myresource.phoenix.edumyresource.phoenix.edu/secure/resource/MAT117R7/mat117...CALCULATOR CORNER...

1111Quadratic Equationsand Functions

11.1 The Basics of Solving Quadratic Equations

11.2 The Quadratic Formula

11.3 Applications Involving Quadratic Equations

11.4 More on Quadratic Equations

11.5 Graphing

11.6 Graphing

11.7 Mathematical Modeling withQuadratic Functions

11.8 Polynomial and Rational Inequalities

f�x� � ax2 � bx � c

f�x� � a�x � h�2 � k

Real-World ApplicationThe base of a triangular sail is 9 m less than its height. The area is Find the base and the height of the sail.

This problem appearsas Exercise 5 inSection 11.3.

56 m2.

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Introductory and Intermediate Algebra, Third Edition, by Marvin L.Bittinger and Judith A.Beecher.Published by Addison Wesley.Copyright ©2007 by Pearson Education, Inc.

754

CHAPTER 11: Quadratic Equations and Functions

ALGEBRAIC – GRAPHICAL CONNECTION

Let’s reexamine the graphical connections to the algebraic equation-solving concepts we have studied before.

In Chapter 7, we introduced the graph of a quadratic function:

, .

For example, the graph of the function and its x-intercepts are shown below.

The x-intercepts are and . These pairs are also thepoints of intersection of the graphs of and (the x-axis). We will analyze the graphs of quadratic functions in greaterdetail in Sections 11.5– 11.7.

In Chapter 5, we solved quadratic equations like using factoring, as here:

Factoring

Using the principle of zero products

.

We see that the solutions of , �4 and �2, are the firstcoordinates of the x-intercepts, and , of the graph of

.

Do Exercise 1.

We now extend our ability to solve quadratic equations.

The Principle of Square Roots

The quadratic equation

is said to be in standard form. The quadratic equation

is equivalent to the preceding, but it is not in standard form.

5x2 � 2 � 8x

5x2 � 8x � 2 � 0

f �x� � x2 � 6x � 8��2, 0��4, 0�

0 � x2 � 6x � 8

x � �4 or x � �2

x � 4 � 0 or x � 2 � 0

�x � 4� �x � 2� � 0

x2 � 6x � 8 � 0

x2 � 6x � 8 � 0

g�x� � 0f �x� � x2 � 6x � 8��2, 0��4, 0�

x-intercepts:(�4, 0), (�2, 0)

f (x) � x 2 � 6x � 8

�1 2

�4

4

�5�6�7 �3 1 3

�3

3

2

1

�2

�1x

y

f �x� � x2 � 6x � 8

a � 0f �x� � ax2 � bx � c

11.111.1 THE BASICS OF SOLVINGQUADRATIC EQUATIONS

1. Consider solving the equation

.

Below is the graph of

.

a) What are the x-intercepts ofthe graph?

b) What are the solutions of?

c) What relationship existsbetween the answers to parts (a) and (b)?

Answers on page A-46

x2 � 6x � 8 � 0

f (x) � x 2 � 6x � 8

�2 �1 2 4

4

�3 1 3 5 6 7

3

2

1

5

6

7

8

9

�1x

y

f �x� � x2 � 6x � 8

x2 � 6x � 8 � 0

ObjectivesSolve quadratic equationsusing the principle ofsquare roots and find the x-intercepts of the graph of arelated function.

Solve quadratic equations bycompleting the square.

Solve applied problems usingquadratic equations.

AG

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QUADRATIC EQUATION

An equation of the type , where a, b, and c are real-number constants and , is called the standard form of aquadratic equation.

To find the standard form of the quadratic equation ,we find an equivalent equation by multiplying by �1 on both sides:

. Writing in standard form

In Section 5.8, we studied the use of factoring and the principle of zeroproducts to solve certain quadratic equations. Let’s review that procedure andintroduce a new one.

EXAMPLE 1

a) Solve: .

b) Find the x-intercepts of .

a) We first find standard form and then factor:

Subtracting 25

Factoring


.

The solutions are 5 and �5.

b) The x-intercepts of are and . The solutions of the equation

are the first coordinates of the x-intercepts of the graph of .

EXAMPLE 2 Solve: .

We factor and use the principle of zero products:

.

The solutions are 0 and . The check is left to the student.

Do Exercises 2 and 3.

52

x � 0 or x � 52

3x � 0 or 2x � 5 � 0

3x�2x � 5� � 0

6x2 � 15x � 0

6x2 � 15x � 0

f �x� � x2 � 25x2 � 25

�5, 0��5, 0�f �x� � x2 � 25

x � 5 or x � �5

x � 5 � 0 or x � 5 � 0

�x � 5� �x � 5� � 0

x2 � 25 � 0

f �x� � x2 � 25

x2 � 25

5x2 � 4x � 7 � 0

�1��5x2 � 4x � 7� � �1�0�

�5x2 � 4x � 7 � 0

a � 0ax2 � bx � c � 0

2. a) Solve: .

b) Find the x-intercepts of.

3. a) Solve: .



? ?

f (x)� 4x 2 � 14x

x

�5

�10

�15

5

5 10

10

�10 �5

y

f �x� � 4x2 � 14x

4x2 � 14x � 0

f (x) � x 2 � 16

x

�5

�10

�15

5

5

? ?

10

10

�10 �5

y

f �x� � x2 � 16

x2 � 16

755


�4 �2 2 4

5

�5

x

y

�10

�15

�20

�25

(5, 0)(�5, 0)

f (x) � x2 � 25

�1 1 32

�8

�4

10

x

y

(0, 0)

f (x) � 6x2 � 15x

(e, 0)CALCULATORCORNER

Solving QuadraticEquations Use the ZERO

feature to solve the equationsin Example 2 and MarginExercise 3. See the CalculatorCorner on p. 371 to review theprocedure.

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EXAMPLE 3

a) Solve: .


a) We first find standard form. Then we factor and use the principle of zeroproducts.

Adding x and subtracting 2 to get the standard form

Factoring


Check: For :

TRUE 43 43

3 �49 63 �

23

3�23�2 ? 2 � �2

3� 3x2 � 2 � x

23

x � 23 or x � �1

3x � 2 or x � �1

3x � 2 � 0 or x � 1 � 0

�3x � 2� �x � 1� � 0

3x2 � x � 2 � 0

3x2 � 2 � x

f �x� � 3x2 � x � 2

3x2 � 2 � x

4. a) Solve: .



f (x) � 5x 2 � 8x � 3

�1 2

2

3

1

3

4

x

y

? ?

f �x� � 5x2 � 8x � 3

5x2 � 8x � 3

756


For �1:

TRUE 3 3 3 � 1 2 � 1

3��1�2 ? 2 � ��1�

3x2 � 2 � x

The solutions are �1 and .

b) The x-intercepts of are and . The solutions of the equation are the first coordinates of the x-intercepts of the graph of .

Do Exercise 4.

SOLVING EQUATIONS OF THE TYPE Consider the equation again. We know from Chapter 10 that the num-ber 25 has two real-number square roots, namely, 5 and �5. Note that theseare the solutions of the equation in Example 1. This exemplifies the principleof square roots, which provides a quick method for solving equations of thetype .

THE PRINCIPLE OF SQUARE ROOTS

The solutions of the equation are and .

When the solutions are two real numbers.

When the only solution is 0.

When the solutions are two imaginary numbers.d � 0,

d � 0,

d � 0,

��d�dx2 � d

x2 � d

x2 � 25

x2 � d

f �x� � 3x2 � x � 2

3x2 � 2 � x�2

3 , 0��1, 0�f �x� � 3x2 � x � 2

23

�2 1 2

2

1

�1

x

y

f (x) � 3x2 � x � 2

(�1, 0) (s, 0)

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EXAMPLE 4 Solve: . Give the exact solutions and approximatethe solutions to three decimal places.

We have

.

We often use the symbol to represent both of the solutions.

Check: For :

TRUE 6 3 � 2

3��2 �2 ? 6

3x2 � 6

�2

��2

x � �2 or x � ��2

x2 � 2

3x2 � 6

3x2 � 6 5. Solve: . Give the exactsolution and approximate thesolutions to three decimalplaces.

6. Solve: . Give theexact solution and approximatethe solutions to three decimalplaces.


�3x2 � 8 � 0

5x2 � 15

757


The solutions are and , or , which are about 1.414 and �1.414when rounded to three decimal places.

Do Exercise 5.

Sometimes we rationalize denominators to simplify answers.

EXAMPLE 5 Solve: . Give the exact solutions and approxi-mate the solutions to three decimal places.

Check: We check both numbers at once,since there is no x-term in the equation.We could have checked both numbers at once in Example 4 as well.

TRUE

The solutions are and , or . We can use a calculator for

approximations:

.�

�105

� �0.632

�

�105

�

�105

�105

0 �2 � 2

�5�1025� � 2

�5��

�105 �2

� 2 ? 0

�5x2 � 2 � 0

x ��10

5 or x � �

�105

Rationalizing thedenominators x � � 2

5�

55

or x � �� 25

�55

Using the principleof square roots x � � 2

5 or x � �� 2

5

Subtracting 2 anddividing by �5

x2 �25

�5x2 � 2 � 0

�5x2 � 2 � 0

��2��2�2

For :

TRUE 6 3 � 2

3��2 �2 ? 6

3x2 � 6

��2

�4 �2 2 4

4

2

x

y

f (x) � 3x2 � 6

(�2, 0)(��2, 0)

�2 �1 1 2

�2

2

�1

x

y

f (x) � �5x2 � 2

��105

, 0 ��105

, 0

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Do Exercise 6 on the preceding page.

Sometimes we get solutions that are imaginary numbers.

EXAMPLE 6 Solve: .

Subtracting 9 and dividing by 4

Using the principle of square roots

Simplifying

Check:

TRUE

The solutions are and , or .We see that the graph of does not cross the x-axis. This

is true because the equation has imaginary complex-numbersolutions.

Do Exercise 7.

SOLVING EQUATIONS OF THE TYPE The equation can also be solved using the principle of squareroots.

EXAMPLE 7

a) Solve: .


a) We have


.

The solutions are and , or .

b) The x-intercepts of are and .

Do Exercise 8.

�2 � �7, 0��2 � �7, 0�f �x� � �x � 2�2 � 7

2 � �72 � �72 � �7

x � 2 � �7 or x � 2 � �7

x � 2 � �7 or x � 2 � ��7

�x � 2�2 � 7

f �x� � �x � 2�2 � 7

�x � 2�2 � 7

�x � 2�2 � 7

�x � c�2 � d

4x2 � 9 � 0f �x� � 4x2 � 9

� 32 i�

32 i3

2 i

0 �9 � 9

4��

94� � 9

4��

32

i�2

� 9 ? 0

4x2 � 9 � 0

x �32

i or x � �

32

i

x � ��

94

or x � ��

94

x2 � �

94

4x2 � 9 � 0

4x2 � 9 � 0

7. Solve: .

8. a) Solve: .



f (x) � (x � 1)2 � 5

�1 2

�4

�5

�6

4

�2�3 1 4 5 6

�3

3

2

1

�1x

y

? ?

f �x� � �x � 1�2 � 5

�x � 1�2 � 5

2x2 � 1 � 0

758


�2 �1 1 2

20

16

12

8

4

x

y

f (x) � 4x2 � 9No x-intercepts

�4 �2 2 4 6

�6

�4

2

�2

x

y

f (x) � (x � 2)2 � 7

(2 � �7, 0)(2 � �7, 0)

CALCULATORCORNER

Imaginary Solutions ofQuadratic EquationsWhat happens when you usethe ZERO feature to solve theequation in Example 6? Explainwhy this happens.

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If we can express the left side of an equation as the square of a binomial,we can proceed as we did in Example 7.

EXAMPLE 8 Solve: .

We have

The left side is the square of a binomial.


.

The solutions are and , or .

Do Exercise 9.

Completing the Square

We can solve quadratic equations like and by using the principle of square roots. We can also solve an equation such as

in like manner because the expression on the left side is the square of a binomial, . This second procedure is the basis fora method called completing the square. It can be used to solve any qua-dratic equation.

Suppose we have the following quadratic equation:

.

If we could add on both sides of the equation a constant that would make theexpression on the left the square of a binomial, we could then solve the equa-tion using the principle of square roots.

How can we determine what to add to to construct the square of a binomial? We want to find a number a such that the following equation is satisfied:

.

Thus a is such that . Solving, we get . That is, a is half of the coefficient of x in . Since , we add 49 to our origi-nal expression:

is the square of ;

that is,

.

COMPLETING THE SQUARE

When solving an equation, to complete the square of an expressionlike , we take half the x-coefficient, which is , and squareit. Then we add that number, , on both sides of the equation.�b2�2

b2x2 � bx

x2 � 14x � 49 � �x � 7�2

x � 7x2 � 14x � 49

a2 � �142 �2 � 72 � 49x2 � 14x

a � 72a � 14

x2 � 14x � a2 � �x � a� �x � a� � x2 � 2ax � a2

x2 � 14x

x2 � 14x � 4

�x � 3�2x2 � 6x � 9 � 2

�x � 2�2 � 73x2 � 6

�3 � �2�3 � �2�3 � �2

x � �3 � �2 or x � �3 � �2

x � 3 � �2 or x � 3 � ��2

�x � 3�2 � 2

x2 � 6x � 9 � 2

x2 � 6x � 9 � 2

9. Solve: .

Answer on page A-46

x2 � 16x � 64 � 11

759


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Returning to solving our original equation, we first add 49 on both sidesto complete the square on the left. Then we solve:

Original equation

Adding 49:


.

The solutions are .We have seen that a quadratic equation can be solved using

the principle of square roots. Any equation, such as , can beput in this form by completing the square. Then we can solve as before.

EXAMPLE 9 Solve: .

We have

. Subtracting 8

We take half of �6 and square it, to get 9. Then we add 9 on both sides of theequation. This makes the left side the square of a binomial, . We havenow completed the square.

Adding 9:


The solutions are 2 and 4.


EXAMPLE 10 Solve by completing the square.

We have

Adding 7

Adding 4:


.

The solutions are .

Do Exercise 12.

�2 � �11

x � �2 � �11 or x � �2 � �11

x � 2 � �11 or x � 2 � ��11

�x � 2�2 � 11

�42�2 � �2�2 � 4 x2 � 4x � 4 � 7 � 4

x2 � 4x � 7

x2 � 4x � 7 � 0

x2 � 4x � 7 � 0

x � 4 or x � 2

x � 3 � 1 or x � 3 � �1

�x � 3�2 � 1

��62 �2 � ��3�2 � 9 x2 � 6x � 9 � �8 � 9

x � 3

x2 � 6x � �8

x2 � 6x � 8 � 0

x2 � 6x � 8 � 0

x2 � 6x � 8 � 0�x � c�2 � d

�7 � �53

x � �7 � �53 or x � �7 � �53

x � 7 � �53 or x � 7 � ��53

�x � 7�2 � 53

�142 �2 � 72 � 49 x2 � 14x � 49 � 4 � 49

x2 � 14x � 4

Solve.

10.

11.

12. Solve by completing the square:

.


x2 � 6x � 1 � 0

x2 � 8x � 20 � 0

x2 � 6x � 8 � 0

760


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When the coefficient of is not 1, we can make it 1, as shown in thefollowing example.


We have

Multiplying and simplifying

Adding

Finding a common denominator

The solutions are


�2

�4

�6

2 4�4 x�6 �2

4

2

6

6

yf(x) � 3x2 � 7x � 2

(� � ��, 0)��673

��76 (� � ��, 0)��

673

��76

�

76

��73

6.

x � �

76

��73

6 or x � �

76

��73

6.

x �76

��73

6 or x �

76

� �

�736

Using the principle ofsquare roots x �

76

� �7336

or x �76

� ��7336

�x �76�2

�7336

�x �76�2

�2436

�4936

� 12

�73 2

�4936

4936

: x2 �73

x �4936

�23

�4936

x2 �73

x �23

13

�3x2 � 7x� �13

� 2

3x2 � 7x � 2

3x2 � 7x � 2

x2 Solve by completing the square.

13.

14.


3x2 � 2x � 7

2x2 � 6x � 5

761


Multiplying by to make thecoefficient 1x2-

13

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Subtracting 3x

Multiplying and simplifying

Finding a common denominator

The solutions are .

We see that the graph of does not cross the x-axis.This is true because the equation has nonreal complex-numbersolutions.

Do Exercise 15.

SOLVING BY COMPLETING THE SQUARE

To solve an equation by completing the square:

1. If , multiply by so that the -coefficient is 1.2. If the -coefficient is 1, add or subtract so that the equation is in

the form

, or if step (1) has been applied.

3. Take half of the x-coefficient and square it. Add the result on bothsides of the equation.

4. Express the side with the variables as the square of a binomial.5. Use the principle of square roots and complete the solution.

x2 �ba

x � �

ca

x2 � bx � �c

x2x21aa � 1

ax2 � bx � c � 0

2x2 � 3x � 7f �x� � 2x2 � 3x � 7

�2 �1 1 2

10

6

4

2

x

y

No x-interceptsf (x) � 2x2 � 3x � 7

34

� i

�474

x �34

�i�47

4 or x �

34

�i�47

4

��1 � i x �34

� i�4716

or x �34

� �i�4716

Using the principleof square roots x �

34

� ��

4716

or x �34

� ��

4716

�x �34�2

� �

4716

�x �34�2

� �

5616

�9

16

x2 �32

x �9

16� �

72

�9

16

x2 �32

x � �72

12

�2x2 � 3x� �12

� ��7�

2x2 � 3x � �7

2x2 � 3x � 7

2x2 � 3x � 715. Solve by completing the square:

Answer on page A-46

3x2 � 2x � 1.

762


Adding : � 12 ��

32� 2

� ��

34 2

�9

169

16

Multiplying by to make the-coefficient 1x2

12

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Completing the square provides a base for proving the quadratic formulain Section 11.2.

Applications and Problem Solving

EXAMPLE 13 Hang Time. One of the most exciting plays in basketball isthe dunk shot. The amount of time T that passes from the moment a playerleaves the ground, goes up, makes the shot, and arrives back on the ground iscalled the hang time. A function relating an athlete’s vertical leap V, in inches,to hang time T, in seconds, is given by

.

a) Hall-of-Famer Michael Jordan has a hang time of about 0.889 sec. What ishis vertical leap?

b) Although his height is only 5 ft 7 in., Spud Webb, formerly of the Sacra-mento Kings, had a vertical leap of about 44 in. What is his hang time?Source: Peter Brancazio, “The Mechanics of a Slam Dunk,” Popular Mechanics, November 1991.Courtesy of Professor Peter Brancazio, Brooklyn College.

a) To find Jordan’s vertical leap, we substitute 0.889 for T in the function andcompute V :

in.

Jordan’s vertical leap is about 37.9 in. Surprisingly, Jordan does not have thevertical leap most fans would expect.

b) To find Webb’s hang time, we substitute 44 for V and solve for T :

Substituting 44 for V

Solving for

Hang time is positive.

Using a calculator

Webb’s hang time is 0.957 sec. Note that his hang time is greater than Jordan’s.


0.957 � T.

�0.916 � T

0.916 � T 2

T 2 4448

� T 2

44 � 48T 2

V�0.889� � 48�0.889�2 � 37.9

V

V�T � � 48T 2

16. Hang Time. Carl Landry, ofPurdue University, has a hangtime of about 0.835 sec. What ishis vertical leap?

17. Hang Time. The record for thegreatest vertical leap in the NBAis held by Darryl Griffith of theUtah Jazz. It was 48 in. What washis hang time?


763


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764


Co

pyr

igh

t ©

200

7 P

ears

on

Ed

uca

tio

n, I

nc.

11.111.1 Student’sSolutionsManual

Digital VideoTutor CD 6

Videotape 11

Math TutorCenter

InterActMath

MyMathLabMathXLEXERCISE SET For Extra Help

1. a) Solve:.

b) Find the x-intercepts of

.

�4 4 x

y

f (x) � 6x2 � 30�30

�20

�10

5

? ?

f �x� � 6x2 � 30

6x2 � 302. a) Solve:

.b) Find the

x-intercepts of.

�4 2 4 x

y

f (x) � 5x2 � 35�30

�20

�10

5

f �x� � 5x2 � 35

5x2 � 353. a) Solve:

.b) Find the

x-intercepts of.

�2 �1 1 2 x

y

10

20

30

40

f (x) � 9x2 � 25

f �x� � 9x2 � 25

9x2 � 25 � 04. a) Solve:

.b) Find the

x-intercepts of.

�1 1 x

y100

80

60

40

20

f (x) � 36x2 � 49

f �x� � 36x2 � 49

36x2 � 49 � 0

Solve. Give the exact solution and approximate solutions to three decimal places, when appropriate.

5. 6. 7. 8. �x � 1�2 � 6�x � 2�2 � 493x2 � 7 � 02x2 � 3 � 0

9. 10. 11. 12. �x � 9�2 � 34�x � 11�2 � 7�x � 3�2 � 9�x � 4�2 � 16

13. 14. 15. 16. �t � 2�2 � 25�x � 9�2 � 81�x � 1�2 � �9�x � 7�2 � �4

17. 18. 19. x2 � 6x � 9 � 64� y �34�2 � 17

16�x �32�2 � 7

2

20. 21. 22. p2 � 8p � 16 � 1y2 � 14y � 49 � 4x2 � 10x � 25 � 100

Solve by completing the square. Show your work.

23. 24. 25. 26. x2 � 18x � 10x2 � 22x � 11x2 � 2x � 5x2 � 4x � 2

27. x2 � x � 1 28. x2 � x � 3 29. t 2 � 5t � 7 30. y2 � 9y � 8

31. x2 �32 x � 3 32. x2 �

43 x � 2

3 33. m2 �92 m � 3

2 34. r 2 �25 r � 4

5

35. x2 � 6x � 16 � 0 36. x2 � 8x � 15 � 0

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765

Exercise Set 11.1

37. 38. 39. 40. x2 � 10x � 4 � 0x2 � 10x � 4 � 0x2 � 18x � 74 � 0x2 � 22x � 102 � 0

41. a) Solve: .

b) Find the x-intercepts of

�8 �6 �4 �2

4

�4

x

y

�12

? ?

f (x) � x2 � 7x � 2

f �x� � x2 � 7x � 2.

x2 � 7x � 2 � 042. a) Solve:

.b) Find the

x-intercepts of

2 864

�8

�4

x

y

�12

f (x) � x2 � 7x � 2

f �x� � x2 � 7x � 2.

x2 � 7x � 2 � 043. a) Solve:

.b) Find the

x-intercepts of

�4 �2 2 4

8

4

x

y

f (x) � 2x2 � 5x � 8

f �x� � 2x2 � 5x � 8.

2x2 � 5x � 8 � 044. a) Solve:

.b) Find the

x-intercepts of

�1 1 32

8

4

x

y

16

f (x) � 2x2 � 3x � 9

f �x� � 2x2 � 3x � 9.

2x2 � 3x � 9 � 0

Solve by completing the square. Show your work.

45. 46. 47. 48. 2x2 � 3x � 1 � 02x2 � 3x � 17 � 0x2 �32 x � 2 � 0x2 �

32 x �

12 � 0

49. 50. 51. 52. x2 � x � 1 � 0x2 � x � 2 � 03x2 � 4x � 3 � 03x2 � 4x � 1 � 0

53. x2 � 4x � 13 � 0 54. x2 � 6x � 13 � 0

Hang Time. For Exercises 55 and 56, use the hang-time function , relating vertical leap to hang time.V�T � � 48T 2

55. The NBA’s Vince Carter, of the New Jersey Nets, has avertical leap of about 36 in. What is his hang time?

56. Tracy McGrady, of the Houston Rockets, has a verticalleap of 40 in. What is his hang time?

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59. Reaching 745 ft above the water, the towers ofCalifornia’s Golden Gate Bridge are the world’s tallestbridge towers. How long would it take an object to fallfreely from the top?Source: The Guinness Book of Records

60. The Gateway Arch in St. Louis is 640 ft high. How longwould it take an object to fall freely from the top?

61. Explain in your own words a sequence of steps thatyou might follow to solve any quadratic equation.

62. Write a problem involving hang time for aclassmate to solve. (See Example 13.) Devise theproblem so that the solution is “The player’s hang timeis about 0.935 sec.”

DWDW

63. Tattoo Removal. The following table lists dataregarding the number of people who visited a doctor for tattoo removal in 1996 and in 2000. [7.5e]

a) Use the two data points to find a linear functionthat fits the data.

b) Use the function to estimate the number of people who will visit a doctor in 2008 to have a tattoo removed.

c) In what year will 1,085,000 people visit a doctor fortattoo removal?

T�t� � mt � b

Free-Falling Objects. The function is used to approximate the distance s, in feet, that an object falls freely from restin t seconds. Use the formula for Exercises 57–60.

s�t� � 16t2

SKILL MAINTENANCE

NUMBER OF PEOPLE WHO VISITED A DOCTOR FOR TATTOO REMOVAL

(in thousands)NUMBER OF YEARS

SINCE 1995

Source: Star-Tribune, Minneapolis–St. Paul

57. Suspended 1053 ft above the water, the bridge overColorado’s Royal Gorge is the world’s highest bridge.How long would it take an object to fall freely from thebridge?Source: The Guinness Book of Records

58. The CN Tower in Toronto, at 1815 ft, is the world’s tallestself-supporting tower (no guy wires). How long would ittake an object to fall freely from the top?Source: The Guinness Book of Records

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767

Exercise Set 11.1

Graph. [7.1c], [7.4a]

64. 65. 66. 67.

x

y

f �x� � �5 � 2x�

x

y

2x � 5y � 10

x

y

f �x� � 5 � 2x

x

y

f �x� � 5 � 2x2

68. Simplify: . [10.3a] 69. Rationalize the denominator: [10.5a]� 25

.�88

74. Use a graphing calculator to solve each of thefollowing equations.

a)b)c)

75. Problems such as those in Exercises 17, 21, and 25can be solved without first finding standard form byusing the INTERSECT feature on a graphing calculator. Welet the left side of the equation and the rightside. Use a graphing calculator to solve Exercises 17, 21,and 25 in this manner.

y2 �y1 �

2.101x � 3.121 � 0.97x2�0.0644x2 � 0.0936x � 4.56 � 025.55x2 � 1635.2 � 0

Find b such that the trinomial is a square.

76. 77. x2 � bx � 64x2 � bx � 75

Solve.

78. 79. x�2x2 � 9x � 56� �3x � 10� � 0�x �13� �x �

13� � �x �

13� �x �

29� � 0

80. Boating. A barge and a fishing boat leave a dock at the same time, traveling at right angles to each other.The barge travels 7 slower than the fishing boat.After 4 hr, the boats are 68 km apart. Find the speed ofeach vessel.

68 km

kmh

SYNTHESIS

Solve. [10.6a, b]

70. 71. �4x � 4 � �x � 4 � 1�5x � 4 � �13 � x � 7

72. 73. �35 � �2x � 5�7x � 5 � �4x � 7

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There are at least two reasons for learning to complete the square. One isto enhance your ability to graph certain equations that are needed to solveproblems in Section 11.7. The other is to prove a general formula for solvingquadratic equations.

Solving Using the Quadratic Formula

Each time you solve by completing the square, the procedure is the same.When we do the same kind of procedure many times, we look for a formula tospeed up our work. Consider

Note that if we can get an equivalent form with by first multiply-ing by �1.

Let’s solve by completing the square. As we carry out the steps, comparethem with Example 12 in the preceding section.

Multiplying by

Subtracting

Half of is The square is We add on both sides:

Adding

.

Since so we can simplify as follows:

Thus,

We now have the following.

THE QUADRATIC FORMULA

The solutions of are given by

x ��b � �b2 � 4ac

2a.

ax2 � bx � c � 0

x � �

b2a

��b2 � 4ac

2a, or x �

�b � �b2 � 4ac2a

.

x �b

2a�

�b2 � 4ac2a

or x �b

2a� �

�b2 � 4ac2a

.

�4a2 � 2a,a � 0,

Using the principleof square rootsx �

b2a

� �b2 � 4ac4a2

or x �b

2a� ��b2 � 4ac

4a2

�x �b

2a�2

�b2 � 4ac

4a2

Factoring the left side and finding acommon denominator on the right �x �

b2a�2

� �

4ac4a2 �

b2

4a2

b2

4a2 x2 �ba

x �b2

4a2 � �

ca

�b2

4a2

b2

4a2b2

4a2 .b

2a.

ba

ca

x2 �ba

x � �

ca

1a

x2 �ba

x �ca

� 0

a � 0a � 0,

ax2 � bx � c � 0, a � 0.

11.211.2 THE QUADRATIC FORMULAObjectiveSolve quadratic equationsusing the quadratic formula,and approximate solutionsusing a calculator.

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The formula also holds when A similar proof would show this, butwe will not consider it here.

ALGEBRAIC–GRAPHICAL CONNECTION

The Quadratic Formula (Algebraic). The solutions of are given by

The Quadratic Formula(Graphical).The x-intercepts of the graph of the function

if they exist, are given by

EXAMPLE 1 Solve using the quadratic formula.

We first find standard form and determine a, b, and c:

We then use the quadratic formula:

Substituting

The solutions are and �1.� 35

x � �

35

or x � �1.

x ��610

or x ��1010

x ��8 � 2

10or x �

�8 � 210

x ��8 � 2

10

x ��8 � �4

10

x ��8 � �64 � 60

10

x ��8 � �82 � 4 � 5 � 3

2 � 5

x ��b � �b2 � 4ac

2a

a � 5, b � 8, c � 3.

5x2 � 8x � 3 � 0;

5x2 � 8x � �3

��b � �b2 � 4ac2a

, 0�.

a � 0,f �x� � ax2 � bx � c,

x ��b � �b2 � 4ac

2a.

a � 0,ax2 � bx � c � 0,

a � 0.

769


x

y f (x) � ax 2 � bx � c,a . 0

�b � �b 2 � 4ac2a

, 0�� b � �b 2 � 4ac2a

, 0��

AG

�1.5 0.5

�4

4

�2

�4

4

�2

x

y

(�1, 0)

(�E, 0)

f (x) � 5x2 � 8x � 3

Be sure to write the fractionbar all the way across.

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It turns out that we could have solved the equation in Example 1 moreeasily by factoring as follows:

To solve a quadratic equation:

1. Check for the form or If it is in this form, usethe principle of square roots as in Section 11.1.

2. If it is not in the form of step (1), write it in standard formwith a and b nonzero.

3. Then try factoring.4. If it is not possible to factor or if factoring seems difficult, use the

quadratic formula.

The solutions of a quadratic equation cannot always be found byfactoring. They can always be found using the quadratic formula.

The solutions to all the exercises in this section could also be found bycompleting the square. However, the quadratic formula is the preferredmethod when solving applied problems because it is faster.

Do Exercise 1.

We will see in Example 2 that we cannot always rely on factoring.

EXAMPLE 2 Solve: Give the exact solution and approxi-mate the solutions to three decimal places.

We first find standard form and determine a, b, and c:

We then use the quadratic formula,

Substituting

Caution!

To avoid a common error in simplifying, remember to factor the numeratorand the denominator and then remove a factor of 1.

We can use a calculator to approximate the solutions:

4 � �315

1.914;4 � �31

5 �0.314.

�8 � 2�31

10�

2�4 � �31 �2 � 5

�22

�4 � �31

5�

4 � �315

.

�8 � �64 � 60

10�

8 � �12410

�8 � �4 � 31

10

x ��8� � ��8�2 � 4 � 5 � ��3�

2 � 5

x ��b � �b2 � 4ac

2a:

a � 5, b � �8, c � �3.

5x2 � 8x � 3 � 0;

5x2 � 8x � 3.

ax2 � bx � c � 0

�x � c�2 � d.x2 � d

x � � 35 or x � �1.

5x � �3 or x � �1

5x � 3 � 0 or x � 1 � 0

�5x � 3� �x � 1� � 0

5x2 � 8x � 3 � 0

1. Consider the equation

a) Solve using the quadraticformula.

b) Solve by factoring.


2x2 � 4 � 7x.

770


CALCULATORCORNER

Approximating Solutions of Quadratic Equations InExample 2, we found that thesolutions of the equation

are

and We can use a

calculator to approximate these solutions. To

approximate we

press ( 4 a F +

3 1 ) ) d 5

[. To approximate

we press ( 4

c F + 3 1 )

) d 5 [. We see that the solutions areapproximately 1.914 and�0.314.

Exercises: Use a calculatorto approximate the solutions ineach of the following. Round tothree decimal places.

1. Example 4

2. Margin Exercise 2

3. Margin Exercise 4

1.913552873

�.3135528726

(4�√ (31))/5

(4�√ (31))/5

4 � �315

,

4 � �315

,

4 � �315

.

4 � �315

5x2 � 8x � 3

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Check: Checking the exact solutionscan be quite cumbersome. It

could be done on a calculator or by using theapproximations. Here we check 1.914; thecheck for �0.314 is left to the student.

For 1.914:

APPROXIMATELY TRUE

We do not have a perfect check due to the rounding error. But our checkseems to confirm the solutions.

Do Exercise 2.

Some quadratic equations have solutions that are nonreal complexnumbers.

EXAMPLE 3 Solve:

We have We use the quadratic formula:

The solutions are

The solutions can also be expressed in the form

Do Exercise 3.

EXAMPLE 4 Solve: Give the exact solution and approximate

solutions to three decimal places.

We first find standard form:

. Subtracting 5 2x2 � 7x � 5 � 0

2x2 � 7x � 5

x2�2 �7x � � x2 �

5x2

2 �7x

�5

x2 .

�12

� i�3

2and �

12

� i�3

2.

�1 � i�32

and�1 � i�3

2.

��1 � i�3

2.

��1 � ��3

2

��1 � �1 � 4

2

x ��1 � �12 � 4 � 1 � 1

2 � 1

c � 1.b � 1,a � 1,

x2 � x � 1 � 0.

3.00498 5�3.663396� � 15.312 5�1.914�2 � 8�1.914� ? 3

5x2 � 8x � 3

�4 � �31 �52. Solve using the quadratic

formula:

Give the exact solution andapproximate solutions to threedecimal places.

3. Solve:


x2 � x � 2 � 0.

3x2 � 2x � 7.

771


�1 1 32

�6

�4

4

2

�2

x

y

f (x) � 5x2 � 8x � 3

(�0.314, 0) (1.914, 0)

�4 �2 2 4

�4

4

2

�2

x

y

No x-intercepts

f (x) � x2 � x � 1

Multiplying by to clear fractions,noting that x � 0

x2

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Then

Substituting

The quadratic formula always gives correct results when we begin withthe standard form. In such cases, we need check only to detect errors in sub-stitution and computation. In this case, since we began with a rational equa-tion, which was not in standard form, we do need to check. We cleared thefractions before obtaining standard form, and this step could introduce num-bers that do not check in the original equation. At the very least, we need toshow that neither of the numbers makes a denominator 0. Since neither ofthem does, the solutions are

We can use a calculator to approximate the solutions:

Do Exercise 4.

�7 � �89

4 �4.108.

�7 � �89

4 0.608;

�7 � �894

and�7 � �89

4.

x ��7 � �89

4or x �

�7 � �894

.

x ��7 � �49 � 40

4�

�7 � �894

x ��7 � �72 � 4 � 2 � ��5�

2 � 2

a � 2, b � 7, c � �5

4. Solve:

Give the exact solution andapproximate solutions to threedecimal places.

Answer on page A-47

3 �5x

�4

x2 .

772


�8 �4 4 8

4

2

x

y

(�4.108, 0)

(0.608, 0)

f (x) � 2 � �7x

5x2

CALCULATOR CORNER

Visualizing Solutions of Quadratic Equations The graph of in Example 2 has twox-intercepts. This indicates that the equation (or, equivalently, ) has two real-number solutions. The graph of in Example 3 has no x-intercepts. This indicates that the equation

has no real-number solutions.

Exercises:1. Explain how the graph of shows that the equation has no real-number

solutions.

2. Use a graph to determine whether the equation has real-number solutions. If so, how manyare there?

3. Use a graph to determine whether the equation has real-number solutions. If so, howmany are there?

4. Use a graph to determine whether the equation has real-number solutions. If so, howmany are there?

4x2 � 1 � 4x

x2 � x � 2 � 0

x2 � x � 1

x2 � x � 2 � 0y � x2 � x � 2

x2 � x � 1 � 0f �x� � x2 � x � 1

5x2 � 8x � 35x2 � 8x � 3 � 0f �x� � 5x2 � 8x � 3

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CALCULATOR CORNER

Solving Quadratic Equations A quadratic equation written with 0 on one side of the equals sign can be solvedusing the ZERO feature of a graphing calculator. See the Calculator Corner on p. 371 for the procedure.

We can also use the INTERSECT feature to solve a quadratic equation. Consider the equation in Exercise 19 inExercise Set 11.2: First, we enter and on theequation-editor screen and graph the equations in a window that shows the point(s) of intersection of the graphs.We use the window We see that there are two points of intersection, so the equation has twosolutions.

We use the INTERSECT feature to find the coordinates of the left-hand point of intersection. (See the CalculatorCorner on p. 550 for the procedure.) The first coordinate of this point, �2, is one solution of the equation. We usethe INTERSECT feature again to find the other solution, �1.

Note that we could use the ZERO feature to solve this equation if we first write it with 0 on one side:

Exercises: Solve.

1.

2. 2x2 � 15 � 7x

5x2 � �11x � 12

5x�x � 1� � 2 � 0.4x�x � 2� �

Y�2

3�5

4

IntersectionX��1

y1 � 4x(x � 2) �5x(x � 1), y2 � 2

�2Y�2

3�5

4

IntersectionX��2

y1 � 4x(x � 2) �5x(x � 1), y2 � 2

�2

��5, 3, �2, 4�.

y2 � 2y1 � 4x�x � 2� � 5x�x � 1�4x�x � 2� � 5x�x � 1� � 2.

3.

4. �x � 1� �x � 4� � 3�x � 4�6�x � 3� � �x � 3� �x � 2�

Study Tips BEGINNING TO STUDY FOR THE FINAL EXAM (PART 2)

The best scenario forpreparing for a final

exam is to do so over aperiod of at least two

weeks. Work in adiligent, disciplined

manner, doingsome final-exam

preparation each day.Here is a detailed planthat many find useful.

1. Begin by browsing through each chapter, reviewing the highlighted or boxed informationregarding important formulas in both the text and the Summary and Review. Theremay be some formulas that you will need to memorize.

2. Retake each chapter test that you took in class, assuming your instructor has returned it.Otherwise, use the chapter test in the book. Restudy the objectives in the text thatcorrespond to each question you missed.

3. If you are still missing questions, use the supplements for extra review. For example,you might check out the videotapes or audio recordings, the Student’s Solutions Manual,the InterAct Math Tutorial Web site, or MathXL.

4. For remaining difficulties, see your instructor, go to a tutoring session, or participate in astudy group.

5. Take the Final Examination in the text during the last couple of days before the final.Set aside the same amount of time that you will have for the final. See how much of thefinal exam you can complete under test-like conditions.

“The door of opportunity won’t open unless you do some pushing.”

Anonymous

773


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Videotape 11

Math TutorCenter

InterActMath


1. 2. 3. 4. 3u2 � 18u � 63p2 � �8p � 1x2 � 6x � 4 � 0x2 � 8x � 2 � 0

5. 6. 7. 8. x2 � 13 � 6xx2 � 13 � 4xx2 � x � 2 � 0x2 � x � 1 � 0

9. 10. 11. 12. 1 �5

x2 �2x

1 �2x

�5

x2 � 0h2 � 4 � 6hr 2 � 3r � 8

13. a) Solve:b) Find the x-intercepts of



f �x� � 11x2 � 3x � 5.

11x2 � 3x � 5 � 0.

f �x� � 4x � x�x � 3�.

4x � x�x � 3� � 0.

f �x� � 3x � x�x � 2�.

3x � x�x � 2� � 0.




f �x� � 49x2 � 14x � 1.

49x2 � 14x � 1 � 0.

f �x� � 25x2 � 20x � 4.

25x2 � 20x � 4.

f �x� � 7x2 � 8x � 2.

7x2 � 8x � �2.

Solve.

19. 20. 3x�x � 1� � 7x�x � 2� � 64x�x � 2� � 5x�x � 1� � 2

21. 22. 11�x � 2� � �x � 5� � �x � 2� �x � 6�14�x � 4� � �x � 2� � �x � 2� �x � 4�

23. 5x2 � 17x � 2 24. 15x � 2x2 � 16 25. x2 � 5 � 4x 26. x2 � 5 � 2x

27. 28.3x

�x3

�52

x �1x

�136

29.1y

�1

y � 2�

13

30.1x

�1

x � 4�

17

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775

Exercise Set 11.2

31. 32. 33. �x � 2�2 � �x � 1�2 � 02y2 � � y � 2� � y � 3� � 12�2t � 3�2 � 17t � 15

37. x2 � 6x � 4 � 0 38. x2 � 4x � 7 � 0 39. 40. x2 � 4x � 1 � 0x2 � 6x � 4 � 0

41. 2x2 � 3x � 7 � 0 42. 3x2 � 3x � 2 � 0 43. 5x2 � 3 � 8x 44. 2y2 � 2y � 3 � 0

34. 35.(Hint : Factor the difference ofcubes. Then use the quadraticformula.)

36. x3 � 27 � 0x3 � 1 � 0�x � 3�2 � �x � 1�2 � 0

Solve. Give the exact solution and approximate solutions to three decimal places.

45. The list of steps on p. 770 does not mentioncompleting the square as a method of solving quadraticequations. Why not?

46. Given the solutions of a quadratic equation, is itpossible to reconstruct the original equation? Why orwhy not?

DWDW

Solve. [10.6a, b]

47. 48. 49. �x � 2 � �2x � 8x � �15 � 2xx � �x � 2

50. 51. 52. �2x � 6 � 11 � 2�x � 5 � �7�x � 1 � 2 � �3x � 1

53. 54. �4 3x � 1 � 2�3 4x � 7 � 2

SKILL MAINTENANCE

55. Use a graphing calculator to solve the equa-tions in Exercises 3, 16, 17, and 37 using the INTERSECT

feature, letting the left side and the right side. Then solve

56. Use a graphing calculator to solve the equations inExercises 9, 27, and 30. Then solve 5.33x2 � 8.23x � 3.24.

2.2x2 � 0.5x � 1 � 0.y2 �y1 �

Solve.

57. 58.5x

�x4

�117

2x2 � x � �5 � 0

59. 60. �3x2 � 6x � �3 � 0ix2 � x � 1 � 0

61. 62. �1 � �3 �x2 � �3 � 2�3 �x � 3 � 0x

x � 1� 4 �

13x2 � 3

63. Let Find all inputs x such that 64. Let Find all inputs x such thatf�x� � 36.

f�x� � x2 � 14x � 49.f�x� � 13.

f�x� � �x � 3�2.

SYNTHESIS

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776


Applications and Problem Solving

Sometimes when we translate a problem to mathematical language, the resultis a quadratic equation.

EXAMPLE 1 Box Construction. An open box is to be made from a 10-ft by20-ft rectangular piece of cardboard by cutting a square from each corner. Thearea of the bottom of the box is to be What is the length of the sides ofthe squares that are cut from the corners?

1. Familiarize. We first make a drawing and label it with the knowninformation. Since we do not know the length of the sides of the squaresthat are cut from the corners, we have called that length x.

2. Translate. Remember, the area of a rectangle is lw (length timeswidth). We find two expressions for the area of the bottom of the box.First, we know from the statement of the problem that

the area of the bottom of the box is

Second, the width of the bottom of the box is 10 ft minus twice the lengthof the squares cut from the corners, or The length of the bottomof the box is 20 ft minus twice the length of the squares cut from the cor-ners, or Then we can multiply the expressions to find anotherexpression for the area of the bottom of the box:

These expressions give us the equation

3. Solve. We solve the equation:

Using FOIL on the left

Obtaining 0 on the right andcollecting like termsDividing by 4

Factoring

Using the principle of zeroproducts

4. Check. We check in the original problem. We see that 13 is not a solutionbecause 13 is longer than the width of the piece of cardboard.

If the length of a side of the squares is 2, then the bottom of the boxwill have width or 6 ft. The length will be or 16 ft.The area is or

5. State. The length of the sides of the squares is 2 ft.

Do Exercise 1.

96 ft2.6 � 16,20 � 2 � 2,10 � 2 � 2,

x � 2 or x � 13.

�x � 2� �x � 13� � 0

x2 � 15x � 26 � 0

4x2 � 60x � 104 � 0

200 � 20x � 40x � 4x2 � 96

�10 � 2x� �20 � 2x� � 96

�10 � 2x� �20 � 2x� � 96.

�10 � 2x� �20 � 2x�.

20 � 2x.

10 � 2x.

96 ft2.

10 ft

x x

x x

x

x

x

x

20 ft

10 � 2x

20 � 2x

96 ft2

96 ft2.

11.311.3 APPLICATIONS INVOLVINGQUADRATIC EQUATIONS

1. Landscaping. A rectangulargarden is 60 ft by 80 ft. Part ofthe garden is torn up to install asidewalk of uniform widtharound it. The area of the newgarden is of the old area. Howwide is the sidewalk?

Answer on page A-47

60

Oldgarden

Newgarden

Sidewalk

80

xx

xx x

x

x

x80 � 2x

60 � 2x

12

ObjectivesSolve applied problemsinvolving quadraticequations.

Solve a formula for a givenletter.

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EXAMPLE 2 Town Planning. Three towns A, B, and C are situated asshown. The roads at A form a right angle. The distance from A to B is 2 mi lessthan the distance from A to C. The distance from B to C is 10 mi. Find the dis-tance from A to B and the distance from A to C.

1. Familiarize. We first make a drawing and label it. We let the dis-tance from A to C. Then the distance from A to B is

2. Translate. We see that a right triangle is formed. We can use thePythagorean equation, which we studied in Chapter 10: .In this problem, we have

3. Solve. We solve the equation:

Squaring

Finding standard form

Multiplying by or dividing by 2

Factoring

Using the principle of zeroproducts

4. Check. We know that �6 cannot be a solution because distances are notnegative. If then and

Since the distance 8 mi checks.

5. State. The distance from A to C is 8 mi, and the distance from A to Bis 6 mi.

Do Exercise 2.

EXAMPLE 3 Town Planning. Three towns A, B, and C are situated asshown in Example 2. The roads at A form a right angle. The distance from Ato B is 2 mi less than the distance from A to C. The distance from B to C is8 mi. Find the distance from A to B and the distance from A to C. Find exactand approximate answers to the nearest hundredth of a mile.

102 � 100,

d2 � �d � 2�2 � 82 � 62 � 64 � 36 � 100.

d � 2 � 6,d � 8,

d � 8 or d � �6.

d � 8 � 0 or d � 6 � 0

�d � 8� �d � 6� � 0

12 , d2 � 2d � 48 � 0

2d2 � 4d � 96 � 0

100 � d2 � d2 � 4d � 4

102 � d2 � �d � 2�2

102 � d2 � �d � 2�2.

c2 � a2 � b2

B

A Cd

d � 210 mi

d � 2.d �

A

B

C

2. Ladder Location. A ladderleans against a building, asshown below. The ladder is 20 ftlong. The distance to the top ofthe ladder is 4 ft greater than thedistance d from the building.Find the distance d and thedistance to the top of the ladder.

Answer on page A-47

20 ft

d

d � 4

777


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Using the same reasoning that we did in Example 2, we translate theproblem to the equation

We solve as follows. Note that the quadratic equation we get is not easilyfactored, so we use the quadratic formula:

Squaring

Finding standard form

Multiplying by or dividing by 2

Then

Since and it follows that Using a calculator, we find that and that Thus the distance from A to C is about 6.57 mi, and the distance from A to Bis about 4.57 mi.

Do Exercise 3.

EXAMPLE 4 Motorcycle Travel. Karin’s motorcycle traveled 300 mi at acertain speed. Had she gone 10 mph faster, she could have made the trip in 1 hr less time. Find her speed.

1. Familiarize. We first make a drawing, labeling it with known and unknown information. We can also organize the information, in a table,as we did in Section 6.8. We let the speed, in miles per hour, and the time, in hours.

Recalling the motion formula and solving for r, we get .From the rows of the table, we obtain

and .r � 10 �300

t � 1r �

300t

r � d�td � rt

300 miles

300 miles

t hours

t – 1 hours

r mph

r + 10 mph

t �r �

d � 2 � 4.57 mi.d � 1 � �31 � 6.57 mi,d � 1 � �31.1 � �31 � 0,1 � �31 � 0

�2 � �124

2�

2 � �4�31�2

�2 � 2�31

2� 1 � �31.

Substituting 1 for a, �2 for b,and �30 for c �

��2� � ��2�2 � 4�1� ��30�2�1�

d ��b � �b2 � 4ac

2a

12 , d2 � 2d � 30 � 0.

2d2 � 4d � 60 � 0

64 � d2 � d2 � 4d � 4

82 � d2 � �d � 2�2.

3. Ladder Location. Refer toMargin Exercise 2. Suppose thatthe ladder has length 10 ft. Findthe distance d and the distance

Answer on page A-47

d � 4.

778


300 r t

300 t � 1r � 10

DISTANCE SPEED TIME

r �300

t

r � 10 �300

t � 1

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2. Translate. We substitute for r from the first equation into the secondand get a translation:

.

3. Solve. We solve as follows:

Standard form

Dividing by 10

Factoring

. Using the principleof zero products

4. Check. Since negative time has no meaning in this problem, we try 6 hr.Remembering that , we get mph.

To check, we take the speed 10 mph faster, which is 60 mph, and seehow long the trip would have taken at that speed:

hr.

This is 1 hr less than the trip actually took, so we have an answer.

5. State. Karin’s speed was 50 mph.

Do Exercise 4.

Solving Formulas

Recall that to solve a formula for a certain letter, we use the principles for solv-ing equations to get that letter alone on one side.

EXAMPLE 5 Period of a Pendulum.The time T required for a pendulum of length L to swing back and forth (complete one period) is given by the formula , where g is the gravitational constant. Solve for L.

Clearing fractions

. Multiplying by 1

4�2 gT 2

4�2 � L

gT 2 � 4�2L

T 2 � 22�2 Lg

Principle of powers(squaring) T 2 � �2�� L

g �2

This is a radicalequation (seeSection 10.6).

T � 2�� Lg

T � 2��L�g

t �dr

�30060

� 5

r � 300�6 � 50r � d�t

t � 6 or t � �5

�t � 6� �t � 5� � 0

t2 � t � 30 � 0

10t 2 � 10t � 300 � 0

300�t � 1� � 10�t 2 � t� � 300t

t�t � 1� �300

t� t�t � 1� � 10 � t�t � 1� �

300t � 1

Multiplying bythe LCM t�t � 1� 300

t� 10 � t�t � 1� �

300t � 1

300

t� 10 �

300t � 1

300t

� 10 �300

t � 1

4. Marine Travel. Two shipsmake the same voyage of 3000nautical miles. The faster shiptravels 10 knots faster than theslower one (a knot is 1 nauticalmile per hour). The faster shipmakes the voyage in 50 hr lesstime than the slower one. Findthe speeds of the two ships.

Complete this table to help withthe familiarization.

5. Solve for .


w2A � �w1

w2

779


L

DIS

TA

NC

ESP

EE

DT

IME

Fast

erS

hip

3000

Slo

wer

Shi

p30

00r

t

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We now have L alone on one side and L does not appear on the other side, sothe formula is solved for L.


In most formulas, variables represent nonnegative numbers, so we do notneed to use absolute-value signs when taking square roots.

EXAMPLE 6 Hang Time. An athlete’s hang time is the amount of time thatthe athlete can remain airborne when jumping. A formula relating an athlete’svertical leap V, in inches, to hang time T, in seconds, is . (See Exam-ple 13 in Section 11.1.) Solve for T.

We have

Using the principle of square roots; note that .

.

Do Exercise 6.

EXAMPLE 7 Falling Distance. An object that is tossed downward withan initial speed (velocity) of will travel a distance of s meters, where

and t is measured in seconds. Solve for t.

s

s � 4.9t 2 � v0tv0

��3V

2 � 2 � 3�

�3V12

T � � V2 � 2 � 2 � 2 � 3

�33

T � 0 T � � V48

T 2 �V48

48T 2 � V

V

V � 48T 2

6. Solve for r.

(Volume of a right circularcylinder)

7. Solve for t.


s � gt � 16t 2

r

h

V � �r 2h

780


Multiplying by to get aloneT 2

148

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Since t is squared in one term and raised to the first power in the otherterm, the equation is quadratic in t. The variable is t ; and s are treated asconstants.

We have

Writing standard form

, ,

Since the negative square root would yield a negative value for t, we use onlythe positive root:

.

The steps listed in the margin should help you when solving formulas fora given letter. Try to remember that when solving a formula, you do the samethings you would do to solve any equation.


EXAMPLE 8 Solve for a.

In this case, we could either clear the fractions first or use the principle ofpowers first. Let’s clear the fractions. Multiplying by , we have

.

Now we square both sides and then continue:

Squaring

Getting all -terms together

Factoring out

Dividing by

Taking the square root

Simplifying

You need not rationalize denominators in situations such as this.

Do Exercise 8.

tb

�1 � t 2� a.

� t 2b2

1 � t 2 � a

1 � t 2 t 2b2

1 � t 2 � a2

a2 t 2b2 � a2�1 � t 2�a2 t 2b2 � a2 � t 2a2

t 2a2 � t 2b2 � a2

t 2�a2 � b2� � a2

�t�a2 � b2 �2 � a2

t�a2 � b2 � a

�a2 � b2

t �a

�a2 � b2

t ��v0 � �v 0

2 � 19.6s9.8

t ��v0 � �v 0

2 � 19.6s9.8

.

t ��v0 � �v 0

2 � 4�4.9� ��s�2�4.9�

c � �sb � v0a � 4.9

4.9t 2 � v0t � s � 0

4.9t 2 � v0t � s

v0

8. Solve for b.

Answer on page A-47

b

�a2 � b2� t

781


Using the quadratic formula:

t ��b � �b2 � 4ac

2a

⎫ ⎪ ⎬ ⎪ ⎭

Caution!

Don’t forget to square both t and �a2 � b2.

To solve a formula for aletter, say, b:

1. Clear the fractions anduse the principle ofpowers, as needed, until bdoes not appear in anyradicand or denominator.(In some cases, you mayclear the fractions first,and in some cases youmay use the principle ofpowers first.)

2. Collect all terms with in them. Also collect allterms with b in them.

3. If does not appear, youcan finish by using justthe addition andmultiplication principles.

4. If appears but b doesnot, solve the equation for

. Then take square rootson both sides.

5. If there are termscontaining both b and ,write the equation instandard form and usethe quadratic formula.

b2

b2

b2

b2

b2

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Translate each word problem to anequation or a system of equationsand select a correct translationfrom equations A–O.

A.

B.

C.

D.

E.

F.

G.

H.

I.

J.

K.

L.

M.

N.

O.


800x

�800

x � 2� 10

x � �x � 1� � �x � 2� � 537

752 � 782 � x2

3x � 2

�3

x � 2� 4

752 � x2 � 782

x2 � �x � 1�2 � 72

4x � 2

�4

x � 2� 3

2.24 � 18% � 2.24 � x

x � y � 1170.10x � 0.25y � 26.95,

x � �x � 7� �12 x � 537

2x � 2�x � 7� � 537

x � y � 117x � 25y � 26.95,

x � 18% � x � 2.24

800x

� 10 �800

x � 2

13 � 80 � 100�80 � 2x� �100 � 2x� �

Translatingfor Success

6. Gasoline Prices. One day theprice of gasoline was increased18% to a new price of $2.24 pergallon. What was the originalprice?

7. Triangle Dimensions. Thehypotenuse of a right triangle is7 ft. The length of one leg is 1 ftlonger than the other. Find thelengths of the legs.

8. Rectangle Dimensions. Theperimeter of a rectangle is 537 ft.The width of the rectangle is 7 ftshorter than the length. Find thelength and the width.

9. Guy Wire. A guy wire is 78 ftlong. It is attached to the top ofa 75-ft cellphone tower. How faris it from the base of the pole tothe point where the wire isattached to the ground?

10. Landscaping. A rectangulargarden is 80 ft by 100 ft. Part ofthe garden is torn up to install asidewalk of uniform widtharound it. The area of the newgarden is of the old area. Howwide is the sidewalk?

13

1. Car Travel. Sarah drove her car800 mi to see her friend. Thereturn trip was 2 hr faster at aspeed that was 10 mph more.Find her return speed.

2. Coin Mixture. A collection ofdimes and quarters is worth$26.95. There are 117 coins inall. How many of each coin arethere?

3. Wire Cutting. A 537-in. wire iscut into three pieces. Thesecond piece is 7 in. shorterthan the first. The third is half aslong as the first. How long iseach piece?

4. Marine Travel. The ColumbiaRiver flows at a rate of 2 mph forthe length of a popular boatingroute. In order for a motorizeddinghy to travel 3 mi upriverand return in a total of 4 hr, howfast must the boat be able totravel in still water?

5. Locker Numbers. The numberson three adjoining lockers areconsecutive integers whose sumis 537. Find the integers.

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Solve.

783

Exercise Set 11.3

EXERCISE SET For Extra Help11.311.3 Student’sSolutionsManual


Videotape 11

Math TutorCenter

InterActMath

MyMathLabMathXL

1. Flower Bed. The width of a rectangular flower bed is 7 ft less than the length. The area is 18 . Find thelength and the width.

2. Feed Lot. The width of a rectangular feed lot is 8 m lessthan the length. The area is 20 . Find the length andthe width.

m2ft2

3. Parking Lot. The length of a rectangular parking lot istwice the width. The area is 162 . Find the length andthe width.

4. Computer Part. The length of a rectangular computerpart is twice the width. The area is 242 . Find thelength and the width.

cm2yd2

5. Sailing. The base of a triangular sail is 9 m less than itsheight. The area is 56 . Find the base and the heightof the sail.

6. Parking Lot. The width of a rectangular parking lot is50 ft less than its length. Determine the dimensions ofthe parking lot if it measures 250 ft diagonally.

l

l – 50

Area =56 m2

m2

7. Sailing. The base of a triangular sail is 8 ft less than itsheight. The area is 56 . Find the base and the height ofthe sail.

8. Parking Lot. The width of a rectangular parking lot is51 ft less than its length. Determine the dimensions ofthe parking lot if it measures 250 ft diagonally.

ft2

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9. Picture Framing. The outside of a picture framemeasures 12 cm by 20 cm; 84 of picture shows. Find the width of the frame.

10. Picture Framing. The outside of a picture framemeasures 14 in. by 20 in.; 160 of picture shows. Find the width of the frame.

20 in.

x

x x

x

14 in.

160 in2

in2cm2

11. Landscaping. A landscaper is designing a flowergarden in the shape of a right triangle. She wants 10 ft of a perennial border to form the hypotenuse of the triangle, and one leg is to be 2 ft longer than the other. Find the lengths of the legs.

12. The hypotenuse of a right triangle is 25 m long. Thelength of one leg is 17 m less than the other. Find the lengths of the legs.

10 ft

x + 2x

13. Page Numbers. A student opens a literature book totwo facing pages. The product of the page numbers is 812. Find the page numbers.

14. Page Numbers. A student opens a mathematics bookto two facing pages. The product of the page numbers is 1980. Find the page numbers.

x � 1

x

17. Page Dimensions. The outside of an oversized bookpage measures 14 in. by 20 in.; 100 of printed textshows. Find the width of the margin.

18. Picture Framing. The outside of a picture framemeasures 13 cm by 20 cm; 80 of picture shows. Find the width of the frame.

cm2in2

Solve. Find exact and approximate answers rounded to three decimal places.

15. The width of a rectangle is 4 ft less than the length. Thearea is 10 . Find the length and the width.

16. The length of a rectangle is twice the width. The area is328 . Find the length and the width.cm2ft2

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785

Exercise Set 11.3

19. The hypotenuse of a right triangle is 24 ft long. Thelength of one leg is 14 ft more than the other. Find the lengths of the legs.

20. The hypotenuse of a right triangle is 22 m long. Thelength of one leg is 10 m less than the other. Find the lengths of the legs.

21. Car Trips. During the first part of a trip, Meira’s Hondatraveled 120 mi at a certain speed. Meira then drove another 100 mi at a speed that was 10 mph slower. If the total time for Meira’s trip was 4 hr, what was her speed on each part of the trip?

22. Canoeing. During the first part of a canoe trip, Tim covered 60 km at a certain speed. He then traveled 24 km at a speed that was 4 slower. If the total time for Tim’s trip was 8 hr, what was his speed on each part of the trip?

km�h

23. Car Trips. Petra’s Plymouth travels 200 mi at a certainspeed. If the car had gone 10 mph faster, the trip wouldhave taken 1 hr less. Find Petra’s speed.

24. Car Trips. Sandi’s Subaru travels 280 mi at a certainspeed. If the car had gone 5 mph faster, the trip wouldhave taken 1 hr less. Find Sandi’s speed.

25. Air Travel. A Cessna flies 600 mi at a certain speed. ABeechcraft flies 1000 mi at a speed that is 50 mph faster,but takes 1 hr longer. Find the speed of each plane.

26. Air Travel. A turbo-jet flies 50 mph faster than a super-prop plane. If a turbo-jet goes 2000 mi in 3 hr less timethan it takes the super-prop to go 2800 mi, find thespeed of each plane.

27. Bicycling. Naoki bikes the 40 mi to Hillsboro at acertain speed. The return trip is made at a speed that is 6 mph slower. Total time for the round trip is 14 hr.Find Naoki’s speed on each part of the trip.

28. Car Speed. On a sales trip, Gail drives the 600 mi toRichmond at a certain speed. The return trip is made at a speed that is 10 mph slower. Total time for theround trip is 22 hr. How fast did Gail travel on each partof the trip?

29. Navigation. The current in a typical Mississippi Rivershipping route flows at a rate of 4 mph. In order for abarge to travel 24 mi upriver and then return in a totalof 5 hr, approximately how fast must the barge be ableto travel in still water?

30. Navigation. The Hudson River flows at a rate of 3 mph. A patrol boat travels 60 mi upriver and returnsin a total time of 9 hr. What is the speed of the boat instill water?

DISTANCE SPEED TIME DISTANCE SPEED TIME

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Solve the formula for the given letter. Assume that all variables represent nonnegative numbers.

31. , for s(Surface area of a cube)

s s

s

A � 6s2 32. , for r(Surface area of a sphere)

r

A � 4�r 2 33. , for rF �Gm1m2

r 2

34. for s

(Number of phone calls betweentwo cities)

35. , for c(Einstein’s energy–massrelationship)

36. , for s(Volume of a pyramid)

ss

h

V � 13 s2hE � mc2N �

kQ1Q 2

s2 ,

37. , for b(Pythagorean formula in two dimensions)

a2 � b2 � c2 38. , for c(Pythagorean formula in three dimensions)

a2 � b2 � c2 � d2 39. , for k

(Number of diagonals of a polygonof k sides)

N �k2 � 3k

2

40. , for t

(A motion formula)

41. , for r(Surface area of a cylinder)

42. , for r(Surface area of a cone)

s

r

A � �r 2 � �rs

r

h

A � 2�r 2 � 2�rhs � v0t �gt 2

2

43. , for g

(A pendulum formula)

44. , for L

(An electricity formula)

45. , for H

(Body mass index; see Exercises 71and 72 of Exercise Set 9.1)

I �704.5W

H 2W �� 1LC

T � 2�� Lg

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787

Exercise Set 11.3

46. , for R 47. , for v

(A relativity formula)

48. Solve the formula given in Exercise 47 for c.

m �m0

�1 �v 2

c2

N � p �6.2A2

pR2

49. Explain how Exercises 1–30 can be solvedusing a calculator but without using factoring,completing the square, or the quadratic formula.

50. Explain how the quadratic formula can be used tofactor a quadratic polynomial into two binomials. Use it to factor .5x2 � 8x � 3

DWDW

Add or subtract.

51. [6.4a]1

x � 1�

1x2 � 3x � 2

52. [6.5a]x � 1x � 1

�x � 1

x2 � x � 153. [6.5b]

2x � 3

�x

x � 1�

x2 � 2x2 � 2x � 3

54. Multiply and simplify: . [10.3a] 55. Express in terms of i: . [10.8a]��20�3x2 �3x3

Simplify. [6.6a]

56. 57.

4a2b

3a

�4

b2

3x � 1

1x � 1

�2

x � 1

58. Solve: . 59. Find a when the reciprocal of is .a � 1a � 14

2x � i�

1x � i

�2

x � i

60. Bungee Jumping. Jesse is tied to one end of a 40-melasticized (bungee) cord. The other end of the cord istied to the middle of a train trestle. If Jesse jumps off the bridge, for how long will he fall before the cordbegins to stretch? (See Example 7 and let .)

61. Surface Area. A sphere is inscribed in a cube as shownin the figure below. Express the surface area of thesphere as a function of the surface area S of the cube.

40 m

v0 � 0

62. Pizza Crusts. At Pizza Perfect, Ron can make 100 largepizza crusts in 1.2 hr less than Chad. Together they cando the job in 1.8 hr. How long does it take each to do the job alone?

63. The Golden Rectangle. For over 2000 yr, theproportions of a “golden” rectangle have beenconsidered visually appealing. A rectangle of width wand length l is considered “golden” if

.

Solve for l.

wl

�l

w � l

SKILL MAINTENANCE

SYNTHESIS

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The Discriminant

From the quadratic formula, we know that the solutions and of a quad-ratic equation are given by

and .

The expression is called the discriminant. When using the quad-ratic formula, it is helpful to compute the discriminant first. If it is 0, therewill be just one real solution. If it is positive, there will be two real solutions.If it is negative, we will be taking the square root of a negative number; hencethere will be two nonreal complex-number solutions, and they will be com-plex conjugates.

If the discriminant is a perfect square, we can solve the equation by fac-toring, not needing the quadratic formula.

EXAMPLE 1 Determine the nature of the solutions of .

We have

, , .

We compute the discriminant:

.

There is just one solution, and it is a real number. Since 0 is a perfect square,the equation can be solved by factoring.


We have

, , .

We compute the discriminant:

.

Since the discriminant is negative, there are two nonreal complex-numbersolutions. The equation cannot be solved by factoring because �7 is not aperfect square.

� �7

� 25 � 32

b2 � 4ac � 52 � 4 � 1 � 8

c � 8b � 5a � 1

x2 � 5x � 8 � 0

� 0

� 144 � 144

b2 � 4ac � ��12�2 � 4 � 9 � 4

c � 4b � �12a � 9

9x2 � 12x � 4 � 0

b2 � 4ac

x2 ��b � �b2 � 4ac

2ax1 �

�b � �b2 � 4ac2a

x2x1

11.411.4 MORE ON QUADRATIC EQUATIONSObjectivesDetermine the nature of the solutions of aquadratic equation.

Write a quadratic equationhaving two numbersspecified as solutions.

Solve equations that arequadratic in form.

0 Only one solution; it is a real number Only one

Positive Two different real-number solutions Two different

Negative Two different nonreal complex-number solutions (complex conjugates)

None

DISCRIMINANT b2 � 4ac NATURE OF SOLUTIONS x-INTERCEPTS

Study Tips

TEST PREPARATION:SYSTEMATIC VS. INTENSE STUDY

Let’s consider two ways toprepare for an examscheduled for November 15.

1. Systematic Study : You beginstudying on November 1and continue every dayuntil the day of the test.

2. Intense Study : You waituntil November 14 to beginstudying and “cram” all dayand through the night.

Which of these methodswould produce a better resulton the exam? Research hasshown that best results comefrom a combination of thesetwo methods: Studysystematically well ahead ofthe test, but do intense studythe day before. (This works solong as you don’t stay up allnight.) Try this on your nextfew exams.

788


1�0.5

4

3

2

1

x

y

One x-intercept

f (x) � 9x2 � 12x � 4

�4 �2 2 4

8

4

2

x

y

No x-interceptsf (x) � x2 � 5x � 8

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We have

, , ;

.

Since the discriminant is positive, there are two solutions, and they are real numbers. The equation can be solved by factoring since the discriminant is a perfect square.

The discriminant, , tells us how many real-number solutions theequation has, so it also indicates how many x-intercepts thegraph of has. Compare the following.

Do Exercises 1–3.

Writing Equations from Solutions

We know by the principle of zero products that has solu-tions 2 and �3. If we know the solutions of an equation, we can write theequation, using this principle in reverse.

EXAMPLE 4 Find a quadratic equation whose solutions are 3 and .

We have

Getting the 0’s on one side

Clearing the fraction

Using the principle of zero products in reverse

. Using FOIL 5x2 � 13x � 6 � 0

�x � 3� �5x � 2� � 0

x � 3 � 0 or 5x � 2 � 0

x � 3 � 0 or x �25 � 0

x � 3 or x � �25

�25

�x � 2� �x � 3� � 0

�4 4

4

6�6

6

�2

�4

�2 2 �4 4

4

6�6

2

6

�2

�4

�2 2 x

y

�4 4

4

6�6

2

6

�4

�2 2

f (x) � x 2 � 4x � 4b 2 � 4ac � 0

One real solutionOne x-intercept

f (x) � x 2 � 4x � 6b 2 � 4ac � �8 � 0No real solutionsNo x-intercepts

f (x) � x 2 � 2b 2 � 4ac � 8 � 0

Two real solutionsTwo x-intercepts

x

y

x

y

f �x� � ax2 � bx � cax2 � bx � c � 0

b2 � 4ac

b2 � 4ac � 52 � 4 � 1 � 6 � 1

c � 6b � 5a � 1

x2 � 5x � 6 � 0 Determine the nature of thesolutions without solving.

1.

2.

3.


3x2 � 2x � 1 � 0

9x2 � 6x � 1 � 0

x2 � 5x � 3 � 0

789


�4 �2 2 4

8

6

2

x

y

Two x-interceptsf (x) � x2 � 5x � 6

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EXAMPLE 5 Write a quadratic equation whose solutions are 2i and �2i.

We have


Using the principle of zero products in reverse

Using

.

EXAMPLE 6 Write a quadratic equation whose solutions are and.

We have



Using FOIL

. Collecting like terms

Do Exercises 4–7.

Equations Quadratic in Form

Certain equations that are not really quadratic can still be solved as quadratic.Consider this fourth-degree equation.

Thinking of as

To make this clearer, write u instead of .

The equation can be solved by factoring or by the quadraticformula. After that, we can find x by remembering that . Equationsthat can be solved like this are said to be quadratic in form, or reducible toquadratic.

EXAMPLE 7 Solve: .

Let . Then we solve the equation found by substituting u for :

Factoring


. u � 8 or u � 1

u � 8 � 0 or u � 1 � 0

�u � 8� �u � 1� � 0

u2 � 9u � 8 � 0

x2u � x2

x4 � 9x2 � 8 � 0

x2 � uu2 � 9u � 8 � 0

x2 u2 � 9u � 8 � 0

�x2�2x4 �x2�2 � 9�x2� � 8 � 0

x4 � 9x2 � 8 � 0

x2 � �3 x � 6 � 0

x2 � 2�3 x � �3 x � 2��3 �2 � 0

�x � �3 � �x � 2�3 � � 0

x � �3 � 0 or x � 2�3 � 0

x � �3 or x � �2�3

�2�3�3

x2 � 4 � 0

x2 � 4��1� � 0

x2 � 4i2 � 0

�A � B� �A � B� � A2 � B2 x2 � �2i �2 � 0

�x � 2i � �x � 2i � � 0

x � 2i � 0 or x � 2i � 0

x � 2i or x � �2i

Find a quadratic equation havingthe following solutions.

4. 7 and �2

5. �4 and

6. 5i and �5i

7. and


�2�2�2

53

790


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Next, we substitute for u and solve these equations:

.

Note that when a number and its opposite are raised to an even power, theresults are the same. Thus we can make one check for and one for �1.

Check:

For :

TRUE 0 64 � 9 � 8 � 8

��2�2 �4 � 9��2�2 �2 � 8 ? 0

x4 � 9x2 � 8 � 0

�2�2

�2�2

x � �2�2 or x � �1

x � ��8 or x � �1

x2 � 8 or x2 � 1

x2 8. Solve: .

9. Solve . Be sureto check.


x � 3�x � 10 � 0

x4 � 10x2 � 9 � 0

791


For :

TRUE 0 1 � 9 � 8

��1�4 � 9��1�2 � 8 ? 0

x4 � 9x2 � 8 � 0

�1

The solutions are 1, �1, , and .

Caution!

A common error is to solve for u and then forget to solve for x. Rememberthat you must find values for the original variable!

Do Exercise 8.

Solving equations quadratic in form can sometimes introduce numbersthat are not solutions of the original equation. Thus a check by substitutionin the original equation is necessary.

EXAMPLE 8 Solve: .

Let . Then we solve the equation found by substituting u for and for x.


.

Squaring the first equation, we get . Squaring the second equation, weget . We check both solutions.

Check:

For 16:

TRUE 0 16 � 12 � 4

16 � 3 � 4 � 4 16 � 3�16 � 4 ? 0

x � 3�x � 4 � 0

x � 1x � 16

�x � 4 or �x � �1

�x

u � 4 or u � �1

�u � 4� �u � 1� � 0

u2 � 3u � 4 � 0

u2�xu � �x

x � 3�x � 4 � 0

�2�22�2

For 1:

FALSE �6 1 � 3 � 1 � 4 1 � 3�1 � 4 ? 0

x � 3�x � 4 � 0

The solution is 16.

Do Exercise 9.

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EXAMPLE 9 Solve: .

Let . Then we solve the equation found by substituting u for and for :

.

Next, we substitute or for u and solve these equations:

.

Solving, we get

or .

The numbers and �1 both check. They are the solutions.

Do Exercise 10.

EXAMPLE 10 Find the x-intercepts of the graph of

.

The x-intercepts occur where so we must have

.

Let . Then we solve the equation found by substituting u for:

.


.

The numbers , , and 0 check. They are the solutions of . Thus the x-intercepts of the graph of are ,

, and .

Do Exercise 11.

(�√3, 0)

2

4

6

�2

(0, 0) (√3, 0)

f (x) � (x 2 � 1)2 � (x 2 � 1) � 2

y

�2 �1 1 2 x

��3, 0��0, 0��3, 0�f �x��x2 � 1� � 2 � 0

�x2 � 1�2 ��3�3

x � ��3 or x � 0

x2 � 3 or x2 � 0

x2 � 1 � 2 or x2 � 1 � �1

x2 � 1

u � 2 or u � �1

�u � 2� �u � 1� � 0

u2 � u � 2 � 0

x2 � 1u � x2 � 1

�x2 � 1�2 � �x2 � 1� � 2 � 0

f �x� � 0

�x2 � 1� � 2f �x� � �x2 � 1�2 �

12

y �1

��1�� 1y �

12

1y

� 2 or1y

� �1

1�yy�1

u � 2 or u � �1

�u � 2� �u � 1� � 0

u2 � u � 2 � 0

y�2u2y�1u � y�1

y�2 � y�1 � 2 � 010. Solve: .

11. Find the x-intercepts of

.


f�x� � �x2 � x�2 � 14�x2 � x� � 24

x�2 � x�1 � 6 � 0

792


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Determine the nature of the solutions of the equation.

793

Exercise Set 11.4



Videotape 11

Math TutorCenter

InterActMath

MyMathLabMathXL

1. 2. 3. 4. x2 � 6 � 0x2 � 1 � 0x2 � 12x � 36 � 0x2 � 8x � 16 � 0

5. 6. 7. 8. 4x2 � 8x � 5 � 04x2 � 12x � 9 � 0x2 � 3 � 0x2 � 6 � 0

9. 10. 11. 12. 4m2 � 7m � 09t2 � 3t � 0x2 � 3x � 4 � 0x2 � 2x � 4 � 0

13. 14. y2 �94

� 4yy2 �12

y �35

15. 16. 6y2 � 2�3 y � 1 � 04x2 � 4�3 x � 3 � 0

Write a quadratic equation having the given numbers as solutions.

17. �4 and 4 18. �11 and 9 19. �2 and �7

20. 3 and 10 21. 8, only solution[Hint: It must be a doublesolution, that is,

.]

22. �3, only solution

�x � 8� �x � 8� � 0

23. and 24. and 25. and m4

k3

�12

�14

65

�25

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26. and 27. and 28. and 3�2�22�3��3d2

c2

Solve.

29. 30. 31. x � 10�x � 9 � 0x4 � 7x2 � 12 � 0x4 � 6x2 � 9 � 0

32. 33. 34. �x2 � 5x�2 � 2�x2 � 5x� � 24 � 0�x2 � 6x�2 � 2�x2 � 6x� � 35 � 02x � 9�x � 4 � 0

35. 36. 37. �1 � �x �2 � �1 � �x � � 6 � 03x�2 � x�1 � 14 � 0x�2 � 5x�1 � 36 � 0

38. 39. 40. �2t2 � t�2 � 4�2t2 � t� � 3 � 0� y2 � 5y�2 � 2� y2 � 5y� � 24 � 0�2 � �x �2 � 3�2 � �x � � 10 � 0

41. 42. 43. 2x�2 � x�1 � 1 � 0w 4 � 29w 2 � 100 � 0t 4 � 10t 2 � 9 � 0

44. 45. 46. 6x4 � 17x2 � 5 � 06x4 � 19x2 � 15 � 0m�2 � 9m�1 � 10 � 0

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795

Exercise Set 11.4

47. 48. x2/3 � 2x1/3 � 8 � 0x2/3 � 4x1/3 � 5 � 0

49. 50. �x � 3x � 3�2

� �x � 3x � 3� � 6 � 0�x � 4

x � 1�2

� 2�x � 4x � 1� � 35 � 0

51. 52. 16�x � 1x � 8�2

� 8�x � 1x � 8� � 1 � 09�x � 2

x � 3�2

� 6�x � 2x � 3� � 1 � 0

53. 54. � y2 � 1y �2

� 4� y2 � 1y � � 12 � 0�x2 � 2

x �2

� 7�x2 � 2x � � 18 � 0

57. 58. f�x� � x2/5 � x1/5 � 6f�x� � �x2 � 3x�2 � 10�x2 � 3x� � 24

Find the x-intercepts of the function.

55. 56. f �x� � 3x � 10�x � 8f �x� � 5x � 13�x � 6

59. Describe a procedure that could be used to writean equation having the first seven natural numbers as solutions.

60. Describe a procedure that could be used to writean equation that is quadratic in and has real-number solutions.

3x2 � 1DWDW

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SKILL MAINTENANCE

SYNTHESIS

Solve. [8.4a]

61. Coffee Beans. Twin Cities Roasters has Kenyan coffeeworth $6.75 per pound and Peruvian coffee worth$11.25 per pound. How many pounds of each kindshould be mixed in order to obtain a 50-lb mixture that is worth $8.55 per pound?

62. Solution Mixtures. Solution A is 18% alcohol andsolution B is 45% alcohol. How many liters of eachshould be mixed in order to get 12 L of a solution that is 36% alcohol?

Multiply and simplify. Assume that all expressions under radicals represent nonnegative numbers. [10.3a]

63. 64. 65. 66. �5 16 �5 64�4 9a2 �4 18a3�3 x2 �3 27x4�8x �2x

Graph. [7.1c], [7.4a, c]

67. 68. 69. 70. f�x� � �x � 3y � 45x � 2y � 8f�x� � �35 x � 4

71. Use a graphing calculator to check your answersto Exercises 30, 32, 34, and 37.

72. Use a graphing calculator to solve each of thefollowing equations.

a)b)c) x4 � x3 � 13x2 � x � 12 � 0

�x4 � �2x2 � �99.36.75x � 35�x � 5.26 � 0

For each equation under the given condition, (a) find k and (b) find the other solution.

73. ; one solution is �3. 74. ; one solution is 3.kx2 � 17x � 33 � 0kx2 � 2x � k � 0

75. Find a quadratic equation for which the sum of thesolutions is and the product is 8.

76. Find k given that and theproduct of the solutions is 3.

kx2 � 4x � �2k � 1� � 0�3

77. The graph of a function of the form

is a curve similar to the one shown below. Determine a,b, and c from the information given.

78. While solving a quadratic equation of the formwith a graphing calculator, Shawn-

Marie gets the following screen.

How could the discriminant help her check the graph?

�10

�10

10

10

ax2 � bx � c � 0

�4 �2 2 4

�2

2

4

�5 �1 3 5

�5

�3

�1

1

3

5

x

y

f�x� � ax2 � bx � c

Solve.

79. 80. 81. �x � 3 � �4 x � 3 � 12x

x � 3� 24 � 10� x

x � 3x

x � 1� 6� x

x � 1� 40 � 0

82. 83. 84. x6 � 7x3 � 8 � 0x6 � 28x3 � 27 � 0a3 � 26a3/2 � 27 � 0

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797

11.5 Graphing f �x� � a�x � h�2 � k

In this section and the next, we develop techniques for graphing quad-ratic functions.

Graphs of

The most basic quadratic function is

EXAMPLE 1 Graph:

We choose some values for x and compute for each. Then we plot theordered pairs and connect them with a smooth curve.

All quadratic functions have graphs similar to the one in Example 1. Suchcurves are called parabolas. They are cup-shaped curves that are symmetricwith respect to a vertical line known as the parabola’s line of symmetry, oraxis of symmetry. In the graph of shown above, the y-axis (or theline ) is the line of symmetry. If the paper were to be folded on this line,the two halves of the curve would match. The point is the vertex of thisparabola.

Let’s compare the graphs of and with the graph ofWe choose x-values and plot points for both functions.

Note the symmetry: For equal increments to the left and right ofthe vertex, the y-values are the same.

f �x� � x2.h�x� � 2x2g �x� � 1

2 x2

�0, 0�x � 0

f �x� � x2,

�4 �2 2 4

�2

2

4

�5 �3 �1 3 5�1

3

5

6

7

8

9

1

1

(1, 1)

(2, 4)

(3, 9)(�3, 9)

(�2, 4)

(�1, 1)

Vertex: (0, 0)

f (x) � x 2

x

y

Line ofsymmetry:

x � 0

f �x�f �x� � x2.

f �x� � x2.

f�x� � ax2

11.511.5 GRAPHING f�x� � a�x � h�2 � kObjectivesGraph quadratic functions ofthe type andthen label the vertex and theline of symmetry.

Graph quadratic functions ofthe type and then label the vertexand the line of symmetry.

Graph quadratic functions of the type

finding thevertex, the line of symmetry,and the maximum orminimum y-value.

a�x � h�2 � k,f �x� �

f �x� � a�x � h�2

f �x� � ax2

�4 �2 2 4

2

4

�5 �3 �1 3 5

3

5

6

7

8

9

10

11

12

13

14

15

16

17

18

1

1

h(x) � 2x 2

f (x) � x 2

g(x) � qx 2

x

y

Vertex: (0, 0)

Line of symmetry:x � 0

CALCULATORCORNER

Graphing QuadraticFunctions Use a graphingcalculator to make a table ofvalues for (See theCalculator Corner on p. 230 forthe procedure.)

f �x� � x2.

�3 9�2 4�1 1

0 01 12 43 9 �3, 9�

�2, 4��1, 1��0, 0��1, 1��2, 4��3, 9�

x (x, f(x))f(x) � x2

�3�2 2�1

0 012 23 9

2

12

12

92

x g(x) � 12 x2

�3 18�2 8�1 2

0 01 22 83 18

x h(x) � 2x2

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Note that the graph of is a wider parabola than the graph ofand the graph of is narrower. The vertex and the line of

symmetry, however, remain and respectively.

When we consider the graph of we see that the parabolaopens down and is the same shape as the graph of

GRAPHS OF

The graph of or is a parabola with as itsline of symmetry; its vertex is the origin.

For the parabola opens up; for the parabola opensdown.

If is greater than 1, the parabola is narrower than

If is between 0 and 1, the parabola is wider than

Do Exercises 1–3.

y � x2.�a�y � x2.�a�

a � 0,a � 0,

x � 0y � ax2,f �x� � ax2,

f �x� � ax 2

�4 �2 2 4

�4

�2

�5 �3 3 5

�5

�3

�1

1

k(x) � �qx 2

x

y

g�x� � 12 x2.

k�x� � � 12 x2,

x � 0,�0, 0�h�x� � 2x2f �x� � x2,g�x� � 1

2 x2Graph.

1.

2.

3.


�8 �4 4 8

�8

�4

4

8

�10�12 �6 �2 2 6 10 12

�10

�12

�6

�2

2

6

10

12

x

y

f�x� � �2x2

�8 �4 4 8

�8

�4

4

8

�10�12 �6 �2 2 6 10 12

�10

�12

�6

�2

2

6

10

12

x

y

f�x� � 3x2

�4 �2 2 4

�4

�2

�5 �3 �1 1 3 5

�5

�3

�1

2

4

1

3

5

x

y

f�x� � �

13

x2

798


CALCULATOR CORNER

Graphs of Quadratic Functions1. Graph each of the following equations in the viewing window

Determine a rule that describes the effect of the constant a onthe graph of where and where Graphsome other equations to test your rule.

2. Graph each of the following equations in the viewing window

Determine a rule that describes the effect of the constant a onthe graph of where and where Graph some other equations to test your rule.

�1 � a � 0.a � �1y � ax2,

y1 � �x2, y2 � �4x2, y3 � �

23

x2.

��5, 5 �10, 10�:

0 � a � 1.a � 1y � ax2,

y1 � x2, y2 � 3x2, y3 �13

x2.

��5, 5, �10, 10�:

�3�2 �2�1

0 012 �23 �

92

� 12

� 12

� 92

x k(x) � �12 x2

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Graphs of

It would seem logical now to consider functions of the type

We are heading in that direction, but it is convenient to first consider graphsof and then where a, h, and k areconstants.

EXAMPLE 2 Graph:

We choose some values for x and compute Then we plot the pointsand draw the curve.

First note that for an x-value of 3, As we increase x-valuesfrom 3, note that the corresponding y-values increase. Then as we decreasex-values from 3, note that the corresponding y-values increase again. The line

is a line of symmetry. Equal distances of x-values to the left and right ofthe vertex produce the same y-values.

EXAMPLE 3 Graph:

We choose some values for x and compute Then we plot the pointsand draw the curve.

First note that for an x-value of As we increasex-values from �3, note that the corresponding y-values increase. Then as wedecrease x-values from note that the y-values increase again. The line

is a line of symmetry.x � �3�3,

t��3� � ��3 � 3�2 � 0.�3,

(�4, 1)

(�5, 4)

(�6, 9) (0, 9)

(�1, 4)

(�2, 1)

Vertex: (�3, 0)

4

2

6

8t (x) � (x � 3)2

x � �3 y

�4 �2 2 4 x

t�x�.t�x� � �x � 3�2.

x � 3

g�3� � �3 � 3�2 � 0.

�2 2 4

�2

2

4

�3 �1 5 6 7 8 9

�3

�1

3

5

6

7

8

9

1

1

Vertex: (3, 0)

(4, 1)(2, 1)

(1, 4)(5, 4)

(6, 9)(0, 9)

x � 3

g(x) � (x � 3)2

x

y

g�x�.g�x� � �x � 3�2.

f �x� � a�x � h�2 � k,f �x� � a�x � h�2

f �x� � ax2 � bx � c.

f�x� � a�x � h�2

799


3 04 15 46 92 11 40 9

�1 16

x g(x) � �x � 3�2

�3 0�2 1�1 4

0 9�4 1�5 4�6 9�7 16

x t(x) � �x � 3�2

Vertex

Vertex

CALCULATORCORNER

Graphing QuadraticFunctions Use a graphingcalculator to graph theequations in MarginExercises 1–3.

CALCULATORCORNER

Graphs of QuadraticFunctions Graph eachof the following equations inthe viewing window

Determine a rule that describesthe effect of the constant h onthe graph ofGraph some other equations totest your rule.

y � a�x � h�2.

y3 � 7�x � 2�2.

y2 � 7�x � 1�2,

y1 � 7x2,

��5, 5, �5, 5�:

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The graph of in Example 2 looks just like the graph ofin Example 1, except that it is moved, or translated, 3 units to the

right. Comparing the pairs for with those for we see that when aninput for is 3 more than an input for the outputs match.

The graph of in Example 3 looks just likethe graph of in Example 1, except that it is moved, or translated,3 units to the left. Comparing the pairs for with those for we see thatwhen an input for is 3 less than an input for the outputs match.

GRAPHS OF

The graph of has the same shape as the graph of

If h is positive, the graph of is shifted h units to the right.

If h is negative, the graph of is shifted to the left.

The vertex is and the axis of symmetry is

EXAMPLE 4 Graph:

We first rewrite the equation as In this case,and so the graph looks like that of translated

3 units to the left and, since the graph opens down. The vertex isand the line of symmetry is Plotting points as needed, we

obtain the graph shown below.


�4 2 4

�4

�2

2

4

�5�6�7�8 �1 3

�5

�6

�7

�3

�1

3

1

1

x � �3

Vertex: (�3, 0)

f (x) � �2(x � 3)2

g(x) � 2x 2

x

y

x � �3.��3, 0�,�2 � 0,

g�x� � 2x2h � �3,a � �2f �x� � �2�x � ��3��2.

f �x� � �2�x � 3�2.

x � h.�h, 0��h� unitsy � ax2

y � ax2

y � ax2.f �x� � a�x � h�2

f �x� � a�x � h�2

f �x�,t�x�f �x�,t�x�

f �x� � x2t�x� � �x � 3�2 � �x � ��3��2

Vertex: (3, 0)

(4, 1)(2, 1)

(1, 4)

(5, 4)

(6, 9)(0, 9)

x � 3

g(x) � (x � 3)2

f (x) � x 2

�4 �2 2 4

2

4

�3 �1 5 6�1

3

5

6

7

8

9

1

1 x

y

f �x�,g�x�f �x�,g�x�

f �x� � x2g�x� � �x � 3�2Graph. Find and label the vertex and

the line of symmetry.

4.

5.


�8 �4 4 8

�8

�4

4

8

�10�12 �6 �2 2 6 10 12

�10

�12

�6

�2

2

6

10

12

x

y

f �x� � �

12

�x � 4�2

�8 �4 4 8

�8

�4

4

8

�10�12 �6 �2 2 6 10 12

�10

�12

�6

�2

2

6

10

12

x

y

f �x� �12

�x � 4�2

800


x � �3

(�4, 1)

(�5, 4)

(�6, 9)(0, 9)

(�2, 1)

Vertex: (�3, 0)

4

2

6

8

y

�4 �2 2 4 x

t (x) � (x � 3)2

� [x � (�3)]2

f (x) � x 2

(�1, 4)

�3 0�2 �2�1 �8

0 �18�4 �2�5 �8�6 �18�7 �32

x f (x) � �2�x � 3�2

Vertex

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Graphs of

Given a graph of what happens if we add a constant k?Suppose that we add 2. This increases each function value by 2, so thecurve is moved up. If k is negative, the curve is moved down. The line of sym-metry for the parabola remains but the vertex will be at or equiv-alently,

Note that if a parabola opens up the function value, or y-value, atthe vertex is a least, or minimum, value. That is, it is less than the y-value atany other point on the graph. If the parabola opens down the func-tion value at the vertex is a greatest, or maximum, value.

GRAPHS OF

The graph of has the same shape as the graphof

If k is positive, the graph of is shifted k units up.

If k is negative, the graph of is shifted down.

The vertex is and the line of symmetry is

For k is the minimum function value. For k is themaximum function value.

EXAMPLE 5 Graph and find the minimum functionvalue.

The graph will look like that of (see Example 2) but trans-lated 5 units down. You can confirm this by plotting some points. Forinstance,

whereas in Example 2,

Note that the vertex is so we begin calculating points onboth sides of The line of symmetry is and the minimum functionvalue is �5.

x � 3,x � 3.�h, k� � �3, �5�,

�4 � 3�2 � 1.g�4� �

f �4� � �4 � 3�2 � 5 � �4,

g�x� � �x � 3�2

f �x� � �x � 3�2 � 5,

a � 0,a � 0,

x � h.�h, k�,�k� unitsy � a�x � h�2

y � a�x � h�2

y � a�x � h�2.f �x� � a�x � h�2 � k

f �x� � a�x � h�2 � k

Maximum: k(h, k)

x � h

k

x

y

(h, k)Minimum: k

x � h

k

h hx

y

�a � 0�,

�a � 0�,�h, f �h��.

�h, k�,x � h,

f �x�f �x� � a�x � h�2,

f�x� � a�x � h�2 � k

801


CALCULATORCORNER

Graphs of QuadraticFunctions Graph eachof the following equations inthe viewing window

Determine a rule thatdescribes the effect of theconstant k on the graph of

Graphsome other equations to testyour rule.

y � a�x � h�2 � k.

y3 � 7�x � 1�2 � 3.

y2 � 7�x � 1�2 � 2,

y1 � 7�x � 1�2,

��5, 5, �5, 5�:

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EXAMPLE 6 Graph and find the minimum functionvalue.

The graph looks just like that of but moved 3 units to the rightand 5 units up. The vertex is and the line of symmetry is We draw

and then shift the curve over and up. The minimum function valueis 5. By plotting some points, we have a check.

EXAMPLE 7 Graph Find the vertex, the line ofsymmetry, and the maximum or minimum value.

We first express the equation in the equivalent form

The graph looks like that of translated 3 units to the left and5 units up. The vertex is and the line of symmetry is Since

we know that the graph opens down so 5, the second coordinate ofthe vertex, is the maximum y-value.

We compute a few points as needed. The graph is shown on the follow-ing page.

�2 � 0,x � �3.��3, 5�,

g�x� � �2x2

f �x� � �2�x � ��3��2 � 5.

f �x� � �2�x � 3�2 � 5.

f �x� � 12 x2

x � 3.�3, 5�,f �x� � 1

2 x2

t�x� � 12 �x � 3�2 � 5,

�2 2 4

�4

�2

2

4

�3 �1 6 7

�5

�6

�3

�1

3

5

6

7

8

9

1

Vertex: (3, �5)

Minimum: �5

x � 3

g(x) � (x � 3)2

f (x) � (x � 3)2 � 5

x

yGraph. Find the vertex, the line ofsymmetry, and the maximum orminimum y-value.

6.

7.


�8 �4 4 8

�8

�4

4

8

�10�12 �6 �2 2 6 10 12

�10

�12

�6

�2

2

6

10

12

x

y

f �x� � �2�x � 5�2 � 3

�8 �4 4 8

�8

�4

4

8

�10�12 �6 �2 2 6 10 12

�10

�12

�6

�2

2

6

10

12

x

y

f �x� �12

�x � 2�2 � 4

802


3 �54 �45 �16 42 �41 �10 4

�1 11

x f (x) � �x � 3�2 � 5

Vertex

3 545 76

21 70 9

12

5 12

9 12

5 12

x t(x) � 12�x � 3�2 � 5

Vertex

�4 �2 2 4

2

4

�5 �3 �1 3 5 6 7 8

3

5

6

7

8

9

1

1

Vertex (3, 5)

Minimum: 5

x � 3

t(x) � q(x � 3)2 � 5

f (x) � qx 2

x

t

�3 5�2 3�1 �3�4 3�5 �3

x f (x) � �2�x � 3�2 � 5

Vertex

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Do Exercises 6 and 7 on the preceding page.

�4 �2 2 4

�2

2

4

�5�6�7�8�9 3 5

�3

�1

3

5

6

7

8

9

1

1

Vertex (�3, 5)

Maximum: 5

f (x) � �2(x � 3)2 � 5

x � �3

g(x) � �2x 2

x

y

Study Tips THE 20 BEST JOBS

Although this does notqualify as a Study Tip,

you can use theinformation to

motivate your study ofmathematics. The book

Best Jobs for the 21stCentury, 3rd ed.,

developed by MichaelFarr with database

work by LaurenceShatkin, Ph.D., ranks

500 jobs, on the basis ofoverall scores for pay,growth rate through

2010, and number ofannual openings. Note

that the use ofmathematics is

significant in most ofthe 20 top jobs.

1. Computer software engineers, applications $70,210 100.0% 28,000

2. Computer systems analysts 61,990 59.7 34,000

3. Computer and information systems managers 82,480 47.9 28,000

4. Teachers, postsecondary 52,115 23.5 184,000

5. Management analysts 57,970 28.9 50,000

6. Registered nurses 46,670 25.6 140,000

7. Computer software engineers, systems software 73,280 89.7 23,000

8. Medical and health services managers 59,220 32.3 27,000

9. Sales agents, financial services 59,690 22.3 55,000

10. Sales agents, securities and commodities 59,690 22.3 55,000

11. Securities, commodities, and financial services sales agents 59,690 22.3 55,000

12. Computer support specialists 38,560 97.0 40,000

13. Sales managers 71,620 32.8 21,000

14. Computer security specialists 53,770 81.9 18,000

15. Network and computer systems administrators 53,770 81.9 18,000

16. Financial managers 70,210 18.5 53,000

17. Financial managers, branch or department 70,210 18.5 53,000

18. Treasurers, comptrollers, and chief financial officers 70,210 18.5 53,000

19. Accountants 45,380 18.5 100,000

20. Accountants and auditors 45,380 18.5 100,000

JOBANNUAL

EARNINGSANNUAL

OPENINGSGROWTH RATETHROUGH 2010

Source: Based on “The 20 Best Jobs,” from Best Jobs for the 21st Century, 3rd ed. Indianapolis: Jist Works, 2004, p. 16.

803


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Videotape 11

Math TutorCenter

InterActMath


1.

Vertex: ,Line of symmetry:

2.

Vertex: ,Line of symmetry: x �

��

�4 �2 2 4

�4

�2

2

4

�5 �3 �1 1 3 5

�5

�3

�1

1

3

5

x

y

f �x� � 5x2

x ��

�4 �2 2 4

�4

�2

2

4

�5 �3 �1 1 3 5

�5

�3

�1

1

3

5

x

y

f �x� � 4x2

3.


4.


��

�4 �2 2 4

�4

�2

2

4

�5 �3 �1 1 3 5

�5

�3

�1

1

3

5

x

y

f �x� � 14 x2

x ��

�4 �2 2 4

�4

�2

2

4

�5 �3 �1 1 3 5

�5

�3

�1

1

3

5

x

y

f �x� � 13 x2

5.


��

�4 �2 2 4

�4

�2

2

4

�5 �3 �1 1 3 5

�5

�3

�1

1

3

5

x

y

f �x� � � 12 x2 6.


��

�4 �2 2 4

�4

�2

2

4

�5 �3 �1 1 3 5

�5

�3

�1

1

3

5

x

y

f �x� � � 14 x2

0

1

2

�1

�2

x f �x�

0

1

2

�1

�2

x f �x�

0

1

2

�1

�2

x f �x�

x f �x�

0

1

2

�1

�2

x f �x�

x f �x�

, Graph. Find and label the vertex and the line of symmetry.

Ch11pgs797-807 1/19/06 12:48 PM Page 804

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805

Exercise Set 11.5

�3

�2

�1

�4

�5

x f �x�

9.


10.


��

�4 �2 2 4

�4

�2

2

4

�5 �3 �1 1 3 5

�5

�3

�1

1

3

5

x

y

f �x� � �x � 1�2

x ��

�4 �2 2 4

�4

�2

2

4

�5 �3 �1 1 3 5

�5

�3

�1

1

3

5

x

y

f �x� � �x � 3�2

11.


��

�4 �2 2 4

�4

�2

2

4

�5 �3 �1 1 3 5

�5

�3

�1

1

3

5

x

y

f �x� � 2�x � 4�2 12.


��

�4 �2 2 4

�4

�2

2

4

�5 �3 �1 1 3 5

�5

�3

�1

1

3

5

x

y

f �x� � 4�x � 1�2

7.


8.


��

�4 �2 2 4

�4

�2

2

4

�5 �3 �1 1 3 5

�5

�3

�1

1

3

5

x

y

f �x� � �3x2

x ��

�4 �2 2 4

�4

�2

2

4

�5 �3 �1 1 3 5

�5

�3

�1

1

3

5

x

y

f �x� � �4x2

x f �x�

�1

0

1

�2

�3

x f �x�

x f �x�

x f �x� x f �x�

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19.

Vertex: ,Line of symmetry:Minimum value:

x ��

�4 �2 2 4

�4

�2

2

4

�5 �3 �1 1 3 5

�5

�3

�1

1

3

5

x

y

f �x� � �x � 3�2 � 1 20.

Vertex: ,Line of symmetry:Minimum value:

x ��

�4 �2 2 4

�4

�2

2

4

�5 �3 �1 1 3 5

�5

�3

�1

1

3

5

x

y

f �x� � �x � 2�2 � 3 21.

Vertex: ,Line of symmetry:Maximum value:

x ��

�4 �2 2 4

�4

�2

2

4

�5 �3 �1 1 3 5

�5

�3

�1

1

3

5

x

y

f �x� � �3�x � 4�2 � 1 22.

Vertex: ,Line of symmetry:Maximum value:

x ��

x

y

f �x� � � 12 �x � 1�2 � 3

Graph. Find and label the vertex and the line of symmetry. Find the maximum or minimum value.

15.

�4 �2 2 4

�4

�2

2

4

�5 �3 �1 1 3 5

�5

�3

�1

1

3

5

x

y

f �x� � 3�x � 1�2 16.

�4 �2 2 4

�4

�2

2

4

�5 �3 �1 1 3 5

�5

�3

�1

1

3

5

x

y

f �x� � 4�x � 2�2 17.

�4 �2 2 4

�4

�2

2

4

�5 �3 �1 1 3 5

�5

�3

�1

1

3

5

x

y

f �x� � � 32 �x � 2�2 18.

�4 �2 2 4

�4

�2

2

4

�5 �3 �1 1 3 5

�5

�3

�1

1

3

5

x

y

f �x� � � 52 �x � 3�2

13.


14.


��

�4 �2 2 4

�4

�2

2

4

�5 �3 �1 1 3 5

�5

�3

�1

1

3

5

x

y

f �x� � �2�x � 4�2

x ��

�4 �2 2 4

�4

�2

2

4

�5 �3 �1 1 3 5

�5

�3

�1

1

3

5

x

y

f �x� � �2�x � 2�2


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807

Exercise Set 11.5

23.

Vertex:Line of symmetry:

value:

24.


value:x �

� , �

�4 �2 2 4

�4

�2

2

4

�5 �3 �1 1 3 5

�5

�3

�1

1

3

5

x

y

f �x� � �2�x � 5�2 � 3

x �� , �

�4 �2 2 4

�4

�2

2

4

�5 �3 �1 1 3 5

�5

�3

�1

1

3

5

x

y

f �x� � 12 �x � 1�2 � 4

25.


value:

26.


value:x �

� , �

�4 �2 2 4

�4

�2

2

4

�5 �3 �1 1 3 5

�5

�3

�1

1

3

5

x

y

f �x� � 3�x � 4�2 � 2

x �� , �

�4 �2 2 4

�4

�2

2

4

�5 �3 �1 1 3 5

�5

�3

�1

1

3

5

x

y

f �x� � ��x � 1�2 � 2

27. Explain, without plotting points, why the graph of looks like the graph of

translated 3 units to the left.

28. Explain, without plotting points, why the graph of looks like the graph of

translated 3 units to the left and 4 units down.f �x� � x2f �x� � �x � 3�2 � 4

DWf �x� � x2f �x� � �x � 3�2

DW

Solve. [10.6a, b]

35. Use the TRACE and/or TABLE features on a graphing calculator to confirm the maximum or minimum values given inExercises 23, 24, and 26.

29. x � 5 � �x � 7 30. �2x � 7 � �5x � 4 31. �x � 4 � �11 32. x � 7 � 2�x � 1

SKILL MAINTENANCE

SYNTHESIS

Multiply and simplify. Assume that all expressions under radicals represent nonnegative numbers. [10.3a]

33. 34. �9a3 �16ab4�4 5x3y5 �4 125x2y3

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Graphing and Analyzing

By completing the square, we can begin with any quadratic polynomialand find an equivalent expression . This allows

us to combine the skills of Sections 11.1 and 11.5 to graph and analyze anyquadratic function .

EXAMPLE 1 For , find the vertex, the line of symmetry,and the minimum value. Then graph.

We first find the vertex and the line of symmetry. To do so, we find the equivalent form by completing the square, beginning as follows:

.

We complete the square inside the parentheses, but in a different mannerthan we did before. We take half the x-coefficient, , and squareit: . Then we add 0, or , inside the parentheses. (Because weare using function notation, instead of adding on both sides of anequation, we add and subtract it on the same side, effectively adding 0 andnot changing the value of the expression.)

Adding 0

Substituting for 0

Using the associative law of addition to regroup

. Factoring and simplifying

(This equation was graphed in Example 5 of Section 11.5.) The vertex is, and the line of symmetry is . The coefficient of is 1, which

is positive, so the graph opens up. This tells us that �5 is a minimum. Weplot the vertex and draw the line of symmetry. We choose some x-values onboth sides of the vertex and graph the parabola. Suppose we compute thepair :

.

We note that it is 2 units to the right of the line of symmetry. There will also be a pair with the same y-coordinate on the graph 2 units to the left ofthe line of symmetry. Thus we get a second point, , without makinganother calculation.

(3, �5)

Minimum: �5

x � 3

f (x) � x 2 � 6x � 4� (x � 3)2 � 5

�2 2 4

�4

�2

2

4

�3 �1 6 7 8 9

�5

�6

�3

�1

1

3

x

y

(4, �4)(2, �4)

(1, �1) (5, �1)

(6, 4)(0, 4)

�1, �1�

f �5� � 52 � 6�5� � 4 � 25 � 30 � 4 � �1

�5, �1�

x2x � 3�3, �5�

� �x � 3�2 � 5

� �x2 � 6x � 9� � ��9 � 4�9 � 9 � �x2 � 6x � 9 � 9� � 4

f �x� � �x2 � 6x � 0� � 4

�b�2�29 � 9��3�2 � 9

�6�2 � �3

f �x� � x2 � 6x � 4 � �x2 � 6x � � 4

a�x � h�2 � k

f �x� � x2 � 6x � 4

f �x� � ax2 � bx � c

a�x � h�2 � kax2 � bx � c

f �x� � ax

2 � bx � c

11.611.6 GRAPHING f �x� � ax

2 � bx � c

1. For , find thevertex, the line of symmetry, and the minimum value. Then graph.

Vertex: ,Line of symmetry: Minimum value:

Answer on page A-50

�8 �4 4 8

�8

�4

4

8

�10�12 �6 �2 2 6 10 12

�10

�12

�6

�2

2

6

10

12

x

y

x ��

f�x� � x2 � 4x � 7

ObjectivesFor a quadratic function, findthe vertex, the line ofsymmetry, and the maximumor minimum value, and graphthe function.

Find the intercepts of aquadratic function.

3 �54 �45 �16 42 �41 �10 4

x f �x�

Vertex

x f �x�

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EXAMPLE 2 For , find the vertex, the line of sym-metry, and the minimum value. Then graph.

Since the coefficient of is not 1, we factor out 3 from only the firsttwo terms of the expression. Remember that we want to get to the form

:

. Factoring 3 out of the first two terms

Next, we complete the square inside the parentheses:

.

We take half the x-coefficient, , and square it: . Then we add 0,or , inside the parentheses:

Adding 0

Substituting for 0

. Factoring and simplifying

The vertex is , and the line of symmetry is . The coefficient ofis 3, so the graph is narrow and opens up. This tells us that 1 is a minimum.

We choose a few x-values on one side of the line of symmetry, compute y-values, and use the resulting coordinates to find more points on the otherside of the line of symmetry. We plot points and graph the parabola.

Do Exercise 2.

�4 2

�2

2

4

�5�6�7�8�9 �3 �1 3

�3

�1

3

5

6

7

1

1

(�2, 1)

Minimum: 1(�3, 4)

(�1, 4)

x � �2

� 3(x � 2)2 � 1f (x) � 3x 2 � 12x � 13

x

y

x2x � �2��2, 1�

� 3�x � 2�2 � 1

� 3�x2 � 4x � 4� � 3��4� � 13

� 3�x2 � 4x � 4 � 4� � 13

4 � 4 � 3�x2 � 4x � 4 � 4� � 13

f �x� � 3�x2 � 4x � 0� � 13

4 � 422 � 41

2 � 4 � 2

f �x� � 3�x2 � 4x � � 13

� 3�x2 � 4x� � 13

f �x� � 3x2 � 12x � 13

f �x� � a�x � h�2 � k

x2

f �x� � 3x2 � 12x � 13

2. For , find the vertex, the line of symmetry, and the minimum value. Then graph.

Vertex: ,Line of symmetry: Minimum value:

Answer on page A-50

x

y

�8 �4 4 8

�8

�4

4

8

�10�12 �6 �2 2 6 10 12

�10

�12

�6

�2

2

6

10

12

x ��

f�x� � 3x2 � 24x � 43

809

11.6 Graphing f �x� � ax2 � bx � c

⎫⎪⎬⎪⎭ ⎫⎪⎬⎪⎭

Using the distributivelaw to separate �4from the trinomial

�2 1�1 4�3 4

0 13�4 13

x f �x�

Vertex

x f �x�

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EXAMPLE 3 For , find the vertex, the line of sym-metry, and the maximum value. Then graph.

Again, the coefficient of is not 1. We factor out �2 from only the firsttwo terms of the expression. This makes the coefficient of inside theparentheses 1:

.

Next, we complete the square as before:

.

We take half the x-coefficient, , and square it: . Thenwe add 0, or , inside the parentheses:

Adding 0, or

. Factoring andsimplifying

The vertex is , and the line of symmetry is . The coefficient of is �2, so the graph is narrow and opens down. This tells us that is a maxi-mum. We choose a few x-values on one side of the line of symmetry, computey-values, and use the resulting coordinates to find more points on the otherside of the line of symmetry. We plot points and graph the parabola.

Do Exercise 3.

�4 �2 2 4

�4

�2

2

4

�5�6 �3 �1 3 5 6 7 8 9

�5

�6

�7

�3

�1

3

5

6

1

1

(e, y)

Maximum: y

x � e

f (x) � �2x 2 � 10x � 7

� �2(x � e)2 � y

x

y

(0, �7)

(1, 1) (4, 1)

(2, 5) (3, 5)

(5, �7)

112

x2x � 52�5

2 , 112 �

� �2�x �52�2 �

112

� �2�x2 � 5x �254 � �

252 � 7

� �2�x2 � 5x �254 � � ��2� ��

254 � � 7

� �2�x2 � 5x �254 �

254 � � 7

254 �

254 f �x� � �2�x2 � 5x �

254 �

254 � � 7

254 �

254

��52�2 � 25

412 ��5� � �

52

f �x� � �2�x2 � 5x � � 7

� �2�x2 � 5x� � 7

f �x� � �2x2 � 10x � 7

x2x2

f �x� � �2x2 � 10x � 73. For , findthe vertex, the line of symmetry,and the maximum value. Then graph.

Vertex: ,Line of symmetry: Maximum value:

Answer on page A-50

�8 �4 4 8

�8

�4

4

8

�10�12 �6 �2 2 6 10 12

�10

�12

�6

�2

2

6

10

12

x

y

x ��

f�x� � �4x2 � 12x � 5

810


⎫⎪⎬⎪⎭

⎫⎪⎪⎬⎪⎪⎭

Using the distributivelaw to separate the from the trinomial

�254

or 3 54 15 �72 51 10 �7

5 12

112

,52

x f �x�

Vertex

x f �x�

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The method used in Examples 1–3 can be generalized to find a formulafor locating the vertex. We complete the square as follows:

.Factoring a out of the first two terms.Check by multiplying.

Half of the x-coefficient, , is . We square it to get and add

inside the parentheses. Then we distribute the a:

.

Thus we have the following.

VERTEX; LINE OF SYMMETRY

The vertex of a parabola given by is

, or .

The x-coordinate of the vertex is . The line of symmetry is. The second coordinate of the vertex is easiest to find by

computing .

Let’s reexamine Example 3 to see how we could have found the vertexdirectly. From the formula above,

the x-coordinate of the vertex is .

Substituting into , we find the second coordinate ofthe vertex:

.

The vertex is . The line of symmetry is .We have developed two methods for finding the vertex. One is by com-

pleting the square and the other is by using a formula. You should check withyour instructor about which method to use.

Do Exercises 4–6.

x � 52�5

2 , 112 �

� �252 � 18 � �

252 �

362 � 11

2

� �2�254 � � 25 � 7

f �52� � �2�5

2�2 � 10�52� � 7

f �x� � �2x2 � 10x � 752

�b

2a� �

102��2�

�52

f��b

2a�x � �b��2a�

�b��2a�

��b

2a, f��

b2a��

b2a

,4ac � b2

4a �f �x� � ax2 � bx � c

� a�x � ��b

2a�2

�4ac � b2

4a

� a�x �b

2a�2

��b2

4a�

4ac4a

� a�x2 �ba

x �b2

4a2� � a��b2

4a2� � c

f �x� � a�x2 �ba

x �b2

4a2 �b2

4a2� � c

b2

4a2 �b2

4a2b2

4a2b

2aba

� a�x2 �ba

x� � c

f �x� � ax2 � bx � c

Find the vertex of the parabola usingthe formula.

4.

5.

6.


f�x� � �4x2 � 12x � 5

f�x� � 3x2 � 24x � 43

f�x� � x2 � 6x � 4

811

11.6 Graphing f �x� � ax2 � bx � c

⎫⎪⎪⎬⎪⎪⎭

⎫⎪⎪⎪⎬⎪⎪⎪⎭

Using thedistributive law

Factoring and finding acommon denominator

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Finding the Intercepts of a Quadratic Function

The points at which a graph crosses an axis are called intercepts. We deter-mine the y-intercept by finding . For , the y-interceptis

To find the x-intercepts, we look for values of x for which . For, we solve

.

EXAMPLE 4 Find the intercepts of .

The y-intercept is . Since , the y-intercept is . To find the x-intercepts, we solve

.

Using the quadratic formula gives us . Thus the x-intercepts areand , or, approximately, and .

Do Exercises 7–9.

x

�4

�2

4

6

2

4 62�4 �2

y

f (x) � x 2 � 2x � 2

(1 � �, 0)3(1 � �, 0)3

(0, �2)

�2.732, 0��0.732, 0��1 � 3, 0��1 � 3, 0�x � 1 � 3

0 � x2 � 2x � 2

�0, �2�f �0� � 02 � 2 � 0 � 2 � �2�0, f �0��

f �x� � x2 � 2x � 2

x

y

x-intercepts

y-intercept

(0, c)

f (x) � ax 2 � bx � c

0 � ax2 � bx � c

f �x� � ax2 � bx � cf �x� � 0

�0, c�.f �x� � ax2 � bx � cf �0�

Find the intercepts.

7.

8.

9.


f�x� � x2 � 4x � 1

f�x� � x2 � 8x � 16

f�x� � x2 � 2x � 3

812


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�1

�2

�3

�4

�5

1

2

3

4

5y

A

�4 �2 2 4�5 �3 �1 1 3 5 x

�1

�2

�3

�4

�5

1

2

3

4

5y

B

�4 �2 2 4�5 �3 �1 1 3 5 x

�1

�2

�3

�4

�5

1

2

3

4

5y

C

�2 5 7�1 4321 6 8 x

�1

�2

�3

�4

�5

1

2

3

4

5y

D

�4 �2 2�5�6�7�8 �3 �1 1 x

�1

�2

�3

�4

�5

1

2

3

4

5y

E

�4 �2 2 4�5 �3 �1 1 3 5 x

Match each equation or inequalitywith its graph.

1.

2.

3.

4.

5.

6.

7.

8.

9.

10.


y � ��x � 5�2 � 2

y � 5x

5x � 2y � �10

5x � 2y � 10

y � x2 � 5

y � �x � 5�2 � 2

2x � 5y � 10

5x � 2y � 10

2x � 5y � 10

y � ��x � 5�2 � 2

�1

�2

�3

�4

�5

1

2

3

4

5y

�2 5 7�1 4321 6 8 x

F

�1

�2

�3

�4

�5

1

2

3

4

5y

G

�4 �2 2 4�5 �3 �1 1 3 5 x

�1

�2

�3

�4

�5

1

2

3

4

5y

H

�4 �2 2 4�5 �3 �1 1 3 5 x

�1

�2

�3

�4

�5

1

2

3

4

5y

I

�4 �2 2 4�5 �3 �1 1 3 5 x

�1

�2

�3

�4

�5

1

2

3

4

5y

J

�4 �2 2 4�5 �3 �1 1 3 5 x

Visualizingfor Success

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814


For each quadratic function, find (a) the vertex, (b) the line of symmetry, and (c) the maximum or minimum value.Then (d) graph the function.

Co

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igh

t ©

200

7 P

ears

on

Ed

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nc.



Videotape 11

Math TutorCenter

InterActMath


3.


value:

4.


value:x �

��

�4 �2 2 4

�4

�2

2

4

�5 �3 �1 1 3 5

�5

�3

�1

1

3

5

x

y

f�x� � �x2 � 4x � 1

x ��

�4 �2 2 4

�4

�2

2

4

�5 �3 �1 1 3 5

�5

�3

�1

1

3

5

x

y

f�x� � �x2 � 4x � 2

1.


value:

2.


value:x �

��

x

y

�4 �2 2 4

�4

�2

2

4

�5 �3 �1 1 3 5

�5

�3

�1

1

3

5

f�x� � x2 � 2x � 5

x ��

�4 �2 2 4

�4

�2

2

4

�5 �3 �1 1 3 5

�5

�3

�1

1

3

5

x

y

f�x� � x2 � 2x � 3



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815

Exercise Set 11.6

5.


value:

6.


value:x �

��

�4 �2 2 4

�4

�2

2

4

�5 �3 �1 1 3 5

�5

�3

�1

1

3

5

x

y

f�x� � 4x2 � 8x � 1

x ��

�4 �2 2 4

�4

�2

2

4

�5 �3 �1 1 3 5

�5

�3

�1

1

3

5

x

y

f�x� � 3x2 � 24x � 50

7.


value:

8.


value:x ��

�4 �2 2 4

�4

�2

2

4

�5 �3 �1 1 3 5

�5

�3

�1

1

3

5

x

y

f �x� � �2x2 � 2x � 1

x ��

�4 �2 2 4

�4

�2

2

4

�5 �3 �1 1 3 5

�5

�3

�1

1

3

5

x

y

f �x� � �2x2 � 2x � 3



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816


Co

pyr

igh

t ©

200

7 P

ears

on

Ed

uca

tio

n, I

nc.

9.


value:

10.


value:x �

��

x

y

�4 �2 2 4

�4

�2

2

4

�5 �3 �1 1 3 5

�5

�3

�1

1

3

5

f�x� � x2 � 3x

x ��

�4 �2 2 4

�4

�2

2

4

�5 �3 �1 1 3 5

�5

�3

�1

1

3

5

x

y

f�x� � 5 � x2

11.


value:

12.


value:x �

��

x

y

�4 �2 2 4

�4

�2

2

4

�5 �3 �1 1 3 5

�5

�3

�1

1

3

5

f�x� � �4x2 � 7x � 2

x ��

x

y

�4 �2 2 4

�4

�2

2

4

�5 �3 �1 1 3 5

�5

�3

�1

1

3

5

f�x� � 2x2 � 5x � 2

Find the x- and y-intercepts.



13. f�x� � x2 � 6x � 1 14. f�x� � x2 � 2x � 12 15. f�x� � �x2 � x � 20 16. f�x� � �x2 � 5x � 24

17. f�x� � 4x2 � 12x � 9 18. f�x� � 3x2 � 6x � 1 19. f�x� � 4x2 � x � 8 20. f�x� � 2x2 � 4x � 1

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817

Exercise Set 11.6

21. Does the graph of every quadratic function have ay-intercept? Why or why not?

22. Is it possible for the graph of a quadratic functionto have only one x-intercept if the vertex is off the x-axis? Why or why not?

DWDW

Solve. [6.9a, b]

23. Determining Medication Dosage. A child’s dosage D, in milligrams, of a medication varies directly as thechild’s weight w, in kilograms. To control a fever, adoctor suggests that a child who weighs 28 kg be given 420 mg of Tylenol.

a) Find an equation of variation.b) How much Tylenol would be recommended for a

child who weighs 42 kg?

24. Calories Burned. The number C of calories burnedwhile exercising varies directly as the time t, in minutes,spent exercising. Harold exercises for 24 min on aStairMaster and burns 356 calories.

a) Find an equation of variation.b) How many calories would he burn if he were to

exercise for 48 min?

Find the variation constant and an equation of variation in which y varies inversely as x and the following are true. [6.9c]

25. when 26. when x � 125y � 2x � 2y � 125

Find the variation constant and an equation of variation in which y varies directly as x and the following are true. [6.9a]

27. when 28. when x � 125y � 2x � 2y � 125

29. Use the TRACE and/or TABLE features of a graphingcalculator to estimate the maximum or minimumvalues of the following functions.

a)b)

30. Use the TRACE and/or TABLE features of a graphingcalculator to confirm the maximum or minimum valuesgiven in Exercises 8, 11, and 12.

f �x� � �18.8x2 � 7.92x � 6.18f �x� � 2.31x2 � 3.135x � 5.89

Graph.

31. 32. 33. 34. f �x� � �2�x � 3�2 � 5�f �x� � �x2 � 3x � 4�f �x� � �x2 � 6x � 4�f �x� � �x2 � 1�

35. A quadratic function has as one of its interceptsand as its vertex. Find an equation for thefunction.

36. A quadratic function has as one of its interceptsand as its vertex. Find an equation for thefunction.

��1, 7��4, 0�

�3, �5��1, 0�

37. Consider

.

Find the vertex, the line of symmetry, and the maximumor minimum value. Then draw the graph.

38. Use only the graph in Exercise 37 to approximate thesolutions of each of the following equations.

a) b)

c)x2

8�

x4

�38

� 2

x2

8�

x4

�38

� 1x2

8�

x4

�38

� 0f �x� �

x2

8�

x4

�38

Use the INTERSECT feature of a graphing calculator to find the points of intersection of the graphs of each pair of functions.

39. , 40. , g�x� � �2x2 � 4x � 1f �x� � x2 � 2x � 1g�x� � 2 � xf �x� � x2 � 4x � 2

SKILL MAINTENANCE

SYNTHESIS

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818


We now consider some of the many situations in which quadratic func-tions can serve as mathematical models.

Maximum–Minimum Problems

We have seen that for any quadratic function , the valueof at the vertex is either a maximum or a minimum, meaning that eitherall outputs are smaller than that value for a maximum or larger than thatvalue for a minimum.

There are many types of applied problems in which we want to find amaximum or minimum value of a quantity. If a quadratic function can beused as a model, we can find such maximums or minimums by finding coor-dinates of the vertex.

EXAMPLE 1 Fenced-In Land. A farmer has 64 yd of fencing. What are thedimensions of the largest rectangular pen that the farmer can enclose?

1. Familiarize. We first make a drawing and label it. We let the lengthof the pen and the width. Recall the following formulas:

Perimeter: ;

Area: .

To become familiar with the problem, let’s choose some dimensions(shown at left) for which and then calculate the corre-sponding areas. What choice of l and w will maximize A?

2l � 2w � 64

l � w

2l � 2w

w �l �

(x, f (x))

(x, f (x))

a � 0 a � 0

f (x) at the vertexa minimum

f (x) at the vertexa maximum

xx

yy

f �x�f �x� � ax2 � bx � c

11.711.7 MATHEMATICAL MODELING WITHQUADRATIC FUNCTIONS

ObjectivesSolve maximum–minimumproblems involving quadraticfunctions.

Fit a quadratic function to a set of data to form amathematical model, andsolve related appliedproblems.

l

w

22 10 22020 12 24018 14 25218.5 13.5 249.7512.4 19.6 243.0415 17 255

l w A

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2. Translate. We have two equations, one for perimeter and one for area:

,

.

Let’s use them to express A as a function of l or w, but not both. To expressA in terms of w, for example, we solve for l in the first equation:

Substituting for l, we get a quadratic function , or just A:

.

3. Carry out. Note here that we are altering the third step of our five-stepproblem-solving strategy to “carry out” some kind of mathematical ma-nipulation, because we are going to find the vertex rather than solve anequation. To do so, we complete the square as in Section 11.6:

This is a parabola opening down,so a maximum exists.

Factoring out �1

; .We add 0, or .

Using the distributive law

.

The vertex is . Thus the maximum value is 256. It occurs whenand .

4. Check. We note that 256 is larger than any of the values found in theFamiliarize step. To be more certain, we could make more calculations.We leave this to the student. We can also use the graph of the functionto check the maximum value.

5. State. The largest rectangular pen that can be enclosed is 16 yd by 16 yd;that is, a square.

Do Exercise 1.

A(w) � �(w � 16)2 � 256

(16, 256)

Maximum: 256

30

100

10 20 40

50

0

150

300

250

200

w

y

l � 32 � w � 32 � 16 � 16w � 16�16, 256�

� ��w � 16�2 � 256

� �1�w 2 � 32w � 256� � ��1� ��256�

256 � 256��16�2 � 2561

2 ��32� � �16 � �1�w 2 � 32w � 256 � 256� � �1�w 2 � 32w�

A � �w 2 � 32w

A � lw � �32 � w�w � 32w � w2 � �w 2 � 32w

A�w�32 � w

� 32 � w.

l �64 � 2w

2

2l � 64 � 2w

2l � 2w � 64

A � l � w

2l � 2w � 64

1. Fenced-In Land. A farmer has100 yd of fencing. What are thedimensions of the largestrectangular pen that the farmer can enclose?

To familiarize yourself with theproblem, complete the following table.

Answer on page A-50

819


⎫⎪⎪⎬⎪⎪⎭

12 38 45615 3524 2625 2526.2 23.8

l w A

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820


CALCULATOR CORNER

Maximum and Minimum Values We can use a graphing calculator to find the maximum or minimum value of aquadratic function. Consider the quadratic function in Example 1, . First, we replace w with x and Awith y and graph the function in a window that displays the vertex of the graph. We choose , with

and . Now, we press F m 4 or F m e e e [ to select the MAXIMUM featurefrom the CALC menu. We are prompted to select a left bound for the maximum point. This means that we mustchoose an x-value that is to the left of the x-value of the point where the maximum occurs. This can be done byusing the left- and right-arrow keys to move the cursor to a point to the left of the maximum point or by keying inan appropriate value. Once this is done, we press [. Now, we are prompted to select a right bound. We move thecursor to a point to the right of the maximum point or key in an appropriate value.

We press [ again. Finally, we are prompted to guess the x-value at which the maximum occurs. We move thecursor close to the maximum or key in an x-value. We press [ a third time and see that the maximum functionvalue of 256 occurs when . (One or both coordinates of the maximum point might be approximations of theactual values, as shown with the x-value below, because of the method the calculator uses to find these values.)

To find a minimum value, we select item 3, “minimum,” from the CALC menu by pressing F m 3 or Fm e e [.

Exercises: Use the maximum or minimum feature on a graphing calculator to find the maximum or minimumvalue of the function.

1.

2.

3.

4. y � �4x2 � 5x � 1

y � �x2 � 4x � 2

y � 2x2 � x � 5

y � 3x2 � 6x � 4

Y�256MaximumX�15.999999

400

300y � �x 2 � 32x

00 Y�255.93481Guess ?X�15.744681

Y1��X2 � 32X

40

300y � �x 2 � 32x

0

x � 16

0

Yscl�20Xscl�5

Y�232.46718RightBound?X�20.851064

Y1��X2 � 32X

40

300y � �x 2 � 32x

00 Y�227.25215

LeftBound?X�10.638298

Yscl�20Xscl�5

Y1��X2 � 32X

40

300y � �x 2 � 32x

0

Yscl � 20Xscl � 5�0, 40, 0, 300�

A � �w 2 � 32w

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Fitting Quadratic Functions to Data

As we move through our study of mathematics, we develop a library of func-tions. These functions can serve as models for many applications. Some ofthem are graphed below. We have not considered the cubic or quartic func-tions in detail other than in the Calculator Corners (we leave that discussionto a later course), but we show them here for reference.

Now let’s consider some real-world data. How can we decide which typeof function might fit the data of a particular application? One simple way is tograph the data and look for a pattern resembling one of the graphs above. Forexample, data might be modeled by a linear function if the graph resembles astraight line. The data might be modeled by a quadratic function if the graphrises and then falls, or falls and then rises, in a curved manner resembling aparabola. For a quadratic, it might also just rise or fall in a curved manner asif following only one part of the parabola.

Quartic function:f (x) � ax 4 � bx 3 � cx 2 � dx � e, a � 0

Cubic function:f (x) � ax 3 � bx 2 � cx � d, a � 0

f (x) � �x �Absolute-value function:

f (x) � ax 2 � bx � c, a � 0Quadratic function:

f (x) � ax 2 � bx � c, a � 0Quadratic function:

f (x) � mx � bLinear function:

821


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Let’s now use our library of functions to see which, if any, might fit certaindata situations.

EXAMPLES Choosing Models. For the scatterplots and graphs below, de-termine which, if any, of the following functions might be used as a model forthe data.

Linear, ;

Quadratic, , ;

Quadratic, , ;

Polynomial, neither quadratic nor linear

2.

The data rise and then fall in a curved manner fitting a quadratic function, .

3.

The data seem to fit a linear function .

4.

The data rise in a manner fitting the right side of a quadratic function, .

5.

The data fall and then rise in a curved manner fitting a quadratic function, .a � 0f �x� � ax2 � bx � c

Ages

Driver Fatalities by Age

Nu

mb

er o

f dri

ver

dea

ths

per

100

,000

15–24 25–39 40–69 70–79 80+

28

15

10

15

25

Source: National Highway Traffic Safety Administration

0

5

10

15

20

25

30

Number of licensed drivers per 100,000 who died in motorvehicle accidents in 1990. The fatality rates for both the 70–79 group and the 80+ age group were lower than for the 15- to 24-year olds.

a � 0f �x� � ax2 � bx � c

020

Yearx

y

4 6

5

10

Po

pu

lati

on

(in

mil

lio

ns)

f �x� � mx � b

050 10

Year

4

8

Po

pu

lati

on

(in

mil

lio

ns)

x

y

a � 0f �x� � ax2 � bx � c

20Year

x

y

4 6 80

8

4

Po

pu

lati

on

(in

mil

lio

ns)

a � 0f �x� � ax2 � bx � c

a � 0f �x� � ax2 � bx � c

f �x� � mx � b

Choosing Models. For thescatterplots and graphs in MarginExercises 2–5, determine which, ifany, of the following functions mightbe used as a model for the data.

Linear, ;

Quadratic, ,;

Quadratic, ,;

Polynomial, neither quadraticnor linear

2.

3.

4.

5.


121086420 x

y

020406080

100

Shoe size

Lif

e ex

pec

tan

cyfo

r w

om

en(i

n y

ears

)

Source: Orthopedic Quarterly

020

Yearx

y

4 6

10

20

Sale

s(i

n m

illi

on

s)

50 10Year

x

y

0

8

4Sale

s(i

n m

illi

on

s)

050 10

Year

4

8

Sale

s(i

n m

illi

on

s)

x

y

a � 0f�x� � ax2 � bx � c

a � 0f�x� � ax2 � bx � c

f�x� � mx � b

822


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6.

The data fall, then rise, then fall again. They do not appear to fit a linearor quadratic function but might fit a polynomial function that is neitherquadratic nor linear.

Do Exercises 2–5 on the preceding page.

Whenever a quadratic function seems to fit a data situation, that functioncan be determined if at least three inputs and their outputs are known.

EXAMPLE 7 River Depth. The drawing below shows the cross section ofa river. Typically rivers are deepest in the middle, with the depth decreasingto 0 at the edges. A hydrologist measures the depths D, in feet, of a river atdistances x, in feet, from one bank. The results are listed in the table atright.

x � distance from left bank (in feet)

D(x) � depth of river (in feet)

823


70

60

50

40

30

20

10

0

Source: Centers for Disease Control and Prevention

U.S. Birth Rate for Women Ages 15 –19

Nu

mb

er o

f bir

ths

per

100

0 w

om

en

1970s 1990s1980s

0 0

15 10.2

25 17

50 20

90 7.2

100 0

DISTANCE xFROM THE

RIVERBANK(in feet)

DEPTH DOF THE RIVER

(in feet)

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a) Make a scatterplot of the data.

b) Decide whether the data seem to fit a quadratic function.

c) Use the data points , , and to find a quadratic functionthat fits the data.

d) Use the function to estimate the depth of the river at 75 ft.

a) The scatterplot is as follows.

b) The data seem to rise and fall in a manner similar to a quadratic function.The dashed black line in the graph represents a sample quadratic functionof fit. Note that it may not necessarily go through each point.

c) We are looking for a quadratic function

.

We need to determine the constants a, b, and c. We use the three datapoints , , and and substitute as follows:

,

,

.

After simplifying, we see that we need to solve the system

,

,

.

Since , the system reduces to a system of two equations in two variables:

(1)

. (2)

We multiply equation (1) by �2, add, and solve for a (see Section 8.3):

,

Adding

Solving for a

. �0.008 � a

�405000

� a

�40 � 5000a

0 � 10,000a � 100b �40 � �5,000a � 100b

0 � 10,000a � 100b

20 � 2,500a � 50b,

c � 0

0 � 10,000a � 100b � c

20 � 2,500a � 50b � c

0 � c

0 � a � 1002 � b � 100 � c

20 � a � 502 � b � 50 � c

0 � a � 02 � b � 0 � c

�100, 0��50, 20��0, 0�

D�x� � ax2 � bx � c

DRiver Depth

Distance from the river bank (in feet)

Dep

th (

in fe

et)

40 80

10

20 60 100

5

0

15

20

x

�100, 0��50, 20��0, 0�

6. Ticket Profits. ValleyCommunity College ispresenting a play. The profit P,in dollars, after x days is given inthe following table. (Profit canbe negative when costs exceedrevenue. See Section 8.7.)

a) Make a scatterplot of the data.

b) Decide whether the data canbe modeled by a quadraticfunction.

c) Use the data points ,, and to

find a quadratic function thatfits the data.

d) Use the function to estimatethe profits after 225 days.


�360, 548��180, 872��0, �100�

200

400

600

800

1000

Days

Pro

fit

x

P

1200

$1400

�200400 500300200100

824


0 $�100

90 560

180 872

270 870

360 548

450 �100

DAYS x PROFIT P

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Next, we substitute �0.008 for a in equation (2) and solve for b:

.

This gives us the quadratic function:

.

d) To find the depth 75 ft from the riverbank, we substitute:

.

At a distance of 75 ft from the riverbank, the depth of the river is 15 ft.


10020 40 60 80

20

15

10

5

0x

y

75

D(75) � 15

D(x) � �0.008x 2 � 0.8x

D�75� � �0.008�75�2 � 0.8�75� � 15

D�x� � �0.008x2 � 0.8x

0.8 � b

80 � 100b

0 � �80 � 100b

0 � 10,000��0.008� � 100b

825


CALCULATOR CORNER

Mathematical Modeling: Fitting a Quadratic Function to Data We can use the quadratic regression featureon a graphing calculator to fit a quadratic function to a set of data. The following table shows the average number oflive births for women of various ages.

a) Make a scatterplot of the data and verify that the data can be modeled with a quadratic function.b) Fit a quadratic function to the data using the quadratic regression feature on a graphing calculator.c) Use the function to estimate the average number of live births per 1000 women of age 20 and of age 30.

(continued )

16 34

18.5 86.5

22 111.1

27 113.9

32 84.5

37 35.4

42 6.8

AGE

AVERAGE NUMBEROF LIVE BIRTHS

PER 1000 WOMEN

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a) We enter the data on the STAT list editor screen, turn on Plot 1, and plot the points as described in the CalculatorCorner on p. 534. The data points rise and then fall in a manner consistent with a quadratic function.

b) To fit a quadratic function to the data, we press K g 5 O g 1 1 [. The first three keystrokesselect QuadReg from the STAT CALC menu and display the coefficients a, b, and c of the regression equation

. The keystrokes O g 1 1 copy the regression equation to the equation-editor screen as. We see that the regression equation is . We can press

B 9 to see the regression equation graphed with the data points.

c) Since the equation is entered on the equation-editor screen, we can use a table set in ASK mode to estimate theaverage number of live births per 1000 women of age 20 and of age 30. We see that when , , andwhen , , so we estimate that there are about 85.8 live births per 1000 women of age 20 andabout 101.8 live births per 1000 women of age 30.

Remember to turn off the STAT PLOT as described on p. 536 before you graph other equations.

Exercise:1. Consider the data in the table in Example 7.

a) Use a graphing calculator to make a scatterplot of the data.b) Use a graphing calculator to fit a quadratic function to the data. Compare this function with the one

found in Example 7.c) Graph the quadratic function with the scatterplot.d) Use the function found in part (b) to estimate the depth of the river 75 ft from the riverbank.

Compare this estimate with the one found in Example 7.

85.769101.84

X

X�

Y1

2030

y � 101.8x � 30y � 85.8x � 20

Plot1 Plot2 Plot3\Y1 � �.48684650353607X^2�25.949851822606X��238.48921925252\Y2 �\Y3 �\Y4 �

QuadReg y�ax2�bx�c a��.4868465035 b �25.94985182 c��238.4892193

y � �0.4868465035x 2 � 25.94985182x � 238.4892193y1

y � ax 2 � bx � c

L2

L2(7)�6.8

L3L11618.52227323742

3486.5111.1113.984.535.46.8

2

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Solve.

827

Exercise Set 11.7



Videotape 11

Math TutorCenter

InterActMath

MyMathLabMathXL

1. Architecture. An architect is designing the floors for ahotel. Each floor is to be rectangular and is allotted 720 ft of security piping around walls outside the rooms. What dimensions of the atrium will allow anatrium at the bottom to have maximum area?

2. Stained-Glass Window Design. An artist is designing arectangular stained-glass window with a perimeter of 84 in. What dimensions will yield the maximum area?

w

l

3. Molding Plastics. Economite Plastics plans to producea one-compartment vertical file by bending the longside of an 8-in. by 14-in. sheet of plastic along two linesto form a U shape. How tall should the file be in order tomaximize the volume that the file can hold?

4. Patio Design. A stone mason has enough stones toenclose a rectangular patio with 60 ft of perimeter,assuming that the attached house forms one side of the rectangle. What is the maximum area that themason can enclose? What should the dimensions of the patio be in order to yield this area?

8 in.

14 in.x

5. Minimizing Cost. Aki’s Bicycle Designs has determinedthat when x hundred bicycles are built, the average costper bicycle is given by

,

where is in hundreds of dollars. How many bicyclesshould the shop build in order to minimize the averagecost per bicycle?

6. Corral Design. A rancher needs to enclose twoadjacent rectangular corrals, one for sheep and one forcattle. If a river forms one side of the corrals and 180 ydof fencing is available, what is the largest total area thatcan be enclosed?

180 – 3x

x

x

x

C�x�C�x� � 0.1x2 � 0.7x � 2.425

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igh

t ©

200

7 P

ears

on

Ed

uca

tio

n, I

nc.

7. Garden Design. A farmer decides to enclose arectangular garden, using the side of a barn as one side of the rectangle. What is the maximum area that the farmer can enclose with 40 ft of fence?What should the dimensions of the garden be in order to yield this area?

8. Composting. A rectangular compost container is to beformed in a corner of a fenced yard, with 8 ft of chickenwire completing the other two sides of the rectangle. Ifthe chicken wire is 3 ft high, what dimensions of thebase will maximize the volume of the container?

9. Ticket Sales. The number of tickets sold each day foran upcoming performance of Handel’s Messiah is given by

,

where x is the number of days since the concert wasfirst announced. When will daily ticket sales peak andhow many tickets will be sold that day?

10. Stock Prices. The value of a share of R. P. Mugahti, indollars, can be represented by ,where x is the number of months after January 2003.What is the lowest value will reach, and when didthat occur?

V�x�

V�x� � x2 � 6x � 13

N�x� � �0.4x2 � 9x � 11

Maximizing Profit. Recall (Section 8.7) that total profit P is the difference between total revenue R and total cost C. Given thefollowing total-revenue and total-cost functions, find the total profit, the maximum value of the total profit, and the value of xat which it occurs.

11. , 12. ,C�x� � 5000 � 8xR�x� � 200x � x2

C�x� � 3000 � 20xR�x� � 1000x � x2

13. What is the maximum product of two numbers whosesum is 22? What numbers yield this product?

14. What is the maximum product of two numbers whosesum is 45? What numbers yield this product?

15. What is the minimum product of two numbers whosedifference is 4? What are the numbers?

16. What is the minimum product of two numbers whosedifference is 6? What are the numbers?

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829

Exercise Set 11.7

17. What is the maximum product of two numbers that addto �12? What numbers yield this product?

18. What is the minimum product of two numbers thatdiffer by 9? What are the numbers?

Choosing Models. For the scatterplots and graphs in Exercises 19–26, determine which, if any, of the followingfunctions might be used as a model for the data: Linear, ; quadratic, , ;quadratic, , ; polynomial, neither quadratic nor linear.a � 0f �x� � ax2 � bx � c

a � 0f �x� � ax2 � bx � cf �x� � mx � b

19. 20. 21.

0

5

10

15

20

25

x

y

30

35

64 5321

Year

Valley Community College

Ave

rage

cla

ss s

ize

0

20

40

60

80

100

x

y

120

140

54321

Years since 1995

Growth ofWorld Wide Web Sites

Nu

mb

er (

in m

illi

on

s)0 5

8.6

8.8

9.0

9.2

9.4

Years since 1990

Media Usage

Ho

urs

per

day

x

y

9.6

9.8

10

22.

0

5

10

15

20

25

x

y

30

35

64 5321

Year


Ave

rage

cla

ss s

ize

23.

0

10

x

y

20

30

40

50

60

64 5321

Year


Ave

rage

cla

ss s

ize

24.

0

10

20

30

40

50

x

y

60

70

64 5321

Quantity

Demand for Earphones

Pri

ce p

er u

nit

25. 26.

Year

Sony Electronics, Inc.

Gro

ss p

rofi

t (i

n b

illi

on

s)

6420

1997 1998 1999 2000 2001 2002

12.3

15.2 1513.8

15

6.8810121416

$18

Source: The New York Stock Exchange

Age

Average Number of Live Births per 1000 Women

Bir

ths

per

100

0 w

om

en

60

40

20

016 18.5 22 27 32 37 42

34

86.5

111.1 113.9

84.5

35.4

6.8

80

100

120

140


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igh

t ©

200

7 P

ears

on

Ed

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tio

n, I

nc.

Find a quadratic function that fits the set of data points.

27. , , 28. , , ��2, 16��1, 6��1, 4��2, 13��1, �2��1, 4�

29. , , 30. , , �6, 6��3, 0��3, �30��12, �5��4, 3��2, 0�

31. Nighttime Accidents.

a) Find a quadratic function that fits the following data.

b) Use the function to estimate the number of nighttimeaccidents that occur at 50 .

32. Daytime Accidents.

a) Find a quadratic function that fits the following data.

b) Use the function to estimate the number of daytimeaccidents that occur at 50 .km�hkm�h

33. Archery. The Olympic flame tower at the 1992 Summer Olympics was lit at a height of about 27 m by a flaming arrow that was launched about 63 m fromthe base of the tower. If the arrow landed about 63 mbeyond the tower, find a quadratic function thatexpresses the height h of the arrow as a function of the distance d that it traveled horizontally.

34. Pizza Prices. Pizza Unlimited has the following pricesfor pizzas.

Is price a quadratic function of diameter? It probablyshould be, because the price should be proportional tothe area, and the area is a quadratic function of thediameter. (The area of a circular region is given by

or .)

a) Express price as a quadratic function of diameterusing the data points , , and .

b) Use the function to find the price of a 14-in. pizza.�16, 11.50��12, 8.50��8, 6�

��4� � d2A � �r 2

63 m

27 m

63 m

60 400

80 250

100 250

TRAVEL SPEED(in kilometers

per hour)

NUMBER OF NIGHTTIME ACCIDENTS(for every 200 million

kilometers driven)

60 100

80 130

100 200

TRAVEL SPEED(in kilometers

per hour)

NUMBER OF DAYTIME ACCIDENTS(for every 200 million

kilometers driven)

8 in. $ 6.00

12 in. $ 8.50

16 in. $11.50

DIAMETER PRICE

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831

Exercise Set 11.7

35. Explain the restrictions that should be placed onthe domains of the quadratic functions found inExercises 11 and 31 and why such restrictions areneeded.

36. Explain how the leading coefficient of a quadraticfunction can be used to determine whether a maximumor minimum function value exists.

DWDW

45. Sony Electronics, Inc. Use the REGRESSION featureon your graphing calculator to fit a quartic function tothe data in Exercise 26.

46. The sum of the base and the height of a triangle is 38 cm. Find the dimensions for which the area is amaximum, and find the maximum area.

SYNTHESIS

SKILL MAINTENANCE

37. In the expression the symbol is calleda(n) and is called the

. [10.1a]

38. When a system of equations has infinitely many solutions,the equations are . [8.1a]

39. The degree of a term of a polynomial is theof the exponents of the variables. [4.3g]

40. A consistent system of equations has solution. [8.1a]

41. The equation where k is a positive constant, is anequation of variation. [6.9c]

42. When a system of equations has one solution or nosolutions, the equations are . [8.1a]

43. If the exponents in a polynomial decrease from left to right,the polynomial is written in order. [4.3f]

44. A(n) is a point [7.4a]�a, 0�.

y � k�x,

2x � 9�5�2x � 9 � 3, at least one

no

dependent

independent

ascending

descending

direct

inverse

sum

product

x-intercept

y-intercept

radical

radicand

i VOCABULARY REINFORCEMENT

In each of Exercises 37–44, fill in the blank with the correct word(s) from the given list. Some of the choices maynot be used.

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832


Quadratic and Other Polynomial Inequalities

Inequalities like the following are called quadratic inequalities:

In each case, we have a polynomial of degree 2 on the left. We will solve suchinequalities in two ways. The first method provides understanding and thesecond yields the more efficient method.

The first method for solving a quadratic inequality, such as is by considering the graph of a related function,

EXAMPLE 1 Solve:

Consider the function and its graph. The graphopens up since the leading coefficient is positive. We find the x-intercepts by setting the polynomial equal to 0 and solving:

Values of y will be positive to the left and right of the intercepts, as shown.Thus the solution set of the inequality is

Do Exercise 1.

We can solve any inequality by considering the graph of a related functionand finding x-intercepts as in Example 1. In some cases, we may need to usethe quadratic formula to find the intercepts.

�x � x � �5 or x � 2�, or ��, �5� � �2, ��.

x-values:�x�x � �5�, or (��, �5)

givepositive

y-values.

x-values:�x�x � 2�, or (2, �)

givepositive

y-values.

f (x) � x 2 � 3x � 10

Positive y-valuesPositive y-values

�8 4 8

�8

�4

4

8

�10�12 �6 �2 6 10 12

�10

�6

�2

2

6

10

12

x

y

x � �5 or x � 2.

x � 5 � 0 or x � 2 � 0

�x � 5� �x � 2� � 0

x2 � 3x � 10 � 0

�a � 1�f �x� � x2 � 3x � 10

x2 � 3x � 10 � 0.

ax2 � bx � c.f �x� �bx � c � 0,ax2 �

x2 � 3x � 10 � 0, 5x2 � 3x � 2 � 0.

11.811.8 POLYNOMIAL AND RATIONAL INEQUALITIES

1. Solve by graphing:

Answer on page A-51

�4 �2 2 4

�4

�2

2

4

�5 �3 �1 3 5

�5

�3

�1

3

5

1

1 x

y

x2 � 2x � 3 � 0.

ObjectivesSolve quadratic and otherpolynomial inequalities.

Solve rational inequalities.

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EXAMPLE 2 Solve:

Looking again at the graph of or at least visualizingit tells us that y-values are negative for those x-values between �5 and 2.

That is, the solution set is

When an inequality contains � or �, the x-values of the x-interceptsmust be included. Thus the solution set of the inequality is , or .

Do Exercises 2 and 3. (Exercise 3 is on the following page.)

��5, 2��x � �5 � x � 2�x2 � 3x � 10 � 0

�x � �5 � x � 2�, or ��5, 2�.

x-values:�x ��5 � x � 2�, or (�5, 2)

givenegativey-values.

f (x) � x 2 � 3x � 10

Negative y-values

�8 4 8

�8

�4

4

8

�10�12 �6 �2 6 10 12

�10

�6

�2

2

6

10

12

x

y

f �x� � x2 � 3x � 10

x2 � 3x � 10 � 0. 2. Solve by graphing:

Answer on page A-51

�4 �2 2 4

�4

�2

2

4

�5 �3 �1 3 5

�5

�3

�1

3

5

1

1 x

y

x2 � 2x � 3 � 0.

CALCULATOR CORNER

Solving Polynomial Inequalities We can solve polynomial inequalities graphically. Consider the inequality in Example 2, We first graph the function Then we use the ZERO

feature to find the solutions of the equation or The solutions, �5 and 2, divide the number line

into three intervals, andSince we want to find the values of x for which we look

for the interval(s) on which the function values are negative. That is, we note where the graph lies below the x-axis. This occurs in the interval

so the solution set is or If we were solving the inequality we would look for the intervals on which the graph lies

above the x-axis. We can see that for or If theinequality symbol were � or �, we would include the endpoints of the intervals as well.

Exercises: Solve graphically.

1.

2. x2 � x � 6 � 0

x2 � 3x � 4 � 0

��, �5� � �2, ��.�x � x � �5 or x � 2�,x2 � 3x � 10 � 0x2 � 3x � 10 � 0,

��5, 2�.�x � �5 � x � 2�,��5, 2�,

f �x� � 0,�2, ��.��5, 2�,��, �5�,

x2 � 3x � 10 � 0.f �x� � 0,

f �x� � x2 � 3x � 10.x2 � 3x � 10 � 0.

833


10y � x 2 � 3x � 10

10�10

�5

�15

2

3.

4. x3 � 16x � 0

6x3 � 9x2 � 6x � 0

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We now consider a more efficient method for solving polynomialinequalities. The preceding discussion provides the understanding for thismethod. In Examples 1 and 2, we see that the x-intercepts divide the numberline into intervals.

If a function has a positive output for one number in an interval, it will bepositive for all the numbers in the interval. The same is true for negativeoutputs. Thus we can merely make a test substitution in each interval to solvethe inequality. This is very similar to our method of using test points to grapha linear inequality in a plane.

EXAMPLE 3 Solve:

We set the polynomial equal to 0 and solve. The solutions of or are �5 and 2. We then locate them on a num-

ber line as follows. Note that the numbers divide the number line into threeintervals, which we will call A, B, and C.

We choose a test number in interval A, say �7, and substitute �7 for x in thefunction

Thus,

Note that so the function values will be positive for any number in in-terval A.

Next, we try a test number in interval B, say 1, and find the correspondingfunction value:

Thus,

Note that so the function values will be negative for any number ininterval B.

�6 � 0,

f �1� � 0. � 1 � 3 � 10 � �6.

f �1� � 12 � 3�1� � 10

18 � 0,

f ��7� � 0. � 49 � 21 � 10 � 18.

f ��7� � ��7�2 � 3��7� � 10

f �x� � x2 � 3x � 10:

�5 2

A B C

�x � 5� �x � 2� � 0,10 � 0,x2 � 3x �

x2 � 3x � 10 � 0.

f (x) � x 2 � 3x � 10

�2 6

�8

�10

�14

�6�8 4�4 2 8

�6

�4

6

4

2

�2x

y

x 2 � 3x � 10 � 0Positive y-values

x 2 � 3x � 10 � 0Positive y-values

x 2 � 3x � 10 � 0Negative y-values

3. Solve by graphing:

Answer on page A-51

�4 �2 2 4

�4

�2

2

4

�5 �3 �1 3 5

�5

�3

�1

3

5

1

1 x

y

x2 � 2x � 3 � 0.

834


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Next, we try a test number in interval C, say 4, and find the correspondingfunction value:

Thus,

Note that so the function values will be positive for any number ininterval C.

We are looking for numbers x for which Thusany number x in interval B is a solution. If the inequality had been �, it wouldhave been necessary to include the intercepts �5 and 2 in the solution set aswell. The solution set is or the interval

To solve a polynomial inequality:

1. Get 0 on one side, set the expression on the other side equal to 0,and solve to find the x-intercepts.

2. Use the numbers found in step (1) to divide the number line intointervals.

3. Substitute a number from each interval into the related function.If the function value is positive, then the expression will bepositive for all numbers in the interval. If the function value isnegative, then the expression will be negative for all numbers inthe interval.

4. Select the intervals for which the inequality is satisfied and writeset-builder or interval notation for the solution set.


EXAMPLE 4 Solve:

The solutions of or are �3, 0, and 2. Theydivide the real-number line into four intervals, as shown below.

We try test numbers in each interval:

A: Test �5,

B: Test �2,

C: Test 1,

D: Test 3,

�� 3 20

A B C D

f �3� � 5�3� �3 � 3� �3 � 2� � 90 � 0.

f �1� � 5�1� �1 � 3� �1 � 2� � �20 � 0.

f ��2� � 5��2� ��2 � 3� ��2 � 2� � 40 � 0.

f ��5� � 5��5� ��5 � 3� ��5 � 2� � �350 � 0.

�3 20

A B C D

5x�x � 3� �x � 2� � 0,f �x� � 0,

5x�x � 3� �x � 2� � 0.

��5, 2�.�x � �5 � x � 2�,

f �x� � x2 � 3x � 10 � 0.

�5 2

��

A B C

18 � 0,

f �4� � 0. � 16 � 12 � 10 � 18.

f �4� � 42 � 3�4� � 10

Solve using the method ofExample 3.

4.

5.


x2 � 3x � 4

x2 � 3x � 4

835


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The expression is positive for values of x in intervals B and D. Since theinequality symbol is �, we need to include the x-intercepts. The solution setof the inequality is

or

We visualize this with the graph below.

Do Exercise 6.

Rational Inequalities

We adapt the preceding method when an inequality involves rational expres-sions. We call these rational inequalities.

EXAMPLE 5 Solve:

We write a related equation by changing the � symbol to �:

Then we solve this related equation. First, we multiply both sides of the equa-tion by the LCM, which is

With rational inequalities, we also need to determine those numbers forwhich the rational expression is not defined—that is, those numbers thatmake the denominator 0. We set the denominator equal to 0 and solve:

or Next, we use the numbers �11 and �4 to divide thenumber line into intervals, as shown below.

We try test numbers in each interval to see if each satisfies the originalinequality.

�11 �4

A B C

x � �4.x � 4 � 0,

�11 � x.

x � 3 � 2x � 8

�x � 4� x � 3x � 4

� �x � 4� 2

x � 4:

x � 3x � 4

� 2.

x � 3x � 4

� 2.

f (x) � 5x(x � 3)(x � 2)

�1 3

40

�2�4�5 1 4 5

10

�10

x

y

�20

�30

�40

f (x) � 0

f (x) � 0

f (x) � 0

f (x) � 0

��3, 0� � �2, ��.�x � �3 � x � 0 or 2 � x�,

6. Solve:

Answer on page A-51

6x�x � 1� �x � 1� � 0.

836


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A: Test �15,

FALSE

Since the inequality is false for the number �15 is not a solution ofthe inequality. Interval A is not part of the solution set.

B: Test �8,

TRUE

Since the inequality is true for the number �8 is a solution of theinequality. Interval B is part of the solution set.

C: Test 1,

FALSE

Since the inequality is false for the number 1 is not a solution of theinequality. Interval C is not part of the solution set.

The solution set includes the interval B. The number �11 is also includedsince the inequality symbol is � and �11 is a solution of the related equation.The number �4 is not included; it is not an allowable replacement because itresults in division by 0. Thus the solution set of the original inequality is

To solve a rational inequality:

1. Change the inequality symbol to an equals sign and solve therelated equation.

2. Find the numbers for which any rational expression in theinequality is not defined.

3. Use the numbers found in steps (1) and (2) to divide the numberline into intervals.

4. Substitute a number from each interval into the inequality. If thenumber is a solution, then the interval to which it belongs is partof the solution set.

5. Select the intervals for which the inequality is satisfied and writeset-builder or interval notation for the solution set.


�11 �4

A B C

�x � �11 � x � �4�, or ��11, �4�.

x � 1,

�25

1 � 31 � 4

? 2

x � 3x � 4

� 2

x � �8,

114

�8 � 3�8 � 4

? 2

x � 3x � 4

� 2

x � �15,

1811

�15 � 3�15 � 4

? 2

x � 3x � 4

� 2Solve.

7.

8.


xx � 5

� 2

x � 1x � 2

� 3

837


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838


Solve algebraically and verify results from the graph.

Co

pyr

igh

t ©

200

7 P

ears

on

Ed

uca

tio

n, I

nc.



Videotape 11

Math TutorCenter

InterActMath


1. 2. 3. 4.

�4 �2 2 4

8

6

4

2

x

y

9 � x2 � 0

�4 4

�4

2

�2

x

y

4 � x2 � 0

�2 2 64

�8

8

4

�4

x

y

�x � 5� �x � 1� � 0

2 4

�12

�16

4

�4

x

y

�x � 6� �x � 2� � 0

Solve.

5. 6. 7. 8. x2 � x � 2 � 0x2 � x � 2 � 0�x � 7� �x � 3� � 03�x � 1� �x � 4� � 0

9. 10. 11. 12. x2 � 12 � 4xx2 � 8 � 6xx2 � 6x � 9 � 0x2 � 2x � 1 � 0

13. 14. 15. �x � 9� �x � 4� �x � 1� � 05x�x � 1� �x � 1� � 03x�x � 2� �x � 2� � 0

16. 17. 18. �x � 2� �x � 3� �x � 1� � 0�x � 3� �x � 2� �x � 1� � 0�x � 1� �x � 8� �x � 2� � 0

Solve.

19. 20. 21. 22.x � 2x � 5

� 0x � 1x � 3

� 01

x � 4� 0

1x � 6

� 0

23. 24. 25. 26.x � 1

2x � 3� 1

x � 1x � 2

� 35 � 2x4x � 3

� 03x � 2x � 3

� 0

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839

Exercise Set 11.8

27. 28. 29. 30.x

x � 2� 0

x � 3x

� 0�x � 4� �x � 1�

x � 3� 0

�x � 2� �x � 1�x � 5

� 0

31. 32. 33. 34.x � 2

�x � 2� �x � 7�� 0

x � 1�x � 3� �x � 4�

� 0x � 5

x� 1

xx � 1

� 2

35. 36. 37. 38.x2 � 11x � 30x2 � 8x � 9

� 0�x � 1� �x � 2��x � 3� �x � 4�

� 01x

� 23 �1x

39. Describe a method that could be used to createquadratic inequalities that have no solution.

40. Describe a method that could be used to createquadratic inequalities that have all real numbers assolutions.

DWDW

Simplify. [10.3b]

49. Use a graphing calculator to solve Exercises 11, 22,and 25 by graphing two curves, one for each side of aninequality.

50. Use a graphing calculator to solve each of thefollowing.

a) b)

c) 13 x3 � x �

23 � 0

x � �x � 0x �1x

� 0

41. 42. 43. 44. �3 27c5

343d3�16a3

b4�254a2�3 125

27

Add or subtract. [10.4a]

45. 46. 7�45 � 2�203�8 � 5�2

47. 48. 3�10 � 8�20 � 5�805�3 16a4 � 7�3 2a

Solve.

51. 52. 53. x4 � 2x2 � 0x2 � 2x � 4x2 � 2x � 2

56. Total Profit. A company determines that its total profiton the production and sale of x units of a product isgiven by

a) A company makes a profit for those nonnegativevalues of x for which Find the values of x forwhich the company makes a profit.

b) A company loses money for those nonnegative valuesof x for which Find the values of x for whichthe company loses money.

57. Height of a Thrown Object. The function

gives the height H of an object thrown from a cliff1920 ft high, after time t seconds.

a) For what times is the height greater than 1920 ft?b) For what times is the height less than 640 ft?

H�t� � �16t2 � 32t � 1920

P�x� � 0.

P�x� � 0.

P�x� � �x2 � 812x � 9600.

SKILL MAINTENANCE

SYNTHESIS

54. x4 � 3x2 � 0 55. �x � 2x � 1� � 3

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IMPORTANT PROPERTIES AND FORMULAS

Principle of Square Roots: has solutions and .

Quadratic Formula: ; Discriminant :

The vertex of the graph of is , or .

The line of symmetry of the graph of is .x � �

b2a

f �x� � ax2 � bx � c

��

b2a

, f��

b2a��

b2a

,4ac � b2

4a �f �x� � ax2 � bx � c

b2 � 4acx ��b � �b2 � 4ac

2a

��d�dx2 � d

Review Exercises

1. a) Solve: . [11.1a]b) Find the x-intercepts of .f �x� � 2x2 � 7

2x2 � 7 � 0

The review that follows is meant to prepare you for a chapter exam. It consists of three parts. The first part, ConceptReinforcement, is designed to increase understanding of the concepts through true/false exercises. The second part isa list of important properties and formulas. The third part is the Review Exercises. These provide practice exercises forthe exam, together with references to section objectives so you can go back and review. Before beginning, stop andlook back over the skills you have obtained. What skills in mathematics do you have now that you did not have beforestudying this chapter?

840


Summary and Review1111

i CONCEPT REINFORCEMENT

Determine whether the statement is true or false. Answers are given at the back of the book.

1. Every quadratic equation has exactly two real-number solutions.

2. The quadratic formula can be used to find all the solutions of any quadraticequation.

3. The graph of opens downward.

4. The graph of is a translation downward of the graph of

5. The graph of is a translation to the right of the graph of

6. If the graph of a quadratic equation crosses the x-axis, then it has two real-number solutions.

f �x� � �3x2 � 5.f �x� � �3�x � 2�2 � 5

g�x� � �x2 � 8x.f �x� � �x2 � 8x � 3

f �x� � ��x2 � 8x � 3�

Solve. [11.2a]

2. 3. x2 � 12x � 27 � 014x2 � 5x � 0

4. 4x2 � 3x � 1 � 0

5. x2 � 7x � 13 � 0

6. 4x�x � 1� � 15 � x�3x � 4�

7. . Give exact solutions and approximatesolutions to three decimal places.x2 � 4x � 1 � 0

8.x

x � 2�

4x � 6

� 0

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841

Summary and Review: Chapter 11

9.x4

�4x

� 2 10. 15 �8

x � 2�

6x � 2

11. Solve by completing the square. Showyour work. [11.1b]

x2 � 4x � 1 � 0

12. Hang Time. Use the function . Abasketball player has a vertical leap of 39 in. What is his hang time? [11.1c]

V�T � � 48T 2

13. DVD Player Screen. The width of a rectangular screenon a portable DVD player is 5 cm less than the length.The area is 126 . Find the length and the width.[11.3a]

cm2

14. Picture Matting. A picture mat measures 12 in. by 16 in.; 140 of picture shows. Find the width of the mat. [11.3a]

x

x

12 in.

16 in.

in2

15. Motorcycle Travel. During the first part of a trip, amotorcyclist travels 50 mi at a certain speed. The ridertravels 80 mi on the second part of the trip at a speedthat is 10 mph slower. The total time for the trip is 3 hr.What is the speed on each part of the trip? [11.3a]

Determine the nature of the solutions of the equation.[11.4a]

16. x2 � 3x � 6 � 0 17. x2 � 2x � 5 � 0

Write a quadratic equation having the given solutions.[11.4b]

18. , � 35

15 19. �4, only solution

Solve for the indicated letter. [11.3b]

20. , for pN � 3�� 1p

21. , for T2A �3BT 2

Solve. [11.4c]

22. x4 � 13x2 � 36 � 0 23. 15x�2 � 2x�1 � 1 � 0

24. �x2 � 4�2 � �x2 � 4� � 6 � 0

25. x � 13�x � 36 � 0

For each quadratic function in Exercises 26–28, find andlabel (a) the vertex, (b) the line of symmetry, and (c) themaximum or minimum value. Then (d) graph the function.[11.5c], [11.6a]

26.

Vertex: , Line of symmetry:

value:x ��

�4 �2 2 4

2

4

�5 �3 �1 1 3 5

�4

�2

�5

�3

�1

1

3

5

x

y

f �x� � �12 �x � 1�2 � 3

27.


value:x ��

�8 �4 4 8

4

8

�10 �6 �2 2 6 10

�8

�4

�10

�6

�2

2

6

10

x

y

f �x� � x2 � x � 6

x f �x�

x f �x�

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842


28.


value:x ��

�4 �2 2 4

2

4

�5 �3 �1 1 3 5

�4

�2

�5

�3

�1

1

3

5

x

y

f �x� � �3x2 � 12x � 8 a) Use the data points , , and to fit a quadratic function to the data.

�37, 35.4��27, 113.9��16, 34�

b) Use the quadratic function to estimate the number oflive births per 1000 women of age 30.

Solve. [11.8a, b]

33. �x � 2� �x � 1� �x � 2� � 0

34.�x � 4� �x � 1�

�x � 2�� 0

35. Explain as many characteristics as you can of thegraph of the quadratic function [11.5a, b, c], [11.6a, b]

f �x� � ax2 � bx � c.DW

36. Explain how the x-intercepts of a quadraticfunction can be used to help find the vertex of the function. What piece of information would still bemissing? [11.6a, b]

DW

29. Find the x- and y-intercepts: [11.6b]

.f �x� � x2 � 9x � 14

30. What is the minimum product of two numbers whosedifference is 22? What numbers yield this product?[11.7a]

31. Find the quadratic function that fits the data points, , and . [11.7b]�3, 7��1, 3��0, �2�

32. Live Births by Age. The average number of live birthsper 1000 women rises and falls according to age, as seenin the following bar graph. [11.7b]

Age

Average Number of Live Births per 1000 Women

Bir

ths

per

100

0 w

om

en

60

40

20

016 18.5 22 27 32 37 42

34

86.5

111.1 113.9

84.5

35.4

6.8

80

100

120

140


x f �x�

SYNTHESIS

37. A quadratic function has x-intercepts and and y-intercept . Find an equation for

the function. What is its maximum or minimum value? [11.6a, b]

�0, �7��5, 0��3, 0�

38. Find h and k such that , the sum of the solutions is 20, and the product of the solutions is 80. [11.2a], [11.7b]

3x2 � hx � 4k � 0

39. The average of two numbers is 171. One of the numbersis the square root of the other. Find the numbers.[11.3a]

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1. a) Solve: .b) Find the x-intercepts of .f �x� � 3x2 � 4

3x2 � 4 � 0

843

Test: Chapter 11

Chapter Test1111

Solve.

2. 3. x � 8�x � 7 � 0x2 � x � 1 � 0

4. 5. x4 � 5x2 � 5 � 04x�x � 2� � 3x�x � 1� � �18

6. . Give exact solutions and approximatesolutions to three decimal places.

7.1

4 � x�

12 � x

�34

x2 � 4x � 2

8. Solve by completing the square. Show your work.x2 � 4x � 1 � 0

9. Free-Falling Objects. The Peachtree Plaza in Atlanta,Georgia, is 723 ft tall. Use the function toapproximate how long it would take an object to fallfrom the top.

10. Marine Travel. The Columbia River flows at a rate of 2 mph for the length of a popular boating route. Inorder for a motorized dinghy to travel 3 mi upriver andthen return in a total of 4 hr, how fast must the boat beable to travel in still water?

s�t� � 16t2

11. Memory Board. A computer-parts company wants tomake a rectangular memory board that has a perimeterof 28 cm. What dimensions will allow the board to havea maximum area?

12. Hang Time. Use the function . AnferneeHardaway of the New York Knicks has a vertical leap of36 in. What is his hang time?

V�T � � 48T 2

13. Determine the nature of the solutions of the equation.

14. Write a quadratic equation having the solutions and .3�3�3x2 � 5x � 17 � 0

15. Solve for T.V � 48T 2

Work It Out!Chapter Test Video

on CD

For Extra Help

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Co

pyr

igh

t ©

200

7 P

ears

on

Ed

uca

tio

n, I

nc.

For each quadratic function in Exercises 16 and 17, find and label (a) the vertex, (b) the line of symmetry, and (c) the maximumor minimum value. Then (d) graph the function.

16.


value:

17.


value:x ��

x

y

�8 �4 4 8

�8

�4

�10�12 �6 �2 2 6 10 12

�10

�12

�6

�2

4

6

10

12

8

2

f �x� � 4x2 � 24x � 41

x ��

x

y

�8 �4 4 8

�8

�4

�10�12 �6 �2 2 6 10 12

�10

�12

�6

�2

4

6

10

12

8

2

f �x� � �x2 � 2x


18. Find the x- and y-intercepts:

.

19. What is the minimum product of two numbers whosedifference is 8? What numbers yield this product?f �x� � �x2 � 4x � 1

20. Find the quadratic function that fits the data points , , and .�5, 2��3, 0��0, 0�

21. Consumer Credit. The graph at right showsoutstanding consumer credit in the United States forvarious years. It appears that the graph might be fit byone part or side of a quadratic function.Source: Board of Governors of the Federal Reserve System, FederalReserve Bulletin (monthly); www.federalreserve.gov

a) Use the data points and to fit a quadratic function to thedata, where C is the outstanding consumer credit tyears after 1980 and corresponds to 1980.

b) Use the quadratic function to predict the outstandingconsumer credit in 2007 and in 2010.

Outstanding Consumer Credit, 1980–2003(in billions of dollars)

1980

$349 $593

$780

$1095

$1692$1987

1985 1990 1995 2000 2003

t � 0

C�t� � at2 � bt � c�20, 1692��10, 780�,�0, 349�,

25. A quadratic function has x-intercepts and and y-intercept . Find an equation for the function.What is its maximum or minimum value?

�0, 8��7, 0��2, 0� 26. One solution of is �2. Find the other

solution.kx2 � 3x � k � 0

27. Solve: .x8 � 20x4 � 64 � 0

SYNTHESIS

22. 23. 24.�x � 2�

�x � 3� �x � 1�� 0

x � 5x � 3

� 0x2 � 6x � 7

Solve.

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