273

About the Chapter Project

The Chapter Project, Out of This World, extendsthe idea presented in the Portfolio Activities.By using different accelerations caused by thegravity on other planets, the height of the ballafter a vertical toss on different planets can becompared with its height after a similar toss on Earth.

After completing the Chapter Project, you will beable to do the following:

● Use the function h(t) = �12�gt2 + v0t + h0 to

model the vertical motion of a basketball.● Compare and contrast algebraic models of the

form h(t) = �12�gt2 + v0t + h0 for the vertical

motion of the basketball on different planets.

About the Portfolio Activities

Throughout the chapter, you will be givenopportunities to complete Portfolio Activitiesthat are designed to support your work on theChapter Project.

● Finding a reasonable model for the basketballdata is included in the Portfolio Activity onpage 280.

● Using various algebraic methods to answerquestions about the height of the basketballalong its path is in the Portfolio Activities onpages 298, 306, and 313.

● Comparing the algebraic model from physicswith the quadratic regression model for thebasketball data is included in the PortfolioActivity on page 329.

● Solving quadratic inequalities to answerquestions about the height of the basketballalong its path is in the Portfolio Activity onpage 337.

A basketball game is about to begin. The referee tosses the ball verticallyinto the air. A video camera follows the motion of the ball as it rises toits maximum height and then begins to fall.

The following table and graph represent the height of the ball in feet at0.1-second intervals. After 1.1 seconds, one of the players makes contactwith the ball by tapping it out of its vertical path to a teammate.

Time, x Height, y

0.0 6.00

0.1 7.84

0.2 9.36

0.3 10.61

0.4 11.47

0.5 12.00

0.6 12.30

0.7 12.19

0.8 11.83

0.9 11.12

1.0 9.98

1.1 8.64

8

10

12

6

4

2

y

x0.2 1.00.4 0.6 0.8 1.2

Time (s)

Hei

ght (

ft)

330 CHAPTER 5

Objectives

● Write, solve, and grapha quadratic inequality in one variable.

● Write, solve, and grapha quadratic inequalityin two variables. Katie makes and sells T-shirts. A consultant found that her monthly costs, C,

are related to the selling price, p, of the shirts by the function C(p) = 75p + 2500.The revenue, R, from the sale of the shirts is represented by R(p) = −25p2 + 700p.Her profit, P, is the difference between the revenue and the costs each month.

P(p) = R(p) − C(p)= −25p2 + 700p − (75p + 2500)= −25p2 + 625p − 2500

For what range of prices can Katie sell the shirts in order to make a profit?That is, for what values of p will −25p2 + 625p − 2500 > 0? You will answer thisquestion in Example 2.

Exploring Quadratic Inequalities

You will need: a graphics calculator

The display at right shows the values off(x) = x2 − 2x − 3 for integer values of xbetween −2 and 4 inclusive.

The table suggests the following three cases:• When x = −1 or x = 3, f(x) = 0.• When x < −1 or x > 3, f(x) > 0.• When −1 < x < 3, f(x) < 0.

A P P L I C A T I O N

SMALL BUSINESS

One-Variable Quadratic Inequalities

0−2−3 2 4 51−1 3

f(x) > 0f(x) < 0f(x) > 0

f(x) = 0 f(x) = 0

x

TECHNOLOGYGRAPHICS

CALCULATOR

Keystroke Guide, page 351

You can solve many real-worldproblems, such as those involvingbusiness profits related to revenue andcost, by solving quadratic inequalities.

WhyWhy

Solving QuadraticInequalities

331LESSON 5.8 SOLVING QUADRATIC INEQUALITIES

E X A M P L E 1

CHECKPOINT ✔

TRY THIS

1. Copy and complete the table below. What values of x satisfy eachequation or inequality?

2. Repeat Step 1 for the functions in the table below.

3. Repeat Step 1 for the functions in the table below.

4. a. If the graph of a quadratic function crosses the x-axis at 2 distinctpoints, the graph separates the x-axis into ? distinct interval(s).

b. If the graph of a quadratic function crosses or touches the x-axis at 1 point, the graph separates the x-axis into ? distinct interval(s).

c. If the graph of a quadratic function does not cross the x-axis, thegraph separates the x-axis into ? distinct interval(s).

You can determine the solution to a given inequality by finding the roots ofthe related quadratic equation or by using the graph of the related quadraticequation.

Solve x2 − 2x − 15 ≥ 0. Graph the solution on a number line.

SOLUTION

The graph of y = x2 − 2x − 15 indicates thatthe solution has two parts.

x ≤ smaller root or x ≥ larger root

x2 − 2x − 15 = 0(x + 3)(x − 5) = 0x = −3 or x = 5

Therefore, the solution to the giveninequality is x ≤ −3 or x ≥ 5.

Solve x2 − 8x + 12 ≤ 0. Graph the solution on a number line.

Number ofFunctionx-intercepts

f(x) = 0 f(x) > 0 f(x) < 0

f(x) = x2 1

f(x) = −x2

Number ofFunctionx-intercepts

f(x) = 0 f(x) > 0 f(x) < 0

f(x) = −x2 + x − 1 0

f(x) = x2 + x + 3

Number ofFunctionx-intercepts

f(x) = 0 f(x) > 0 f(x) < 0

f(x) = x2 − 4 2

f(x) = −x2 + 2x + 3

TECHNOLOGYGRAPHICS

CALCULATOR

Keystroke Guide, page 351

2 4 6 7−2−3−4−5 0 3 5−1 1

x is greater thanor equal to thegreater root.

x is less thanor equal to thelesser root.

332 CHAPTER 5

Refer to Katie’s T-shirt business from the beginning of the lesson.

At what price range can Katie sell her T-shirts in order to make a profit?

SOLUTION

Use the quadratic formula to find the roots of −25x2 + 625x − 2500 = 0.

The graph of y = −25x2 + 625x − 2500 indicatesthat the profit is positive between the roots ofthe related equation. If Katie sells her shirts at a price between $5 and $20, she will make a profit.

Special types of solutions to quadratic inequalities are shown in Example 3.

Solve each inequality.a. (x − 2)2 ≥ 0 b. (x + 3)2 < 0 c. 2(x + 1)2 > 0

SOLUTION

Find all solutions of −(x + 1)2 > 0, if any exist.

E X A M P L E 2

A P P L I C A T I O N

SMALL BUSINESS

p = �−b ± �2ba

2� −� 4�ac��

p =

p = 5 or p = 20

−625 ± �62�52� −� 4�(−�25�)(�−�25�00�)�����2(−25)

TRY THIS

TECHNOLOGYGRAPHICS

CALCULATOR

Keystroke Guide, page 351

a. The square of everyreal number isgreater than or equalto 0. Therefore, thesolution is all realnumbers.

The graph ofy = (x − 2)2 indicatesthat all y-values ofy = (x − 2)2 aregreater than or equalto zero for all realnumbers x.

b. The square of a realnumber cannot benegative. Therefore,there is no solution.

The graph ofy = (x + 3)2 indicatesthat no y-values ofy = (x + 3)2 are lessthan zero.

c. The square of anynonzero real numberis greater than 0.Thus, 2(x + 1)2 > 0 isalways true unlessx + 1 = 0, or x = −1.

The graph ofy = 2(x + 1)2

indicates that the y-values ofy = 2(x + 1)2 aregreater than zero forall real numbers xexcept x = −1.

−3 2

y = 2(x + 1)2

−1

2

1

3

4

5

y

x

(−1, 0)−4−5 −3 −2

y = (x + 3)2

−1

2

1

3

4

5

y

x1 2 3 4

y = (x − 2)2

1

2

3

4

5

y

x

E X A M P L E 3

(5, 0) (20, 0)

333LESSON 5.8 SOLVING QUADRATIC INEQUALITIES

Two-Variable Quadratic Inequalities

A quadratic inequality in two variables is an inequality that can be written inone of the forms below, where a, b, and c are real numbers and a ≠ 0.

y ≥ ax2 + bx + c y > ax2 + bx + cy ≤ ax2 + bx + c y < ax2 + bx + c

Example 4 shows how to graph a quadratic inequality in two variables.

Graph the solution to y ≥ (x − 2)2 + 1.

SOLUTION

1. Graph the related equation, y = (x − 2)2 + 1. Use a solid curve because theinequality symbol is ≥.

2. Test (0, 0) to see if this point satisfies thegiven inequality.

y ≥ (x − 2)2 + 10 ≥

? (0 − 2)2 + 10 ≥ 5 False

Test a point inside the parabola,such as (2, 3).

y ≥ (x − 2)2 + 13 ≥

? (2 − 2)2 + 13 ≥ 1 True

3. Shade the region inside the graph of y = (x − 2)2 + 1 because this regioncontains the test point that satisfies y ≥ (x − 2)2 + 1.

Graph the solution to y < (x + 2)2 − 3.

Explain why you should not use (0, 0) as a test point when graphing thesolution to an inequality such as y > x2 + 2x.

On most graphics calculators, you can graph a quadratic inequality in twovariables. The graph of y ≥ (x − 2)2 + 1 from Example 4 is shown below. Thekeystrokes for a TI-83 model are also given.

Use viewing window [−2, 6] by [−2, 6].

( Y1=) 2

1 GRAPH+x2

)–X,T,O,n(ENTERENTERY=

E X A M P L E 4

y

Test points

–4 –2 2 4x

4

2

–2

–4

TRY THIS

CRITICAL THINKING

TECHNOLOGYGRAPHICS

CALCULATOR

334 CHAPTER 5

1. Explain how to solve x2 − 2x − 8 > 0.

2. Explain how a graph can assist you when solving x2 − 2x − 8 > 0.

3. Explain how to graph an inequality such as y < (x − 2)2 + 2.

4. Explain how to test whether the correct area has been shaded in the graphof an inequality.

5. Explain how to determine the possible solutions to (x + 7)2 < 0 withoutsolving the inequality.

6. Solve x2 − 7x + 12 ≥ 0. Graph the solution on a number line.(EXAMPLE 1)

7. For what integer values of x is −2x2 + 25x − 72 > 0 true? (EXAMPLE 2)

Solve each inequality. (EXAMPLE 3)

8. (x − 3)2 < 0 9. (x − 5)2 > 0 10. (x − 1)2 < 0

11. Graph the solution to y ≤ 2(x − 3)2 − 2. (EXAMPLE 4)

Solve each inequality. Graph the solution on a number line.

12. x2 − 1 ≥ 0 13. −x2 + 5x − 6 > 0 14. x2 − 8x + 12 ≤ 0

15. x2 − 4x − 5 < 0 16. x2 − 7x + 10 ≤ 0 17. 50 − 15x > –x2

18. x2 ≤ �34� + x 19. x2 − x − 12 ≤ 0 20. −x2 + �

43�x − �

59� > 0

21. x2 − 4x − 12 > 0 22. x2 − 2x − 99 > 0 23. x2 + x − 6 ≤ 0

24. −x2 −x + 20 < 0 25. x2 ≤ 7x − 6 26. x2 + 35 > −12x

27. 10 − x2 ≥ 9x 28. x2 + 10x + 25 > 0 29. x2 + 3x − 18 > 0

30. x2 − 2 > x 31. x2 + 6x ≥ 7 32. 15 − 8x ≤ −x2

33. −x2 + 3x + 6 < 0 34. 4x − 1 > 8 − x2 35. x2 + 5x − 7 < 4x

Sketch the graph of each inequality. Then decide which of the given

points are in the solution region.

36. y ≥ (x − 1)2 + 5; A(4, 1), B(4, 14), C(4, 20)

37. y > −(x − 3)2 + 8; A(5, 1), B(5, 4), C(5, 6)

38. y < (x − 2)2 + 6; A(3, 1), B(3, 7), C(3, 10)

39. y ≤ −(x − 4)2 + 7; A(6, 1), B(6, 3), C(6, 5)

ExercisesExercises

Communicate

Guided Skills Practice

Practice and Apply

Activities OnlineGo To: go.hrw.comKeyword:MB1 Videos

Graph each inequality.

40. y ≤ (x − 2)2 + 2 41. y ≥ (x + 2)2 42. y < (x − 5)2 + 1

43. y > 2(x + 3)2 − 5 44. y ≤ (x − �12�)

2+ 1 45. y ≥ (x + 1)2 + 2

46. y ≤ x2 + 2x + 1 47. y < x2 − 3x + 2 48. y ≥ 2x2 + 5x + 1

49. y > x2 + 4x + 2 50. y − 3 ≤ x2 − 6x 51. y − 1 < x2 − 4x

52. y ≤ (x − π)2 + 1 53. y ≤ − (x − �57�)

2+ 2 54. y + 3 < (x − 1)2

55. y > x2 + 12x + 35 56. x + y > x2 − 6 57. y − 2x ≤ x2 − 8

58. Create a quadratic function in which f(x) ≥ 0 for values of x between 2 and 6 inclusive.

59. Write a quadratic inequality whose solution is x < 3 or x > 7.

60. MAXIMUM/MINIMUM Jon is a sales representative for a winter sportsequipment wholesaler. The price per snowboard varies based on thenumber of snowboards purchased in each order. Beginning with a price of$124 for one snowboard, the price for each snowboard is reduced by $1when additional snowboards arepurchased.a. Copy and complete the table.b. What is the function for the

revenue?c. What is the maximum revenue

per order?d. How many snowboards must be

sold per order to attain themaximum revenue?

e. Assume that it costs thewholesaler $68 to produce eachsnowboard and that John spendsan average of $128 in fixed costs(travel expenses, phone calls, and so on) per order. Based on these twofactors alone, what is the function for the costs? Is the function linear orquadratic?

f. Graph the revenue and cost functions on the same coordinate plane. In order for the revenue to be greater than the costs, how many snowboards does Jon need to sell?

g. What is the function for profit per order?h. Graph the profit function on the same

coordinate plane as the functions for revenue per order and cost per order.How many snowboards does Jon need to sell per order to make a profit?

i. What is the maximum profit per order? How many snowboards must be sold to earn the maximum profit per order?

C H A L L E N G E S

C O N N E C T I O N

Number of Price Revenuesnowboards per per orderpurchased board ($) ($)

1 124 124

2 123 246

3 122 366

4 121 484

5 120 600

� � �

x

LESSON 5.8 SOLVING QUADRATIC INEQUALITIESLESSON 5.8 SOLVING QUADRATIC INEQUALITIES 335335

HomeworkHelp OnlineGo To: go.hrw.comKeyword:MB1 Homework Help

for Exercises 40–57

336 CHAPTER 5

61. SPORTS At the beginning of a basketball game, the referee tosses the ballvertically into the air. Its height, h, in feet after t seconds is given by h(t) = − 16t2 + 24t + 5. During what time interval (to the nearest tenth of a second) is the height of the ball greater than 9 feet?

62. SMALL BUSINESS Suppose that the profit, p, for selling x bumper stickers isgiven by p(x) = −0.1x2 + 8x − 50.a. What is the minimum number of bumper stickers that must be sold to

make a profit?b. Is it possible for the profit to be greater than $100? Justify your answer

algebraically and graphically.

63. BUSINESS A camping suppliescompany has determined cost andrevenue information for theproduction of their backpacks.

The cost is given by C(x) = 50 + 30x,and the revenue is given by R(x) = 5x(40 − x), where x is thenumber of backpacks sold inthousands. Both the cost andrevenue are given in thousands of dollars.

The profit is given byP(x) = R(x) − C(x).

Use the graph at right to giveapproximate answers to thequestions below.a. To make a profit, revenue must

be greater than cost. What is therange for the quantity ofbackpacks that the companymust sell in order to make aprofit?

b. At what number of backpacks sold is the revenue maximized?c. Is there a greatest cost? Why or why not?d. Graph the profit function.

e. What is the range for the quantity ofbackpacks that the company must sell to makea profit? Compare this answer to your answerfrom part a. Would you expect the answers tobe the same? Why or why not?

f. For what number of backpacks sold is theprofit maximized? Compare this answer toyour answer from part b. Would you expectthe answers to be the same? Why or why not?

g. At what point will the company start to losemoney by producing too many backpacks?

PHYSICS The approximate length of a pendulum, l, in feet is related to thetime, t, in seconds required for one complete swing as given by the formula l ≈ 0.81t2.

64. For what values of t is l > 2? 65. For what values of t is l < 5?

A P P L I C A T I O N S

400

200

600

800

1000

1200

1400

1600

2000

Number of backpacks sold (in thousands)

Nu

mb

er o

f dol

lars

(in

thou

san

ds)

2010 30 40x

y

Revenue

Cost

1800

PortfolioExtensionGo To: go.hrw.comKeyword:MB1 QInequalities

PHYSICS An object is dropped from a height of 1000 feet. Its height, h, in feetafter t seconds is given by the function h(t) = −16t2 + 1000.

66. For approximately how long, to the nearest hundredth of a second, will theheight of the object be above 500 feet?

67. About how long, to the nearest hundredth of a second, does it take theobject to fall from 500 feet to the ground?

SMALL BUSINESS A small company canproduce up to 200 handmade sandals amonth. The monthly cost, C, of producingx sandals is C(x) = 1000 + 5x. The monthlyrevenue, R, is given by R(x) = 75x − 0.4x2.

68. For what values of x is the revenuegreater than the cost?

69. At what production levels will thecompany make a profit?

70. At what production levels will thecompany lose money?

Graph each equation, and state whether y is a function of x.

(LESSON 2.3)

71. y = |x| 72. x = |y| 73. y = −|x| 74. x = y2

Solve each equation. Give exact solutions. (LESSON 5.2)

75. −2x2 = −16 76. −3x2 + 15 = −6 77. 32 = 2x2 − 4

Simplify each expression, where i = �−�1�. (LESSON 5.6)

78. (8 − 2i)(6 + 3i) 79. (1 + 5i) − (2 + i)

80. �22

++

3ii� 81. (3 + 2i) + (3 − i)

82 Let f(x) = x3 − 2x2 + 3x + d. If the graph of f contains the point (1, 9), findthe value of d. Verify your answer by substituting this value for d in thefunction and graphing it.

Look Back

Look Beyond

A C T I V

I TY

PO

RTFOLIO

SPORTS Refer to the basketball data on page 273.

1. At what point(s) in time is the basketball at a height of 10 feet?

2. During what period(s) of time is the height of the basketball above 10 feet?

3. During what period(s) of time is the height of the basketball below 10 feet?

4. Use a graph to show that your answers to Steps 1–3 are correct.

337LESSON 5.8 SOLVING QUADRATIC INEQUALITIES

A P P L I C A T I O N S

340 CHAPTER 5

Chapter Review and Assessment

Multiply linear binomials, and identify and

graph a quadratic function.

f(x) = (x + 4)(x − 1)= x(x − 1) + 4(x − 1)= x2 + 3x − 4

The function f is a quadratic function because itcan be written in the form f(x) = ax2 + bx + c,where a = 1, b = 3, and c = −4.

Since a > 0 in f(x) = x2 + 3x − 4, the parabolaopens up and the vertex contains the minimumvalue of the function.

The coordinates of the vertex are (−1.5, −6.25).

The equation of the axis of symmetry is x = −1.5.

Show that each function is a quadratic

function by writing it in the form

f(x) = ax2 + bx + c and identifying a, b, and c.

1. f(x) = −(x + 1)(x − 4)

2. f(x) = 5(2x − 1)(3x + 2)

Graph each function and give the approximate

coordinates of the vertex.

3 f(x) = −x2 + 3x − 1

4 f(x) = 5x2 − x − 12

State whether the parabola opens up or down

and whether the y-coordinate of the vertex is

the maximum or the minimum value of the

function.

5. f(x) = −x2 − x − 1

6. f(x) = (x − 3)(x + 2)

Key Skills Exercises

LESSON 5.1

Key Skills & Exercises

absolute value of a complex number . . . . . . . . 318

axis of symmetry ofa parabola . . . . . . . . . 276, 310

completing the square . . . . . 300complex number . . . . . . . . . . 316complex plane . . . . . . . . . . . 318conjugate of a complex

number . . . . . . . . . . . . . . . . 318difference of two squares . . 293discriminant . . . . . . . . . . . . . 314double root . . . . . . . . . . . . . . 314factoring . . . . . . . . . . . . . . . . 290imaginary axis . . . . . . . . . . . 318imaginary number . . . . . . . . . 316

imaginary part of acomplex number . . . . . . . . 316

imaginary unit . . . . . . . . . . . . 315maximum value . . . . . . . . . . . 277minimum value . . . . . . . . . . . 277parabola . . . . . . . . . . . . . . . . . 276perfect-square trinomials . . 293principal square root . . . . . . 281Product Property of Square

Roots . . . . . . . . . . . . . . . . . 281Pythagorean Theorem . . . . . 284quadratic expression . . . . . . 275quadratic formula . . . . . . . . . 308quadratic function . . . . . . . . 275

quadratic inequality in two variables . . . . . . . . 333

Quotient Property of Square Roots . . . . . . . . . . . 281

rationalizing the denominator . . . . . . . . 318

real axis . . . . . . . . . . . . . . . . . 318real part of a

complex number . . . . . . . . 316standard form of a

quadratic equation . . . . . . 294vertex form . . . . . . . . . . . . . . 302vertex of a parabola . . . . . . . 276zero of a function . . . . . . . . . 294Zero-Product Property . . . . . 294

VOCABULARY

Use factoring to solve a quadratic equation

and to find the zeros of a quadratic function.

6x2 + 9x = 66x2 + 9x − 6 = 0

3(2x2 + 3x − 2) = 03(2x − 1)(x + 2) = 0

x = �12� or x = −2

The zeros of the related quadratic function,

f(x) = 6x2 + 9x − 6, are �12� and −2.

Factor each expression.

23. 7x2 − 21x 24. 6n − 4n2

25. x2 + 7x + 10 26. x2 + 11x + 28

27. t2 − 5t − 24 28. x2 − 7x + 12

29. x2 − 8x − 20 30. y2 − 6y − 27

31. x2 + x − 20 32. x2 + 4x − 21

33. 3y2 − y − 2 34. 2x2 − 5x − 25

35. 16 − 9x2 36. 4x2 − 49

37. x2 − 16x + 64 38. 4a2 + 4a + 1

Use the Zero-Product Property to find the

zeros of each function.

39. f(x) = x2 − 10x + 24

40. g(x) = 2x2 − 3x −2

41. h(t) = 6t2 + 11t − 10

341CHAPTER 5 REVIEW

Solve quadratic equations by taking square

roots.

16(x + 3)2 = 81

(x + 3)2 = �81

16�

x + 3 = ±��81�

16��

x = −3 + �94� or x = −3 − �

94�

x = −�34� x = −5�

14�

Use the Pythagorean Theorem to solve

problems involving right triangles.

Find the unknown length, b, in right triangle ABC.

a2 + b2 = c2

72 + b2 = 92

b = �92� −� 7�2�b = �32�b ≈ 5.66

Solve each equation, giving both exact

solutions and approximate solutions to the

nearest hundredth.

7. x2 = 8 8. 3x2 = 60

9. x2 − 3 = 46 10. x2 + 4 = 9

11. (x − 3)2 = 64 12. (x − 5)2 = 48

13. 7(x + 1)2 = 54 14. 6(x + 2)2 = 30

Find the unknown length in right triangle

ABC. Give your answers to the nearest tenth.

15. a = 4, b = 5

16. c = 4, a = 1

17. b = 7, c = 12

18. a = 12, c = 15

19. c = 25, b = 5

20. b = 6, a = 6

21. a = 0.2, c = 0.75

22. b = 3.2, c = 5.8

Key Skills Exercises

LESSON 5.2

C

A

B

b

a = 7

c = 9

Key Skills Exercises

LESSON 5.3

(−2, 0) (0.5, 0)

C

A

B

b

c

a

342 CHAPTER 5

Use completing the square to solve a

quadratic equation.

3x2 − 8x = 48x2 − �

83�x = 16

x2 − �83�x + (�

86�)

2= 16 + (�

86�)

2

(x − �86�)

2= �

169

0�

x = �86� ± ��1

36�0��

x = �4 + �

316�0�� or x = �

4 − �3

16�0��

The coordinates of the vertex of the graph of a quadratic function in vertex form,y = a(x − h)2 + k, are (h, k).

Solve each quadratic equation by completing

the square.

42. x2 − 6x = 27 43. 5x2 = 2x + 1

44. x2 − 10x + 21 = 0 45. x2 + 5x = 84

46. x2 − 7x − 8 = 0 47. 2x2 + 7x = 4

48. 4x2 + 4 = 17x 49. 2x + 8 = 3x2

Write each function in vertex form, and

identify the coordinates of the vertex.

50. y = 2x2 − 16x + 33 51. y = −3x2 − 6x − 7

52. y = −x2 − 5x − 2 53. y = 4x2 − 9x + 2

Key Skills Exercises

LESSON 5.4

Use the quadratic formula to find the real

roots of quadratic equations.

Solve 2x2 + x = 10.2x2 + x − 10 = 0 → a = 2, b = 1, c = −10

x = �−b ± �2ba

2� −� 4�ac��

x =

x = �−1 ±

4�81��

x = 2 or x = −�140�

The coordinates of the vertex of the graph of

f(x) = ax2 + bx + c are (−�2ba�, f(−�2

ba�)).

Use the quadratic formula to solve each

equation.

54. 2x + 1 = 2x2 55. 6x = 2 − 5x2

56. x2 − 7x = −10 57. x2 + 6x = −8

58. 11x = 5x2 − 3 59. x = 6x2 − 3

60. 3 = x2 + 5x 61. x2 = 1 − x

For each function, find the coordinates of the

vertex of the graph.

62. f(x) = x2 + 7x + 6 63. f(x) = x2 − x −12

64. g(x) = x2 + 2x −3 65. f(n) = n2 + 12n + 5

−1 ± �12� −� 4�(2�)(�−�10�)����2(2)

Key Skills Exercises

LESSON 5.5

Find and classify all roots of a quadratic

equation.

Solve x2 + 8 = 0. a = 1, b = 0, and c = 8

b2 − 4ac = 02 − 4(1)(8) = −32

Because the discriminant is less than zero, thesolutions are imaginary.

x2 + 8 = 0x = ±�−�8�x = ±i�8�, or ±2i�2�

Determine the number of real solutions for

each equation by using the discriminant.

66. 4x2 − 20x = −25 67. 9x2 + 12x = −2

68. x2 = 21x − 110 69. −x2 + 6x = 10

Solve each equation. Write your answers in

the form a + bi.

70. x2 − 6x + 25 = 0 71. x2 + 10x + 34 = 0

72. x2 + 8x + 20 = 0 73. x2 − 6x + 11 = 0

74. 4x2 = 2x − 1 75. 3x2 + 2 = 2x

Key Skills Exercises

LESSON 5.6

344 CHAPTER 5

Applications

113. AVIATION A crate of blankets and clothing is dropped without aparachute from a helicopter hovering at a height of 125 feet. Thealtitude of the crate, a, in feet, is modeled by a(t) = −16t 2 + 125, wheret is the time, in seconds, after it is released. How long will it take forthe crate to reach the ground?

114. RECREATION Students are designing an archery target with one ringaround the bull’s-eye. The bull’s-eye has a radius of 6 inches. The areaof the outer ring should be 5 times that of the bull’s-eye. What shouldbe the radius of the outer circle?

Solve and graph quadratic inequalities in one

variable.

x2 − x < 12x2 − x − 12 < 0

(x + 3)(x − 4) < 0

The roots of the related equation, x2 − x − 12 = 0,are −3 and 4.

The graph of y = x2 − x − 12 indicates thatx2 − x −12 is negative when x is between the roots.The solution is −3 < x < 4.

Solve and graph quadratic inequalities in two

variables.

To graph y ≤ x2 − 2x − 3, graph y = x2 − 2x − 3,and test a point, such as (0, 0).

y ≤ x2 − 2x −3

0 ?≤ 02 − 2(0) − 3

0 ≤ −3 False

Solve each quadratic inequality, and graph the

solution on a number line.

99. x2 − 8x + 12 > 0

100. x2 − 3x − 10 < 0

101. x2 + 7x + 10 ≥ 0

102. 2x2 + x < 15

103. 4x2 > 9x + 9

104. 2x2 ≥ 9x

105. 4x2 ≤ 10x

106. −x2 + 6x ≥ 8

Graph each quadratic inequality on a

coordinate plane.

107. y > x2 − 6x + 8

108. y ≤ x2 + 3x − 10

109. y − 2x2 < − x − 1

110. y − 5x ≤ 2x2 − 3

111. y + 4x + 21 ≥ x2

112. y − 5x > 6x2 + 1

Key Skills Exercises

LESSON 5.8

2

y

−4

−6

x2 4−2

−2

The point (0, 0)is not in thesolution region.

(−3, 0) (4, 0)

345CHAPTER 5 TEST

Chapter Test

345

Write each function in the form

f(x) � ax2 � bx � c and identify a, b, and c.

State whether the parabola opens up or down

and whether the y-coordinate of the vertex is

the maximum or the minimum.

1. f(x) � (x � 3)(x � 4)

2. f(x) � �5(x � 1)(x � 7)

3. f(x) � �2(x � 3)(3x)

Solve each equation giving both exact and

approximate solutions to the nearest

hundredth.

4. 3x2 � 81 5. (x � 7)2 � 12

Find the unknown length in the right triangle

ABC to the nearest tenth.

6. a � 7, b � 9

7. a � 2, c � 4

8. b � 8.4, c � 9.2

Use factoring and the Zero-Product Property

to find the zeros of each quadratic function.

9. f(x) � x2 � 9x

10. f(x) � 4x2 � 64

11. f(x) � 4x2 � 4x � 1

12. f(x) � x2 � 3x � 10

13. NUMBER THEORY The product of two numbersis 90. One number is 3 more than twice theother number. Model these numbers with aquadratic equation. Solve the equation byfactoring and using the Zero-ProductProperty.

Solve each quadratic equation by completing

the square.

14. x2 � 8x � 4 � 0

15. 2x2 � 11x � 5 � 0

16. GEOMETRY The area of a triangle is 24 squareinches. The height is 4 inches shorter than the

base. Find the height and the base of thetriangle.

Use the quadratic formula to solve each

equation.

17. x2 � 2x � 5 � 0

18. �3x2 � 15 � 12x

For each quadratic function, write the

equation for the axis of symmetry, and find

the coordinates of the vertex.

19. y � x2 � 7x � 10

20. y � 3x2 � 18x � 6

Use the discriminant to determine the

number of real solutions.

21. x2 � 2x � 5 � 0

22. �3x2 � 5 � 3x

23. 4x2 � 27

Perform the indicated operations.

24. (3 � 2i) � (5 � 7i) 25. (2 � i) � (6 � 3i)

26. 3i(7 � 3i) 27. (�2 � i)(3 � 4i)

28. 29. �5 � 12i�Find a quadratic function that fits each set of

data points exactly.

30. (�1, �6), (2, 3), (1, �2)

31. (2, �11), (3, 9), (�1, �23)

Solve each quadratic inequality. Graph the

solution on a number line.

32. x2 � x � 12 > 0

33. 15x2 � 2x � 8 ≥ 0

Graph each quadratic inequality on a

coordinate plane.

34. y ≤ x2 � 4x � 5

35. y � 1 > x2 � 2x � 7

2 � 3i1 � i

A

BC

b c

a

89. 1 91. �1.�00�01�

93. �13� 95. 1

97. (3, 1), (−2, −4), (−3, 3)99. rectangle 101.

103.

105. (f ° f −1)(x) = x 107. 5 framed prints

LESSON 5.7

TRY THIS (p. 323)f(x) = –x2 + 9x – 17

Exercises3. f(x) ≈ 2x2 – 11x + 64a.

The first differences for the number of handshakesare 2, 3, 4, and 5; the second differences are all equalto 1, indicating a quadratic relationship.

b. h(n) = 0.5n2 – 0.5n c. 45 handshakes5a.

b. yes; d ≈ 0.06x2 + 1.10x + 0.067. y = x2 + 3x – 5 9. y = –x2 + 6x + 1011. y = 3x2 – 2x + 4 13. y = 2x2 – 9x + 1515. y = 2x2 – 18x + 49 17. y = –0.5x2 + 4x + 719. y = 0.5x2 − 0.5x + 4 21. y = 2x2 – 5x + 723. h(t) = –6t2 + 20t + 5 25. 1�

23� seconds 27. about

3.57 seconds 29. 0.69 second, 2.64 seconds31. h(t) ≈ –15.11t2 + 15.19t + 1.6533. yes 35. yes 37. yes 39. −4 41. 8; yes

43. – �112�

; yes

LESSON 5.8

TRY THIS (p. 331)2 ≤ x ≤ 6

TRY THIS (p. 332)no solution

TRY THIS (p. 333)

Exercises6. x ≤ 3 or x ≥ 4

7. 5, 6, and 7 8. no solution 9. all real numbersexcept 5 10. no solution11.

13. 2 < x < 3

−6 −4−5 −3 −2 −1 0 1 2 3 654

x

y

6−2

−2

4

6

2

(2, 0) (4, 0)

−4 −2−3 −1 0 1 2 3 4 5 6 7 8

x2–2–4

y

–2

4

2

−4 −2−3 −1 0 1 2 3 4 5 6 7 8

0 80

400

0

x

y

42

−4

−4

4

2

x

y

42−2

−2

4

2

Real

Imaginary

a + bi

a − bi

−a + bi

−(a + bi)

i1

Imaginary

1

1

Real

+ 13 3

2

2 + 3i

2

1

3

Imaginary

2 31Real

13

1 + 0.01i0.01

Imaginary

1Real

1.0001Real

Imaginary

1 2−1

−2

−2 −1

2

1 0 + i

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Number of Number ofpeople, n handshakes, h

2 1

3 3

4 6

5 10

6 15

15. −1 < x < 5

17. x < 5 or x > 10

19. −3 ≤ x ≤ 4

21. x < −2 or x > 6

23. −3 ≤ x ≤ 2

25. 1 ≤ x ≤ 6

27. −10 ≤ x ≤ 1

29. x < −6 or x > 3

31. x ≤ −7 or x ≥ 1

33. x < �32� – �

�233�� or x > �

32� + �

�233��

35. – �12� – �

�229�� < x < – �

12� + �

�229��

37. 39.

C A and B

41. 43.

45. 47.

49. 51.

53. 55.

57.

59. x2 – 10x + 21 > 0 61. 0.2 ≤ t ≤ 1.3 seconds63a. from 0 to 33,000 b. 20,000 c. No; the costincreases as the number of backpacks increases.d. e. from 1000 to

33,000; same aspart af. 17,000; yesg. after 33,000

Nu

mb

er o

f dol

lars

(in

thou

san

ds)

Number of backpacks(in thousands)

500

1500

1000

y

20 3010x

x

y

84−4

−8

8

4

x

y

84−4

−8

−4−8

8

4x

y

42−2

−4

−6

−2−4

2

x

y

4 62−2

−4

−2

2

x

y

42−2

−4

−2−4

4

2

x

y

42−2

−4

−2−4

2

x

y

42−2

−2−4

4

6

x

y

2−2

−4

−2−6

4

2

x

y

2−2

−2−4−6

4

6

x

y

4 62

4

6

2

x

y

42

4

6

2

−6 −4−5 −3 −2 −1 0 1 2 3 654

−−12 2

29 +− 12 2

29

−6 −4−5 −3 −2 −1 0 1 2 3 654

+32 2

33−32 2

33

−10 −8−9 −7 −6 −5 −4 −3 −2 −1 210

−6 −4−5 −3 −2 −1 0 1 2 3 654

−10 −8−9 −7 −6 −5 −4 −3 −2 −1 210

−6 −4−5 −3 −2 −1 0 1 2 3 654

−6 −4−5 −3 −2 −1 0 1 2 3 654

−6 −4−5 −3 −2 −1 0 1 2 3 654

−6 −4−5 −3 −2 −1 0 1 2 3 654

−12 −8−10 −6 −4 −2 0 2 4 6 12108

5

−6 −4−5 −3 −2 −1 0 1 2 3 654

1029SELECTED ANSWERS

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65. approximately 0 ≤ t < 2.48 67. about 2.32seconds 69. between 16 and 159 pairs, inclusive71.yes 73. yes 75. x = ±�8� 77. x = ±�18�79. −1 + 4i 81. 6 + i

CHAPTER REVIEW AND ASSESSMENT

1. f(x) = –x2 + 3x + 4; a = −1, b = 3, and c = 43. (1.5, 1.25) 5. opens down; maximum7. x = ±�8�; x ≈ ±2.83 9. x ≈ ±7 11. x = –5 or x = 11

13. x = –1 ± ��

754��; x ≈ –3.78 or x ≈ 1.78 15. c ≈ 6.4

17. a ≈ 9.7 19. a ≈ 24.5 21. b ≈ 0.7 23. 7x(x – 3)25. (x + 5)(x + 2) 27. (t + 3)(t – 8)29. (x + 2)(x – 10) 31. (x + 5)(x – 4)33. (3y + 2)(y – 1) 35. (4 + 3x)(4 – 3x) 37. (x – 8)2

39. x = 4 or x = 6 41. t = – �52� or t = �

23�

43. x = �15� ± �

�56�

45. x = −12 or x = 7

47. x = –4 or x = �12� 49. x = – �

43� or x = 2

51. y = –3(x + 1)2 – 4; (−1, −4)

53. y = 4�x – �98��2

– �41

96�

; ��98� , – �

41

96�� 55. x = – �

35� ± �

�1706�

�

57. x = –4 or x = –2 59. x = �112�

± ��1723�

�

61. x = – �12� ± �

�25�� 63. ��

12�, – �

449�� 65. (−6, −31)

67. 2 real solutions 69. no real solutions

71. x = −5 − 3i or x = −5 + 3i 73. x = 3 ± �i�

28�

�

75. x = �13� ± �

i�620�� 77. 1 − 7i 79. −1 81. 6 − 9i

83. 2 − 4i

85. �10� 87. �2�

89. f(x) = 5x2 + 2x – 9 91. f(x) = –5x2 – x + 693. f(x) = 3x2 – 3x – 3 95. f(x) = 7x2 – 197. f(x) ≈ –0.7x2 + 1.5x – 3.199. x < 2 or x > 6

101. x ≤ –5 or x ≥ –2

103. x < – �34� or x > 3

105. 0 ≤ x ≤ �52�

107. 109.

111.

113. about 2.8 seconds

Chapter 6

LESSON 6.1

TRY THIS (p. 356)188,700,000; 194,400,000

TRY THIS (p. 357)192 milligrams, 62.9 milligrams

Exercises5. 1.055 6. 1.0025 7. 0.97 8. 0.995 9. 8 10. 135011. 0.512 12. 42.1875 13. 31,400,000 14. 31.0milligrams and 27.3 milligrams 15. 1.07 17. 0.9419. 1.065 21. 0.9995 23. 1.00075 25. 1.927. 1,638,400 29. 37.9 31. 394.0 33. 941,013.735. 32.6 37a. 8000 bacteria b. 32,000 bacteria39a. 1200 bacteria b. 4800 bacteria 41a. 6975bacteria b. 62,775 bacteria 43. exponential45. linear 47. 278,700,000 49. 1,359,600,000 and1,399,800,000 51a. 1,002,600,000 and 1,140,800,000b. 13.79% c. 1,107,500,000

53. 310,000 gallons 55. 2 57. 7 59. �nm

1

3

5� 61.

63. reflection across y-axis and horizontalcompression by a factor of �

12� 65. reflection across

x-axis, horizontal stretch by a factor of 2, and verticaltranslation 3 units up 67. reflection across x-axis,vertical stretch by a factor of 5, horizontal translation2 units to the right, and vertical translation 4 unitsdown 69. opens down; maximum value

LESSON 6.2

TRY THIS (p. 364) TRY THIS (p. 366)a. exponential growth; �

13� 7.2%

b. exponential decay; �14�

1�12xy 2

x

y

2010−10

−20

−10−20

x

y

42−2

−4

−2−4

4

2

x

y

84–4

–8

–4–8

8

4

–6 –4–5 –3 –2 –1 0 1 2 3 654

52

–4 –2–3 –1

–

0 1 2 3 4 5 6 7 8

34

–4 –2–3–6 –5 –1 0 1 2 3 4 5 6

–4 –2–3 –1 0 1 2 3 4 5 6 7 8

1

Imaginary

1Real

1 + i

√2

–6

–4

1

Imaginary

–3 – i–6

Real

√10

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