UNIT 5 • QUADRATIC FUNCTIONS Lesson 3: Creating Quadratic Equations...

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

Instruction

CCGPS Analytic Geometry Teacher Resource U5-156

© Walch Education

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

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

Key Concepts

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

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

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

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

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

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

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

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

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

Prerequisite Skills

This lesson requires the use of the following skills:

• identifying the standard form of a quadratic function

• finding the vertex of a quadratic function

• solving equations by taking the square root


Instruction

CCGPS Analytic Geometry Teacher Resource © Walch EducationU5-157

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

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

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

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

ax2 + bx + c

a xb

ax

b

a

b

ac2

2 2

2 2+ +

−

+

a xb

aa

b

ac+

−

+2 2

2 2

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

the equation a xb

ak+

+2

2

, where k ab

ac= −

+2

2

.

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

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

Common Errors/Misconceptions

• forgetting to make sure the coefficient of the quadratic term, x2, is 1 before completing the square


Instruction


© Walch Education

Example 1

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

26 22 , identify the vertex and determine whether it is a

minimum or maximum.

1. Identify the vertex.

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

f x x( )= +( ) −1

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

function is (–6, –2).

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

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

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

f x x( )= +( ) −1

26 22 is in vertex form, so a =

1

2.

1

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

Guided Practice 5.3.3


Instruction


Example 2

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

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

f(x) = a(x – h)2 + k Vertex form

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

f(x) = a(x + 4)2 – 8 Simplify.

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

f(x) = a(x + 4)2 – 8 Equation

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

–5 = a(2)2 – 8 Simplify.

–5 = 4a – 8

3 = 4a

3

4= a

3. Substitute a into the equation from step 1.

f(x) = a(x + 4)2 – 8

f x x( )= +( ) −3

44 82

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

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

44 82 .


Instruction


© Walch Education

Example 3

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

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

g(x) = –7x2 + 14x Original function

g(x) = –7(x2 – 2x) Factor out –7.

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

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

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

2. Summarize your result.

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

Example 4

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

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

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

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

The axis of symmetry extends through the vertex.

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


Instruction


2. Find the y-intercept.

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

Substitute 0 for x to find y.

y = (x + 3)2 – 8 Original equation

y = (0 + 3)2 – 8 Substitute 0 for x.

y = 32 – 8 Simplify.

y = 1

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

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

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

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

In this case, let’s use x = 1.

y = (x + 3)2 – 8 Original equation

y = (1 + 3)2 – 8 Substitute 1 for x.

y = 42 – 8 Simplify.

y = 8

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

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

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


Instruction


© Walch Education

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

x

y

–8 –6 –4 –2 0 2 4 6 8

–10

–8

–6

–4

–2

0

2

4

6

10

8Axis of symmetryx = –3(–7, 8) (1, 8)

(–6, 1) (0, 1)

V (–3, –8)

UNIT 5 • QUADRATIC FUNCTIONS Lesson 3: Creating Quadratic Equations...

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