exploring quadratic graphs

Exploring Quadratic GraphsALGEBRA 1 LESSON 10-1

Evaluate each expression for h = 3, k = 2, and j = –4.

1. hkj 2. kh2 3. hk2 4. kj 2 + h

(For help, go to Lessons 1-2 and 5-3.)

Graph each equation.

5. y = 2x – 1 6. y = | x | 7. y = x2 + 2

10-1


1. hkj for h = 3, k = 2, j = –4: (3)(2)(–4) = 6(–4) = –24

2. kh2 for h = 3, k = 2: 2(32) = 2(9) = 18

3. hk2 for h = 3, k = 2: 3(22) = 3(4) = 12

4. kj 2 + h for h = 3, k = 2, j = –4: 2(–4)2 + 3 = 2(16) + 3 = 32 + 3 = 35

5. y = 2x – 1 6. y = | x |

7. y = x2 + 2

Solutions

10-1

Identify the vertex of each graph. Tell whether it is a minimum or a maximum.

Exploring Quadratic Graphs

a.

The vertex is (1, 2).

b.

The vertex is (2, –4).

It is a maximum. It is a minimum.

ALGEBRA 1 LESSON 10-1

10-1

Make a table of values and graph the quadratic

function y = x2.13


x y = x2 (x, y)13

0 (0)2 = 0 (0, 0)13

2 (2)2 = 1 (2, 1 )13

13

13

3 (3)2 = 3 (3, 3)13

10-1

Use the graphs below. Order the quadratic functions

(x) = –x2, (x) = –3x2, and (x) = x2 from widest to narrowest graph.

So, the order from widest to narrowest is (x) = x2, (x) = –x2,(x) = –3x2.

12

12

(x) = –x2 (x) = x2 12

Of the three graphs, (x) = x2 is the widest and (x) = –3x2 is the narrowest.

12

(x) = –3x2


10-1

Graph the quadratic functions y = 3x2 and y = 3x2 – 2. Compare the graphs.

The graph of y = 3x2 – 2 has the same shape as the graph of y = 3x2, but it is shifted down 2 units.


x y = 3x2 y = 3x2 – 2

2 12 10

1 3 1

0 0 2

–1 3 1

2 12 10

10-1

Height h is dependent on time t.

Graph t on the x-axis and h on the y-axis.

Use positive values for t.


A monkey drops an orange from a branch 26 ft above the ground. The force of gravity causes the orange to fall toward Earth. The function h = –16t2 + 26 gives the height of the orange, h, in feet after t seconds. Graph this quadratic function.

t h = –16t2 + 260 261 102 –38

10-1

9.

10. y = x2, y = 3x2, y = 4x2

11. ƒ(x) = x2, ƒ(x) = x2, ƒ(x) = 5x2

12. y = – x2, y = – x2, y = 5x2

13. ƒ(x) = – x2, ƒ(x) = –2x2,

ƒ(x) = –4x2

14.


pages 513–516 Exercises

1. (2, 5); max.

2. (–3, –2); min.

3. (2, 1); min.

4.

5.

6.

7.

8.

12

13

14

12

23

10-1


15.

16.

17.

18.

19.

20.

21. E

22. A

23. F

24. B

25. C

26. D

27. The graph of y = 2x2 is narrower.

28. The graph of y = –x2 opens downward.

29. The graph of y = 1.5x2 is narrower.

30. The graph of y = x2 is wider.

12

10-1


31.

32.

33.

34.

35.

36.

37.

38. a.

b. 184 ftc. 56 ft

39. a. 0 < r < 6b. 0 < A < 36 113.1c.

10-1


40. K, L

41. M

42. K

43. M

44. Answers may vary. Sample:a. y = 5x2 b. y = –5x2 c. y = 3x2

45. a.

b. 16 ftc. No; the apple falls 48 ft

from t = 1 to t = 2, because it is accelerating.

46. a. c 0 and a and c have opp. signs.b. c 0 and a and c have the same signs.

47. a.

b. 0 < x < 12; the side length of the square garden must be less than the width of the patio.

c. 96 < A < 240; as the side length of the garden increases from 0 to 12, the area of the patio decreases from 240 to 96.

d. about 6 ft

10-1

=/=/


48. a. a > 0b. |a| > 1

49. a.

b.

50. B

51. G

52. D

53. [4] a.

b. 3.5 s[3] estimate incorrect or missing[2] error in table or graph[1] table OR reasonable graph only

t h0 2001 1842 1363 564 –56

10-1


54. (x2 + 2)(x – 4)

55. (3a2 – 2)(5a – 6)

56. (7b2 + 1)(b + 2)

57. (y + 2)(y – 2)(y + 3)

58. 2n(n + 3)(n – 4)

59. 3m(2m + 3)(5m + 1)

60. 15x2 – 20x

61. 9n2 – 63n

62. –12t 3 + 22t 2

63. 12m6 – 4m5 + 20m2

64. –15y6 – 10y4 + 20y

65. –12c5 + 21c4 – 24c3

66. 15 balloons

10-1

1. a. Graph y = – x2 – 1.

b. Identify the vertex. Tell whether it is a

maximum or a minimum.

c. Compare this graph to the graph of y = x2.

2. a. Graph y = 4x2 + 3.

b. Identify the vertex. Tell whether it is a

maximum or a minimum.

c. Compare this graph to the graph of y = x2.


12

(0, -1); maximum

This graph is wider, opens downward, and is shifted one unit down.

(0, 3); minimum

This graph is narrower and shifted 3 units up.

3. Order the quadratic functions y = –4x2, y = x2, and y = 2x2 from widest to narrowest graph.

14

y = x2, y = 2x2, y = –4x214

10-1

Quadratic Functions

(For help, go to Lessons 1-6 and 10-1.)ALGEBRA 1 LESSON 10-2

Evaluate the expression for the following values of a and b.

1. a = –6, b = 4 2. a = 15, b = 20

3. a = –8, b = –56 4. a = –9, b = 108

Graph each function.

5. y = x2 6. y = –x2 + 2 7. y = x2 – 1

–b2a

12

10-2

Quadratic FunctionsALGEBRA 1 LESSON 10-2

Solutions b2a for a = –6, b = 4:1.

b2a for a = 15, b = 20:2.

b2a for a = –8, b = –56:3.

b2a for a = –9, b = 108:4.

–42(–6) = –4

–12 = 13

–202(15) = –20

30 = 23–

–(–56)2(–8) = 56

–16 = 72– = –3 1

2

–

–

–

– –1082(–9) = –108

–18 = 6

5. y = x2 6. y = –x2 + 2 7. y = x2 – 112

10-2


Graph the function y = 2x2 + 4x – 3.

Step 1: Find the equation of the axis of symmetry and thecoordinates of the vertex.

Find the equation of the axis of symmetry.x = b2a– = –4

2(2) = – 1

The x-coordinate of the vertex is –1.

10-2

y = 2x2 + 4x – 3

y = 2(–1)2 + 4(–1) – 3

= –5

To find the y-coordinate of the vertex, substitute –1 for x.

The vertex is (–1, –5).


(continued)

Step 2: Find two other points.Use the y-intercept.For x = 0, y = –3, so one point is (0, –3).

10-2

Choose a value for x on the same side of the vertex.Let x = 1y = 2(1)2 + 4(1) – 3 = 3For x = 1, y = 3, so another point is (1, 3).

Find the y-coordinate for x = 1.


(continued)

10-2

Step 3: Reflect (0, –3) and (1, 3) across the axis of symmetry to get two more points.

Then draw the parabola.


Aerial fireworks carry “stars,” which are made of a sparkler-like material, upward, ignite them, and project them into the air in fireworks displays. Suppose a particular star is projected from an aerial firework at a starting height of 610 ft with an initial upward velocity of 88 ft/s. How long will it take for the star to reach its maximum height? How far above the ground will it be?

The equation h = –16t2 + 88t + 610 gives the height of the star h in feet at time t in seconds.

Since the coefficient of t 2 is negative, the curve opens downward, and the vertex is the maximum point.

10-2


(continued)

Step 2: Find the h-coordinate of the vertex.h = –16(2.75)2 + 88(2.75) + 610 Substitute 2.75 for t.h = 731 Simplify using a calculator.

The maximum height of the star will be about 731 ft.

Step 1: Find the x-coordinate of the vertex.

After 2.75 seconds, the star will be at its greatest height.

b2a– = –88

2(–16) = 2.75

10-2

Graph the boundary curve, y = –x2 + 6x – 5.

Use a dashed line because the solution of the inequality y > –x2 + 6x – 5 does not include the boundary.


Graph the quadratic inequality y > –x2 + 6x – 5.

Shade above the curve.

10-2



1. x = 0, (0, 4)

2. x = –1, (–1, –7)

3. x = 4, (4, –25)

4. x = 1.5, (1.5, –1.75)

5. B

6. E

7. C

8. F

9. A

10. D

11.

12.

13.

14.

15. a. 20 ftb. 400 ft2

16. a. 1.25 s b. 31 ft

17.

10-2


18.

19.

20.

21.

22.

23.

24.

25.

26.

10-2


27.

28.

29.

30.

31.

32–34. Answers may vary. Samples are given.

32. y = 2x2 – 8x + 1

33. y = –3x2

34. y = 2x2 + 4

35. a. 1.2 mb. 7.2 m

36. a. y < –0.1x2 + 12b.

c. Yes; when x = 6, y = 8.4, so the camper will fit.

37. a. $12.50b. $10,000

38. 26 units2

39. 26 units2

10-2

40. Answers may vary. Sample: If the coefficient of the squared term is pos., the vertex point is a min.; if it is neg., the vertex point is a max.

41. Answers may vary. Sample: a affects whether the parabola opens up or down, b affects the axis of symmetry, and c affects the y-intercept.

42. a. w = 13 – b. A = – 2 + 13c. (6.5, 42.25)d. 6.5 ft by 6.5 ft


43. a. 0.4 sb. No; after 0.6 s, the ball will

have a height of about 2.23 m but the net has a height of 2.43 m.

44.

45. a. 0.4 sb. No; it takes about 0.8 s to return to

h = 0.5 m, so it will take more time to reach the ground.

10-2


10-2

46. a. (0, 2)b. x = –2.5c. 5d. y = x2 + 5x + 2e. Answers may vary.

Sample: Test (–4, –2).–2 (–4)2 + 5(–4) + 2–2 16 – 20 + 2–2 = –2

f. No; you would not be able to determine the b value using the vertex formula.

47. A

48. I

49. B

50. [2] axis of symmetry:

x = = 16.7

maximum height:y –0.009(16.7)2 + 0.3(16.7) + 4.5y 7 ft

[1] appropriate methods, but with a minor computational error

51. C

52. A

53. F

54. D

55. B

56. E

–b2a

–0.3 2(–0.009)


57. c2 – 5c – 36

58. 2x2 + 7x – 30

59. 20t 2 + 17t + 3

60. 21n4 – 62n2 + 16

61. 2a3 + 9a2 – a + 20

62. 6r 3 + 9r 2 – 20r + 7

10-2


Graph each relation. Label the axis of symmetry and the vertex.

2. ƒ(x) = –x2 + 4x – 2

14

<3. y – x2 – 2x – 6

1. y = x2 – 8x + 15

x = –4

10-2

Finding and Estimating Square Roots


Simplify each expression.

1. 112 2. (–12)2 3. –(12)2 4. 1.52

5. 0.62 6. 2 7. 2 8. 212

23– 4

5

10-3

1. 112 = 11 • 11 = 121 2. (–12)2 = (–12)(–12) = 144

3. (–12)2 = –(12)(12) = –144 4. 1.52 = (1.5)(1.5) = 2.25

5. 0.62 = (0.6)(0.6) = 0.36 6. 2 = • =

7. 2 = = 8.

2 = • =

Solutions

12

23– 4

5

12

12

14

23–2

3– 45

45

1625

Finding and Estimating Square RootsALGEBRA 1 LESSON 10-3

10-3

49

a. 25



c. – 64

d. –49

e. ± 0

b. ± 925

10-3

positive square root= 5

35

35

35The square roots are and – .= ±

negative square root= –8

For real numbers, the square root of a negative number is undefined.

is undefined

There is only one square root of 0. = 0

a. ± 144 = ± 12


Tell whether each expression is rational or irrational.

c. – 6.25 = –2.5

e. 7 = 2.64575131 . . .

b. – = –0.44721359 . . .15

d. = 0.319

10-3

rational

irrational

rational

rational

irrational

Between what two consecutive integers is 28.34 ?


28.34 is between 5 and 6.

28.34 is between the two consecutive perfect squares 25 and 36.25 < 28.34 < 36

The square roots of 25 and 36 are 5 and 6, respectively.28.345 < < 6

10-3

Find 28.34 to the nearest hundredth.


28.34 5.323532662 Use a calculator.

5.32 Round to the nearest hundredth.

10-3

Suppose a rectangular field has a length three times its width x. The formula d = x2 + (3x)2 gives the distance of the diagonal of a rectangle. Find the distance of the diagonal across the field if x = 8 ft.


d = x2 + (3x)2

d = 82 + (3 • 8)2 Substitute 8 for x.

The diagonal is about 25.3 ft long.

Use a calculator. Round to the nearest tenth.d 25.3

10-3

Simplify.d = 64 + 576

d = 640



1. 13

2. 20

3.

4. 30

5. 0.5

6.

7. –1.1

8. 1.4

9. 0.6

10. –12

11.

13

67

54

12. ± 0.1

13. irrational

14. rational

15. irrational

16. rational

17. 5 and 6

18. 5 and 6

19. –12 and –11

20. 13 and 14

21. 3.46

22. –14.25

23. 107.47

24. –12.25

25. 0.93

26. ± 20

27. 0

28. ± 25

29. ±

30. ± 1.3

31. ±

32. ± 27

33. ± 1.5

34. ± 16

35. ± 0.1

37

19

10-3

=/

54. false; 1 = 1

55. true

56. true

57. False; answers may vary. Sample: 4 + 9 4 + 9.

58. False; answers may vary. Sample: 12 and 3 are irrational but 36 is rational.

59. a. 4 units2

b. unit2

c. 2 units2

d. 2 units


36. ±

37. ± 202

38. 1

39. a. 8450 kmb. 7684 km

40. 21

41. –

42. 1.41

43. 1.26

44. –5.48

45. –33

46. –0.8

8 11

25

47. 6.40

48. 8.66

49. Answers may vary. Sample: The first expression means the neg. square root of 1 and the second expression means the pos. square root of 1.

50. Answers may vary. Sample: 3 and 4

51. 452. a. 5 s

b. 10 sc. No; the object takes

twice as long to fall.

53. False; zero has one square root.

12

10-3


60. 1.3

61. 6

62.

63. 9

64. 11

65. 128

66.

29

67.

68.

69.

70.

71.

72. d 2 – 81

73. 9t 2 – 25

74. 8075

10-3


75. x2 + 26x + 169

76. 16y2 – 56y + 49

77. 10,201

78. 813,604

79. 36k2 – 49

80. 144b2 – 168b + 49

81. –2

82.

83. –

84. –972

85.

86.

3 16

34

1 12814

10-3

1. Simplify each expression.

a. 196 b. ±

2. Tell whether each expression is rational, irrational, or undefined.

a. ± b. –25 c. – 2.25

3. Between what two consecutive integers is – 54?

4. The formula s = 13.5d estimates the speed s in miles per hour that a car was traveling, when it applied its brakes and left a skid mark d feet long on a wet road. Estimate the speed of a car that left a 120 foot long skid mark.


425

35

14 ± 25

irrational undefined rational

–8 and –7

about 40.25 mph

10-3

Solving Quadratic EquationsALGEBRA 1 LESSON 10-4

(For help, go to Lesson10-3.)


1. 36 2. – 81 3. ± 121

4. 1.44 5. 0.25 6. ± 1.21

7. 8. ± 9. ±14

19

49100

10-4


Solutions

13

14

19

49100

12

710

10-4

1. 36 = 6 2. – 81 = –9

3. ± 121 = ±11 4. 1.44 = 1.2

5. 0.25 = 0.5 6. ± 1.21 = ±1.1

7. = 8. ± = ±

9. ± =


Solve each equation by graphing the related function.

a. 2x2 = 0 b. 2x2 + 2 = 0 c. 2x2 – 2 = 0

There is one solution, x = 0.

There is no solution. There are two solutions, x = ±1.

10-4

Graph y = 2x2 Graph y = 2x2 + 2 Graph y = 2x2 – 2

x = ± 25 Find the square roots.


Solve 3x2 – 75 = 0.

3x2 – 75 + 75 = 0 + 75 Add 75 to each side.

3x2 = 75

x2 = 25 Divide each side by 3.

x = ± 5 Simplify.

10-4


S = 4 r 2

315 = 4 r 2 Substitute 315 for S.

Put in calculator ready form.315 (4) = r 2

= r 2315 (4)

Find the principle square root.

5.00668588 r Use a calculator.

The radius of the sphere is about 5 ft.

10-4

A museum is planning an exhibit that will contain a large globe. The surface area of the globe will be 315 ft2. Find the radius of the sphere producing this surface area. Use the equation S = 4 r2, where S is the surface area and r is the radius.



1.

±32.

no solution3.

0

4.

±25.

no solution6.

±3

7.

no solution

8.

0

9.

±2

10-4


10. ± 7

11. ± 21

12. ± 15

13. 0

14. no solution

15. ±

16. ±

17. ± 2

18. ± 27

19. x2 = 256; 16 m

20. x2 = 90; 9.5 ft

21. r 2 = 80; 5.0 cm

14

52

22. a. 6.0 in.b. The length of a radius

cannot be negative.

23. none

24. two

25. one

26. 10.4 in. by 10.4 in.

27. a. 11.3 ftb. 16.0 ftc. No; the radius increases

by about 1.4 times.

28. no solution

29. ± 37

30. ±

31. ± 2.8

32. ± 0.4

33. ± 3.5

34. 3.5 s

35. 121

36. a. n > 0b. n = 0c. n < 0

37. Answers may vary. Sample: Michael subtracted 25 from the left side of the equation but added 25 to the right side.

16

10-4


38. a. 2, –2; 2, –2

b. The first equation multiplied by 2 on both sides equals the second equation.

39. a. square: 4r 2, circle: r 2

b. 4r 2 – r 2 = 80c. 9.7 in., 19.3 in.

40. Answers may vary. Sample:a. 5x2 + 10 = 0, no solutionb. 2x2 + 0 = 0, x = 0c. –20x2 + 80 = 0, x = ± 2

41. 6.3 ft

42. 11.0 cm

43. a. 0.2 mb. 2.5 sc. 3.0 sd. Shorten; as decreases, t decreases.

44. a. –7b. (–7, 0)c. Answers may vary.

Sample: h = 5, –5, (–5, 0)d. (4, 0); the vertex is at (–h, 0).

45. 28 cm

46. B

47. I

48. B

10-4


49. [2]

x-intercepts 1.5, –1.5

[1] minor error in table OR incorrect graph

x y–2 5–1 –40 –71 –42 5

50. [4] a. 96 = 6s2

s2 = 16s = 4, so side is 4 ft.

b. 6(8)2 = 6 • 64 = 384, so surface area is 384 ft2. The surface area is quadrupled.

[3] appropriate methods, but with one computational error

[2] part (a) done correctly[1] no work shown

51. 3

52. –13

53. 40

54. 15

55. 0.2

10-4


56. –1.6

57.

58.

59. (x + 4)(x + 1)

60. (y – 13)(y – 2)

61. (a + 5)(a – 2)

62. (z – 12)(z + 6)

63. (c – 12d)(c – 2d)

64. (t + 2u)(t – u)

65. 3.6135 106

66. 3.48 10–5

67. –8.12 100

68. 31,000

69. 701,000

70. 0.00062

5879

10-4


1. Solve each equation by graphing the related function. If the equation has no solution, write no solution.

a. 2x2 – 8 = 0

b. x2 + 2 = –2

2. Solve each equation by finding square roots.

a. m2 – 25 = 0

b. 49q2 = 9

3. Find the speed of a 4-kg bowling ball with a kinetic energy of 160 joules. Use the equation E = ms2, where m is the object’s mass in kg, E is its kinetic energy, and s is the speed in meters per second.

±2

no solution

±537±

12

about 8.94 m/s

10-4

Factoring to Solve Quadratic EquationsALGEBRA 1 LESSON 10-5


Solve and check each equation.

1. 6 + 4n = 2 2. – 9 = 4 3. 7q + 16 = –3

Factor each expression.

4. 2c2 + 29c + 14 5. 3p2 + 32p + 20 6. 4x2 – 21x – 18

a8

10-5


Solutions1. 6 + 4n = 2

4n = –4 n = –1Check: 6 + 4(–1) = 6 + (–4) = 2

2. – 9 = 4

= 13

a = 104

Check: – 9 = 13 – 9 = 4

3. 7q + 16 = –3 7q = –19

q = –2

Check: 7 (–2 ) + 16 = 7(– ) + 16 = –19 + 16 = –3

a8 a

8

1048

57

57

197

10-5


Solutions (continued)

4. 2c2 + 29c + 14 = (2c + 1)(c + 14)Check: (2c + 1)(c + 14) = 2c2 + 28c + c + 14 = 2c2 + 29c + 14

5. 3p2 + 32p + 20 = (3p + 2)(p + 10)Check: (3p + 2)(p + 10) = 3p2 + 30p + 2p + 20 = 3p2 + 32p + 20

6. 4x2 – 21x – 18 = (4x + 3)(x – 6)Check: (4x + 3)(x – 6) = 4x2 – 24x + 3x – 18 = 4x2 – 21x – 18

10-5


Solve (2x + 3)(x – 4) = 0 by using the Zero Product Property.

(2x + 3)(x – 4) = 0

2x + 3 = 0 or x – 4 = 0 Use the Zero-Product Property.

2x = –3 Solve for x.

x = – 32 or x = 4

Check: Substitute – for x.32 Substitute 4 for x.

(2x + 3)(x – 4) = 0 (2x + 3)(x – 4) = 0

[2(– ) + 3](– – 4) 032

32 [2(4) + 3](4 – 4) 0

(0)(– 5 ) = 012 (11)(0) = 0

10-5


Solve x2 + x – 42 = 0 by factoring.

x2 + x – 42 = 0

(x + 7)(x – 6) = 0 Factor using x2 + x – 42

x + 7 = 0 or x – 6 = 0 Use the Zero-Product Property.

x = –7 or x = 6 Solve for x.

10-5


Solve 3x2 – 2x = 21 by factoring.

3x2 – 2x = 21 Subtract 21 from each side.

(3x + 7)(x – 3) = 0 Factor 3x2 – 2x – 21.

3x + 7 = 0 or x – 3 = 0 Use the Zero-Product Property

3x = –7 Solve for x.

x = – or x = 373

10-5


The diagram shows a pattern for an open-top box. The total area of the sheet of materials used to make the box is 130 in.2. The

height of the box is 1 in. Therefore, 1 in. 1 in. squares are cut from each corner. Find the dimensions of the box.

Define: Let x = width of a side of the box.Then the width of the material = x + 1 + 1 = x + 2The length of the material = x + 3 + 1 + 1 = x + 5

Relate: length width = area of the sheet

10-5


(continued)

Write: (x + 2) (x + 5) = 130

x2 + 7x + 10 = 130 Find the product (x + 2) (x + 5).

x2 + 7x – 120 = 0 Subtract 130 from each side.

(x – 8) (x + 15) = 0 Factor x2 + 7x – 120.

x – 8 = 0 or x + 15 = 0 Use the Zero-Product Property.

x = 8 or x = –15 Solve for x.

The only reasonable solution is 8. So the dimensions of the box are 8 in. 11 in. 1 in.

10-5

12. 3, –4

13. –3, –5

14. –4, 7

15. 0, 6

16. 1, 2.5

17. –5, –

18. –2.5, 2.5

19. , –4

20. 2, 3

21. , –4

22. 5 cm

23. 5

25

32



1. 3, 7

2. –4, 4.5

3. 0, –1

4. 0, 2.5

5. – , –

6. , –

7. 1, –4

8. –2, 7

9. 0, 8

10. 5, 11

11. –2, 5

27

45

74

83

24. 6 ft 15 ft

25. base: 10 ftheight: 22 ft

26. 2 and 3 or 7 and 8

27. 2q2 + 22q + 60 = 0; –6, –5

28. 6n2 – 5n – 4 = 0; , –

29. 4y2 + 12y + 9 = 0; –

30. a2 + 6a + 9 = 0; –3

31. 2t 2 + 11t + 12 = 0; –1.5, –4

32. x2 – 10x + 24 = 0; 4, 6

33. 2k2 + 11k – 63 = 0; , –9

34. 20y2 + 41y – 9 = 0; , –

7215

94

43

12

32

10-5

13


35. 8 in. 10 in.

36. a. 2 sb. about 19 ft

37. Answers may vary. Sample: To solve a quadratic equation, write the equation in standard form, factor the quadratic expression, use the Zero-Product Property, and solve for the variable.

x2 + 8x = –15x2 + 8x + 15 = 0(x + 3)(x + 5) = 0x + 3 = 0 or x + 5 = 0x = –3 or x = –5

38. Answers may vary. Sample: x = 6, a = 2, b = 1; x = 3, a = 1, b = 11

39. Answers may vary. Sample:x2 – 2x – 8 = 0

(x – 4)(x + 2) = 0x – 4 = 0 or x + 2 = 0 x = 4 or x = –2

40. a. 0, 1; –1, 0b. 0

41. 0, 4, 6

42. 0, 1, 4

43. 0, 3

44. 0, 7, –10

45. 0, 1, 9

46. 0, 4, –5

47. 4

10-5


48. Answers may vary. Samples:a. x2 – 3x – 40 = 0b. x2 – x – 6 = 0c. 2x2 + 19x – 10 = 0d. 21x2 + x – 10 = 0

49. –1, 1, –5

50. –2, 2, –1

51. D

52. H

53. A

54. C

55. D

56. [2] 3x2 + 20x + 1 = 83x2 + 20x – 7 = 0(3x – 1)(x + 7) = 0x = , x = –7


57. x2 = 320; 17.9 ft

58. 38 = r 2; 3.5 ft

59. (2x + 3)(x + 5)

60. (3y – 1)(y – 3)

61. (4t – 3)(t + 2)

62. (3n – 1)(2n + 3)

63. 5a(3a + 2)(a – 4)

64. –2b(3b – 2)(3b – 5)

10-5

13


1. Solve (2x – 3)(x + 2) = 0.

Solve by factoring.

2. 6 = a2 – 5a 3. 12x + 4 = –9x2 4. 4y2 = 25

–2, 32

–1, 6 – 23 ± 5

2

10-5

Completing the SquareALGEBRA 1 LESSON 10-6


Find each square.

1. (d – 4)2 2. (x + 11)2 3. (k – 8)2

Factor.

4. b2 + 10b + 25 5. t2 + 14t + 49 6. n2 – 18n + 81

10-6


1. (d – 4)2 = d2 – 2d(4) + 42 = d2 – 8d + 16

2. (x + 11)2 = x2 + 2x(11) + 112 = x2 + 22x + 121

3. (k – 8)2 = k2 – 2k(8) + 82 = k2 – 16k + 64

4. b2 + 10b + 25 = b2 + 2b(5) + 52 = (b + 5)2

5. t2 + 14t + 49 = t2 + 2t(7) + 72 = (t + 7)2

6. n2 – 18n + 81 = n2 – 2n(9) + 92 = (n – 9)2

Solutions

10-6


Find the value of c to complete the square for x2 – 16x + c.

The value of b in the expression x2 – 16x + c is –16.

The term to add to x2 – 16x is or 64. –162

2

10-6


Solve the equation x2 + 5 = 50 by completing the square.

Step 1: Write the left side of x2 + 5 = 50 as a perfect square.

x2 + 5 = 50

x2 + 5 + = 50 +52

2 52

2 Add , or , to each side of the

equation.

52

2 254

52

2x + 200

4 + 254= Write x2 + 5x + as a square. 5

22

Rewrite 50 as a fraction with denominator 4.

52

2x + = 225

4

10-6


(continued)

Step 2: Solve the equation.

Find the square root of each side.52

2x + = 225

4±

52

x + = 152

± Simplify.

52

x + = 152 or

52x + = 15

2– Write as two equations.

x = 5 or x = –10 Solve for x.

10-6


Solve x2 + 10x – 16 = 0 by completing the square. Round to the nearest hundredth.

Step 1: Rewrite the equation in the form x2 + bx = c and complete the square.

x2 + 10x – 16 = 0

x2 + 10x = 16 Add 16 to each side of the equation.

(x + 5)2 = 41 Write x2 + 10x +25 as a square.

x2 + 10x + 25 = 16 + 25 Add , or 25, to each side of the equation.102

2

10-6


(continued)

Step 2: Solve the equation.

x + 5 = ± 41 Find the square root of each side.

Use a calculator to find 41x + 5 ± 6.40

x + 5 6.40 or x + 5 –6.40 Write as two equations.

Subtract 5 from each side.x 6.40 – 5 or x –6.40 – 5

x 1.40 or x –11.40 Simplify

10-6


Suppose you wish to section off a soccer field as shown in the diagram to run a variety of practice drills. If the area of the field is 6000 yd2, what is the value of x?

Define: width = x + 10 + 10 = x + 20length = x + x + 10 + 10 = 2x + 20

Relate: length width = area

Write: (2x + 20)(x + 20) = 60002x2 + 60x + 400 = 6000

Step 1: Rewrite the equation in the form x2 + bx = c.

2x2 + 60x + 400 = 60002x2 + 60x = 5600 Subtract 400 from each side.x2 + 30x = 2800 Divide each side by 2.

10-6


(continued)

Step 2: Complete the square.

x2 + 30x + 255 = 2800 + 225 Add , or 225, to each side.302

2

(x + 15)2 = 3025 Write x2 + 30x + 255 as a square.

Step 3: Solve each equation.

(x + 15) = ± 3025 Take the square root of each side.

x + 15 = ± 55 Use a calculator.

x + 15 = 55 or x + 15 = –55x = 40 or x = –70 Use the positive answer for this problem.

The value of x is 40 yd.

10-6



1. 49

2. 16

3. 400

4. 9

5. 144

6. 324

7. 4, –12

8. 13.06, –3.06

9. –5, –17

10. 1.24, –7.24

11. 9, –29

12. 19, –17

13. 7, –5

14. –2.17, –7.83

15. 11, 1

16. 1.19, –4.19

17. 4.82, –5.82

18. 22, –31

19. 1

20. 4

21.

22. 2.16, –4.16

23. 5, –1

24. 7, –2

25. a. (2x + 1)(x + 1)b. 2x2 + 3x + 1 = 28c. 3

26. –0.27, –3.73

27. –3, –4

28. 4, –10

29. 6, 2

30. 8.32, 1.68

31. no solution

32. 9.37, –1.87

33. 8.12, –0.12

34. –4, –5

10-6

81 100


35. a. = 50 – 2wb. w(50 – 2w) = 150; 21.5, 3.5c. 7 ft 21.5 ft or 43 ft 3.5 ftd. No; the answers in part

(b) were rounded.

36. The student did not divide each side of the equation by 4.

37. Answers may vary. Sample: Add 1 to each side of the equation, and then complete the square by adding 225 to each side of the equation. Write x2 + 30x + 225 as the square (x + 15)2 and add 1 and 225 to get 226. Then take square roots and solve the resulting equations.

38. Answers may vary. Sample:x2 + 10x – 50 = 0

x2 + 10x = 50x2 + 10x + 25 = 50 + 25

(x + 5)2 = ± 75x + 5 = ± 8.7x + 5 ± 8.7

x + 5 8.7 or x + 5 –8.7 x 3.7 or x –13.7

39. 5.16, –1.16

40. 6.83, 1.17

41. 5.6 ft by 14.2 ft

42. a. 6x2 + 28xb. 6x2 + 28x = 384c. 13 in. 6 in. 6 in.

10-6


43. a. A = x2 + 5x + 1

b. 6.86c. 207.5 ft2

44. a. 3 ± 5 b. (3, –5)c. Answers may vary.

Sample: p is the x-coordinate of the vertex.

45. B

46. I

47. D

48. [2] (x)(x + x + 4) = 200

(x)(2x + 4) = 200

(x)(x + 2) = 200

x2 + 2x = 200

x2 + 2x + 1 = 201

(x + 1)2 = 201

x + 1 ±14.18

x 13.18 x –15.18

The value of x is about 13.18 cm.


72

12

12

10-6


49. [4] a. (8 + x)(12 + x) = 2 • (8 • 12)x2 + 20x – 96 = 0

b. x2 + 20x = 96x2 + 20x + 100 = 196

(x + 10)2 = 196x + 10 = ±14

x = 4c. 12 ft by 16 ft


[2] part (c) not done[1] no work shown

50. –3, 7

51. –6, –5

52. 0, 5

53. – ,

54. –

55. – ,

56. (x + 2)2

57. (t – 11)2

58. (b + 5)(b – 5)

59. (4c + 3)2

60. (7s + 13)(7s – 13)

61. 2m(2m + 3)(2m – 3)

62. (5m + 12)2

63. (20k + 3)(20k – 3)

64. (16g – 11)(16g + 11)

83

83

3216

52

65. r 12

66. p13

67. –y

68.

69. –

70. t 29

1 m40

1w

10-6


1. x2 + 14x = –43

2. 3x2 + 6x – 24 = 0

3. 4x2 + 16x + 8 = 40

Solve each equation by completing the square. If necessary, round to the nearest hundredth.

–9.45, –4.55

–4, 2

–5.46, 1.46

10-6

Using the Quadratic Formula

(For help, go to Lesson 10-6.)ALGEBRA 1 LESSON 10-7

Find the value of c to complete the square for each expression.

1. x2 + 6x + c 2. x2 + 7x + c 3. x2 – 9x + c

Solve each equation by completing the square.

4. x2 – 10x + 24 = 0 5. x2 + 16x – 36 = 0

6. 3x2 + 12x – 15 = 0 7. 2x2 – 2x – 112 = 0

10-7

Using the Quadratic FormulaALGEBRA 1 LESSON 10-7

1. x2 + 6x + c; c = 2 = 32 = 9 2. x2 + 7x + c; c = 2 =

3. x2 – 9x + c; c = 2 =

4. x2 – 10x + 24 = 0;5. x2 + 16x – 36 = 0;

c = 2 = (–5)2 = 25 c = 2 = 82 = 64

x2 – 10x = –24 x2 + 16x = 36x2 – 10x + 25 = –24 + 25 x2 + 16x + 64 = 36 + 64 (x – 5)2 = 1 (x + 8)2 = 100 (x – 5) = ±1 (x + 8) = ±10x – 5 = 1 or x – 5 = –1 x + 8 = 10 or x + 8 = –10 x = 6 or x = 4 x = 2 or x = –18

62

72

494

–92

814

–102

162

Solutions

10-7



6. 3x2 + 12x – 15 = 0 7. 2x2 – 2x – 112 = 0

3(x2 + 4x – 5) = 0 2(x2 – x – 56) = 0

x2 + 4x – 5 = 0; x2 – x – 56 = 0;

c = 2 = 22 = 4 c = 2 =

x2 + 4x = 5 x2 – x = 56

x2 + 4x + 4 = 5 + 4 x2 – x + = 56 +

(x + 2)2 = 9 (x – )2 =

(x + 2) = ±3 x – = ±

x + 2 = 3 or x + 2 = –3 x – = or x – =

x = 1 or x = –5 x = 8 or x = –7

42

–12

14

14

14

12

2254

12

152

12

152

12

–152

10-7


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

x2 + 3x + 2 = 0 Add 3x to each side and write in standard form.

x = –b ± b2 – 4ac2a Use the quadratic formula.

x = –3 ± (–3)2 – 4(1)(2)2(1) Substitute 1 for a, 3 for b, and 2 for c.

x = –3 ± 12 Simplify.

x = –3 + 12 x = –3 – 1

2or Write two solutions.

x = –1 or x = –2 Simplify.

10-7


(continued)

Check: for x = –1 for x = –2

(–1)2 + 3(–1) + 2 0 (–2)2 + 3(–2) + 2 0

1 – 3 + 2 0 4 – 6 + 2 0

0 = 0 0 = 0

10-7


Solve 3x2 + 4x – 8 = 0. Round the solutions to the nearest hundredth.

x = –b ± b2 – 4ac2a Use the quadratic formula.

x = –4 ± 42 – 4(3)(–8)2(3) Substitute 3 for a, 4 for b, and –8 for c.

–4 ± 1126x =

Use a calculator.x –4 + 10.5830052446 x –4 – 10.583005244

6or

x 1.10 or x –2.43 Round to the nearest hundredth.

10-7

x = or x = Write two solutions.–4 + 1126

–4 – 1126


A child throws a ball upward with an initial upward velocity of 15 ft/s from a height of 2 ft. If no one catches the ball, after how many seconds will it land? Use the vertical motion formula h = –16t2 + vt + c, where h = 0, v = velocity, c = starting height, and t = time to land. Round to the nearest hundredth of a second.

Step 1: Use the vertical motion formula.h = –16t2 + vt + c0 = –16t2 + 15t + 2 Substitute 0 for h, 15 for v, and 2 for c.

Step 2: Use the quadratic formula.

x = –b ± b2 – 4ac2a

10-7


(continued)

t = –15 ± 152 – 4(–16)(2)2(–16) Substitute –16 for a, 15 for b, 2 for c, and t for x.

t = –15 + 18.79–32 or t = –15 – 18.79

–32 Write two solutions.

t –0.12 or t 1.06 Simplify. Use the positive answer because it is the only reasonable answer in this situation.

The ball will land in about 1.06 seconds.

–15 ± 225 + 128–32t = Simplify.

–15 ± 353–32t =

10-7


Which method(s) would you choose to solve each equation? Justify your reasoning.

10-7

a. 5x2 + 8x – 14 = 0 Quadratic formula; the equation cannot be factored easily.

b. 25x2 – 169 = 0 Square roots; there is no x term.

c. x2 – 2x – 3 = 0 Factoring; the equation is easily factorable.

d. x2 – 5x + 3 = 0 Quadratic formula, completing the square, or graphing; the x2 term is 1, but the equation is not factorable.

e. 16x2 – 96x + 135 = 0 Quadratic formula; the equation cannot be factored easily and the numbers are large.



1. –1, –1.5

2. 2.8, –6

3. 1.5

4. –0.67, –15

5. 6.67, –0.25

6. –4, –9

7. 2.67, –16

8. 13, –8.5

9. 16, –2.4

10. 0.07, –2.67

11. 10.42, 1.58

13. 1.14, –0.7714. 2.20, –3.0315. 3.84, –0.1716. a. 0 = –16t 2 + 10t + 3

b. t 0.8; 0.8 s17. a. 0 = –16t 2 + 50t + 3.5

b. t 3.2; 3.2 s18. Completing the square or graphing;

the x2 term is 1 but the equation is not factorable.19. Factoring or square roots; the equation

is easily factorable and there is no x term.20. Quadratic formula; the equation

cannot be factored.21. Quadratic formula; the equation

cannot be factored.22. Factoring; the equation is easily factorable.

10-7


23. Quadratic formula; the equation cannot be factored.

24. 6, –6

25. 0.87, –1.54

26. 1.41, –1.41

27. 1.28, –2.61

28. 2

29. 3, –3

30. 1.72, –0.39

31. 1.4, –1

32. 2.23, –1.43

33. a. 7 ft 8 ftb. x(x + 1) = 60, 7.26 ft 8.26 ft

34. About 2.1s35. Answers may vary. Sample: You solve

the linear equation using transformations and you solve the quadratic equation using the quadratic formula.

36. 7.40 ft and 5.40 ft

37. 13.44 cm and 7.44 cm

38. Answers may vary. Sample: A rectangle has length x. Its width is 5 feet longer than three times the length. Find the dimensions if its area is 182 ft2.

7 ft 26 ft

10-7


39. if the expression b2 – 4ac equals zero

40. about 1.9 s

41. a. Check students’ work.b. 356.9 millionc. 2007

42. a. s = –b. 6.5

ab

46. [2]

–1.8, 3.7[1] correct substitution into

quadratic formula, with one calculation error

47. 1.54, 8.46

48. –1, –2

49. 0.1, –6.1

50. (2c + 5)(c + 3)

51. (3z – 2)(z + 4)

52. (5n + 2)(n – 7)

11 ± (112) – 4(6)(–40)2(6)

53. (6v – 5)(2v + 7)

54. (2x – 1)(3x – 5)

55. (5t + 3)(3t + 2)

x =

10-7

43. a. 7.62 107 lbb. 3.3 104 tonsc. 6.27 s

44. D

45. F


1. Solve 2x2 – 11x + 12 = 0 by using the quadratic formula.

2. Solve 4x2 – 12x = 64. Round the solutions to the nearest hundredth.

3. Suppose a model rocket is launched from a platform 2 ft above the ground with an initial upward velocity of 100 ft/s. After how many seconds will the rocket hit the ground? Round the solution to the nearest hundredth.

1.5, 4

–2.77, 5.77

6.27 seconds

10-7

Using the Discriminant


Evaluate b2 – 4ac for the given values of a, b, and c.

1. a = 3, b = 4, c = 8 2. a = –2, b = 0, c = 9

3. a = 11, b = –5, c = 7

Solve using the quadratic formula. If necessary, round to the nearest hundredth.

4. 3x2 – 7x + 1 = 0 5. 4x2 + x – 1 = 0 6. x2 – 12x + 35

= 0

10-8

Using the DiscriminantALGEBRA 1 LESSON 10-8

Solutions

1. b2 – 4ac for a = 3, b = 4, c = 8: 42 – 4(3)(8) = 16 – 96 = –80

2. b2 – 4ac for a = –2, b = 0, c = 9: 02 – 4(–2)(9) = 0 + 72 = 72

3. b2 – 4ac for a = 11, b = –5, c = 7: (–5)2 – 4(11)(7) = 25 – 308 = –283

4. 3x2 – 7x + 1 = 0; x = for a = 3, b = –7, c = 1:

x = = =

x = 2.18 or x = 0.15

–b ± b2 – 4ac2a

–(–7) ± (–7)2 – 4(3)(1)2(3)

7 ± 49 – 126

7 ± 376

7 + 376

7 – 376

10-8



5. 4x2 + x – 1 = 0; x = for a = 4, b = 1, c = –1:

x = = =

x = 0.39 or x = –0.64

–b ± b2 – 4ac2a

–1 ± 12 – 4(4)(–1)2(4)

–1 ± 1 + 168

–1 ± 178

6. x2 – 12x + 35 = 0; x = for a = 1, b = –12, c = 35:

x = =

x = =

x = = = 7 or x = = = 5

–1 – 178

–1 + 178

–b ± b2 – 4ac2a

–(–12) ± (–12)2 – 4(1)(35)2(1)

12 ± 144 – 1402

12 ± 42

12 ± 22

12 + 22

142

12 – 22

102

10-8


Find the number of solutions of x2 = –3x – 7 using the discriminant.

x2 + 3x + 7 = 0 Write in standard form.

b2 – 4ac = 32 – 4(1)(7) Evaluate the discriminant. Substitute for a, b, and c.

= 9 – 28 Use the order of operations.

= –19 Simplify.

Since –19 < 0, the equation has no solution.

10-8


A football is kicked from a starting height of 3 ft with an initial upward velocity of 40 ft/s. Will the football ever reach a height of 30 ft?

h = –16t2 + vt + c Use the vertical motion formula.

30 = –16t2 + 40t + 3 Substitute 30 for h, 40 for v, and 3 for c.

0 = –16t2 + 40t – 27 Write in standard form.

b2 – 4ac = (40)2 – 4 (–16)(–27) Evaluate the discriminant.

= 1600 – 1728 Use the order of operations.

= –128 Simplify.

The discriminant is negative. The football will never reach a height of 30 ft.

10-8



1. A

2. C

3. B

4. 0

5. 1

6. 2

7. 2

8. 2

9. 2

10. 0

11. 2

12. 2

13. 1

14. 2

15. 0

16. none

17. No; the discriminant is negative.

18. a. yesb. noc. nod. no

19. 0

20. 0

21. 2

22. 2

23. 0

24. 2

25. a. S = –0.75p2 + 54pb. noc. $36d. If a product is too expensive,

fewer people will buy it.

26. a. k > 4b. k = 4c. k < 4

27. a. A2 ^ 2 – 4;A2 ^ 2 – 8

b. |b| < 2

10-8


30. a. 16; 5, 1b. 81; 4, –5c. 73; 3.89, –0.39d. Rational; the square root

of a discriminant that is a perfect square is a pos. integer.

31. no

32. no

33. yes; 1, –1.25

34. yes; –1,

35. no

36. yes; 2.5, –1

37. Answers may vary. Sample: Use values for a, b, and c such that the discriminant is positive.

38. never

39. sometimes

40. always

41. 2; since the parabola crosses the x-axis once, it must cross again.

42. y = 2x2 + 8x + 10 has a vertex closer to the x-axis; its discriminant is closer to zero.

43. C

44. G

45. B

46. D

47. A

10-8

23


48. [2] x(25 – x) = 136x2 – 25x + 136 = 0

x =

= = 17 or 8

yes, 17 cm by 8 cm[1] no work shown OR appropriate

methods with one computational error

49. 0.5, –1.5

50. 1.83, –3.83

51. 0.61, –0.27

52. 1.79, –2.79

53. 2.54, 0.13

54. 2, 1.5

–(–25) ± (–25)2 – 4(1)(136)2(1)

25 ± 92

55. $1093.81

56. $312.88

57. $6104.48

58. arithmetic

59. arithmetic

60. geometric

61. arithmetic

10-8


Find the number of solutions for each equation.

1. 3x2 – 4x = 7

2. 4x2 = 4x – 1

3. –3x2 + 2x – 12 = 0

4. A ball is thrown from a starting height of 4 ft with an initial upward velocity of 30 ft/s. Is it possible for the ball to reach a height of 18 ft?

two

one

none

yes

10-8

Choosing a ModelALGEBRA 1 LESSON 10-9

(For help, go to Lessons 6-2, 8-7 and 10-1.)

Graph each function.

1. y = 3x – 1 2. y = x + 2

3. y = 2x 4. y = x

5. y = x2 + 5 6. y = 2x2 – 1

14

13

10-9


1. y = 3x – 1 2. y = x + 2

3. y = 2x 4. y = x

5. y = x2 + 5 6. y = 2x2 – 1

14

13

Solutions

10-9


Graph each set of points. Which model is most appropriate for each set?

a. (–2, 2.25), (0, 3),(1, 4) (2, 6)

b. (–2, –2), (0, 2),(1, 4), (2, 6)

c. (–1, 5), (2, 11), (0, 3),(1, 5), (–2, 11)

exponential model linear modelquadratic model

10-9


a. Which kind of function best models the data below? Write an equation to model the data.

x y0 01 1.42 5.63 12.64 22.4

Step 1: Graph the data

There is a common second difference, 2.8.

Step 2: The data appear to be quadratic. Test for a common second difference.

x y0 01 1.42 5.63 12.64 22.4

+1

+1

+1+1

+ 1.4

+ 4.2+ 7.0+ 9.8

+ 2.8

+ 2.8

+ 2.8

10-9


a. (continued)

Step 3: Write a quadratic model. y = ax2

5.6 = a(2)2 Use a point other than(0, 0) to find a.

5.6 = 4a Simplify.1.4 = a Divide each side

by 4. y = 1.4x2 Write a quadratic

function.

Step 4: Test two points other than (2, 5.6) and (0, 0)

y = 1.4(1)2 y = 1.4(3)2

y = 1.4 • 1 y = 1.4 • 9y = 1.4 y = 12.6(1, 1.4) and (3, 12.6) are both data points.

The equation y = 1.4x2 models the data.

10-9


b. Which kind of function best models the data below? Write an equation to model the data.

x y–1 4 0 2 1 1 2 0.5 3 0.25

Step 1: Graph the data Step 2: The data appear to suggest an exponential model. Test for a common ratio.

x y–1 4 0 2 1 1 2 0.5 3 0.25

+1

+1

+1+1

2 4 = 0.51 2 = 0.50.5 1 = 0.50.25 0.5 = 0.5

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b. (continued)

Step 4: Test two points other than (0, 2).

y = 2 • 0.51 y = 2 • 0.52

y = 2 • 0.5 y = 2 • 0.25

y = 1.0 y = 0.5

(1, 1) and (2, 0.5) are both data points.

Step 3: Write an exponential model.

Relate: y = a • bx

Define: Let a = the initial value, 2.

Let b = the decay factor, 0.5.

Write: y = 2 • 0.5x

The equation y = 2 • 0.5x models the data.

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Suppose you are studying deer that live in an area. The data in the table was collected by a local conservation organization. It indicates the number of deer estimated to be living in the area over a five-year period. Determine which kind of function best models the data. Write an equation to model the data.

Year Estimated Population

0 90 1 69 2 52 3 40 4 31

Step 1: Graph the data to decide which model is most appropriate.

The graph curves, and it does not look quadratic. It may be exponential.

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(continued)

Step 2: Test for a common ratio.

Year Estimated Population

0 90 1 69 2 52 3 40 4 31

+1

+1

+1+1

69 90 0.76652 69 0.75340 52 0.76931 40 0.775

The common ratio is roughly 0.77.

The population of deer is roughly 0.77 times its value the previous year.

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(continued)

Step 3: Write an exponential model.

Relate: y = a • bx

Define: Let a = the initial value, 90.

Let b = the decay factor, 0.77.

Write: y = 90 • 0.77x

Step 4: Test two points other than (0, 90).

y = 90 • 0.771 y = 90 • 0.772

y 69 y 53

The predicted value (1, 69) matches the corresponding data point. The point (2, 53) is close to the data point (2, 52).

The equation y = 90 • 0.77x models the data.

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5.

exponential

6.

linear

7. quadratic; y = 1.5x2

8. linear; y = 2x – 5

9. quadratic; y = 2.8x2



1.

quadratic

2.

linear

3.

exponential

4.

quadratic

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10. exponential; y = 1 • 1.2x

11. exponential; y = 5 • 0.4x

12. linear; y = – x + 2

13. a.

linearb. 65, 64, 64; yesc. 64d. y = 64x – 5

12

14. a. exponentialb. y = 16,500 • 0.88x

15. a. 41, 123, 206b. 82, 83c. d = 41t 2

d. 256.25 cm16. a.

linearb. 5 yearsc. 600, 600, 600; 120, 120, 120d. p = 120t + 5100

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17. a. 5b. 398, 429, 407, 389;

79.6, 85.8, 81.4, 77.8c. 81.2d. p = 81.2t + 4457e. 6893 million,

or about 6.9 billion

18. Answers may vary. Sample: Linear data have a common first difference, quadratic data have a common second difference, and exponential data have a common ratio.

19. y = 0.875x2 – 0.435x + 1.515

20. y = 1.987 • 0.770x

21. y = 2.125x2 – 4.145x + 2.955

22. y = –0.336x2 – 0.219x + 4.666

23. y = –1.1x + 3.5

24. y = 0.102 • 2.582x

25. a. i.

ii.

x y1 –22 13 64 135 22

) 3 ) 2) 5 ) 2) 7 ) 2) 9

x y1 32 123 274 485 75

) 9 ) 6) 15 ) 6) 21 ) 6) 27

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iii.

b. The second common difference is twice the coefficient of x2.

c. When second differences are the same, the data are quadratic. You can determine the coefficient of x2 by dividing the second difference by 2.


x y1 –12 63 214 445 75

) 7 ) 8) 15 ) 8) 23 ) 8) 31

26. Answers may vary. Sample:

27. a. quadratic b. d = 13.6t 2

c. 54.5 ft

28. Check students’ work.

x y0 52 134 296 53

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29. a. 1.85, 1.28, 1.45, 1.43b. 139, 85, 174, 240c. –54, 89, 66d. 1.85; the ratio is much

larger than the other ratios.e. 85; the difference is much

smaller than the other differences.f.

30. B

31. H

32. [2] p = 33,500(1.014)n, 33,500(1.014)10 38,497

[1] correct formula, inaccurate evaluation

33. [4] a. linearb. d = –2.5n + 43.5c. 18


[2] part (c) not answered[1] no work shown

34. 1

35. 0

36. 2

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37.

38.

39.

40.

41.

42.

43. 0.125

44. –32

45.

46. 250

47. 16

48. 30.375

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227


Which kind of function best models the data in each table? Write an equation to model the data.

x y–1 15 0 3 1 0.6 2 0.12 3 0.024

1. x y–1 –5 0 –3 1 –1 2 1 3 3

2. x y–1 2.2 0 0 1 2.2 2 8.8 3 19.8

3.

exponential;y = 3 • 0.2x

linear;y = 2x – 3

quadratic;y = 2.2x2

10-9

Quadratic Equations and FunctionsALGEBRA 1 CHAPTER 10

1. D2. C3. A4. B5. x = 0, (0, –7); min.6. x = 1.5, (1.5, –0.25); min.7. x = 2.5, (2.5, 11.5); max.8. x = –6, (–6, –18); min.9.

10.

11.

12.

13.

14.

15. Answers may vary. Sample: You can tell how wide it is and whether it opens upward or downward.

10-A

Quadratic Equations and FunctionsALGEBRA 1 CHAPTER 10

16. 117. 018. 219. 220.

21.

22. 1.2

23. 40

24.

25. 0.2

26. 5, 6

27. 11, 12

28. 18, 19

29. –3, –2

30. 1

31. no solution

32. 2

33. 2

23

34. 135. 5, –536. 3.33, –137. 2, –838. 1, –239. 0.24, –4.2440. 1, –2.3341. Answers may vary.

Sample: y = –x2 + 4;

42. 140 = 10 r 2, 2.1 ft

43. 800 = 2w 2, w = 20 ft,

= 40 ft

44. quadratic; y = x2

45. exponential; y = (2x)

46. linear; y = x + 2

47. exponential; y = 40(0.5x)

10-A

12

12

exploring quadratic graphs

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