoe

Alg

ebra

2

Enri

chm

ent

Usi

ng

Patt

ern

s to

Facto

r

Stud

y th

e pa

tter

ns b

elow

for

fact

orin

g th

e su

m a

nd t

he d

iffer

ence

of c

ubes

.

a3 + b

3 = (a

+ b

)(a2 -

ab

+ b

2 )a3 -

b3 =

(a -

b)(a

2 + a

b +

b2 )

This

pat

tern

can

be

exte

nded

to

othe

r od

d po

wer

s. S

tudy

the

se e

xam

ples

.

Fa

ctor

a5 +

b5 .

Ext

end

the

first

pat

tern

to

obta

in a

5 + b

5 = (a

+ b

)(a4 -

a3 b

+ a

2 b2 -

ab3 +

b4 ).

Che

ck: (

a +

b)(a

4 - a

3 b +

a2 b

2 - a

b3 + b

4 ) =

a5 -

a4 b

+ a

3 b2 -

a2 b

3 + a

b4

+ a

4 b -

a3 b

2 + a

2 b3 -

ab4 +

b5

= a

5 +

b5

Fa

ctor

a5 -

b5 .

Ext

end

the

seco

nd p

atte

rn t

o ob

tain

a5 -

b5 =

(a -

b)(a

4 + a

3 b +

a2 b

2 + a

b3 + b

4 ).C

heck

: (a

- b

) (a4 +

a3 b

+ a

2 b2 +

ab3 +

b4 )

= a

5 + a

4 b +

a3 b

2 + a

2 b3 +

ab4

-

a4 b

- a

3 b2 -

a2 b

3 - a

b4 - b

5

= a

5 -

b5

In g

ener

al, i

f n is

an

odd

inte

ger,

whe

n yo

u fa

ctor

an +

bn o

r an -

bn ,

one

fact

or w

ill b

e ei

ther

(a

+ b

) or

(a -

b),

depe

ndin

g on

the

sig

n of

the

ori

gina

l exp

ress

ion.

The

oth

er fa

ctor

will

ha

ve t

he fo

llow

ing

prop

erti

es:

• Th

e fir

st t

erm

will

be

an -

1 a

nd t

he la

st t

erm

will

be

bn -

1.

• Th

e ex

pone

nts

of a

will

dec

reas

e by

1 a

s yo

u go

from

left

to

righ

t.•

The

expo

nent

s of

b w

ill in

crea

se b

y 1

as y

ou g

o fr

om le

ft t

o ri

ght.

• Th

e de

gree

of e

ach

term

will

be

n -

1.

• If

the

ori

gina

l exp

ress

ion

was

an +

bn ,

the

term

s w

ill a

lter

nate

ly h

ave

+ a

nd -

sig

ns.

• If

the

ori

gina

l exp

ress

ion

was

an -

bn ,

the

term

s w

ill a

ll ha

ve +

sig

ns.

Use

the

pat

tern

s ab

ove

to f

acto

r ea

ch e

xpre

ssio

n.

1. a

7 + b

7

2. c

9 - d

9

3. f

11 +

g11

To f

acto

r x10

- y

10, c

hang

e it

to

(x5 +

y5 )

(x5 -

y5 )

and

fac

tor

each

bin

omia

l. U

se t

his

appr

oach

to

fact

or e

ach

expr

essi

on.

4. x

10 -

y10

5. a

14 -

b14

Exam

ple

1

Exam

ple

2

(a +

b)(

a6 -

a5 b

+ a

4 b2 -

a3 b

3 + a

2 b4 -

ab

5 + b

6 ) (a

- b

)

(a

6 + a

5 b +

a4 b

2 + a

3 b3 +

a2 b

4 + a

b5 +

b6 )

(c -

d)(

c8 +

c7 d

+ c

6 d2 +

c5 d

3 + c

4 d4 +

c3 d

5 + c

2 d6 +

cd

7 + d

8 )

(f +

g)(f

10 -

f 9 g

+ f

8 g2 -

f 7 g

3 + f

6 g4 -

f 5 g

5 + f

4 g6 -

f 3 g

7 + f

2 g8 -

fg9 +

g10

)

(x +

y)(x

4 - x

3 y +

x2 y

2 - x

y3 +

y4 )(

x -

y)(x

4 + x

3 y +

x2 y

2 + x

y3 +

y4 )

(a +

b)(

a6 -

a5 b

+ a

4 b2 -

a3 b

3 + a

2 b4 -

ab

5 + b

6 )

011_

029_

ALG

2_A

_CR

M_C

04_C

R_6

6078

5.in

dd

2212

/20/

10

9:06

PM


NA

ME

DAT

E

P

ER

IOD

Lesson 4-3

4-3

Ch

ap

ter

4

23

Gle

ncoe

Alg

ebra

2

Grap

hing

Cal

cula

tor

Activ

ityU

sin

g T

ab

les

to F

acto

r b

y G

rou

pin

g

The

TAB

LE

feat

ure

of a

gra

phin

g ca

lcul

ator

can

be

used

to

help

fact

or a

pol

ynom

ial o

f the

fo

rm a

x2 + b

x +

c. (

The

sam

e pr

oble

ms

can

be s

olve

d w

ith

the

List

s an

d Sp

read

shee

t ap

plic

atio

n on

the

TI-

Nsp

ire.

)

Fa

ctor

10x

2 - 4

3x +

28

by g

roup

ing.

Mak

e a

tabl

e of

the

neg

ativ

e fa

ctor

s of

10

� 28

or 2

80. L

ook

for

a pa

ir

of fa

ctor

s w

hose

sum

is -

43.

Ent

er t

he e

quat

ion

y =

280

−

x

in Y

1 to

find

the

fact

ors

of 2

80. T

hen,

find

the

sum

of t

he fa

ctor

s us

ing

y =

280

−

x

+ x

in Y

2. S

et u

p th

e ta

ble

to d

ispl

ay t

he n

egat

ive

fact

ors

of 2

80 b

y se

ttin

g �

Tbl

= t

o -

1.

Exa

min

e th

e re

sult

s.

Key

stro

kes:

Y=

280

÷

E

NT

ER

V

AR

S

EN

TE

R E

NT

ER

+

E

NT

ER

2

nd

[TB

LSE

T] (

–) 1

EN

TE

R (

–) 1

EN

TE

R

2n

d

[TA

BLE

].

The

last

line

of t

he t

able

sho

ws

that

-43

x m

ay b

e re

plac

ed w

ith

-8x

+ (-

35x)

.

10x2 -

43x

+ 2

8 =

10x

2 - 8

x +

(-35

x) +

28

=

2x(

5x -

4) +

(-7)

(5x

- 4

)

= (5

x -

4)(2

x -

7)

Thus

, 10x

2 - 4

3x +

28

= (5

x -

4)(2

x -

7).

Fact

or e

ach

poly

nom

ial.

1. y

2 - 2

0y -

96

2. 4

z2 - 3

3z +

35

3. 4

y2 + y

-18

4.

6a2 +

2a

- 1

5

5. 6

m2 +

17m

+ 1

2 6.

24z

2 - 4

6z +

15

7. 3

6y2 +

84y

+ 4

9 8.

4b2 +

36b

- 4

03

Fa

ctor

12x

2 - 7

x -

12.

Look

at

the

fact

ors

of 1

2(-

12) o

r -

144

for

a pa

ir w

ith

a su

m o

f -7.

E

nter

an

equa

tion

to

dete

rmin

e th

e fa

ctor

s in

Y1

and

an e

quat

ion

to

find

the

sum

of f

acto

rs in

Y2.

Exa

min

e th

e ta

ble

to fi

nd a

sum

of -

7.K

eyst

roke

s: Y

= (

–) 1

44

÷

EN

TE

R

VA

RS

E

NT

ER

EN

TE

R

+

EN

TE

R

2n

d [T

BLS

ET]

1 E

NT

ER

1 E

NT

ER

2

nd

[T

AB

LE].

12x2 -

7x

- 1

2 =

12x

2 + 9

x +

(-16

x) -

12

=

3x(

4x +

3) -

4(4

x +

3)

=

(4x

+ 3

)(3x

- 4

) Th

us, 1

2x2 -

7x

- 1

2 =

(4x

+ 3

)(3x

- 4

).

Exer

cise

s

Exam

ple

1

Exam

ple

2

(

y +

4)(

y -

24)

(

4z -

5)(

z -

7)

(4y +

9)(

y -

2)

pr

ime

(

2m +

3)(

3m +

4)

(1

2z -

5)(

2z -

3)

(

6y +

7)2

(

2b +

31)

(2b

- 1

3)

011_

029_

ALG

2_A

_CR

M_C

04_C

R_6

6078

5.in

dd

2312

/20/

10

9:06

PM



Copyri

ght

© G

lencoe/M

cG

raw

-Hill

, a

div

isio

n o

f T

he

McG

raw

-Hill

Co

mp

an

ies,

Inc.

PDF Pass



NA

ME

DAT

E

P

ER

IOD

4-4

Cha

pter

4

24

Gle

ncoe

Alg

ebra

2

Stud

y Gu

ide

and

Inte

rven

tion

Co

mp

lex

Nu

mb

ers

So

lve

x2 + 5

= 0

.

x2 + 5

= 0

O

rigin

al e

quat

ion.

x

2 = -

5

Sub

tract

5 fr

om e

ach

side

.

x

= ±

√

�

5 i

Squ

are

Roo

t Pro

pert

y.

Exer

cise

sSi

mpl

ify.

1. √

��

-

72

2.

√

��

-

24

3. √

��

-

84

4.

(2 +

i) (

2 -

i)

Solv

e ea

ch e

quat

ion.

5.

5x2 +

45

= 0

6.

4x2 +

24

= 0

7. -

9x2 =

9

8. 7

x2 + 8

4 =

0

Pure

Imag

inar

y N

umbe

rs A

squ

are

root

of a

num

ber

n is

a n

umbe

r w

hose

squ

are

is n

. For

non

nega

tive

rea

l num

bers

a a

nd b

, √

��

ab

= √

�

a �

√

�

b an

d √

�

a −

b =

√

�

a −

√

�

b , b ≠

0.

•

The

imag

inar

y un

it i

is d

efin

ed t

o ha

ve t

he p

rope

rty

that

i2 =

-1.

•

Sim

plifi

ed s

quar

e ro

ot e

xpre

ssio

ns d

o no

t ha

ve r

adic

als

in t

he d

enom

inat

or, a

nd a

ny

num

ber

rem

aini

ng u

nder

the

squ

are

root

has

no

perf

ect

squa

re fa

ctor

oth

er t

han

1.

a. S

impl

ify

√

��

-

48 .

√

��

-

48 =

√

��

��

16

� (-

3)

=

√

��

16

� √

�

3 � √

��

-

1

= 4

i √

�

3

b. S

impl

ify

√

��

-

63 .

√

��

-

63 =

√

��

��

-

1 � 7

� 9

=

√

��

-

1 � √

�

7 .

√

�

9

= 3

i √

�

7

a. S

impl

ify

-3i

� 4i

.-

3i �

4i =

-12

i2

=

-12

(-1)

=

12

b. S

impl

ify

√ �

�

-3

� √

��

-15

.

√

��

-

3 � √

��

-

15 =

i √

�

3 � i

√

��

15

= i

2 √

��

45

= -

1 � √

�

9 � √

�

5

= -

3 √

�

5

Exam

ple

1Ex

ampl

e 2

Exam

ple

3 6i √

� 2

2i √

� 6

2i √

��

21

5

±3i

±i √

� 6

± i

±2i

√ �

3

011_

029_

ALG

2_A

_CR

M_C

04_C

R_6

6078

5.in

dd

2412

/20/

10

9:06

PM


NA

ME

DAT

E

P

ER

IOD

Lesson 4-4

4-4

Cha

pter

4

25

Gle

ncoe

Alg

ebra

2

Stud

y Gu

ide

and

Inte

rven

tion

(con

tinu

ed)

Co

mp

lex

Nu

mb

ers

Ope

rati

ons

wit

h Co

mpl

ex N

umbe

rs

Com

plex

Num

ber

A c

ompl

ex n

umbe

r is

any

num

ber t

hat c

an b

e w

ritte

n in

the

form

a +

bi,

whe

re a

and

b a

re

real

num

bers

and

i is

the

imag

inar

y un

it (i

2 = -

1). a

is c

alle

d th

e re

al p

art,

and

b is

cal

led

the

imag

inar

y pa

rt.

Add

ition

and

S

ubtr

actio

n of

C

ompl

ex N

umbe

rs

Com

bine

like

term

s.

(a +

bi)

+ (c

+ d

i) =

(a +

c) +

(b +

d)i

(a +

bi)

- (c

+ d

i) =

(a -

c) +

(b -

d)i

Mul

tiplic

atio

n of

C

ompl

ex N

umbe

rsU

se th

e de

fi niti

on o

f i2 a

nd th

e FO

IL m

etho

d:

(a +

bi)

(c +

di)

= (a

c -

bd

) + (a

d +

bc)

i

Com

plex

Con

juga

tea

+ b

i an

d a

- b

i ar

e co

mpl

ex c

onju

gate

s. T

he p

rodu

ct o

f com

plex

con

juga

tes

is a

lway

s a

real

num

ber.

To d

ivid

e by

a c

ompl

ex n

umbe

r, fir

st m

ulti

ply

the

divi

dend

and

div

isor

by

the

com

plex

co

njug

ate

of t

he d

ivis

or.

Exer

cise

sSi

mpl

ify.

1. (

-4

+ 2

i) +

(6 -

3i)

2.

(5 -

i) -

(3 -

2i)

3.

(6 -

3i)

+ (4

- 2

i)

4. (

-11

+ 4

i) -

(1 -

5i)

5.

(8 +

4i)

+ (8

- 4

i)

6. (5

+ 2

i) -

(-6

- 3

i)

7. (

2 +

i)(3

- i

) 8.

(5 -

2i)

(4 -

i)

9. (4

- 2

i)(1

- 2

i)

10.

5

−

3 +

i

11. 7

- 1

3i

−

2i

12

. 6 -

5i

−

3i

Si

mpl

ify

(6 +

i) +

(4 -

5i)

.

(6 +

i) +

(4 -

5i)

=

(6 +

4) +

(1 -

5)i

=

10

- 4

i

Si

mpl

ify

(2 -

5i)

� (-

4 +

2i)

.

(2 -

5i)

� (-

4 +

2i)

=

2(-

4) +

2(2

i) +

(-5i

)(-4)

+ (-

5i)(2

i)

= -

8 +

4i

+ 2

0i -

10i

2

=

-8

+ 2

4i -

10(

-1)

=

2 +

24i

Si

mpl

ify

(8 +

3i)

- (6

- 2

i).

(8 +

3i)

- (6

- 2

i)

= (8

- 6

) + [3

- (-

2)]i

=

2 +

5i

Si

mpl

ify

3 -

i

−

2 +

3i .

3 -

i

−

2 +

3i =

3

- i

−

2 +

3i �

2 -

3i

−

2 -

3i

=

6 -

9i

- 2

i +

3 i 2

−

−

4 -

9 i 2

=

3 -

11i

−

13

=

3 −

13 -

11

−

13 i

Exam

ple

1Ex

ampl

e 2

Exam

ple

3Ex

ampl

e 4

2

- i

2

+ i

1

0 -

5i

-

12 +

9i

16

11

+ 5

i

7 +

i18

- 1

3i-

10i

- 13

−

2 -

7 −

2 i-

5 −

3 - 2

i 3 −

2 - 1 −

2 i

011_

029_

ALG

2_A

_CR

M_C

04_C

R_6

6078

5.in

dd

2512

/20/

10

9:06

PM



Co

pyrig

ht ©

Gle

nco

e/M

cG

raw

-Hill, a

div

isio

n o

f Th

e M

cG

raw

-Hill C

om

pa

nie

s, In

c.

PDF Pass



NA

ME

DAT

E

P

ER

IOD

Lesson 4-4

4-4

Cha

pter

4

27

Gle

ncoe

Alg

ebra

2

Prac

tice

Co

mp

lex

Nu

mb

ers

Sim

plif

y.

1. √

��

-

36

2. √

��

-

8 �

√

��

-

32

3. √

��

-

15 �

√

��

-

25

4. (

-3i

) (4i

)(-5i

) 5.

(7i)

2 (6i)

6.

i42

7. i

55

8. i

89

9. (5

- 2

i) +

(-13

- 8

i)

10. (

7 -

6i)

+ (9

+ 1

1i)

11. (

-12

+ 4

8i) +

(15

+ 2

1i)

12. (

10 +

15i

) - (4

8 -

30i

)

13. (

28 -

4i)

- (1

0 -

30i

) 14

. (6

- 4

i) (6

+ 4

i)

15. (

8 -

11i

) (8

- 1

1i)

16. (

4 +

3i)

(2 -

5i)

17

. (7

+ 2

i) (9

- 6

i)

18. 6

+ 5

i −

-2i

-

5 −

2 + 3

i

19.

2 −

7 -

8i

20

. 3 -

i

−

2 -

i

21

. 2 -

4i

−

1 +

3i

Solv

e ea

ch e

quat

ion.

22. 5

n2 + 3

5 =

0

23. 2

m2 +

10

= 0

24. 4

m2 +

76

= 0

25

. -2m

2 - 6

= 0

26. -

5m2 -

65

= 0

27

. 3 −

4 x2 +

12

= 0

Fin

d th

e va

lues

of

ℓ an

d m

tha

t m

ake

each

equ

atio

n tr

ue.

28. 1

5 -

28i

= 3

ℓ +

(4m

)i

29. (

6 -

ℓ) +

(3m

)i =

-12

+ 2

7i

30. (

3ℓ +

4) +

(3 -

m)i

= 1

6 -

3i

31. (

7 +

m) +

(4ℓ

- 1

0)i

= 3

- 6

i

32. E

LECT

RICI

TY T

he im

peda

nce

in o

ne p

art

of a

ser

ies

circ

uit

is 1

+ 3

j ohm

s an

d th

e im

peda

nce

in a

noth

er p

art

of t

he c

ircu

it is

7 -

5j o

hms.

Add

the

se c

ompl

ex n

umbe

rs t

o fin

d th

e to

tal i

mpe

danc

e in

the

cir

cuit

.

33. E

LECT

RICI

TY U

sing

the

form

ula

E =

IZ,

find

the

vol

tage

E in

a c

ircu

it w

hen

the

curr

ent

I is

3 -

j am

ps a

nd t

he im

peda

nce

Z is

3 +

2j o

hms.

2

3 -

14i

7

5 -

24i

-1

- i

±i √

��

19

±i √

� 3

±i √

� 5

±i √

��

13

±4i

5, -

7

4, 6

8 -

2j o

hms 11

+ 3

j vol

ts

6

i -

16

-5 √

��

15

-

60i

-29

4i

-1

-

i i

-

8 -

10i

1

6 +

5i

3 +

69i

-

38 +

45i

1

8 +

26i

5

2 -

57 -

176

i

±i √

� 7

14

−

113 +

16

−

113 i

7 −

5 + 1 −

5 i

1, -

4

18, 9

011_

029_

ALG

2_A

_CR

M_C

04_C

R_6

6078

5.in

dd

2712

/20/

10

9:06

PM


NA

ME

DAT

E

P

ER

IOD

4-4

Cha

pter

4

26

Gle

ncoe

Alg

ebra

2

Skill

s Pr

actic

eC

om

ple

x N

um

bers

Sim

plif

y.

1. √

��

99

2.

√

��

27

−

49

3. √

��

�

52

x3 y5

4.

√

��

�

-

108x

7

5. √

��

�

-

81x6

6. √

��

-

23 �

√

��

-

46

7. (

3i)(-

2i)(5

i)

8. i

11

9. i

65

10. (

7 -

8i)

+ (-

12 -

4i)

11. (

-3

+ 5

i) +

(18

- 7

i)

12. (

10 -

4i)

- (7

+ 3

i)

13. (

7 -

6i)

(2 -

3i)

14

. (3

+ 4

i)(3

- 4

i)

15. 8

- 6

i −

3i

16

.

3i

−

4 +

2i

Solv

e ea

ch e

quat

ion.

17. 3

x2 + 3

= 0

18

. 5x2 +

125

= 0

19. 4

x2 + 2

0 =

0

20. -

x2 - 1

6 =

0

21. x

2 + 1

8 =

0

22. 8

x2 + 9

6 =

0

Fin

d th

e va

lues

of

� a

nd m

tha

t m

ake

each

equ

atio

n tr

ue.

23. 2

0 -

12i

= 5

� +

(4m

)i

24. �

- 1

6i =

3 -

(2m

)i

25. (

4 +

�) +

(2m

)i =

9 +

14i

26

. (3

- m

) + (7

� -

14)

i =

1 +

7i

3 √ �

11

2⎪x

⎪y2 √

��

�

13x

y

6i ⎪ x

3 ⎥ √

��

3x

9 ⎪x

3 ⎥ i

-23

√ �

2

30i

-i

i-

5 -

12i

15 -

2i

3 -

7i

-4

- 3

3i25

±i

±5i

±i √

� 5

±4i

±3i

√ �

2 ±

2i √

� 3

4, -

33,

8

5, 7

3, 2

3 −

10 +

3 −

5 i-

2 -

8 −

3 i

3 √ �

3 −

7

011_

029_

ALG

2_A

_CR

M_C

04_C

R_6

6078

5.in

dd

2612

/20/

10

9:06

PM



Copyri

ght

© G

lencoe/M

cG

raw

-Hill

, a

div

isio

n o

f T

he

McG

raw

-Hill

Co

mp

an

ies,

Inc.

PDF Pass



NA

ME

DAT

E

P

ER

IOD

Lesson 4-4

4-4

Cha

pter

4

29

Gle

ncoe

Alg

ebra

2

Enri

chm

ent

Co

nju

gate

s an

d A

bso

lute

Valu

e

Whe

n st

udyi

ng c

ompl

ex n

umbe

rs, i

t is

oft

en c

onve

nien

t to

rep

rese

nt a

com

plex

nu

mbe

r by

a s

ingl

e va

riab

le. F

or e

xam

ple,

we

mig

ht le

t z

= x

+ y

i. W

e de

note

th

e co

njug

ate

of z

by

−

z . T

hus,

−

z =

x -

yi.

We

can

defin

e th

e ab

solu

te v

alue

of a

com

plex

num

ber

as fo

llow

s.

⎪ z⎥

= ⎪

x +

yi ⎥

= √

��

�

x2 +

y2

Ther

e ar

e m

any

impo

rtan

t re

lati

onsh

ips

invo

lvin

g co

njug

ates

and

abs

olut

e va

lues

of c

ompl

ex n

umbe

rs.

Sh

ow ⎪

z⎥ 2 =

z −

z fo

r an

y co

mpl

ex n

umbe

r z.

Let

z =

x +

yi.

Then

,

zz −

z =

(x +

yi)

(x -

yi)

=

x2 +

y2

=

√

��

��

(x

2 + y

2 )2

=

⎪z⎥

2

Sh

ow

−

z −

⎪ z⎥ 2

is t

he m

ulti

plic

ativ

e in

vers

e fo

r an

y no

nzer

o

com

plex

num

ber

z.

We

know

⎪z⎥

2 = z

−

z . I

f z ≠

0, t

hen

we

have

z (

−

z −

⎪ z⎥

2 )

= 1

.

Thus

, −

z −

⎪ z⎥

2 i

s th

e m

ulti

plic

ativ

e in

vers

e of

z.

Exer

cise

sFo

r ea

ch o

f th

e fo

llow

ing

com

plex

num

bers

, fin

d th

e ab

solu

te v

alue

and

m

ulti

plic

ativ

e in

vers

e.

1. 2

i

2. -

4 -

3i

3.

12

- 5

i

4. 5

- 1

2i

5. 1

+ i

6.

√

�

3 -

i

7. √

�

3 −

3 +

√

�

3 −

3 i

8. √

�

2 −

2 -

√

�

2 −

2 i

9. 1 −

2 - √

�

3 −

2 i

Exam

ple

1

Exam

ple

2

2;

-i

−

2 5

; -

4 +

3i

−

25

1

3; 12

+ 5

i −

16

9

13;

5 +

12i

−

169

√

� 2 ; 1

- i

−

2

2;

√ �

3 +

i −

4

√

� 6

−

3 ; √

� 3 -

i √

� 3

−

2

1;

√ �

2 −

2

+ √

� 2

−

2 i

1;

1 −

2 + √

� 3

−

2 i

011_

029_

ALG

2_A

_CR

M_C

04_C

R_6

6078

5.in

dd

2912

/20/

10

9:06

PM


NA

ME

DAT

E

P

ER

IOD

4-4

Cha

pter

4

28

Gle

ncoe

Alg

ebra

2

Wor

d Pr

oble

m P

ract

ice

Co

mp

lex

Nu

mb

ers

1. S

IGN

ERR

ORS

Jen

nife

r an

d Je

ssic

a co

me

up w

ith

diffe

rent

ans

wer

s to

the

sa

me

prob

lem

. The

y ha

d to

mul

tipl

y (4

+i)

(4 -

i) a

nd g

ive

thei

r an

swer

as

a co

mpl

ex n

umbe

r. Je

nnife

r cl

aim

s th

at

the

answ

er is

15

and

Jess

ica

clai

ms

that

the

ans

wer

is 1

7. W

ho is

cor

rect

? E

xpla

in.

2. C

OM

PLEX

CO

NJU

GAT

ES Y

ou h

ave

seen

tha

t th

e pr

oduc

t of

com

plex

co

njug

ates

is a

lway

s a

real

num

ber.

Show

tha

t th

e su

m o

f com

plex

co

njug

ates

is a

lso

alw

ays

a re

al n

umbe

r.

3. P

YTH

AG

ORE

AN

TRI

PLES

If t

hree

in

tege

rs a

, b, a

nd c

sat

isfy

a2

+b2

=c2 ,

then

the

y ar

e ca

lled

a Py

thag

orea

n tr

iple

. Sup

pose

tha

t a,

b, a

nd c

are

a

Pyth

agor

ean

trip

le. S

how

tha

t th

e re

al

and

imag

inar

y pa

rts

of (a

+bi

)2 , to

geth

er w

ith

the

num

ber

c2 , fo

rm

anot

her

Pyth

agor

ean

trip

le.

4. R

OTA

TIO

NS

Com

plex

num

bers

can

be

used

to

perf

orm

rot

atio

ns in

the

pla

ne.

For

exam

ple,

if (x

, y) a

re t

he c

oord

inat

es

of a

poi

nt in

the

pla

ne, t

hen

the

real

an

d im

agin

ary

part

s of

i(x

+ y

i) a

re

the

hori

zont

al a

nd v

erti

cal c

oord

inat

es

of t

he 9

0° c

ount

ercl

ockw

ise

rota

tion

of

(x, y

) abo

ut t

he o

rigi

n. W

hat

are

the

real

an

d im

agin

ary

part

s of

i(x

+ y

i)?

5. E

LECT

RICA

L EN

GIN

EERI

NG

A

lter

nati

ng c

urre

nt (A

C) i

n an

ele

ctri

cal

circ

uit

can

be d

escr

ibed

by

com

plex

nu

mbe

rs. I

n an

y el

ectr

ical

cir

cuit

, Z, t

he

impe

danc

e in

the

cir

cuit

, is

rela

ted

to

the

volt

age

V a

nd t

he c

urre

nt I

by

the

form

ula

Z =

V

− I . Th

e st

anda

rd e

lect

rica

l vo

ltag

e in

Eur

ope

is 2

20 v

olts

, so

in

thes

e pr

oble

ms

use

V =

220

.

a. F

ind

the

impe

danc

e in

a s

tand

ard

Eur

opea

n ci

rcui

t if

the

curr

ent

is

22 –

11i

am

ps.

b. F

ind

the

curr

ent

in a

sta

ndar

d E

urop

ean

circ

uit

if th

e im

peda

nce

is

10 –

5i

wat

ts.

c. F

ind

the

impe

danc

e in

a s

tand

ard

Eur

opea

n ci

rcui

t if

the

curr

ent

is

20i

amps

.

a

+ b

i and

a -

bi a

re c

ompl

ex

conj

ugat

es a

nd th

eir

sum

is 2

a,

whi

ch is

rea

l.

(

a +

bi)

2 = a

2 - b

2 + 2

ab

i; a

2 - b

2 and

2a

b a

re in

tege

rs a

nd

(a2 -

b2 )2 +

(2a

b)2 =

a

4 - 2

a2 b

2 + b

4 + 4

a2 b

2 =

a4 +

2a

2 b2 +

b4 =

(a2 +

b2 )2 =

(c

2 )2 , so

a2 +

b2 =

c2 a

s de

sire

d.

T

he r

eal p

art i

s -

y an

d im

agin

ary

part

is x

.

18 –

8i a

mps

J

essi

ca is

cor

rect

; (4

+ i

)(4 -

i) =

1

6 +

4i -

4i -

i2

=

16

- (-

1) =

16

+ 1

= 1

7.

8 +

4i

–11

i am

ps

011_

029_

ALG

2_A

_CR

M_C

04_C

R_6

6078

5.in

dd

2812

/20/

10

9:06

PM



Co

pyrig

ht ©

Gle

nco

e/M

cG

raw

-Hill, a

div

isio

n o

f Th

e M

cG

raw

-Hill C

om

pa

nie

s, In

c.

PDF Pass



NA

ME

DAT

E

P

ER

IOD

Lesson 4-5

4-5

Cha

pter

4

31

Gle

ncoe

Alg

ebra

2

Stud

y Gu

ide

and

Inte

rven

tion

(con

tinu

ed)

Co

mp

leti

ng

th

e S

qu

are

Com

plet

e th

e Sq

uare

To

com

plet

e th

e sq

uare

for

a qu

adra

tic

expr

essi

on o

f the

form

x2 +

bx,

fol

low

the

se s

teps

.

1. F

ind

b −

2 .

2.

Squ

are

b −

2 .

3.

Add

( b −

2 ) 2 to

x2 + b

x.

F

ind

the

valu

e of

c t

hat

mak

es x

2 + 2

2x +

c a

pe

rfec

t sq

uare

tri

nom

ial.

The

n w

rite

the

tri

nom

ial a

s th

e sq

uare

of

a bi

nom

ial.

Step

1

b =

22

; b −

2 = 1

1

Step

2 1

12 = 1

21St

ep 3

c

= 1

21

The

trin

omia

l is

x2 + 2

2x +

121

, w

hich

can

be

wri

tten

as

(x +

11)

2 .

So

lve

2x2 -

8x

- 2

4 =

0 b

y co

mpl

etin

g th

e sq

uare

.

2x2 -

8x

- 2

4 =

0

Orig

inal

equ

atio

n

2x

2 - 8

x -

24

−

2

= 0 −

2 D

ivid

e ea

ch s

ide

by 2

.

x2 -

4x

- 1

2 =

0

x2 - 4

x -

12

is n

ot a

per

fect

squ

are.

x2 -

4x

= 1

2 A

dd 1

2 to

eac

h si

de.

x2 -

4x

+ 4

= 1

2 +

4

Since

( 4 −

2 ) 2 = 4

, add

4 to

eac

h si

de.

(x

- 2

)2 =

16

Fact

or th

e sq

uare

.

x

- 2

= ±

4 S

quar

e R

oot P

rope

rty

x =

6 o

r x

= -

2

Sol

ve e

ach

equa

tion.

The

solu

tion

set

is {6

, -2}

.

Exer

cise

s

Fin

d th

e va

lue

of c

tha

t m

akes

eac

h tr

inom

ial a

per

fect

squ

are.

The

n w

rite

the

tr

inom

ial a

s a

perf

ect

squa

re.

1. x

2 - 1

0x +

c

2. x

2 + 6

0x +

c

3. x

2 - 3

x +

c

4. x

2 + 3

.2x

+ c

5.

x2 +

1 −

2 x +

c

6. x

2 - 2

.5x

+ c

Solv

e ea

ch e

quat

ion

by c

ompl

etin

g th

e sq

uare

. 7

. y2 -

4y

- 5

= 0

8.

x2 -

8x

- 6

5 =

0

9. w

2 - 1

0w +

21

= 0

10. 2

x2 - 3

x +

1 =

0

11. 2

x2 - 1

3x -

7 =

0

12. 2

5x2 +

40x

- 9

= 0

13. x

2 + 4

x +

1 =

0

14. y

2 + 1

2y +

4 =

0

15. t

2 + 3

t - 8

= 0

Exam

ple

1Ex

ampl

e 2

2

5; (x

- 5

)2 9

00; (

x +

30)

2 9

− 4 ;

(x -

3 −

2 ) 2

2

.56;

(x +

1.6

)2

1 −

16

; (x

+ 1 −

4 ) 2 1

.562

5; (x

- 1

.25)

2

-

1, 5

-

5, 1

3 3

, 7

1

, 1 −

2 -

1 −

2 , 7

1 −

5 , -

9 −

5

-

2 ±

√ �

3 -

6 ±

4 √

� 2

-3

± √

��

41

−

2

030_

041_

ALG

2_A

_CR

M_C

04_C

R_6

6078

5.in

dd

3112

/20/

10

9:06

PM


NA

ME

DAT

E

P

ER

IOD

4-5

Cha

pter

4

30

Gle

ncoe

Alg

ebra

2

Stud

y Gu

ide

and

Inte

rven

tion

Co

mp

leti

ng

th

e S

qu

are

Squa

re R

oot

Prop

erty

Use

the

Squ

are

Roo

t Pr

oper

ty t

o so

lve

a qu

adra

tic

equa

tion

th

at is

in t

he fo

rm “p

erfe

ct s

quar

e tr

inom

ial =

con

stan

t.”

So

lve

each

equ

atio

n by

usi

ng t

he S

quar

e R

oot

Pro

pert

y. R

ound

to

the

near

est

hund

redt

h if

nec

essa

ry.

a. x

2 - 8

x +

16

= 2

5

x2 - 8

x +

16

= 2

5

(x -

4)2 =

25

x

- 4

= √

��

25

or

x

- 4

= -

√

��

25

x =

5 +

4 =

9 o

r

x =

-5

+ 4

= -

1

Th

e so

luti

on s

et is

{9, -

1}.

b. 4

x2 - 2

0x +

25

= 3

2 4

x2 - 2

0x +

25

= 3

2

(2x

- 5

)2 = 3

22x

- 5

= √

��

32

or

2x -

5 =

- √

��

32

2x

- 5

= 4

√

�

2 or

2x

- 5

= -

4 √

�

2

x

= 5

± 4

√

�

2 −

2

The

solu

tion

set

is {

5 ±

4 √

�

2 −

2

}

.

Exer

cise

sSo

lve

each

equ

atio

n by

usi

ng t

he S

quar

e R

oot

Pro

pert

y. R

ound

to

the

near

est

hund

redt

h if

nec

essa

ry.

1. x

2 - 1

8x +

81

= 4

9 2.

x2 +

20x

+ 1

00 =

64

3. 4

x2 + 4

x +

1 =

16

4. 3

6x2 +

12x

+ 1

= 1

8 5.

9x2 -

12x

+ 4

= 4

6.

25x

2 + 4

0x +

16

= 2

8

7. 4

x2 - 2

8x +

49

= 6

4 8.

16x

2 + 2

4x +

9 =

81

9. 1

00x2 -

60x

+ 9

= 1

21

10. 2

5x2 +

20x

+ 4

= 7

5 11

. 36x

2 + 4

8x +

16

= 1

2 12

. 25x

2 - 3

0x +

9 =

96

Exam

ple

{

-2

± 5

√ �

3 −

5

} {

-2

± √

� 3

−

3 }

{ 3

± 4

√ �

6 −

5

}

{

2, 1

6}

{-

2, -

18}

{ 3 −

2 , -

5 −

2 }

{

-1

± 3

√ �

2 −

6

} {

0, 4 −

3 } {

-4

± 2

√ �

7 −

5

}

{

15

−

2 , -

1 −

2 } {

3 −

2 , -3 }

-

{0.8

, 1.4

}

030_

041_

ALG

2_A

_CR

M_C

04_C

R_6

6078

5.in

dd

3012

/20/

10

9:06

PM



Copyri

ght

© G

lencoe/M

cG

raw

-Hill

, a

div

isio

n o

f T

he

McG

raw

-Hill

Co

mp

an

ies,

Inc.

PDF Pass



NA

ME

DAT

E

P

ER

IOD

4-5

Cha

pter

4

32

Gle

ncoe

Alg

ebra

2

Skill

s Pr

actic

eC

om

ple

tin

g t

he S

qu

are

Solv

e ea

ch e

quat

ion

by u

sing

the

Squ

are

Roo

t P

rope

rty.

Rou

nd t

o th

e ne

ares

t hu

ndre

dth

if n

eces

sary

.

1. x

2 - 8

x +

16

= 1

2.

x2 +

4x

+ 4

= 1

3. x

2 + 1

2x +

36

= 2

5

4. 4

x2 - 4

x +

1 =

9

5. x

2 + 4

x +

4 =

2

6. x

2 - 2

x +

1 =

5

7. x

2 - 6

x +

9 =

7

8. x

2 + 1

6x +

64

= 1

5

Fin

d th

e va

lue

of c

tha

t m

akes

eac

h tr

inom

ial a

per

fect

squ

are.

The

n w

rite

the

tr

inom

ial a

s a

perf

ect

squa

re.

9. x

2 + 1

0x +

c

10. x

2 - 1

4x +

c

11. x

2 + 2

4x +

c

12. x

2 + 5

x +

c

13. x

2 - 9

x +

c

14. x

2 - x

+ c

Solv

e ea

ch e

quat

ion

by c

ompl

etin

g th

e sq

uare

.

15. x

2 - 1

3x +

36

= 0

16

. x2 +

3x

= 0

17. x

2 + x

- 6

= 0

18

. x2 -

4x

- 1

3 =

0

19. 2

x2 + 7

x -

4 =

0

20. 3

x2 + 2

x -

1 =

0

21. x

2 + 3

x -

6 =

0

22. x

2 - x

- 3

= 0

23. x

2 = -

11

24. x

2 - 2

x +

4 =

0

-1,

-3

-1,

2

-11

.87,

-4.

13

2 ±

√ �

�

17

1 −

3 , -1

± i

√ �

11

3, 5 -

1, -

11

25; (

x +

5)2

49; (

x -

7)2

144;

(x +

12)

2

81

−

4 ; (

x -

9 −

2 ) 2

4, 9

0, -

3

2, -

3

-4,

1 −

2

1 ±

i √

� 3

1 ±

√ �

�

13

−

2

-3

± √

��

33

−

2

25

−

4 ; (

x +

5 −

2 ) 2

1 −

4 ; (x

- 1 −

2 ) 2

-3.

41, -

0.59

0.35

, 2.6

5

-1.

24, 3

.24

030_

041_

ALG

2_A

_CR

M_C

04_C

R_6

6078

5.in

dd

3212

/20/

10

9:06

PM


NA

ME

DAT

E

P

ER

IOD

Lesson 4-5

4-5

Cha

pter

4

33

Gle

ncoe

Alg

ebra

2

Prac

tice

Co

mp

leti

ng

th

e S

qu

are

Solv

e ea

ch e

quat

ion

by u

sing

the

Squ

are

Roo

t P

rope

rty.

Rou

nd t

o th

e ne

ares

t hu

ndre

dth

if n

eces

sary

.

1. x

2 + 8

x +

16

= 1

2.

x2 +

6x

+ 9

= 1

3.

x2 +

10x

+ 2

5 =

16

4. x

2 - 1

4x +

49

= 9

5.

4x2 +

12x

+ 9

= 4

6.

x2 -

8x

+ 1

6 =

8

7. x

2 - 6

x +

9 =

5

8. x

2 - 2

x +

1 =

2

9. 9

x2 - 6

x +

1 =

2

Fin

d th

e va

lue

of c

tha

t m

akes

eac

h tr

inom

ial a

per

fect

squ

are.

The

n w

rite

the

tr

inom

ial a

s a

perf

ect

squa

re.

10. x

2 + 1

2x +

c

11. x

2 - 2

0x +

c

12. x

2 + 1

1x +

c

13. x

2 + 0

.8x

+ c

14

. x2 -

2.2

x +

c

15. x

2 - 0

.36x

+ c

16. x

2 + 5 −

6 x +

c

17. x

2 - 1 −

4 x +

c

18. x

2 - 5 −

3 x +

c

Solv

e ea

ch e

quat

ion

by c

ompl

etin

g th

e sq

uare

.

19. x

2 + 6

x +

8 =

0

20. 3

x2 + x

- 2

= 0

21

. 3x2 -

5x

+ 2

= 0

22. x

2 + 1

8 =

9x

23. x

2 - 1

4x +

19

= 0

24

. x2 +

16x

- 7

= 0

25. 2

x2 + 8

x -

3 =

0

26. x

2 + x

- 5

= 0

27

. 2x2 -

10x

+ 5

= 0

28. x

2 + 3

x +

6 =

0

29. 2

x2 + 5

x +

6 =

0

30. 7

x2 + 6

x +

2 =

0

31. G

EOM

ETRY

Whe

n th

e di

men

sion

s of

a c

ube

are

redu

ced

by 4

inch

es o

n ea

ch s

ide,

the

su

rfac

e ar

ea o

f the

new

cub

e is

864

squ

are

inch

es. W

hat

wer

e th

e di

men

sion

s of

the

or

igin

al c

ube?

32. I

NV

ESTM

ENTS

The

am

ount

of m

oney

A in

an

acco

unt

in w

hich

P d

olla

rs a

re in

vest

ed

for

2 ye

ars

is g

iven

by

the

form

ula

A =

P(1

+ r

)2 , w

here

r is

the

inte

rest

rat

e co

mpo

unde

d an

nual

ly. I

f an

inve

stm

ent

of $

800

in t

he a

ccou

nt g

row

s to

$88

2 in

tw

o ye

ars,

at

wha

t in

tere

st r

ate

was

it in

vest

ed?

-

5, -

3 -

4, -

2 -

9, -

1

-

4, -

2 2

− 3 ,

-1

1, 2

− 3

4

, 10

- 1 −

2 , -

5 −

2 4

± 2

√ �

2

3

± √

� 5

1 ±

√ �

2 1

± √

� 2

−

3

3

6; (x

+ 6

)2 1

00; (

x -

10)

2 12

1 −

4

; (x

+ 11

−

2 )

2

0

.16;

(x +

0.4

)2 1

.21;

(x -

1.1

)2 0

.032

4; (x

- 0

.18)

2

25

−

144 ;

(x +

5 −

12

) 2

1 −

64

; (x

- 1 −

8 ) 2 25

−

36

; (x

- 5 −

6 ) 2

6

, 3

7 ±

√ �

�

30

-8

± √

��

71

-

4 ±

√ �

�

22

−

2

-1

± √

��

21

−

2

5 ±

√ �

�

15

−

2

16 in

. by

16 in

. by

16 in

.

5%

-

3 ±

i √

��

15

−

2

-5

± i

√ �

�

23

−

4

-3

± i

√ �

5 −

7

030_

041_

ALG

2_A

_CR

M_C

04_C

R_6

6078

5.in

dd

3312

/20/

10

9:06

PM



Co

pyrig

ht ©

Gle

nco

e/M

cG

raw

-Hill, a

div

isio

n o

f Th

e M

cG

raw

-Hill C

om

pa

nie

s, In

c.

PDF Pass



NA

ME

DAT

E

P

ER

IOD

4-5

Cha

pter

4

34

Gle

ncoe

Alg

ebra

2

Wor

d Pr

oble

m P

ract

ice

Co

mp

leti

ng

th

e S

qu

are

1. C

OM

PLET

ING

TH

E SQ

UA

RESa

man

tha

need

s to

sol

ve t

he e

quat

ion

x2-

12x

= 4

0.

Wha

t m

ust

she

do t

o ea

ch s

ide

of t

he

equa

tion

to

com

plet

e th

e sq

uare

?

2. A

RT T

he a

rea

in s

quar

e in

ches

of t

he

draw

ing

Folia

ge b

y Pa

ul C

ézan

ne is

ap

prox

imat

ed b

y th

e eq

uati

on

y=

x2– 4

0x+

396

. Com

plet

e th

e sq

uare

an

d fin

d th

e tw

o ro

ots,

whi

ch a

re e

qual

to

the

app

roxi

mat

e le

ngth

and

wid

th o

f th

e dr

awin

g.

3. C

OM

POU

ND

INTE

REST

Nik

ki in

vest

ed

$100

0 in

a s

avin

gs a

ccou

nt w

ith

inte

rest

co

mpo

unde

d an

nual

ly. A

fter

tw

o ye

ars

the

bala

nce

in t

he a

ccou

nt is

$12

10.

Use

the

com

poun

d in

tere

st fo

rmul

a A

=P(

1 +

r)t t

o fin

d th

e an

nual

in

tere

st r

ate.

4. R

EACT

ION

TIM

E La

uren

was

eat

ing

lunc

h w

hen

she

saw

her

frie

nd J

ason

ap

proa

ch. T

he r

oom

was

cro

wde

d an

d Ja

son

had

to li

ft h

is t

ray

to a

void

ob

stac

les.

Sud

denl

y, a

glas

s on

Jas

on’s

lunc

h tr

ay t

ippe

d an

d fe

ll of

f the

tra

y. La

uren

lung

ed fo

rwar

d an

d m

anag

ed t

o ca

tch

the

glas

s ju

st b

efor

e it

hit

the

gr

ound

. The

hei

ght

h, in

feet

, of t

he

glas

s t s

econ

ds a

fter

it w

as d

ropp

ed is

gi

ven

by h

=-

16t2

+ 4

.5. L

aure

n ca

ught

th

e gl

ass

whe

n it

was

six

inch

es o

ff th

e gr

ound

. How

long

was

the

gla

ss in

the

ai

r be

fore

Lau

ren

caug

ht it

?

5. P

ARA

BOLA

S A

par

abol

a is

mod

eled

by

y=

x2-

10x

+ 2

8. J

ane’

s ho

mew

ork

prob

lem

req

uire

s th

at s

he fi

nd t

he

vert

ex o

f the

par

abol

a. S

he u

ses

the

com

plet

ing

squa

re m

etho

d to

exp

ress

th

e fu

ncti

on in

the

form

y

= (x

-h)

2+

k, w

here

(h, k

) is

the

vert

ex o

f the

par

abol

a. W

rite

the

fu

ncti

on in

the

form

use

d by

Jan

e.

6. A

UD

ITO

RIU

M S

EATI

NG

The

sea

ts in

an

aud

itor

ium

are

arr

ange

d in

a s

quar

e gr

id p

atte

rn. T

here

are

45

row

s an

d 45

co

lum

ns o

f cha

irs.

For

a s

peci

al c

once

rt,

orga

nize

rs d

ecid

e to

incr

ease

sea

ting

by

addi

ng n

row

s an

d n

colu

mns

to

mak

e a

squa

re p

atte

rn o

f sea

ting

45

+ n

sea

ts o

n a

side

.

a. H

ow m

any

seat

s ar

e th

ere

afte

r th

e ex

pans

ion?

b. W

hat

is n

if o

rgan

izer

s w

ish

to a

dd

1000

sea

ts?

c. I

f org

aniz

ers

do a

dd 1

000

seat

s,

wha

t is

the

sea

ting

cap

acit

y of

th

e au

dito

rium

?

0

.5 s

econ

d

y

= (x

- 5

)2 + 3

n

2 + 9

0n +

202

5

3

025

A

dd 3

6.

1

8 in

ches

by

22 in

ches

1

0%

1

0

030_

041_

ALG

2_A

_CR

M_C

04_C

R_6

6078

5.in

dd

3412

/20/

10

9:06

PM


NA

ME

DAT

E

P

ER

IOD

Lesson 4-5

4-5

Cha

pter

4

35

Gle

ncoe

Alg

ebra

2

Enri

chm

ent

Th

e G

old

en

Qu

ad

rati

c E

qu

ati

on

s

A g

olde

n re

ctan

gle

has

the

prop

erty

tha

t it

s le

ngth

a

a a

b b

a

can

be w

ritt

en a

s a

+ b

, whe

re a

is t

he w

idth

of t

he

rect

angl

e an

d a

+ b

−

a =

a −

b . A

ny g

olde

n re

ctan

gle

can

be

divi

ded

into

a s

quar

e an

d a

smal

ler

gold

en r

ecta

ngle

, as

sho

wn.

The

prop

orti

on u

sed

to d

efin

e go

lden

rec

tang

les

can

be

used

to

deri

ve t

wo

quad

rati

c eq

uati

ons.

The

se a

re

som

etim

es c

alle

d go

lden

qua

drat

ic e

quat

ions

.

Solv

e ea

ch p

robl

em.

1. I

n th

e pr

opor

tion

for

the

gold

en r

ecta

ngle

, let

a e

qual

1. W

rite

the

res

ulti

ng

quad

rati

c eq

uati

on a

nd s

olve

for

b.

2. I

n th

e pr

opor

tion

, let

b e

qual

1. W

rite

the

res

ulti

ng q

uadr

atic

equ

atio

n an

d so

lve

for

a.

3. D

escr

ibe

the

diffe

renc

e be

twee

n th

e tw

o go

lden

qua

drat

ic e

quat

ions

you

fo

und

in e

xerc

ises

1 a

nd 2

.

4. S

how

tha

t th

e po

siti

ve s

olut

ions

of t

he t

wo

equa

tion

s in

exe

rcis

es 1

and

2

are

reci

proc

als.

5. U

se t

he P

ytha

gore

an T

heor

em t

o fin

d a

radi

cal e

xpre

ssio

n fo

r th

e di

agon

al

of a

gol

den

rect

angl

e w

hen

a =

1.

6. F

ind

a ra

dica

l exp

ress

ion

for

the

diag

onal

of a

gol

den

rect

angl

e w

hen

b =

1.

b

2 + b

- 1

= 0

b =

-1

+ √

� 5

−

2

a

2 - a

- 1

= 0

a =

1 +

√ �

5 −

2

T

he s

igns

of t

he fi

rst-

degr

ee te

rms

are

oppo

site

.

(

-1

+ √

� 5

−

2 )

( 1

+ √

� 5

−

2 )

= -

( 1 2 ) +

( √

� 5 ) 2

−

4

= -

1 +

5

−

4 =

1

d

= √

��

��

�

10 -

2 √

� 5

−

2

d =

√ �

��

��

10 +

2 √

� 5

−

2

030_

041_

ALG

2_A

_CR

M_C

04_C

R_6

6078

5.in

dd

3512

/20/

10

9:06

PM



Copyri

ght

© G

lencoe/M

cG

raw

-Hill

, a

div

isio

n o

f T

he

McG

raw

-Hill

Co

mp

an

ies,

Inc.

PDF Pass



NA

ME

DAT

E

P

ER

IOD

4-6

Cha

pter

4

36

Gle

ncoe

Alg

ebra

2

Stud

y Gu

ide

and

Inte

rven

tion

Th

e Q

uad

rati

c F

orm

ula

an

d t

he D

iscri

min

an

t

Qua

drat

ic F

orm

ula

The

Qua

drat

ic F

orm

ula

can

be u

sed

to s

olve

any

qua

drat

ic

equa

tion

onc

e it

is w

ritt

en in

the

form

ax2 +

bx

+ c

= 0

.

Qua

drat

ic F

orm

ula

The

solu

tions

of a

x2 +

bx

+ c

= 0

, with

a ≠

0, a

re g

iven

by

x =

-

b ±

√

��

��

b2 -

4ac

−

2a

.

So

lve

x2 - 5

x =

14

by u

sing

the

Qua

drat

ic F

orm

ula.

Rew

rite

the

equ

atio

n as

x2 -

5x

- 1

4 =

0.

x =

-b

± √

��

��

b

2 - 4

ac

−

2a

Q

uadr

atic

For

mul

a

=

-

(-5)

± √

��

��

��

�

(-5)

2 - 4

(1)(-

14)

−−

2(1)

Rep

lace

a w

ith 1

, b w

ith -

5, a

nd c

with

-14

.

=

5

± √

��

81

−

2

Sim

plify

.

=

5

± 9

−

2

=

7

or

-2

The

solu

tion

s ar

e -

2 an

d 7.

Exer

cise

sSo

lve

each

equ

atio

n by

usi

ng t

he Q

uadr

atic

For

mul

a.

1. x

2 + 2

x -

35

= 0

2.

x2 +

10x

+ 2

4 =

0

3. x

2 - 1

1x +

24

= 0

4. 4

x2 + 1

9x -

5 =

0

5. 1

4x2 +

9x

+ 1

= 0

6.

2x2 -

x -

15

= 0

7. 3

x2 + 5

x =

2

8. 2

y2 + y

- 1

5 =

0

9. 3

x2 - 1

6x +

16

= 0

10. 8

x2 + 6

x -

9 =

0

11. r

2 - 3r

−

5 + 2 −

25 =

0

12. x

2 - 1

0x -

50

= 0

13. x

2 + 6

x -

23

= 0

14

. 4x2 -

12x

- 6

3 =

0

15. x

2 - 6

x +

21

= 0

Exam

ple

5

, -7

-4,

-6

3, 8

1

− 4 ,

-5

- 1 −

2 , -

1 −

7 3

, - 5 −

2

-

2, 1 −

3 5

− 2 ,

-3

4, 4

− 3

-

3 −

2 , 3 −

4 2

− 5 ,

1 −

5 5

± 5

√ �

3

-

3 ±

4 √

� 2

3 ±

6 √

� 2

−

2

3 ±

2i √

� 3

030_

041_

ALG

2_A

_CR

M_C

04_C

R_6

6078

5.in

dd

3612

/20/

10

9:06

PM


NA

ME

DAT

E

P

ER

IOD

Lesson 4-6

4-6

Cha

pter

4

37

Gle

ncoe

Alg

ebra

2

Root

s an

d th

e D

iscr

imin

ant

Dis

crim

inan

tTh

e ex

pres

sion

und

er th

e ra

dica

l sig

n, b

2 - 4

ac, i

n th

e Q

uadr

atic

For

mul

a is

cal

led

the

disc

rimin

ant.

Dis

crim

inan

tTy

pe a

nd N

umbe

r of

Roo

ts

b2 -

4ac

> 0

and

a p

erfe

ct s

quar

e 2

ratio

nal r

oots

b2 -

4ac

> 0

, but

not

a p

erfe

ct s

quar

e2

irrat

iona

l roo

ts

b2 -

4ac

= 0

1 ra

tiona

l roo

t

b2 -

4ac

< 0

2 co

mpl

ex ro

ots

F

ind

the

valu

e of

the

dis

crim

inan

t fo

r ea

ch e

quat

ion.

The

n de

scri

be

the

num

ber

and

type

of

root

s fo

r th

e eq

uati

on.

a. 2

x2 + 5

x +

3Th

e di

scri

min

ant

is

b2 - 4

ac =

52 -

4(2

) (3)

or

1.Th

e di

scri

min

ant

is a

per

fect

squ

are,

so

the

equa

tion

has

2 r

atio

nal r

oots

.

b. 3

x2 - 2

x +

5Th

e di

scri

min

ant

is

b2 - 4

ac =

(-2)

2 - 4

(3) (

5) o

r -

56.

The

disc

rim

inan

t is

neg

ativ

e, s

o th

e eq

uati

on h

as 2

com

plex

roo

ts.

Exer

cise

sC

ompl

ete

part

s a-

c fo

r ea

ch q

uadr

atic

equ

atio

n.a.

Fin

d th

e va

lue

of t

he d

iscr

imin

ant.

b. D

escr

ibe

the

num

ber

and

type

of

root

s.c.

Fin

d th

e ex

act

solu

tion

s by

usi

ng t

he Q

uadr

atic

For

mul

a.

1. p

2 + 1

2p =

-4

2.

9x2 -

6x

+ 1

= 0

3.

2x2 -

7x

- 4

= 0

4. x

2 + 4

x -

4 =

0

5. 5

x2 - 3

6x +

7 =

0

6. 4

x2 - 4

x +

11

= 0

7. x

2 - 7

x +

6 =

0

8. m

2 - 8

m =

-14

9.

25x

2 - 4

0x =

-16

10. 4

x2 + 2

0x +

29

= 0

11

. 6x2 +

26x

+ 8

= 0

12

. 4x2 -

4x

- 1

1 =

0

Exam

ple

Stud

y Gu

ide

and

Inte

rven

tion

(con

tinu

ed)

Th

e Q

uad

rati

c F

orm

ula

an

d t

he D

iscri

min

an

t

2

irra

tiona

l ro

ots;

1

rat

iona

l roo

t; 1 −

3 2

rat

iona

l roo

ts; -

1 −

2 , 4

-6

± 4

√ �

2

2

irra

tiona

l roo

ts;

2 r

atio

nal r

oots

;

-16

0; 2

com

plex

roo

ts;

-

2 ±

2 √

� 2

1 −

5 , 7

1 ±

i √

��

10

−

2

2

rat

iona

l roo

ts;

2 ir

ratio

nal r

oots

; 1

rat

iona

l roo

t; 4 −

5

1, 6

4

± √

� 2

2

com

plex

roo

ts;

2

rat

iona

l roo

ts;

2 ir

ratio

nal r

oots

;

- 5 −

2 ± i

-4,

- 1 −

3

1 −

2 ± √

� 3

030_

041_

ALG

2_A

_CR

M_C

04_C

R_6

6078

5.in

dd

3712

/20/

10

9:06

PM



Co

pyrig

ht ©

Gle

nco

e/M

cG

raw

-Hill, a

div

isio

n o

f Th

e M

cG

raw

-Hill C

om

pa

nie

s, In

c.

PDF Pass



NA

ME

DAT

E

P

ER

IOD

4-6

Cha

pter

4

38

Gle

ncoe

Alg

ebra

2

Skill

s Pr

actic

eTh

e Q

uad

rati

c F

orm

ula

an

d t

he D

iscri

min

an

t

Com

plet

e pa

rts

a-c

for

each

qua

drat

ic e

quat

ion.

a. F

ind

the

valu

e of

the

dis

crim

inan

t.b.

Des

crib

e th

e nu

mbe

r an

d ty

pe o

f ro

ots.

c. F

ind

the

exac

t so

luti

ons

by u

sing

the

Qua

drat

ic F

orm

ula.

1. x

2 - 8

x +

16

= 0

2.

x2 -

11x

- 2

6 =

0

3. 3

x2 - 2

x =

0

4. 2

0x2 +

7x

- 3

= 0

5. 5

x2 - 6

= 0

6.

x2 -

6 =

0

7. x

2 + 8

x +

13

= 0

8.

5x2 -

x -

1 =

0

9. x

2 - 2

x -

17

= 0

10

. x2 +

49

= 0

11. x

2 - x

+ 1

= 0

12

. 2x2 -

3x

= -

2

Solv

e ea

ch e

quat

ion

by u

sing

the

Qua

drat

ic F

orm

ula.

13. x

2 = 6

4 14

. x2 -

30

= 0

15. x

2 - x

= 3

0

16. 1

6x2 -

24x

- 2

7 =

0

17. x

2 - 4

x -

11

= 0

18

. x2 -

8x

- 1

7 =

0

19. x

2 + 2

5 =

0

20. 3

x2 + 3

6 =

0

21. 2

x2 + 1

0x +

11

= 0

22

. 2x2 -

7x

+ 4

= 0

23. 8

x2 + 1

= 4

x

24. 2

x2 + 2

x +

3 =

0

25. P

ARA

CHU

TIN

G I

gnor

ing

win

d re

sist

ance

, the

dis

tanc

e d(

t) in

feet

tha

t a

para

chut

ist

falls

in t

seco

nds

can

be e

stim

ated

usi

ng t

he fo

rmul

a d(

t) =

16t

2 . If

a p

arac

huti

st ju

mps

fr

om a

n ai

rpla

ne a

nd fa

lls fo

r 11

00 fe

et b

efor

e op

enin

g he

r pa

rach

ute,

how

man

y se

cond

s pa

ss b

efor

e sh

e op

ens

the

para

chut

e?

0

; 1 r

atio

nal r

oot;

4

225;

2 r

atio

nal r

oots

; -2,

13

-

3; 2

com

plex

roo

ts;

1 ±

i √ �

3 −

2

-

7; 2

com

plex

roo

ts;

3 ±

i √ �

7 −

4

4

; 2 r

atio

nal r

oots

; 0, 2

− 3

28

9; 2

rat

iona

l roo

ts; -

3 −

5 , 1 −

4

1

20; 2

irra

tiona

l roo

ts; ±

√ �

�

30

−

5

24

; 2 ir

ratio

nal r

oots

; ±

√ �

6

1

2; 2

irra

tiona

l roo

ts; -

4 ±

√ �

3

21; 2

irra

tiona

l roo

ts;

1 ±

√ �

�

21

−

10

7

2; 2

irra

tiona

l roo

ts; 1

± 3

√ �

2 -

196;

2 c

ompl

ex r

oots

; ±7i

±8

-5,

6 2 ±

√ �

�

15

4 ±

√ �

�

33

±5i

± 2

i √ �

3

abou

t 8.3

s

-5

± √

� 3

−

2

1 ±

i −

4

-

1 ±

i √ �

5 −

2

7 ±

√ �

�

17

−

4

± √

��

30 9

− 4 , -

3 −

4

030_

041_

ALG

2_A

_CR

M_C

04_C

R_6

6078

5.in

dd

3812

/20/

10

9:06

PM


NA

ME

DAT

E

P

ER

IOD

Lesson 4-6

4-6

Cha

pter

4

39

Gle

ncoe

Alg

ebra

2

Prac

tice

Th

e Q

uad

rati

c F

orm

ula

an

d t

he D

iscri

min

an

t

Solv

e ea

ch e

quat

ion

by u

sing

the

Qua

drat

ic F

orm

ula.

1. 7

x2 - 5

x =

0

2. 4

x2 - 9

= 0

3. 3

x2 + 8

x =

3

4. x

2 - 2

1 =

4x

5. 3

x2 - 1

3x +

4 =

0

6. 1

5x2 +

22x

= -

8

7. x

2 - 6

x +

3 =

0

8. x

2 - 1

4x +

53

= 0

9. 3

x2 = -

54

10. 2

5x2 -

20x

- 6

= 0

11. 4

x2 - 4

x +

17

= 0

12

. 8x

- 1

= 4

x2

13. x

2 = 4

x -

15

14

. 4x2 -

12x

+ 7

= 0

Com

plet

e pa

rts

a-c

for

each

qua

drat

ic e

quat

ion.

a. F

ind

the

valu

e of

the

dis

crim

inan

t.b.

Des

crib

e th

e nu

mbe

r an

d ty

pe o

f ro

ots.

c. F

ind

the

exac

t so

luti

ons

by u

sing

the

Qua

drat

ic F

orm

ula.

15. x

2 - 1

6x +

64

= 0

16

. x2 =

3x

17. 9

x2 - 2

4x +

16

= 0

18. x

2 - 3

x =

40

19. 3

x2 + 9

x -

2 =

0

20. 2

x2 + 7

x =

0

21. 5

x2 - 2

x +

4 =

0

22. 1

2x2 -

x -

6 =

0

23. 7

x2 + 6

x +

2 =

0

24. 1

2x2 +

2x

- 4

= 0

25

. 6x2 -

2x

- 1

= 0

26

. x2 +

3x

+ 6

= 0

27. 4

x2 - 3

x2 - 6

= 0

28

. 16x

2 - 8

x +

1 =

0

29. 2

x2 - 5

x -

6 =

0

30. G

RAVI

TATI

ON

The

hei

ght h

(t) in

feet

of a

n ob

ject

t se

cond

s af

ter

it is

pro

pelle

d st

raig

ht

up fr

om th

e gr

ound

wit

h an

init

ial v

eloc

ity

of 6

0 fe

et p

er s

econ

d is

mod

eled

by

the

equa

tion

h(

t) =

-16

t2 + 6

0t. A

t wha

t tim

es w

ill th

e ob

ject

be

at a

hei

ght o

f 56

feet

?

31. S

TOPP

ING

DIS

TAN

CE T

he fo

rmul

a d

= 0

.05s

2 + 1

.1s

esti

mat

es t

he m

inim

um s

topp

ing

dist

ance

d in

feet

for

a ca

r tr

avel

ing

s m

iles

per

hour

. If a

car

sto

ps in

200

feet

, wha

t is

the

fa

stes

t it c

ould

hav

e be

en tr

avel

ing

whe

n th

e dr

iver

app

lied

the

brak

es?

1.75

s, 2

s

abou

t 53.

2 m

i/h

0

; 1 r

atio

nal;

8 9

; 2 r

atio

nal;

0, 3

0

; 1 r

atio

nal;

4 −

3

1

69; 2

rat

iona

l; -

5, 8

2

irra

tiona

l; -

9 ±

√

��

10

5 −

6

4

9; 2

rat

iona

l; 0,

- 7 −

2

2

com

plex

; 1

± i

√

�

19

−

5

2

rat

iona

l; 3 −

4 , -

2 −

3 2

com

plex

; -

3 ±

i √

�

5 −

7

2

rat

iona

l; 1 −

2 , -

2 −

3 2

irra

tiona

l; 1

± √

�

7 −

6

2

com

plex

; -

3 ±

i √

�

15

−

2

2

irra

tiona

l; 3

± √

��

10

5 −

8

0

; 1 r

atio

nal;

1 −

4 2

irra

tiona

l; 5

± √

��

73

−

4

± 3 −

2

1 −

3 , 4

7 ±

2i

2 ±

√

�

10

−

5

2 ±

√

�

3 −

2

3 ±

√

�

2 −

2

- 2 −

3 , - 4 −

5

0, 5 −

7

1 −

3 , -3

3 ±

√

�

6

± 3

i √

�

2

2 ±

i √

�

11

-3,

7

1 ±

4i

−

2

030_

041_

ALG

2_A

_CR

M_C

04_C

R_6

6078

5.in

dd

3912

/20/

10

9:06

PM



Copyri

ght

© G

lencoe/M

cG

raw

-Hill

, a

div

isio

n o

f T

he

McG

raw

-Hill

Co

mp

an

ies,

Inc.

PDF Pass



NA

ME

DAT

E

P

ER

IOD

4-6

Cha

pter

4

40

Gle

ncoe

Alg

ebra

2

Wor

d Pr

oble

m P

ract

ice

Th

e Q

uad

rati

c F

orm

ula

an

d t

he D

iscri

min

an

t

1. P

ARA

BOLA

S Th

e gr

aph

of a

qua

drat

ic

equa

tion

of t

he fo

rm y

=ax

2+

bx+

c is

sh

own

belo

w.

y

xO5

-5

Is t

he d

iscr

imin

ant

b2-

4ac

pos

itiv

e,

nega

tive

, or

zero

? E

xpla

in.

2. T

AN

GEN

T K

athl

een

is t

ryin

g to

find

bso

tha

t th

e x-

axis

is t

ange

nt t

o th

e pa

rabo

la y

=x2

+bx

+ 4

. She

find

s on

e va

lue

that

wor

ks, b

= 4

. Is

this

the

onl

y va

lue

that

wor

ks?

Exp

lain

.

3. S

PORT

S In

199

0, A

mer

ican

Ran

dy

Bar

nes

set

the

wor

ld r

ecor

d fo

r th

e sh

ot

put.

His

thr

ow c

an b

e de

scri

bed

by t

he

equa

tion

y=

–16

x2+

368

x. U

se t

he

Qua

drat

ic F

orm

ula

to fi

nd h

ow fa

r hi

s th

row

was

to

the

near

est

foot

.

4. E

XA

MPL

ES G

ive

an e

xam

ple

of a

qu

adra

tic

func

tion

f(x

) tha

t ha

s th

e fo

llow

ing

prop

erti

es.

I. Th

e di

scri

min

ant

of f

is z

ero.

II. T

here

is n

o re

al s

olut

ion

of t

he

equa

tion

f(x

) = 1

0.

Sket

ch t

he g

raph

of x

=f(

x).

yx

O

-2

-2

-4

-6

-8

-4

24

5. T

AN

GEN

TS T

he g

raph

of y

= x

2 is

a pa

rabo

la t

hat

pass

es t

hrou

gh t

he p

oint

at

(1, 1

). Th

e lin

e y

= m

x -

m +

1,

whe

re m

is a

con

stan

t, al

so p

asse

s th

roug

h th

e po

int

at (1

, 1).

a. T

o fin

d th

e po

ints

of i

nter

sect

ion

betw

een

the

line

y =

mx

- m

+ 1

an

d th

e pa

rabo

la y

= x

2 , se

t x2 =

m

x -

m +

1 a

nd t

hen

solv

e fo

r x.

R

earr

angi

ng t

erm

s, t

his

equa

tion

be

com

es x

2 - m

x +

m -

1 =

0. W

hat

is t

he d

iscr

imin

ant

of t

his

equa

tion

?

b. F

or w

hat

valu

e of

m is

the

re o

nly

one

poin

t of

inte

rsec

tion

? E

xpla

in t

he

mea

ning

of t

his

in t

erm

s of

the

co

rres

pond

ing

line

and

the

para

bola

.

N

egat

ive;

the

equa

tion

has

no

real

sol

utio

ns s

o th

e di

scrim

inan

t is

nega

tive.

N

o, b

= -

4 al

so w

orks

; the

x-

axis

will

be

tang

ent w

hen

the

disc

rimin

ant b

2 - 1

6 is

zer

o.

This

hap

pens

whe

n b

= 4

or

-4.

2

3 ft

S

ampl

e an

swer

: f(x

) = -

x2

x

2 - 4

m +

4

m

= 2

; the

par

abol

a y

= x

2 and

th

e lin

e y

= 2

x -

1 h

ave

exac

tly

one

poin

t of i

nter

sect

ion

at

(1, 1

). In

oth

er w

ords

, thi

s lin

e is

tang

ent t

o th

e pa

rabo

la a

t (1

, 1).

030_

041_

ALG

2_A

_CR

M_C

04_C

R_6

6078

5.in

dd

4012

/20/

10

9:06

PM


NA

ME

DAT

E

P

ER

IOD

Lesson 4-6

4-6

Cha

pter

4

41

Gle

ncoe

Alg

ebra

2

Enri

chm

ent

Su

m a

nd

Pro

du

ct

of

Ro

ots

Som

etim

es y

ou m

ay k

now

the

roo

ts o

f a q

uadr

atic

equ

atio

n w

itho

ut k

now

ing

the

equa

tion

it

self.

Usi

ng y

our

know

ledg

e of

fact

orin

g to

sol

ve a

n eq

uati

on, y

ou c

an w

ork

back

war

d to

fin

d th

e qu

adra

tic

equa

tion

. The

rul

e fo

r fin

ding

the

sum

and

pro

duct

of r

oots

is a

s fo

llow

s:

Sum

and

Pro

duct

of R

oots

If th

e ro

ots

of a

x2 + b

x +

c =

0, w

ith a

≠ 0

, are

s1 a

nd

s 2, th

en s

1 + s

2 = -

a −

b and

s1 �

s2 =

c −

a .

W

rite

a q

uadr

atic

equ

atio

n th

at h

as t

he r

oots

3 a

nd -

8.

The

root

s ar

e x

= 3

and

x =

-8.

x

y

O

(–5 – 2, –30

–1 4)

10 –10

–20

–30

24

–2–4

–6–8

3 +

(-8)

= -

5 A

dd th

e ro

ots.

3(

-8)

= -

24

Mul

tiply

the

root

s.

Equ

atio

n: x

2 + 5

x -

24

= 0

Exer

cise

sW

rite

a q

uadr

atic

equ

atio

n th

at h

as t

he g

iven

roo

ts.

1. 6

, -9

2. 5

, -1

3. 6

, 6

4. 4

± √

�

3 5.

- 2 −

5 , 2 −

7 6.

-

2 ±

3 √

�

5 −

7

Fin

d k

such

tha

t th

e nu

mbe

r gi

ven

is a

roo

t of

the

equ

atio

n.

7. 7

; 2x2 +

kx

- 2

1 =

0

8. -

2; x

2 - 1

3x +

k =

0

Exam

ple

x

2 + 3

x -

54

= 0

x

2 - 4

x -

5 =

0

x2 -

12x

+ 3

6 =

0

x

2 - 8

x +

13

= 0

3

5x2 +

4x

- 4

= 0

4

9x2 -

42x

+ 2

05 =

0

-

11

-

30

030_

041_

ALG

2_A

_CR

M_C

04_C

R_6

6078

5.in

dd

4112

/20/

10

9:06

PM



Co

pyrig

ht ©

Gle

nco

e/M

cG

raw

-Hill, a

div

isio

n o

f Th

e M

cG

raw

-Hill C

om

pa

nie

s, In

c.

PDF Pass




NA

ME

DAT

E

P

ER

IOD

4-6

Cha

pter

4

42

Gle

ncoe

Alg

ebra

2

Spre

adsh

eet A

ctiv

ityA

pp

roxi

mati

ng

th

e R

eal

Zero

s o

f P

oly

no

mia

ls

You

have

lear

ned

the

Loca

tion

Pri

ncip

le, w

hich

can

be

used

to

appr

oxim

ate

the

real

zer

os o

f a p

olyn

omia

l.

A1 2 3 4 5

C x f(x)

B 5

G 0 –2

D –5 23

J 5 23

E

–3.3

3333

339.

1111

111

H

1.66

6666

70.

7777

778

I

3.33

3333

39.

1111

111

F

–1.6

6666

670.

7777

778

–5

She

et 1

She

et 2

She

et 3

In t

he s

prea

dshe

et a

bove

, the

pos

itiv

e re

al z

ero

of ƒ

(x) =

x2 -

2 c

an b

e ap

prox

imat

ed in

the

fo

llow

ing

way

. Set

the

spr

eads

heet

pre

fere

nce

to m

anua

l cal

cula

tion

. The

val

ues

in A

2 an

d B

2 ar

e th

e en

dpoi

nts

of a

ran

ge o

f val

ues.

The

val

ues

in D

2 th

roug

h J2

are

val

ues

equa

lly

in t

he in

terv

al fr

om A

2 to

B2.

The

form

ulas

for

thes

e va

lues

are

A2,

A2

+ (B

2 -

A2)

/6, A

2 +

2*

(B2

- A

2)/6

, A2

+ 3

*(B

2 -

A2)

/6, A

2 +

4*(

B2

- A

2)/6

, A2

+ 5

*(B

2 -

A2)

/6, a

nd B

2,

resp

ecti

vely

.

Row

3 g

ives

the

func

tion

val

ues

at t

hese

poi

nts.

The

func

tion

ƒ(x

) = x

2 - 2

is e

nter

ed in

to

the

spre

adsh

eet

in C

ell D

3 as

D2^

2 -

2. T

his

func

tion

is t

hen

copi

ed t

o th

e re

mai

ning

cel

ls

in t

he r

ow.

You

can

use

this

spr

eads

heet

to

stud

y th

e fu

ncti

on v

alue

s at

the

poi

nts

in c

ells

D2

thro

ugh

J2. T

he v

alue

in c

ell F

3 is

pos

itiv

e an

d th

e va

lue

in c

ell G

3 is

neg

ativ

e, s

o th

ere

mus

t be

a

zero

bet

wee

n -

1.66

67 a

nd 0

. Ent

er t

hese

val

ues

in c

ells

A2

and

B2,

res

pect

ivel

y, an

d re

calc

ulat

e th

e sp

read

shee

t. (Y

ou w

ill h

ave

to r

ecal

cula

te a

num

ber

of t

imes

.) Th

e re

sult

is

a ne

w t

able

from

whi

ch y

ou c

an s

ee t

hat

ther

e is

a z

ero

betw

een

1.41

414

and

1.41

4306

. B

ecau

se t

hese

val

ues

agre

e to

thr

ee d

ecim

al p

lace

s, t

he z

ero

is a

bout

1.4

14. T

his

can

be

veri

fied

by u

sing

alg

ebra

.B

y so

lvin

g x2 -

2 =

0, w

e ob

tain

x =

± √

�

2 . T

he p

osit

ive

root

is

x=

± √

�

2 =

1.4

1421

3. .

. , w

hich

ver

ifies

the

res

ult.

Exer

cise

s 1

. Use

a s

prea

dshe

et li

ke t

he o

ne a

bove

to

appr

oxim

ate

the

zero

of ƒ

(x) =

3x

- 2

to

thre

e de

cim

al p

lace

s. T

hen

veri

fy y

our

answ

er b

y us

ing

alge

bra

to fi

nd t

he e

xact

val

ue o

f the

ro

ot.

2. U

se a

spr

eads

heet

like

the

one

abo

ve t

o ap

prox

imat

e th

e re

al z

eros

of f

(x) =

x2 +

2x

+ 0

.5.

Rou

nd y

our

answ

er t

o fo

ur d

ecim

al p

lace

s. T

hen,

ver

ify y

our

answ

er b

y us

ing

the

quad

rati

c fo

rmul

a.

3. U

se a

spr

eads

heet

like

the

one

abo

ve t

o ap

prox

imat

e th

e re

al z

ero

of

ƒ(x)

= x

3 - 3 −

2 x2 -

6x

- 2

bet

wee

n -

0.4

and

-0.

3.

T

he s

prea

dshe

et g

ives

x =

0.6

67. B

y so

lvin

g fo

r x

alge

brai

cally

,

x =

2 −

3 . So,

the

appr

oxim

atio

n is

cor

rect

.

The

proc

ess

give

s -

1.70

71 a

nd -

0.29

29 to

the

near

est

t

en-t

hous

andt

h. T

he q

uadr

atic

form

ula

give

s x

= -

1 ±

√ �

2 −

2 .

-1

- √

� 2

−

2 ≈

-

1.70

71 a

nd -

1 +

√ �

2 −

2

≈ -

0.29

29.

-0.

3781

to th

e ne

ares

t ten

-thou

sand

th

042_

056_

ALG

2_A

_CR

M_C

04_C

R_6

6078

5.in

dd

4212

/20/

10

9:06

PM


NA

ME

DAT

E

P

ER

IOD

Lesson 4-7

4-7

Cha

pter

4

43

Gle

ncoe

Alg

ebra

2

Stud

y Gu

ide

and

Inte

rven

tion

Tran

sfo

rmati

on

s o

f Q

uad

rati

c G

rap

hs

Wri

te Q

uadr

atic

Equ

atio

ns in

Ver

tex

Form

A q

uadr

atic

func

tion

is e

asie

r to

gr

aph

whe

n it

is in

ver

tex

form

. You

can

wri

te a

qua

drat

ic fu

ncti

on o

f the

form

y

= a

x2 + b

x +

c in

ver

tex

from

by

com

plet

ing

the

squa

re.

W

rite

y =

2x2 -

12x

+ 2

5 in

ver

tex

form

. The

n gr

aph

the

func

tion

.

y =

2x2 -

12x

+ 2

5

x

y

O2

2468

46

y =

2(x

2 - 6

x) +

25

y =

2(x

2 - 6

x +

9) +

25

- 1

8y

= 2

(x -

3)2 +

7

The

vert

ex fo

rm o

f the

equ

atio

n is

y =

2(x

- 3

)2 + 7

.

Exer

cise

s

Wri

te e

ach

equa

tion

in v

erte

x fo

rm. T

hen

grap

h th

e fu

ncti

on.

1. y

= x

2 - 1

0x +

32

2. y

= x

2 + 6

x 3.

y =

x2 -

8x

+ 6

x

y

O2468

24

6

x

y

O

-2

-4

-6

-8

-2

-4

-6

x

y

O4

-4

8

8 4

-4

-8

-12

4. y

= -

4x2 +

16x

- 1

1 5.

y =

3x2 -

12x

+ 5

6.

y =

5x2 -

10x

+ 9

x

y

O-

2

-2

-4

2

246

4

x

y

O-

2

-2

-4

-6

2

2

4

x

y

O-

22

246810

4

Exam

ple

y

= -

4(x

- 2

)2 + 5

y

= 3

(x -

2)2 -

7

y=

5(x

- 1

)2 + 4

y

= (x

- 5

)2 + 7

y

= (x

+ 3

)2 - 9

y

= (x

- 4

)2 - 1

0

042_

056_

ALG

2_A

_CR

M_C

04_C

R_6

6078

5.in

dd

4312

/20/

10

9:06

PM


Copyri

ght

© G

lencoe/M

cG

raw

-Hill

, a

div

isio

n o

f T

he

McG

raw

-Hill

Co

mp

an

ies,

Inc.

PDF Pass



NA

ME

DAT

E

P

ER

IOD

4-7

Cha

pter

4

44

Gle

ncoe

Alg

ebra

2

Stud

y Gu

ide

and

Inte

rven

tion

(con

tinu

ed)

Tran

sfo

rmati

on

s o

f Q

uad

rati

c G

rap

hs

Tran

sfor

mat

ions

of

Qua

drat

ic G

raph

s Pa

rabo

las

can

be t

rans

form

ed b

y ch

angi

ng t

he v

alue

s of

the

con

stan

ts a

, h, a

nd k

in t

he v

erte

x fo

rm o

f a q

uadr

atic

equ

atio

n:

y =

a(x

– h

) 2 + k

.

•

The

sign

of a

det

erm

ines

whe

ther

the

gra

ph o

pens

upw

ard

(a >

0) o

r do

wnw

ard

(a <

0).

•

The

abso

lute

val

ue o

f a a

lso

caus

es a

dila

tion

(enl

arge

men

t or

red

ucti

on) o

f the

pa

rabo

la. T

he p

arab

ola

beco

mes

nar

row

er if

⎪a⎥

>1

and

wid

er if

⎪a⎥

< 1

.

•

The

valu

e of

h t

rans

late

s th

e pa

rabo

la h

oriz

onta

lly. P

osit

ive

valu

es o

f h s

lide

the

grap

h to

the

rig

ht a

nd n

egat

ive

valu

es s

lide

the

grap

h to

the

left

.

•

The

valu

e of

k t

rans

late

s th

e gr

aph

vert

ical

ly. P

osit

ive

valu

es o

f k s

lide

the

grap

h up

war

d an

d ne

gati

ve v

alue

s sl

ide

the

grap

h do

wnw

ard.

G

raph

y =

(x

+ 7

)2 + 3

.

•

Rew

rite

the

equ

atio

n as

y =

[x –

(–7)

]2 + 3

.

•

Bec

ause

h =

–7

and

k =

3, t

he v

erte

x is

at

(–7,

3).

The

axis

of

sym

met

ry is

x =

–7.

Bec

ause

a =

1, w

e kn

ow t

hat

the

grap

h op

ens

up, a

nd t

he g

raph

is t

he s

ame

wid

th a

s th

e gr

aph

of y

= x

2 .

•

Tran

slat

e th

e gr

aph

of y

= x

2 sev

en u

nits

to

the

left

and

th

ree

unit

s up

.

Exer

cise

s

Gra

ph e

ach

func

tion

.

1.

y =

–2x

2 +

2

2. y

= –

3(x

– 1)

2 3.

y =

2(x

+ 2

)2 + 3

Exam

ple

x

15 5

-15

-15-5

515

y

-5

x

68 4 2

-6-

4-

8

-6

-4

-2

-8

24

68

y

-2

x

68 4 2

-6-

4-

8

-6

-4

-2

-8

24

68

y

-2

x

68 4 2

-6-

4-

8

-6

-4

-2

-8

24

68

y

-2

042_

056_

ALG

2_A

_CR

M_C

04_C

R_6

6078

5.in

dd

4412

/20/

10

9:06

PM


NA

ME

DAT

E

P

ER

IOD

4-7

Cha

pter

4

45

Gle

ncoe

Alg

ebra

2

Skill

s Pr

actic

eTr

an

sfo

rmati

on

s o

f Q

uad

rati

c G

rap

hs

Wri

te e

ach

quad

rati

c fu

ncti

on in

ver

tex

form

. The

n id

enti

fy t

he v

erte

x, a

xis

of

sym

met

ry, a

nd d

irec

tion

of

open

ing.

1. y

= (x

- 2

)2 2.

y =

-x2 +

4

3. y

= x

2 - 6

4. y

= -

3(x

+ 5

)2 5.

y =

-5x

2 + 9

6.

y =

(x -

2)2 -

18

7. y

= x

2 - 2

x -

5

8. y

= x

2 + 6

x +

2

9. y

= -

3x2 +

24x

Gra

ph e

ach

func

tion

.

10. y

= (x

- 3

)2 - 1

11

. y =

(x +

1)2 +

2

12. y

= -

(x -

4)2 -

4

x

y

O6 4 2

26

4

x

y

O4

-4

6 4 2

-2

2

x

y

O2

–2

–4

–6

46

13. y

= -

1 −

2 (x

+ 2

)2 14

. y =

-3x

2 + 4

15

. y =

x2 +

6x

+ 4

x

y

O

-2

-4

-6

-2

2-

6-

4

x

y

O4 2

24

–2

–4

-4

-2

x

y

O2

2

–2

–4

-4

-2

-6

y

= -

3(x

+ 5

)2 + 0

; y

= -

5(x

- 0

)2 + 9

; y

= (x

- 2

)2 - 1

8;

(

-5,

0);

x =

-5;

dow

n (

0, 9

); x

= 0

; dow

n (

2, -

18);

x =

2; u

p

y

= (x

- 2

)2 + 0

; y

= -

(x -

0)2 +

4;

y =

(x -

0)2 -

6;

(

2, 0

); x

= 2

; up

(0,

4);

x =

0; d

own

(0,

-6)

; x =

0; u

p

y

= (x

- 1

)2 - 6

; y

= (x

+ 3

)2 - 7

; y

= -

3(x

- 4

)2 + 4

8;

(

1, -

6); x

= 1

; up

(-

3, -

7); x

= -

3; u

p (

4, 4

8); x

= 4

; dow

n

042_

056_

ALG

2_A

_CR

M_C

04_C

R_6

6078

5.in

dd

4512

/20/

10

9:06

PM



Co

pyrig

ht ©

Gle

nco

e/M

cG

raw

-Hill, a

div

isio

n o

f Th

e M

cG

raw

-Hill C

om

pa

nie

s, In

c.

PDF Pass



NA

ME

DAT

E

P

ER

IOD

4-7

Cha

pter

4

46

Gle

ncoe

Alg

ebra

2

Prac

tice

Tran

sfo

rmati

on

s o

f Q

uad

rati

c G

rap

hs

Wri

te e

ach

equa

tion

in v

erte

x fo

rm. T

hen

iden

tify

the

ver

tex,

axi

s of

sym

met

ry,

and

dire

ctio

n of

ope

ning

. 1

. y =

-6x

2 -

24

x -

25

2. y

= 2

x2 + 2

3.

y =

-4x

2 + 8

x

4. y

= x

2 + 1

0x +

20

5. y

= 2

x2 + 1

2x +

18

6. y

= 3

x2 - 6

x +

5

7. y

= -

2x2 -

16x

- 3

2 8.

y =

-3x

2 + 1

8x -

21

9. y

= 2

x2 + 1

6x +

29

Gra

ph e

ach

func

tion

.

10. y

= (x

+ 3

)2 - 1

11

. y

= -

x2 + 6

x -

5

12. y

= 2

x2 - 2

x +

1

x

y

O-

2-

4-

6

246

x

y

O-

2

-2

-424

24

6

x

y O-

2-

42

246

4

13. W

rite

an

equa

tion

for

a pa

rabo

la w

ith

vert

ex a

t (1

, 3) t

hat

pass

es t

hrou

gh (-

2, -

15).

14. W

rite

an

equa

tion

for

a pa

rabo

la w

ith

vert

ex a

t (-

3, 0

) tha

t pa

sses

thr

ough

(3, 1

8).

15. B

ASE

BALL

The

hei

ght

h of

a b

aseb

all t

sec

onds

aft

er b

eing

hit

is g

iven

by

h(t)

= -

16t2 +

80t

+ 3

. Wha

t is

the

max

imum

hei

ght

that

the

bas

ebal

l rea

ches

, and

w

hen

does

thi

s oc

cur?

16. S

CULP

TURE

A m

oder

n sc

ulpt

ure

in a

par

k co

ntai

ns a

par

abol

ic a

rc t

hat

st

arts

at

the

grou

nd a

nd r

each

es a

max

imum

hei

ght

of 1

0 fe

et a

fter

a

hori

zont

al d

ista

nce

of 4

feet

. Wri

te a

qua

drat

ic fu

ncti

on in

ver

tex

form

th

at d

escr

ibes

the

sha

pe o

f the

out

side

of t

he a

rc, w

here

y is

the

hei

ght

of a

poi

nt o

n th

e ar

c an

d x

is it

s ho

rizo

ntal

dis

tanc

e fr

om t

he le

ft-h

and

star

ting

poi

nt o

f the

arc

. 10

ft

4 ft

y

= -

6(x

+ 2

)2 - 1

;

y =

2(x

+ 0

)2 + 2

; y

= -

4(x

- 1

)2 + 4

;

(-

2, -

1); x

= -

2; d

own

(0,

2);

x =

0; u

p (

1, 4

); x

= 1

; dow

n

y

= (x

+ 5

)2 - 5

; y

= 2

(x +

3)2 ;

(-3,

0);

y =

3(x

- 1

)2 + 2

;

(-

5, -

5); x

= -

5; u

p x

= -

3; u

p (

1, 2

); x

= 1

; up

y

= -

2(x

+ 4

)2 ; y

= -

3(x

- 3

)2 + 6

; y

= 2

(x +

4)2 -

3;

(

-4,

0);

x =

-4;

dow

n (

3, 6

); x

= 3

; dow

n (

-4,

-3)

; x =

-4;

up

y

= -

2(x

- 1

)2 + 3

y

= 1 −

2 (x +

3)2

103

ft; 2

.5 s

y =

- 5 −

8 (x -

4)2 +

10

042_

056_

ALG

2_A

_CR

M_C

04_C

R_6

6078

5.in

dd

4612

/20/

10

9:06

PM


NA

ME

DAT

E

P

ER

IOD

4-7

Cha

pter

4

47

Gle

ncoe

Alg

ebra

2

Wor

d Pr

oble

m P

ract

ice

Tran

sfo

rmati

on

s o

f Q

uad

rati

c G

rap

hs

1.A

RCH

ES A

par

abol

ic a

rch

is u

sed

as a

br

idge

sup

port

. The

gra

ph o

f the

arc

h is

sh

own

belo

w.

y

xO

5

5

-5

If t

he e

quat

ion

that

cor

resp

onds

to

this

gra

ph is

wri

tten

in t

he fo

rm

y+

a(x

-h)

2+

k, w

hat

are

h an

d k?

2. T

RAN

SLAT

ION

S Fo

r a

com

pute

r an

imat

ion,

Bar

bara

use

s th

e qu

adra

tic

func

tion

f(x)

= -

42(x

- 2

0)2 +

168

00 t

o he

lp h

er s

imul

ate

an o

bjec

t to

ssed

on

anot

her

plan

et. F

or o

ne s

kit,

she

had

to

use

the

func

tion

f(x

+ 5

) - 8

000

inst

ead

of f

(x).

Whe

re is

the

ver

tex

of t

he g

raph

of

y =

f(x

+ 5

) - 8

000?

3. B

RID

GES

The

sha

pe fo

rmed

by

the

mai

n ca

bles

of t

he G

olde

n G

ate

Bri

dge

appr

oxim

atel

y fo

llow

s th

e eq

uati

on

y =

0.0

002x

2 - 0

.23x

+ 2

27. G

raph

the

pa

rabo

la fo

rmed

by

one

of t

he c

able

s.

4.W

ATER

JET

S Th

e gr

aph

show

s th

e pa

th

of a

jet

of w

ater

.

y

xO

5

The

equa

tion

cor

resp

ondi

ng t

o th

is

grap

h is

y=

a(x

-h)

2+

k. W

hat

are

a,

h, a

nd k

?

5. P

ROFI

T A

the

ater

ope

rato

r pr

edic

ts t

hat

the

thea

ter

can

mak

e -

4x2 +

160

x do

llars

per

sho

w if

tic

kets

are

pri

ced

at x

do

llars

.

a. R

ewri

te t

he e

quat

ion

y =

-4x

2 + 1

60x

in t

he fo

rm y

= a

(x -

h) 2 +

k.

b. W

hat

is t

he v

erte

x of

the

par

abol

a an

d w

hat

is it

s ax

is o

f sym

met

ry?

c.G

raph

the

par

abol

a.

y

xO

800

1600

2040

h =

-1

and

k =

5

(15,

880

0)

a =

-2,

h =

2, k

= 6

y =

-4(

x -

20)

2 + 1

600

vert

ex a

t (20

, 160

0); a

xis

is x

= 2

0

Wid

th

Height

x

150

500

1000

y

300

042_

056_

ALG

2_A

_CR

M_C

04_C

R_6

6078

5.in

dd

4712

/20/

10

9:06

PM



Copyri

ght

© G

lencoe/M

cG

raw

-Hill

, a

div

isio

n o

f T

he

McG

raw

-Hill

Co

mp

an

ies,

Inc.

PDF Pass



NA

ME

DAT

E

P

ER

IOD

4-7

Cha

pter

4

48

Gle

ncoe

Alg

ebra

2

Enri

chm

ent

A S

ho

rtcu

t to

Co

mp

lex

Ro

ots

Whe

n gr

aphi

ng a

qua

drat

ic fu

ncti

on, t

he r

eal r

oots

are

sho

wn

in t

he g

raph

. Yo

u ha

ve le

arne

d th

at q

uadr

atic

func

tion

s ca

n al

so h

ave

imag

inar

y ro

ots

that

can

not

be s

een

on t

he g

raph

of t

he fu

ncti

on. H

owev

er, t

here

is a

way

to

grap

hica

lly r

epre

sent

the

com

plex

roo

ts o

f a q

uadr

atic

func

tion

.

F

ind

the

com

plex

roo

ts o

f th

e qu

adra

tic

func

tion

y =

x2 -

4x

+ 5

.

Step

1

Gra

ph t

he fu

ncti

on.

y

xO6

5

Step

2

Ref

lect

the

gra

ph o

ver

the

hori

zont

al li

ne c

onta

inin

g th

e ve

rtex

. In

this

exa

mpl

e,

the

vert

ex is

(2, 1

).

y

xO

Step

3

The

real

par

t of

the

com

plex

roo

t is

the

poi

nt h

alfw

ay b

etw

een

the

x-in

terc

epts

of t

he r

efle

cted

gra

ph a

nd t

he im

agin

ary

part

of t

he

com

plex

roo

ts a

re +

and

- h

alf t

he d

ista

nce

betw

een

the

x-in

terc

epts

of

the

ref

lect

ed g

raph

. So,

in t

his

exam

ple,

the

com

plex

roo

ts a

re

2 +

1i

and

2 -

1i.

Exer

cise

sU

sing

thi

s m

etho

d, f

ind

the

com

plex

roo

ts o

f th

e fo

llow

ing

quad

rati

c fu

ncti

ons.

1. y

= x

2 + 2

x +

5

2. y

= x

2 + 4

x +

8

3. y

= x

2 + 6

x +

13

4. y

= x

2 + 2

x +

17

Exam

ple

-

1 +

2i,

-1

- 2

i -

2 +

2i,

-2

- 2

i

-

3 +

2i,

-3

- 2

i

-1

+ 4

i, -

1 -

4i

042_

056_

ALG

2_A

_CR

M_C

04_C

R_6

6078

5.in

dd

4812

/20/

10

9:06

PM


NA

ME

DAT

E

P

ER

IOD

Lesson 4-8

4-8

Cha

pter

4

49

Gle

ncoe

Alg

ebra

2

Stud

y Gu

ide

and

Inte

rven

tion

Qu

ad

rati

c I

neq

ualiti

es

Gra

ph Q

uadr

atic

Ineq

ualit

ies

To g

raph

a q

uadr

atic

ineq

ualit

y in

tw

o va

riab

les,

use

th

e fo

llow

ing

step

s:

1. G

raph

the

rel

ated

qua

drat

ic e

quat

ion,

y =

ax2 +

bx

+ c

. U

se a

das

hed

line

for

< o

r >

; use

a s

olid

line

for

≤ o

r ≥

.

2. T

est

a po

int

insi

de t

he p

arab

ola.

If i

t sa

tisf

ies

the

ineq

ualit

y, sh

ade

the

regi

on in

side

the

par

abol

a;

othe

rwis

e, s

hade

the

reg

ion

outs

ide

the

para

bola

.

G

raph

the

ineq

uali

ty y

> x

2 + 6

x +

7.

Firs

t gr

aph

the

equa

tion

y =

x2 +

6x

+ 7

. By

com

plet

ing

the

x

y

O2

-224

-4

-2

squa

re, y

ou g

et t

he v

erte

x fo

rm o

f the

equ

atio

n y

= (x

+ 3

)2 - 2

, so

the

ver

tex

is (-

3, -

2). M

ake

a ta

ble

of v

alue

s ar

ound

x =

-3,

an

d gr

aph.

Sin

ce t

he in

equa

lity

incl

udes

>, u

se a

das

hed

line.

Test

the

poi

nt (-

3, 0

), w

hich

is in

side

the

par

abol

a. S

ince

(-

3)2 +

6(-

3) +

7 =

-2,

and

0 >

-2,

(-3,

0) s

atis

fies

the

ineq

ualit

y. Th

eref

ore,

sha

de t

he r

egio

n in

side

the

par

abol

a.

Exer

cise

sG

raph

eac

h in

equa

lity

.

1. y

> x

2 - 8

x +

17

2. y

≤ x

2 + 6

x +

4

3. y

≥ x

2 + 2

x +

2

x

y

O2

46

-2

-4

-6

x

y

O2

-4

-2

-2

-4

-6

2

x

y

O2

246

-2

-4

4. y

< -

x2 + 4

x -

6

5. y

≥ 2

x2 + 4

x 6.

y >

-2x

2 - 4

x +

2

x

y

O2

-2

-4

-6

4-

2

x

y

O2

246

-2

-4

-2

x

y

O2

24

4-

2

-2

-4

-4

Exam

ple

042_

056_

ALG

2_A

_CR

M_C

04_C

R_6

6078

5.in

dd

4912

/20/

10

9:06

PM



Co

pyrig

ht ©

Gle

nco

e/M

cG

raw

-Hill, a

div

isio

n o

f Th

e M

cG

raw

-Hill C

om

pa

nie

s, In

c.

PDF Pass



NA

ME

DAT

E

P

ER

IOD

4-8

Cha

pter

4

50

Gle

ncoe

Alg

ebra

2

Stud

y Gu

ide

and

Inte

rven

tion

(con

tinu

ed)

Qu

ad

rati

c I

neq

ualiti

es

Solv

e Q

uadr

atic

Ineq

ualit

ies

Qua

drat

ic in

equa

litie

s in

one

var

iabl

e ca

n be

sol

ved

grap

hica

lly o

r al

gebr

aica

lly.

Gra

phic

al M

etho

d

To s

olve

ax2 +

bx

+ c

< 0

: Fi

rst g

raph

y =

ax2 +

bx

+ c

. The

sol

utio

n co

nsis

ts o

f the

x-v

alue

s

for w

hich

the

grap

h is

bel

ow th

e x-

axis

.

To s

olve

ax2 +

bx

+ c

> 0

: Fi

rst g

raph

y =

ax2 +

bx

+ c

. The

sol

utio

n co

nsis

ts o

f the

x-v

alue

s fo

r whi

ch th

e gr

aph

is a

bove

the

x-ax

is.

Alg

ebra

ic M

etho

d

Find

the

root

s of

the

rela

ted

quad

ratic

equ

atio

n by

fact

orin

g,

com

plet

ing

the

squa

re; o

r usi

ng th

e Q

uadr

atic

For

mul

a.

2 ro

ots

divi

de th

e nu

mbe

r lin

e in

to 3

inte

rval

s.Te

st a

val

ue in

eac

h in

terv

al to

see

whi

ch in

terv

als

are

solu

tions

.

If t

he in

equa

lity

invo

lves

≤ o

r ≥

, the

roo

ts o

f the

rel

ated

equ

atio

n ar

e in

clud

ed in

the

so

luti

on s

et.

So

lve

the

ineq

uali

ty x

2 - x

- 6

≤ 0

.

Firs

t fin

d th

e ro

ots

of t

he r

elat

ed e

quat

ion

x2 - x

- 6

= 0

. The

x

y

O

equa

tion

fact

ors

as (x

- 3

)(x +

2) =

0, s

o th

e ro

ots

are

3 an

d -

2.

The

grap

h op

ens

up w

ith

x-in

terc

epts

3 a

nd -

2, s

o it

mus

t be

on

or b

elow

the

x-a

xis

for

-2

≤ x

≤ 3

. The

refo

re t

he s

olut

ion

set

is

{x|-

2 ≤

x ≤

3}.

Exer

cise

sSo

lve

each

ineq

uali

ty.

1. x

2 + 2

x <

0

2. x

2 - 1

6 <

0

3. 0

< 6

x -

x2 -

5

4. c

2 ≤ 4

5.

2m

2 - m

< 1

6.

y2 <

-8

7. x

2 - 4

x -

12

< 0

8.

x2 +

9x

+ 1

4 >

0

9. -

x2 + 7

x -

10

≥ 0

10. 2

x2 + 5

x- 3

≤ 0

11

. 4x2 -

23x

+ 1

5 >

0

12. -

6x2 -

11x

+ 2

< 0

13. 2

x2 - 1

1x +

12

≥ 0

14

. x2 -

4x

+ 5

< 0

15

. 3x2 -

16x

+ 5

< 0

Exam

ple

{

x ⎥

-2

< x

< 0

} {

x ⎥

-4

< x

< 4

} {

x ⎥

1 <

x <

5}

{

c ⎥

-2

≤ c

≤ 2

} {

m ⎥

- 1 −

2 < m

< 1

} �

{

x ⎥

-2

< x

< 6

} {

x ⎥

x <

-7

or x

> -

2}

{x

⎥ 2

≤ x

≤ 5

}

{

x⎥

-3

≤ x

≤ 1 −

2 }

{x

⎥ x

< 3 −

4 or

x >

5}

{

x ⎥

x <

-2

or x

> 1 −

6 }

{

x⎥

x <

3 −

2 or

x >

4}

�

{

x⎥

1 −

3 < x

< 5

}

042_

056_

ALG

2_A

_CR

M_C

04_C

R_6

6078

5.in

dd

5012

/20/

10

9:06

PM


NA

ME

DAT

E

P

ER

IOD

Lesson 4-8

4-8

Cha

pter

4

51

Gle

ncoe

Alg

ebra

2

Skill

s Pr

actic

eQ

uad

rati

c I

neq

ualiti

es

Gra

ph e

ach

ineq

uali

ty.

1. y

≥ x

2 - 4

x +

4

2. y

≤ x

2 - 4

3.

y >

x2 +

2x

- 5

x

y

O4

6 4 2

-2

2

x

y

O4 2

–2

–4

-2

-4

24

x

y

O2

-4

4-

2

-2

-4

-6

Solv

e ea

ch in

equa

lity

by

grap

hing

.

4. x

2 - 6

x +

9 ≤

0

5. -

x2 - 4

x +

32

≥ 0

6.

x2 +

x -

10

> 1

0

xO8 6 4 2

2–2

46

y

x

y

O6

-6

15 5

–5

-2

2

–15

Solv

e ea

ch in

equa

lity

alg

ebra

ical

ly.

7. x

2 - 3

x -

10

< 0

8.

x2 +

2x

- 3

5 ≥

0

9. x

2 - 1

8x +

81

≤ 0

10

. x2 ≤

36

11. x

2 - 7

x >

0

12. x

2 + 7

x +

6 <

0

13. x

2 + x

- 1

2 >

0

14. x

2 + 9

x +

18

≤ 0

15. x

2 - 1

0x +

25

≥ 0

16

. -x2 -

2x

+ 1

5 ≥

0

17. x

2 + 3

x >

0

18. 2

x2 + 2

x >

4

19. -

x2 - 6

4 ≤

-16

x 20

. 9x2 +

12x

+ 9

< 0

x

y O2

6

-6

6-

2-

4-

6

183012

{

x ⎥ -

2 <

x <

5}

{x

⎥ x

≤ -

7 or

x ≥

5}

{

x ⎥

x <

-3

or x

> 0

}

{x⎥

x <

-2

or x

> 1

}

{

x ⎥

x <

0 o

r x

> 7

}

{x ⎥

-6

< x

< -

1}

{

all r

eal n

umbe

rs}

{x

⎥ -

5 ≤

x ≤

3}

{

x ⎥

x =

9}

{x

⎥ -

6 <

x <

6}

{

x ⎥

x <

-4

or x

> 3

}

{x ⎥

-6

≤ x

≤ -

3}

{

all r

eal n

umbe

rs}

�

{

x ⎥

x =

3}

{x

⎥ -

8 ≤

x ≤

4}

{x

⎥ x

< -

5 or

x >

4}

042_

056_

ALG

2_A

_CR

M_C

04_C

R_6

6078

5.in

dd

5112

/20/

10

9:06

PM



Copyri

ght

© G

lencoe/M

cG

raw

-Hill

, a

div

isio

n o

f T

he

McG

raw

-Hill

Co

mp

an

ies,

Inc.

PDF Pass



NA

ME

DAT

E

P

ER

IOD

4-8

Cha

pter

4

52

Gle

ncoe

Alg

ebra

2

Prac

tice

Qu

ad

rati

c I

neq

ualiti

es

Gra

ph e

ach

ineq

uali

ty.

1. y

≤ x

2 + 4

2.

y >

x2 +

6x

+ 6

3.

y <

2x2 -

4x

- 2

x

y

O2

268 4

4-

2-

4

x

y

O24

-2

-4

-6

-2

-4

x

y O2

2

4-

2

-2

-4

-4

Solv

e ea

ch in

equa

lity

.

4. x

2 + 2

x +

1 >

0

5. x

2 - 3

x +

2 ≤

0

6. x

2 + 1

0x +

7 ≥

0

7. x

2 - x

- 2

0 >

0

8. x

2 - 1

0x +

16

< 0

9.

x2 +

4x

+ 5

≤ 0

10. x

2 + 1

4x +

49

≥ 0

11

. x2 -

5x

> 1

4 12

. -x2 -

15

≤ 8

x

13. -

x2 + 5

x -

7 ≤

0

14. 9

x2 + 3

6x +

36

≤ 0

15

. 9x

≤ 1

2x2

16. 4

x2 + 4

x +

1 >

0

17. 5

x2 + 1

0 ≥

27x

18

. 9x2 +

31x

+ 1

2 ≤

0

19. F

ENCI

NG

Van

essa

has

180

feet

of f

enci

ng t

hat

she

inte

nds

to u

se t

o bu

ild a

rec

tang

ular

pl

ay a

rea

for

her

dog.

She

wan

ts t

he p

lay

area

to

encl

ose

at le

ast

1800

squ

are

feet

. Wha

t ar

e th

e po

ssib

le w

idth

s of

the

pla

y ar

ea?

20. B

USI

NES

S A

bic

ycle

mak

er s

old

300

bicy

cles

last

yea

r at

a p

rofit

of $

300

each

. The

m

aker

wan

ts t

o in

crea

se t

he p

rofit

mar

gin

this

yea

r, bu

t pr

edic

ts t

hat

each

$20

incr

ease

in

pro

fit w

ill r

educ

e th

e nu

mbe

r of

bic

ycle

s so

ld b

y 10

. How

man

y $2

0 in

crea

ses

in p

rofit

ca

n th

e m

aker

add

in a

nd e

xpec

t to

mak

e a

tota

l pro

fit o

f at

leas

t $1

00,0

00?

a

ll re

als

{x

⎥ x

= -

2}

{x

⎥ x

≤ 0

or

x ≥

3 −

4 }

30 ft

to 6

0 ft

al

l rea

ls

{x ⎥

-1

≤ x

≤ 3

} {x

⎥ x

≤ 2

or

x ≥

5}

{

x ⎥

x <

-4

or x

> 5

} {

x ⎥

2 <

x <

8}

�

a

ll re

als

{x

⎥ x

< -

2 or

x >

7}

{x

⎥ -

5 ≤

x ≤

-3}

{

x⎥

x ≠

- 1 −

2 }

{x

⎥ x

≤ 2 −

5 or

x ≥

5}

{

x⎥

-3

≤ x

≤ -

4 −

9 }

from

5 to

10

042_

056_

ALG

2_A

_CR

M_C

04_C

R_6

6078

5.in

dd

5212

/20/

10

9:06

PM


NA

ME

DAT

E

P

ER

IOD

Lesson 4-8

4-8

Cha

pter

4

53

Gle

ncoe

Alg

ebra

2

Wor

d Pr

oble

m P

ract

ice

Qu

ad

rati

c I

neq

ualiti

es

1.H

UTS

The

spa

ce in

side

a h

ut is

sha

ded

in t

he g

raph

. The

par

abol

a is

des

crib

ed

by t

he e

quat

ion

y=

-

4−

5(x

- 1

)2+

4.

y

xO

Wri

te a

n in

equa

lity

that

des

crib

es t

he

shad

ed r

egio

n.

2. D

ISCR

IMIN

AN

TS C

onsi

der

the

equa

tion

ax2 +

bx

+ c

= 0

. Ass

ume

that

th

e di

scri

min

ant

is z

ero

and

that

a is

po

siti

ve. W

hat

are

the

solu

tion

s of

the

in

equa

lity

ax2 +

bx

+ c

≤ 0

?

3. K

IOSK

S C

aleb

is d

esig

ning

a k

iosk

by

wra

ppin

g a

piec

e of

she

et m

etal

wit

h di

men

sion

s x

+ 5

inch

es b

y 4x

+ 8

in

ches

into

a c

ylin

dric

al s

hape

. Ign

orin

g co

st, C

aleb

wou

ld li

ke a

kio

sk t

hat

has

a su

rfac

e ar

ea o

f at

leas

t 44

80 s

quar

e in

ches

. Wha

t va

lues

of x

sat

isfy

th

is c

ondi

tion

?

4. D

AM

S Th

e H

oove

r D

am is

a c

oncr

ete

arch

dam

des

igne

d to

hol

d th

e w

ater

of

Lake

Mea

d. A

t it

s ce

nter

, the

dam

’s he

ight

is a

ppro

xim

atel

y 72

5 fe

et, a

nd

the

dam

var

ies

from

45

to 6

60 fe

et t

hick

. Th

e da

rk li

ne o

n th

is s

ketc

h of

the

cro

ss-

sect

ion

of t

he d

am is

a p

arab

ola.

y

xTh

ickn

ess

Height

Cro

ss-S

ectio

n of

Hoov

er D

am

Dam

Lake

Mea

d

a. W

rite

an

equa

tion

for

the

Hoo

ver

Dam

par

abol

a. L

et t

he h

eigh

t be

the

y-

valu

e of

the

par

abol

a an

d th

e th

ickn

ess

be t

he x

-val

ue o

f the

pa

rabo

la. (

Hin

t: th

e eq

uati

on w

ill b

e in

the

form

: y =

k(x

– m

axim

um

thic

knes

s) +

max

imum

hei

ght.)

b. U

sing

you

r eq

uati

on, g

raph

the

pa

rabo

la o

f the

Hoo

ver

Dam

for

45 ≤

x ≤

660

. Thic

knes

s

Height

x

250

220

440

660

y

500

750

c. E

stim

ate

to t

he n

eare

st fo

ot t

he

thic

knes

s of

the

dam

whe

n th

e he

ight

is

200

feet

.

y =

–0.

0019

2 (x

– 6

60)2 +

725

0 ≤

y ≤

- 4 −

5 (x

- 1

)2 + 4

x =

- b

−

2a

x >

30

(Not

e th

at th

e va

lues

of

x <

-37

res

ult i

n a

high

er p

rodu

ct,

but n

egat

ive

leng

ths

do n

ot m

ake

sens

e.)

353

feet

042_

056_

ALG

2_A

_CR

M_C

04_C

R_6

6078

5.in

dd

5312

/20/

10

9:06

PM



Co

pyrig

ht ©

Gle

nco

e/M

cG

raw

-Hill, a

div

isio

n o

f Th

e M

cG

raw

-Hill C

om

pan

ies, In

c.

PDF Pass



NA

ME

DAT

E

P

ER

IOD

4-8

Ch

ap

ter

4

54

Gle

ncoe

Alg

ebra

2

Enri

chm

ent

Gra

ph

ing

Ab

solu

te V

alu

e I

neq

ualiti

es

You

can

solv

e ab

solu

te v

alue

ineq

ualit

ies

by g

raph

ing

in m

uch

the

sam

e m

anne

r yo

u gr

aphe

d qu

adra

tic

ineq

ualit

ies.

Gra

ph t

he r

elat

ed a

bsol

ute

func

tion

fo

r ea

ch in

equa

lity

by u

sing

a g

raph

ing

calc

ulat

or. F

or >

and

≥, i

dent

ify t

he

x-va

lues

, if a

ny, f

or w

hich

the

gra

ph li

es b

elow

the

x-a

xis.

For

< a

nd ≤

, ide

ntify

th

e x

valu

es, i

f any

, for

whi

ch t

he g

raph

lies

abo

ve t

he x

-axi

s.

For

each

ineq

uali

ty, m

ake

a sk

etch

of

the

rela

ted

grap

h an

d fi

nd t

he

solu

tion

s ro

unde

d to

the

nea

rest

hun

dred

th.

1. ⎪

x -

3⎥ >

0

2. ⎪

x⎥ -

6 <

0

3. -

⎪ x +

4⎥ +

8 <

0

4. 2

⎪ x +

6⎥ -

2 ≥

0

5. ⎪

3x -

3⎥ ≥

0

6. ⎪

x -

7⎥ <

5

7.⎪ 7

x -

1⎥>

13

8.⎪ x

- 3

.6⎥≤

4.2

9.⎪ 2

x+

5⎥≤

7

x

> 3

or

x <

3

-6 <

x <

6

-12

< x

< 4

x

≤ -

7 or

x ≥

-5

all

real

num

bers

2

< x

< 1

2

x

< -

1.71

or

x >

2

-0.

6 ≤

x ≤

7.8

-

6 ≤

x ≤

1

042_

056_

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NA

ME

DAT

E

P

ER

IOD

Lesson 4-8

4-8

Ch

ap

ter

4

55

Gle

ncoe

Alg

ebra

2

Grap

hing

Cal

cula

tor

Activ

ityQ

uad

rati

c I

neq

ualiti

es

an

d t

he T

est

Men

u

The

ineq

ualit

y sy

mbo

ls, c

alle

d re

lati

onal

ope

rato

rs, i

n th

e T

EST

men

u ca

n be

use

d to

dis

play

th

e so

luti

on o

f a q

uadr

atic

ineq

ualit

y. A

noth

er m

etho

d th

at c

an b

e us

ed t

o fin

d th

e so

luti

on

set

of a

qua

drat

ic in

equa

lity

is t

o gr

aph

each

sid

e of

an

ineq

ualit

y se

para

tely

. Exa

min

e th

e gr

aphs

and

use

the

inte

rsec

t fu

ncti

on t

o de

term

ine

the

rang

e of

val

ues

for

whi

ch t

he

ineq

ualit

y is

tru

e.

Exer

cise

sSo

lve

each

ineq

uali

ty.

1. -

x2 - 1

0x -

21 <

0

2. x

2 - 9

< 0

3.

x2 +

10x

+ 2

5 ≤

0

4. x

2 + 3

x ≤

28

5. 2

x2 + x

≥ 3

6.

4x2 +

12x

+ 9

> 0

7. 2

3 >

-x2 +

10x

8.

x2 -

4x -

13 ≤

0

9. (

x +

1)(x

-3)

> 0

So

lve

x2 +

x ≥

6.

Plac

e th

e ca

lcul

ator

in D

ot m

ode.

Ent

er t

he in

equa

lity

into

Y1.

Th

en t

race

the

gra

ph a

nd d

escr

ibe

the

solu

tion

as

an in

equa

lity.

Key

stro

kes:

Y=

x2

+

2

nd

[TE

ST] 4

6

ZO

OM

4.

Use

TR

AC

E t

o de

term

ine

the

endp

oint

s of

the

seg

men

ts.

Thes

es v

alue

s ar

e us

ed t

o ex

pres

s th

e so

luti

on o

f the

ineq

ualit

y,

{ x |

x ≥

- 3

or

x ≥

2 }.

So

lve

2x2 +

4x -

5 ≤

3.

Plac

e th

e le

ft s

ide

of t

he in

equa

lity

in Y

1 an

d th

e ri

ght

side

in Y

2.

Det

erm

ine

the

poin

ts o

f int

erse

ctio

n. U

se t

he in

ters

ecti

on p

oint

s to

exp

ress

the

sol

utio

n se

t of

the

ineq

ualit

y. B

e su

re t

o se

t th

e ca

lcul

ator

to

Con

nect

ed m

ode.

Key

stro

kes:

Y

= 2

x2

+ 4

—

5 E

NT

ER

3 E

NT

ER

Z

OO

M 6

.

Pres

s 2

nd

[CA

LC] 5

and

use

the

k

ey t

o m

ove

the

curs

or

to t

he le

ft o

f the

firs

t in

ters

ecti

on p

oint

. Pre

ss E

NT

ER

. The

n m

ove

the

curs

or t

o th

e ri

ght

of t

he in

ters

ecti

on p

oint

and

pre

ss E

NT

ER

EN

TE

R. O

ne o

f the

val

ues

used

in t

he s

olut

ion

set

is d

ispl

ayed

. R

epea

t th

e pr

oced

ure

on t

he o

ther

inte

rsec

tion

poi

nt.

The

solu

tion

is {

x | -

3.24

≤ x

≤ 1

.24}

.

Exam

ple

1

Exam

ple

2

[-4.

7, 4

.7] s

cl:1

by

[-3.

1, 3

.1] s

cl:1

[-10

, 10]

scl

:1 b

y [-

10, 1

0] s

cl:1

[-10

, 10]

scl

:1 b

y [-

10, 1

0] s

cl:1

{

x | x

< -

7 or

x >

-3}

{x | -

3 <

x <

3}

{

x | x

= -

5}

{

x | -

7 ≤

x ≤

4 }

{x | x

≤ -

1.5

or x

≥ 1

}

{x |

x ≠

-1.

5}

{

x | x

≤ 3

.59

or x

≥ 6

.41}

{x | -

2.12

≤ x

≤ 6

.12}

{x

| x <

-1

or x

> 3

}

042_

056_

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Chapter 4 Assessment Answer Key

An

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Quiz 1 (Lessons 4-1 and 4-2) Quiz 3 (Lessons 4-5 and 4-6) Mid-Chapter TestPage 59 Page 60 Page 61

1.

2.

3.

4.

5.

xO

f(x)

(0, -3)(-1, -4)

f(x) = x2 + 2x - 3

x = -1

4

2

2 4

–2

–4

-4 -2

-3; x = -1; -1

between 1 and 2;between -6 and -5

A

3, -1

1.

2.

3.

4.

5.

6.

7.

8.

9.

10.

{-5, 2 _ 3 }

{-9, 5}

C

x 2 + 4x - 12 = 0

3 x 2 + 10x - 8 = 0

4i √ � 5

-6 √ � 2

68 + 4i

-11 + 3i

1 _ 2 + 1

_ 2 i

1.

2.

3.

4.

5.

{-10, 2}

1 ± 3 √ � 5

B

2 ± √ � 5

-96; 2 complex roots

xO

y

(2, -1)

x = 24

2

2

–2

–4

-4 -2

y

xO

4

2

2 4

–2

-4 -2

y = 2 (x + 5) 2

1.

2.

3.

4.

5. A

{x ⎪1 < x < 5}

1.

2.

3.

4.

5.

B

H

A

F

D

y

xO

4

2

2 4-4 -2

8.

6.

7.

8.

9.

10.

1, 3

minimum, -9 1 _ 2

{-2, 9}

{0, 1 _ 4 }

- 25

_ 34

+ 15i

_ 34

Quiz 2 (Lessons 4-3 and 4-4)

Page 59

Quiz 4 (Lessons 4-7 and 4-8)

Page 60


Chapter 4 Assessment Answer KeyC

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Vocabulary Test Form 1Page 62 Page 63 Page 64

false; complex conjugates

false; constant termfalse; quadratic

inequality

false; roots

true

false; minimum value

false; quadratic term

true

true

false; discriminant

Sample answer: The value that determines the roots of a quadratic equation.

Sample answer: a quadratic equation written as ax2 + bx + c = 0

1.

2.

3.

4.

5.

6.

7.

8.

9.

10.

11.

12.

B

G

A

H

B

H

B

H

A

F

1.

2.

3.

4.

5.

6.

7.

8.

9.

10. 1 and 7; 14

B

J

D

B

G

F

C

J

B

G

11.

12.

13.

14.

15.

16.

17.

18.

19.

20.

B:



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Form 2A Form 2BPage 65 Page 66 Page 67 Page 68

C

G

A

H

D

G

C

F

D

F

1.

2.

3.

4.

5.

6.

7.

8.

9.

10.

B

G

C

J

D

B

F

C

J

F

Sample answer:16x2 + 3 = 0

11.

12.

13.

14.

15.

16.

17.

18.

19.

20.

B:

B

H

B

J

A

H

A

J

A

H

1.

2.

3.

4.

5.

6.

7.

8.

9.

10.

11.

12.

13.

14.

15.

16.

17.

18.

19.

20.

B:

C

G

D

F

B

G

D

H

A

F

Sample answer:

9x 2 + 2 = 0



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Form 2CPage 69 Page 70

xO

f (x)

(3, 0)

(1, 5)

f(x) = -5x2 + 10x

x = 1

xO

4

2

2 4

–2

–4

-4 -2

y

maximum; 4

2, 4

{-3, 2 _ 5 }

9 in. by 16 in.

6 + 12j ohms

37

_ 17

- 22

_ 17

j amps

4x2 + 21x - 18 = 0

{-8, 2}

1.

2.

3.

4.

5.

6.

7.

8.

9.

y

xO 2 4-2-4

4

2

–2

–4

{-2 ± √ �� 13 }

{-2, 1 _ 2 }

3 ± i √ �� 31

_

10

0; 1 real, rational root

33; 2 real, irrational roots

(-5, -7); x = -5; down

y = 3

_ 2 (x - 2)2 - 1

h(t ) = -16(t - 1.5)2

+ 51; 51ft

y = (x - 3)2 - 1

{x ⎪x ≤ - 1 _ 2 or x ≥ 3}

9x 2 - 7 = 0

10.

11.

12.

13.

14.

15.

16.

17.

18.

19.

20.

B:



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Form 2DPage 71 Page 72

y

xO

4

2

2 4

–2

–4

-4 -2

xO

f(x)

(0, 3)

(2, -1)

x = 2

4

2

2 4

–4

-4 -2

f(x) = x2 - 4x

+ 3

minimum; -17

1, -3

{-1, 4 _ 3 }

8 in. by 18 in.

9 - 6j ohms

23

_ 17

- 27

_ 17

j ohms

2x 2 + 5x - 12 = 0

1.

2.

3.

4.

5.

6.

7.

8.

y

xO

4

2

2 4

–2

-4 -2

{4 ± √ � 2 }

{-1, 2 _ 3 }

{ 9 ± √ �� 41

_

4 }

0; 1 real, rational root

-8; 2 complex roots

(6,-5); x = 6; down

y = - 1 _ 4 (x + 4)2 + 2

y = (x + 2)2 + 4

h(t ) = -16(t - 2)2 + 76; 76 ft

{x ⎪- 3 _

2 < x < 5}

16x 2 - 5 = 0

9.

10.

11.

12.

13.

14.

15.

16.

17.

18.

19.

20.

B:

{ -2 ± √ � 6

_ 3 }



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Form 3Page 73 Page 74

y

x

6

2

2 6-6 -2O

y

x

6

2

2 6-6 -2O

xO

f (x)

(0, 3)(- ),1

383

f(x) = 3x2 + 2x + 3

-13x =

4

2

2 4

–2

-4 -2

minimum; 22

_ 25

$8.00; $6400

3, 6

between -3 and -2;

between 4 and 5

{ 1 _ 2 ,

5

_ 3 }

12x2 - 13x - 14 = 0

4 - 6i

- 1 _ 9 -

4 √ � 5

_ 9 i

1.

2.

3.

4.

5.

6.

7.

8.

9.

y

xO 2 4-4

6

4

2

-2

-2 < k < 2

{-0.35, 0.85}

{ 5 ± i √ �� 39

_

8 }

{-3.5, 1}

6 ± 4 √ � 2

1.2; two real, irrational roots

h(t ) = -9.1(t - 32.5)2

+ 30,000; 30,000 ft

y = - 29 _

200 (x + 9)2 +

29

_ 2

{x ⎪x ≤ - 7 _ 2 or x = 1}

16x 2 + 24x + 29 = 0

y = - 3

_ 5 (x + 7

_ 2 )

2

- 1 _ 2 ;

(- 7 _ 2 , - 1

_ 2 ) ; x = -

7

_ 2 ; down

10.

11.

12.

13.

14.

15.

16.

17.

18.

19.

20.

B:



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Score General Description Specifi c Criteria

4 SuperiorA correct solution that is supported by well-developed, accurate explanations

• Shows thorough understanding of the concepts of graphing, analyzing, and finding the maximum and minimum values of quadratic functions; solving quadratic equations; computing with complex numbers; and solving inequalities.

• Uses appropriate strategies to solve problems. • Computations are correct. • Written explanations are exemplary. • Goes beyond requirements of some of or all problems.

3 SatisfactoryA generally correct solution, but may contain minor flaws in reasoning or computation

• Shows an understanding of the concepts of graphing, analyzing, and finding the maximum and minimum values of quadratic functions; solving quadratic equations; computing with complex numbers; and solving inequalities.

• Uses appropriate strategies to solve problems. • Computations are mostly correct. • Written explanations are effective. • Satisfies all requirements of problems.

2 Nearly SatisfactoryA partially correct interpretation and/or solution to the problem

• Shows an understanding of most of the concepts of graphing, analyzing, and finding the maximum and minimum values of quadratic functions; solving quadratic equations; computing with complex numbers; and solving inequalities.

• May not use appropriate strategies to solve problems. • Computations are mostly correct. • Written explanations are satisfactory. • Satisfies the requirements of most of the problems.

1 Nearly Unsatisfactory A correct solution with no supporting evidence or explanation

• Final computation is correct. • No written explanations or work is shown to substantiate

the final computation. • Satisfies minimal requirements of some of the problems.

0 UnsatisfactoryAn incorrect solution indicating no mathematical understanding of the concept or task, or no solution is given

• Shows little or no understanding of most of the concepts of graphing, analyzing, and finding the maximum and minimum values of quadratic functions; solving quadratic equations; computing with complex numbers; and solving inequalities.

• Does not use appropriate strategies to solve problems. • Computations are incorrect. • Written explanations are unsatisfactory. • Does not satisfy the requirements of problems. • No answer may be given.

Page 75, Extended-Response Test Scoring Rubric



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1. Student responses should indicate that using the Square Root Property, as Mi-Ling’s group did, would take less time than the other method since the equation is already set up as a perfect square set equal to a constant. To solve using the other method, the binomial would need to be expanded and the constant on the right brought to the left side of the equal sign.

2a. Jocelyn had trouble because the problem is impossible. No such parabola exists.

2b. Student responses will vary. One of the three conditions must be omitted or modified. Sample answer: “...that passes through (-1, -12).”

2c. Answers will vary and depend on the answer for part b. For example, for the sample answer in part b above, a possible equation is: y = -2(x + 3)2 - 4.

3a. Answer must be of the form y = a(x - h)2 + 8 where h is any real number and a < 0.

3b. Answers must be of the form y = a[x - (h + n)]2 + 8 where h and a represent the same values as in part a. The student choice is for the value of n. The student should indicate that the graph will shift to the left n units if his or her value of n is negative, but will shift the graph to the right n units if the chosen value of n is positive.

4. Students should indicate that Joseph’s answer is not correct. In Step 2, when he completed the square by adding 9 inside the parentheses, he actually added 2(9) = 18 to the right side of the equation, so he must subtract 18 from the constant on the same side, rather than add 9, to keep the statements equivalent. The correct solution is f(x) = 2(x + 3)2 - 23.

5a. > The graph is strictly above the x-axis for all values of x other than 2.

5b. <; The graph is never below the x-axis.

5c. ≥; The graph is always on or above the x-axis.

In addition to the scoring rubric found on page A33, the following sample answers may be used as guidance in evaluating open-ended assessment items.

Page 75, Extended-Response Test Sample Answers



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Standardized Test PracticePage 76 Page 77

1. A B C D

2. F G H J

3. A B C D

4. F G H J

5. A B C D

6. F G H J

7. A B C D

8. F G H J

9. A B C D

10. F G H J

11. A B C D

12. F G H J

13. A B C D

14. F G H J

15.

9

8

7

6

5

4

3

2

1

0

9

8

7

6

5

4

3

2

1

0

9

8

7

6

5

4

3

2

1

0

9

8

7

6

5

4

3

2

1

0

9

8

7

6

5

4

3

2

1

0

1 5 4

16.

9

8

7

6

5

4

3

2

1

0

9

8

7

6

5

4

3

2

1

0

9

8

7

6

5

4

3

2

1

0

9

8

7

6

5

4

3

2

1

0

9

8

7

6

5

4

3

2

1

0

8 0



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Standardized Test PracticePage 78

y

xO

4

2

2 4

–2

–4

-4 -2

17.

18.

19.

20.

21.

22.

23.

24.

25.

26.

27.

28a.

28b.

28c.

17

- 3 _

4

inconsistent

(-2, 0), (-2, 8), (0, -2), (8, -2)

92

(2, -3)

±2i √ � 5

136 ft; 1.5 s

-1, 3

(-2, 3)

-3, 2 complex roots

y = (

x - 7 _ 2 )

2

- 29

_ 4

(

7 _ 2 , -

29 _

4 )

x = 7 _ 2


Chapter 4 Resource · PDF fileChapter 4 iv Glencoe Algebra 2 Teacher’s Guide to Using...

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Transcript of Chapter 4 Resource · PDF fileChapter 4 iv Glencoe Algebra 2 Teacher’s Guide to Using...