Solutions for Precalculus A Unit Circle Approach 3rd Us Edition by Ratti by 6alsm

Chapter 2 Polynomial and Rational Functions

2.1 Quadratic Functions

2.1 Practice Problems

1. Substitute 1 for h, 5 for k, 3 for x, and 7 for y in the standard form for a quadratic equation to solve for a: 2 7(31)5745 aa 

3 a  . The equation is  2 315.yx Since a = 3 > 0, f has a minimum value of 5 at x = 1.

2. The graph of  2 213fxx is a parabola with a = 2, h = 1 and k = 3. Thus, the vertex is ( 1, 3). The parabola opens down because a < 0. Now, find the x-intercepts:

22 0213213 xx 

2 333 111 222 xxx 

0.22 or 2.22 xx . Next, find the y-intercept:  2 020131 f

. Plot the vertex, the x-intercepts, and the y-intercept, and join them with a parabola.

3. The graph of  2 336fxxx is a parabola with a = 3, b = 3 and c = 6. The parabola opens up because a > 0. Now, find the vertex: 31 22(3)2 b h a

2 11127 336 2224

Thus, the vertex (h, k) is 127 , 24

. Next, find the x-intercepts:

Now, find the y-intercept:

2 0303066 f

Thus, the intercepts are ( 1, 0), (2, 0) and (0, 6). Use the fact that the parabola is symmetric with respect to its axis, 1 2 , x  to locate additional points. Plot the vertex, the x-intercepts, the y-intercept, and any additional points, and join them with a parabola.

4. The graph of  2 361fxxx is a parabola with a = 3, b = 6 and c = 1. The parabola opens up because a > 0. Complete the square to write the equation in standard form:

 22 361321gxxxgxxx     2 2 32113314gxxxx

Thus, the vertex is ( 1, 4). The domain of f is   ,  and the range is   4,.

Next, find the x-intercepts:   22 4 03141 3 xx  2323 112.15 or 33 xxx 

0.15 x  . Now, find the y-intercept:  2 0306011 f  . Use the fact that the parabola is symmetric with respect to its axis, x = 1, to locate additional points. Plot the vertex, the x-intercepts, the y-intercept, and any additional points, and join them with a parabola. (continuedonnextpage)

(continued)

The graph of f is below the x-axis between the x-intercepts, so the solution set for  2 3610gxxx is 2323 1,133 

or 323323 33,.

5.  2 00 2 E g httvth  2 0 32 fts,100 ft, max height244 ft E gh

a. Using the given values, we have  2 0 2 0 32 100 2 16100 httvt tvt 

Using the formula for the vertex of a parabola gives  00 . 21632 vv t  This is the time at which the maximum height h(t) = 244 ft is attained. Thus,

0 2 00 0 222 000 244 32 16100 3232 100100 643264 v hth vv v vvv

Solving for 0 v yields 22 00 2 00 244100144 6464 144641446412896 vv vv

Thus,  2 1696100.httt

b. Using the formula for the vertex of a parabola gives  9696 3. 21632 t 

The ball reached its highest point 3 seconds after it was released.

6. Let x = the length of the playground and y = the width of the playground. Then   21000500 xyxy  500. yx 

The area of the playground is  2 500500. Axxyxxxx 

The vertex for the parabola is (h, k) where  500 250 21 h  and  2 50025025062,500. k 

Thus, the maximum area that can be enclosed is 62,500 ft2. The playground is a square with side length 250 ft.

2.1 Exercises

Concepts and Vocabulary

1. The graph of  2 ,0,fxaxbxca is a parabola.

2. The vertex of the graph of  2 ,0,fxaxhka is (h, k).

3. The graph of  2 ,0,fxaxhka is symmetric with respect to the vertical line x = h.

4. The x-coordinate of the vertex of  2 ,0,fxaxbxca is given by 2 . b a x 

5. True.

6. False. The graph of  2 2 fxx  is the graph of  2 gxx  compressed horizontally by a factor of 2.

7. False. If a > 0, then the parabola opens up and has a minimum.

8. True Building Skills

9. f 10. d 11. a 12. e

13. h 14. b 15. g 16. c

17. Substitute 8 for y and 2 for x to solve for a: 2 8(2)2. aa The equation is 2 2.yx 

18. Substitute 3 for y and –3 for x to solve for a: 2 1 3(3). 3 aa The equation is 2 1 . 3 yx 

19. Substitute 20 for y and 2 for x to solve for a: 2 20(2)5. aa The equation is 2 5.yx 

20. Substitute –6 for y and –3 for x to solve for a: 2 2 6(3). 3 aa The equation is 2 2 . 3 yx 

21. Substitute 0 for h, 0 for k, 8 for y, and –2 for x in the standard form for a quadratic equation to solve for a: 2 8(20)084 aa 

2. a  The equation is 2 2.yx 

22. Substitute 2 for h, 0 for k, 3 for y, and 1 for x in the standard form for a quadratic equation to solve for a: 2 3(12)03. aa 

The equation is 2 3(2).yx

23. Substitute –3 for h, 0 for k, –4 for y, and –5 for x in the standard form for a quadratic equation to solve for a: 2 4(5(3))01. aa 

The equation is 2 (3).yx

24. Substitute 0 for h, 1 for k, 0 for y, and –1 for x in the standard form for a quadratic equation to solve for a: 2 0(10)11. aa 

The equation is 2 1. yx

25. Substitute 2 for h, 5 for k, 7 for y, and 3 for x in the standard form for a quadratic equation to solve for a: 2 7(32)52. aa 

The equation is 2 2(2)5.yx

26. Substitute –3 for h, 4 for k, 0 for y, and 0 for x in the standard form for a quadratic equation to solve for a: 2 4 0(0(3))4. 9 aa 

The equation is 2 4 (3)4. 9 yx

27. Substitute 2 for h, –3 for k, 8 for y, and –5 for x in the standard form for a quadratic equation to solve for a: 2 11 8(52)3. 49 aa 

The equation is 2 11 (2)3. 49 yx

28. Substitute –3 for h, –2 for k, –8 for y, and 0 for x in the standard form for a quadratic equation to solve for a: 2 2 8(0(3))2. 3 aa 

The equation is 2 2 (3)2. 3 yx

29. Substitute 1 2 for h, 1 2 for k, 1 4 for y, and 3 4 for x in the standard form for a quadratic equation to solve for a: 2 1311 12. 4422 aa 

The equation is 2 11 12. 22 yx

30. Substitute 3 2 for h, 5 2 for k, 55 8 for y, and 1 for x in the standard form for a quadratic equation to solve for a: 2 55353 1. 8222 aa

The equation is 2 335 222yx

31. The vertex is (–2, 0), and the graph passes through (0, 3). Substitute 2 for h, 0 for k, 3 for y, and 0 for x in the standard form for a quadratic equation to solve for a: 2 3 3(0(2))0. 4 aa 

The equation is  2 3 2. 4 yx

32. The vertex is (3, 0), and the graph passes through (0, 2). Substitute 3 for h, 0 for k, 2 for y, and 0 for x in the standard form for a quadratic equation to solve for a: 2 2 2(03)0. 9 aa 

The equation is  2 2 3. 9 yx

33. The vertex is (3, –1), and the graph passes through (5, 2). Substitute 3 for h, –1 for k, 2 for y, and 5 for x in the standard form for a quadratic equation to solve for a: 2 3 2(53)1. 4 aa 

The equation is  2 3 31. 4 yx

34. The vertex is (3, –1), and the graph passes through (0, 3). Substitute 3 for h, –1 for k, 3 for y, and 0 for x in the standard form for a quadratic equation to solve for a: 2 4 3(03)194. 9 aaa 

The equation is  2 4 31. 9 yx

35.  2 3 fxx 

Stretch the graph of 2 yx  vertically by a factor of 3.

36.  2 1 3 fxx 

Compress the graph of 2 yx  vertically by a factor of 13.

37.  2 4 gxx

Shift the graph of 2 yx  right four units.

38.  2 3 gxx

Shift the graph of 2 yx  left three units.

39.  2 24fxx

Stretch the graph of 2 yx  vertically by a factor of 2, reflect the resulting graph in the x-axis, then shift the resulting graph down 4 units.

40.  2 3 fxx

Reflect the graph of 2 yx  in the x-axis, then shift the resulting graph up 3 units.

41.  2 32 gxx

Shift the graph of 2 yx  right three units, then shift the resulting graph up two units.

42.  2 13 gxx

Shift the graph of 2 yx  left one unit, then shift the resulting graph down three units.

43.  2 324fxx

Shift the graph of 2 yx  right two units, stretch the resulting graph vertically by a factor of 3, reflect the resulting graph about the x-axis, and then shift the resulting graph up four units.

44.  2 213fxx

Shift the graph of 2 yx  left one unit, stretch the resulting graph vertically by a factor of 2, reflect the resulting graph about the x-axis, and then shift the resulting graph up three units.

45. Complete the square to write the equation in standard form: 2 4 yxx 22 444(2)4.yxxyx  This is the graph of 2 yx  shifted two units left and four units down. The vertex is (–2, –4). The axis of symmetry is 2. x  To find the x-intercepts, let y = 0 and solve 2 0(2)4 x 

2 (2)4 x  224 or 0. xxx  To find the y-intercept, let x = 0 and solve 2 (02)40.yy 

46. Complete the square to write the equation in standard form: 2 22yxx

22 1(21)2(1)1.yxxyx  This is the graph of 2 yx  shifted one unit right and one unit up. The vertex is (1, 1). The axis of symmetry is 1. x  To find the x-intercepts, let y = 0 and solve 2 0(1)1 x 

2 (1)1 x  there is no x-intercept. To find the y-intercept, let x = 0 and solve 2 (01)12.yy 

47. Complete the square to write the equation in standard form:

22 610(610) 9(69)10(3)1. yxxyxx yxxyx

This is the graph of 2 yx  shifted three units right, reflected about the x-axis, and then shifted one unit down. The vertex is (3, –1). The axis of symmetry is 3. x  To find the x-intercepts, let y = 0 and solve 2 0(3)1 x 

2 1(3) x  there is no x-intercept. To find the y-intercept, let x = 0 and solve 2 (03)110.yy

48. Complete the square to write the equation in standard form:

22 83(38)yxxyxx  2 99 38 44yxx

2 341 . 24

This is the graph of 2 yx  shifted 32 units right, reflected about the x-axis, and then shifted 41 4 units up. The vertex is 341 , 24 

. The axis of symmetry is 3 2 x  To find the x-intercepts, let y = 0 and solve 22 341341 0 2424

To find the y-intercept, let x = 0 and solve 2 341 08. 24 yy

49. Complete the square to write the equation in standard form: 22 2892(4)9yxxyxx

22 82(44)92(2)1. yxxyx 

This is the graph of 2 yx  shifted 2 units right, stretched vertically by a factor of 2, and then shifted one unit up. The vertex is (2, 1). The axis of symmetry is 2. x  To find the x-intercepts, let y = 0 and solve

2 02(2)1 x  2 1 (2) 2 x  there is no x-intercept. To find the y-intercept, let x = 0 and solve 2 2(02)19.yy 

50. Complete the square to write the equation in standard form: 22 31273(4)7yxxyxx  22 123(44)73(2)19. yxxyx 

This is the graph of 2 yx  shifted 2 units left, stretched vertically by a factor of 3, and then shifted 19 units down. The vertex is (–2, –19). The axis of symmetry is x = 2. To find the x-intercepts, let y = 0 and solve

2 03(2)19 x 

2 1919 33(2)2 xx 

571

2257. 33 xx  (continuedonnextpage)

(continued)

To find the y-intercept, let x = 0 and solve 2 3(02)197.yy 

51. Complete the square to write the equation in standard form:

22 318113(6)11yxxyxx 

2 273(69)11yxx 

2 3(3)16.yx This is the graph of 2 yx  shifted three units right, stretched vertically by a factor of three, reflected about the x-axis, and then shifted 16 units up. The vertex is (3, 16). The axis of symmetry is x = 3.

To find the x-intercepts, let y = 0 and solve 2 03(3)16yx

2 1643 33(3)3 xx 

33. 3 x  To find the y-intercept, let x = 0 and solve 2 3(03)1611. yy

52. Complete the square to write the equation in standard form:

22 520135(4)13yxxyxx

2 205(44)13yxx

2 5(2)33yx

This is the graph of 2 yx  shifted two units left, stretched vertically by a factor of five, reflected about the x-axis, and then shifted 33 units up. The vertex is (–2, 33). The axis is x = 2. To find the x-intercepts, let y = 0 and solve 22 33 05(2)33(2) 5 xx 

331 22165. 55 xx  To find the y-intercept, let x = 0 and solve 2 5(02)3313.yy 

53. a. a = 1 > 0, so the graph opens up.

b. The vertex is 88 2(1)2(1),(4,1). f

c. The axis of symmetry is x = 4.

d. To find the x-intercepts, let y = 0 and solve 2 08150(3)(5) xxxx  3 or 5. xx To find the y-intercept, let x = 0 and solve 2 08(0)1515.yy 

54. a. a = 1 > 0, so the graph opens up.

b. The vertex is 88 2(1)2(1),(4,3). f

c. The axis of symmetry is x = –4.

2 Polynomial and Rational Functions

d. To find the x-intercepts, let y = 0 and solve 2 2 884(1)(13) 0813 2(1) xxx   812 43. 2 x   To find the y-intercept, let x = 0 and solve 2 08(0)1313. y

55. a. a = 1 > 0, so the graph opens up.

b. The vertex is 11125 2(1)2(1)24,,. f

c. The axis of symmetry is 1 2 x 

d. To find the x-intercepts, let y = 0 and solve 2 060(3)(2)3 xxxxx  or 2. x  To find the y-intercept, let x = 0 and solve 2 0(0)66.yy 

56. a. a = 1 > 0, so the graph opens up.

b. The vertex is 1119 ,,. 2(1)2(1)24 f

c. The axis of symmetry is 1 . 2 x 

d. To find the x-intercepts, let y = 0 and solve 2 020(1)(2) xxxx  1 or x  2. x 

To find the y-intercept, let x = 0 and solve 2 0(0)22.yy 

57. a. a = 1 > 0, so the graph opens up.

b. The vertex is  22 2(1)2(1),1,3. f

c. The axis of symmetry is 1. x 

d. To find the x-intercepts, let y = 0 and solve 2 024 xx  2 (2)(2)4(1)(4) 2(1) x   212 2 x   there are no x-intercepts.

To find the y-intercept, let x = 0 and solve 2 02(0)44.yy 

58. a. a = 1 > 0, so the graph opens up.

b. The vertex is  44 2(1)2(1),2,1. f

c. The axis of symmetry is 2. x 

d. To find the x-intercepts, let y = 0 and solve

2 045 xx 

2 (4)(4)4(1)(5) 2(1) x  

44 2 x   there are no x-intercepts. To find the y-intercept, let x = 0 and solve

2 04(0)55.yy 

59. a. a = –1 < 0, so the graph opens down.

b. The vertex is

c. The axis of symmetry is 1. x 

d. To find the x-intercepts, let y = 0 and solve

2 062 xx  2 (2)(2)4(1)(6) 228 2(1)2 xx   

17. x 

To find the y-intercept, let x = 0 and solve

2 62(0)06.yy 

60. a. a = –3 < 0, so the graph opens down.

b. The vertex is 55549 ,,. 2(3)2(3)612 f

c. The axis of symmetry is 5 . 6 x 

d. To find the x-intercepts, let y = 0 and solve 22 02530(352) xxxx  1 0(31)(2) or 2. 3 xxxx  To find the y-intercept, let x = 0 and solve 2 25(0)3(0)2.yy 

61. a. a = 1 > 0, so the graph opens up and has a minimum value. Find the minimum value by finding the vertex:

44 2(1)2(1),2,1

The minimum value is –1.

b. The range of f is [1,). 

62. a. a = –1 < 0, so the graph opens down and has a maximum value. Find the maximum value by finding the vertex:

66 2(1)2(1),3,1

The maximum value is 1.

b. The range of f is (,1]. 

63. a. a = –1< 0, so the graph opens down and has a maximum value. Find the maximum value by finding the vertex:

44 2(1)2(1),2,0

The maximum value is 0.

b. The range of f is (,0]. 

64.

a. a = 1 > 0, so the graph opens up and has a minimum value. Find the minimum value by finding the vertex:

2(1)2(1),3,0

The minimum value is 0.

b. The range of f is [0,). 

65.

a. a = 2 > 0, so the graph opens up and has a minimum value. Find the minimum value by finding the vertex:

2(2)2(2),2,5

The minimum value is –5.

b. The range of f is [5,). 

66. a. a = 3 > 0, so the graph opens up and has a minimum value. Find the minimum value by finding the vertex:

The minimum value is –17.

b. The range of f is [17,). 

67. a. a = –4 < 0, so the graph opens down and has a maximum value. Find the maximum value by finding the vertex: 12123

2(4)2(4)2,,16 f

The maximum value is 16.

b. The range of f is (,16]. 

68. a. a = –2 < 0, so the graph opens down and has a maximum value. Find the maximum value by finding the vertex:

88 2(2)2(2),2,3

The maximum value is 3.

b. The range of f is (,3]. 

In exercises 69–82, the x-intercepts are the boundaries of the intervals.

69. 2 40 x 

Vertex: (0, –4); x-intercepts: –2, 2

Solution: [–2, 2]

70. 2 40 x  Vertex: (0, 4); x-intercepts: –2, 2

Solution: [–2, 2]

71. 2 430xx

Vertex: (2, –1); x-intercepts: 1, 3

Solution:  1,3

72. 2 20 xx

Vertex: 19 24,,

x-intercepts: –2, 1

Solution: 

73. 2 370xx

Vertex: 337 , 24

To find the x-intercepts, use the quadratic equation to solve 2 370.xx

2 33417 337 212 x

The solution of the inequality 2 370xx is 337337 ,,. 2222

74. 2 80 xx

Vertex: 133 , 24

To find the x-intercepts, use the quadratic equation to solve 2 80. xx

2 11418 133 212 x

The solution of the inequality 2 80 xx is 133133 2222,.

75. 2 490xx

Vertex: (2, 5)

To find the x-intercepts, use the quadratic equation to solve 2 490.xx

2 44419 420 212 x

There are no x-intercepts.

Solution:   , 

76. 2 690xx

Vertex: (3, 0); x-intercept: (3, 0)

Solution: {3}

77. 2 220xx 

Vertex: (1,–1)

Solution: 

78. 2 690xx 

Vertex: (3, 0); x-intercept: (3, 0)

Solution:     ,33,  

79. 2 3100xx 

Vertex: 349 , 24 

x-intercepts: –5, 2

Solution: [–5, 2]

80. 2 120 xx 

Vertex: 149 , 24 

x-intercepts: –4, 3

Solution:    ,43,  

81. 2 61370 xx 

Vertex: 131 , 1224 

x-intercepts: 1, 7 6

Solution:  7 ,1, 6  

82. 2 5940 xx 

Vertex: 91 , 1020    

x-intercepts: 4 ,1 5

Solution: 4 ,1 5

a. Using the graph, we see that the ball traveled approximately 550 ft horizontally.

b. Using the graph, we see that the ball went approximately 140 high.

Thus, the ball traveled approximately 547 ft horizontally. The ball reached its maximum height at the vertex of the function,

a. Using the graph, we see that the maximum revenue from the apartments is approximately $110,000

b. Using the graph, we see that the maximum revenue is generated by about 35 $25 increases.

c. The maximum revenue is at the vertex of the function, 22,. bb

2 34253417003480,000 108,900 R

The maximum revenue of $108,900 is reached at 34 $25 increases.

85. The vertex of the function is 114114 2(3)2(3),(19,1098). f

The revenue is at its maximum when x = 19.

86. 2 2004()2004. pxRxxx 

The vertex of the revenue function is 200200 2(4)2(4),(25,2500).

The revenue is at its maximum when x = 25.

87. 5050 2(1)2(1),(25,425). f

The total cost is minimum when x = 25.

88. 2 100()100. pxRxxx  Profit = revenue – cost

 2 ()100502 Pxxxx 

2 ()9850. Pxxx Now find the vertex of the profit function: 9898 , 2(1)2(1)

  49,2351. So the maximum profit occurs when x = 49. The maximum profit is $2351.

89. Let x = the length of the rectangle. Then 802 40 2 x x  the width of the rectangle.

The area of the rectangle = (40)xx

2 40 xx . Find the vertex to find the maximum value: 4040 2(1)2(1),(20,400).

The rectangle with the maximum area is a square with sides of length 20 units. Its area is 400 square units.

90. The fence encloses three sides of the region. Let x = the width of the region. Then 120 – 2x = the length of the region. The area of the region = 2 (1202)1202. xxxx 

Find the vertex to find the maximum value: 120120 2(2)2(2),(30,1800).

The maximum area that can be enclosed is 1800 square meters.

91. Let x = the width of the fields. Then, 60033 300 22 x x  the length of the two fields together. (Note that there is fencing between the two fields, so there are three “widths.”)

The total area = 2 33 300300. 22 xxxx

Find the vertex to find the dimensions and maximum value: 300300 2(32)2(32),(100,15,000).

So the width of each field is 100 meters. The length of the two fields together is 300 – 1.5(100) = 150 meters, so the length of each field is 150/2 = 75 meters. The area of each field is 100(75) = 7500 square meters.

92. Let x = the amount of high fencing. Then 8x = the cost of the high fencing. The cost of the low fencing = 2400 – 8x, and 24008 6002 4 x x  = the amount of low fencing. This is also the width of the enclosure. So 2(6002)512005 600 222 xxx x  the length of the enclosure.

The area of the entire enclosure is 2 5 (6002)60052700360,000. 2 xxxx

  

Use the vertex to find the dimensions and maximum area: 27002700 2(5)2(5),(270,4500)

     . So the area of the entire enclosure is 4500 square feet. There are 270 feet of high fencing, so the dimensions of the enclosure are 5(270) 60075 2  feet by 6002(270)60  feet. The question asks for the dimensions and maximum area of each half of the enclosure, so each half has maximum area 2250 sq ft with dimensions 37.5 ft by 60 ft.

93. The yield per tree is modeled by the equation of a line passing through (26, 500) where the x-coordinate represents the number of trees planted, and the y-coordinate represents the number of apples per tree. The rate of change is –10; that is, for each tree planted the yield decreases by 10. So, 50010(26)10760.yxyx 

Since there are x trees, the total yield = 2 (10760)10760. xxxx  Use the vertex to find the number of trees that will maximize the yield:

So the maximum yield occurs when 38 trees are planted per acre.

94. Let x = the number of days. The original price is $1.50 per pound and the price decreases $0.02 per pound each day, so the price per pound is (1.5 – 0.02x). The weight of the steer after x days is 300 + 8x. So the selling price = number of pounds times price per pound (3008)(1.50.02). xx The original cost of the steer is 1.5(300) = $450, and the daily cost of the steer is x, so the total cost of the steer after x days is x + 450. The profit is selling price – cost

2 (3008)(1.50.02)(450) (0.166450)(450) xxx xxx

2 0.165. xx

The maximum profit occurs at the x-coordinate of the vertex: 5 15.625. 2(0.16)

The maximum profit occurs after 16 days.

95. If 20 students or less go on the trip, the cost is $72 per students. If more than 20 students go on the trip, the cost is reduced by $2 per the number of students over 20. So the cost per student is a piecewise function based on the number of students, n, going on the trip:

72if 20 () 722(20)1122if 20 fnn nnn

The total revenue is 2 72if 20 () (1122)1122if 20 nn nfn nnnnn

The maximum revenue is either 1440 (the revenue if 20 students go on the trip) or the maximum of 2 1122. nn Find this by using the vertex: 112112 (2)(2)(2)(2),(28,1568).

The maximum revenue is $1568 when 28 students go on the trip.

96. Assume that m bytes per inch can be put on any track. Let x = the radius of the innermost track. Then the maximum number of bytes that can be put on the innermost track is 2. mx So, each track will have 2 mx bytes.

The total number of bytes on the disk is the number of bytes on each track times the number of tracks per inch times the radius (in inches): 2 2((5))2(5) mxpxmpxx  2 210mpxmpx . The maximum occurs at the x-coordinate of the vertex: 1010 2.5 22(2)4 bmp amp    inches.

97.  2 00 2 M g httvth  2 0 1.6ms,5 m, max height25 m M gh

a. Using the given values, we have  22 00 1.6 50.85 2 httvttvt 

Using the formula for the vertex of a parabola gives  00 20.81.6 vv t  This is the time at which the height h(t) = 25 m is attained. Thus,  0 2 00 0 222 000 25 1.6 0.85 1.61.6 55 3.21.63.2 v hth vv v vvv



 (continuedonnextpage)

(continued)

b. Using the formula for the vertex of a parabola gives  88 5. 20.81.6

The ball reached its highest point 5 seconds after it was released.

a. Using the given values, we have

Using the formula for the vertex of a parabola gives  00 20.81.6 vv t  This is the time at which the height h(t) = 52 m is attained. Thus,

Thus,

b. Using the formula for the vertex of a parabola gives

The ball reached its highest point 7.5 seconds after it was released.

99. a. The maximum height occurs at the vertex: 6464 2(16)2(16),(2,64), f

so the maximum height is 64 feet.

b. When the projectile hits the ground, h = 0, so solve 2 0166416(4)0 tttt  0 or 4. tt

The projectile hits the ground at 4 seconds.

100. a. The maximum height occurs at the vertex: 6464 2(16)2(16),(2,464).

The maximum height is 464 feet.

b. When the projectile hits the ground, h = 0, so solve 2 01664400 tt  22 16(425)04250 tttt  2 4(4)4(1)(25) 4116 2(1)2 tt   

2293.39 or 7.39. ttt  Reject the negative solution (time cannot be negative). The projectile hits the ground at 7.39 seconds.

101. Let x = the radius of the semicircle. Then the length of the rectangle is 2x. The circumference of the semicircle is , x  so the perimeter of the rectangular portion of the window is 18. x 

The width of the rectangle = 182 2 xx 

The area of the semicircle is 2 2, x  and the area of the rectangle is 182 2 2 xx x    

22 182. xxx  So the total area is 22 222 182182 22 xx xxxxx    2 182. 2 xx  

The maximum area occurs at the x-coordinate of the vertex: 181818 44 22 2    

(continuedonnextpage)

(continued)

This is the radius of the semicircle. The length of the rectangle is 1836 2 44

ft. The width of the rectangle is 1818 182

102. a.

b. The width of the arch is the difference between the two x-intercepts:

2 00.006354xx  04(0.00158751)0 or xxx 

629.92. x  The arch is 629.92 feet wide.

c. The maximum height occurs at the vertex of the arch. Since we know that the arch is 629.92 feet wide, the vertex occurs at x = 629.92/2 = 314.96. 2 (314.96)0.00635(314.96)4(314.96) f  629.92 feet. 

103. a. 1.181.18 , 2(0.01)2(0.01) (59,36.81)

b. 2 00.011.182 xx  2 1.181.184(0.01)(2) 2(0.01) 1.181.47241.181.21 0.020.02 1.5 or 119.5. Reject the negative answer x 

and round the positive answer to the nearest whole number. The ball hits the ground approximately 120 feet from the punter.

c. The maximum height occurs at the vertex. Since we know that the ball hits the ground at 120, x  the vertex occurs at 60. x  2 (60)0.01(60)1.18(60)236.8. f 

The maximum height is approximately 37 ft.

d. The player is at x = 6 feet. 2 (6)0.01(6)1.18(6)28.72. f 

The player must reach approximately 9 feet to block the ball.

e. 2 2 2 70.011.182 00.011.185

1.181.184(0.01)(5)

2(0.01)

1.181.1924 0.02 4.4 feet or 113.6 feet

104. The maximum height occurs at the vertex: 303015225 2(16)2(16)1616,,.

The maximum height is approximately 14 feet, so it will never reach a height of 16 feet.

105.  2 1.510.220.029 fxxx 

a. We must complete the square to write the function in standard form.

b. Sales were at a minimum at 3.79, x  or during year 4 (2010). This fits the original data.

c. The year 2011 is represented by x = 5.

2 51.510.2250.02951.135

In 2011, there were about 1.14 million vehicles sold.

106.

a. We must complete the square to write the function in standard form.

b. Foreclosure filings were at a maximum during year 4, 2010. This fits the original data.

c. The year 2013 is represented by x = 7.  2 70.8220.89670.1171.7 f 

There were about 1.7 million foreclosure filings in 2013.

Beyond the Basics

107. 22 2 3(2)33(44)3 31215 yxyxx yxx 

108. Substitute the coordinates (1, 5) for x and y and (–3, –2) for h and k into the standard form 2 () yaxhk  to solve for a: 2 7 5(13)2716 16 aaa 

The equation is 2 7 ()(3)2 16 yfxx 2 72131 16816yxx

109. The y-intercept is 4, so the graph passes through (0, 4). Substitute the coordinates (0, 4) for x and y and (1, –2) for h and k into the standard form 2 () yaxhk  to solve for a:

2 4(01)26. aa  The x-coordinate of the vertex is 112. 22(6) bb b a 

The y-intercept = c, so the equation is 2 ()6124. fxxx

110. The x-coordinate of the vertex of the graph is halfway between the x-intercepts:

26 4. 2 h   Substitute the coordinates of one of the x-intercepts and the x-coordinate of the vertex into the standard form 2 () yaxhk  to find an expression relating a and k:

2 0(24)4. akak 

Now substitute the coordinates of the y-intercept, the x-coordinate of the vertex, and the expression for k into the standard form

2 () yaxhk  to solve for a:

2 24(04)4 aa 241642. aaa

Use this value to find k: 4(2)8. k 

The equation is 22 2(4)821624.yxyxx 

111. The x-coordinate of the vertex is 3. Substitute the coordinates of the x-intercept and the xcoordinate of the vertex into the standard form

2 () yaxhk  to find an expression relating a and k: 2 0(73)16. akak  Now substitute the coordinates of the y-intercept, the x-coordinate of the vertex, and the expression for k into the standard form 2 () yaxhk  to solve for a: 2 14(03)162. aaa 

Use this value to find k: 16(2)32. k  The equation is 22 2(3)3221214.yxyxx 

112. First, write 2 22yxx in standard form by completing the square: 22 1(21)2(1)1.yxxyx 

Move the curve three units to the right by subtracting 3 from x + 1; move the curve two units down by subtracting 2 from 1: 22 (13)(12)(2)1yxx  2 43.yxx

113. The x-coordinate of the vertex is 26 2. 2  

Substitute the coordinates of one of the xintercepts and the x-coordinate of the vertex into the standard form 2 () yaxhk  to find an expression relating a and k: 2 0(22)16. akak 

Now substitute the coordinates of the other x-intercept, the x-coordinate of the vertex, and the expression for k into the standard form 2 () yaxhk  to solve for a: 2 0(62)161. aaa  Use this value to find k: 16(1)16. k  So one equation is 22 (2)16412.yxxx  The graph of this equation opens upward. To find the equation of the graph that opens downward, multiply the equation by –1: 2 412.yxx

114. The x-coordinate of the vertex is 35 1. 2  

Substitute the coordinates of one of the x-intercepts and the x-coordinate of the vertex into the standard form 2 () yaxhk  to find an expression relating a and k: 2 0(31)16. akak  Now substitute the coordinates of the other x-intercept, the x-coordinate of the vertex, and the expression for k into the standard form 2 () yaxhk  to solve for a: 2 0(51)161. aaa  Use this value to find k: 16(1)16. k  So one equation is 22 (1)16215.yxxx  The graph of this equation opens upward. To find the equation of the graph that opens downward, multiply the equation by –1: 2 215.yxx

115. The x-coordinate of the vertex is 71 4. 2 

Substitute the coordinates of one of the x-intercepts and the x-coordinate of the vertex into the standard form 2 () yaxhk  to find an expression relating a and k: 2 0(74)9. akak 

Now substitute the coordinates of the other x-intercept, the x-coordinate of the vertex, and the expression for k into the standard form 2 () yaxhk  to solve for a: 2 0(14)91. aaa  Use this value to find k: 9(1)9. k  So one equation is 22 (4)987.yxxx  The graph of this equation opens upward. To find the equation of the graph that opens downward, multiply the equation by –1: 2 87.yxx

116. The x-coordinate of the vertex is 210 6. 2  

Substitute the coordinates of one of the x-intercepts and the x-coordinate of the vertex into the standard form 2 () yaxhk  to find an expression relating a and k: 2 0(26)16. akak  Now substitute the coordinates of the other x-intercept, the x-coordinate of the vertex, and the expression for k into the standard form 2 () yaxhk  to solve for a: 2 0(106)161. aaa  Use this value to find k: 16(1)16. k  So one equation is 22 (6)161220.yxxx  The graph of this equation opens upward. To find the equation of the graph that opens downward, multiply the equation by –1: 2 1220.yxx

117. First, complete the squares on x and y.









Now, consider the two functions  2 21fx and  2 34. gy

The minimum of f is 0 and the minimum of g is 4, so the minimum of f + g = 4.



118. First, complete the squares on x and y 









 22 22 22 22 22 312211 312211 34211 3442111121 3212 xyxy xxyy xxyy xxyy xy

Now consider the two functions  2 32fx and  2 12. gx The maximum of f is 0 and the maximum of g is 2, so the maximum of f + g = 2.

Critical Thinking/Discussion/Writing

   





119. 2 2 2 2 22 22 ()()() ()()() ()() ()() ()() 2 2 422. (We can divide by because 0.)Since 0,2 fhpahpbhpc fhpahpbhpc fhpfhp ahpbhpc ahpbhpc ahahpapbhbpc ahahpapbhbpc ahpbpahb ppaah

   2 b b h a  

120. Write 2 289yxx in standard form to find the axis of symmetry: 22 28982(44)9yxxyxx  2 2(2)1.yx

The axis of symmetry is 2. x 

Using the results of exercise 119, we know that (2)(1)(2). fpffp  21 p  3, so 22(3)5.pp

The point symmetric to the point (–1, 19) across the axis of symmetry is (5, 19).

121. a.  

This is the equation of a parabola. The x-coordinate of the vertex is 

22 2 22 22 or abmahmambh amam bhhb mm

The y-coordinate of the vertex is

2 2 2 2 22 2222 22 22 2 2 2 2242 2 hbhb amabmahm mm abahbk ahbabhhb abahbahk ahahbabababhah abahbahk k

The vertex is ,. hb k m

2 2 22 22 2 2 gfxgaxhk maxhkb maxahxahkb maxmahxmahmkb

This is the equation of a parabola. The x-coordinate of the vertex is 2 2 mah h ma 

The y-coordinate of the vertex is 222 2 mahmahmahmkbmkb 

The vertex is   ,.hmkb 

122. If the discriminant equals zero, there is exactly one real solution. Thus, the vertex of   yfx  lies on the x-axis at 2 b x a 

If the discriminant > 0, there are two unequal real solutions. This means that the graph of   yfx  crosses the x-axis in two places. If a > 0, then the vertex lies below the x-axis and the parabola crosses the x-axis; if a < 0, then the vertex lies above the x-axis and the parabola crosses the x-axis. If the discriminant < 0, there are two nonreal complex solutions. If a > 0, then the vertex lies above the x-axis and the parabola does not cross the x-axis (it opens upward); if a < 0, then the vertex lies below the x-axis and the parabola does not cross the x-axis (it opens downward).

123. Start by writing the function in standard form.

The minimum value of this function is 2 . 2 a

Now, rewrite the function as

This is the same function, so

124. Start by writing the function in standard form.

The maximum value of this function is 2 . 4

Now, rewrite the function as

129. 222 31239 3 4444 xxx

4444 2323 42 3535 11 35 xxxx xxxx xxx

2 492323 xxx

2 2 693xxx

2 1511123453 xxxx

2 14322172 xxxx

2214141 221 xxxxx

2 2 9272527 92527 353527 xxx xx xxx  

2.2 Polynomial Functions

2.2 Practice Problems

1. a.  2 1 1 fxx x   is not a polynomial function because its domain is not   , 

b.  72 2517gxxx is a polynomial function. Its degree is 7, the leading term is 7 2 x , and the leading coefficient is 2.

2.  32 3 23 42517 2517 4 Pxxxx x xxx      

 When |x| is large, the terms 2 25 ,, xx and 3 17 x are close to 0. Therefore, 33 40004. Pxxx

3. Use the leading-term test to determine the end behavior of  42 253yfxxx  . Here n = 4 and 20 n a  . Thus, Case 2 applies. The end behavior is described as as yx and as . yx

4. First group the terms, then factor and solve   0 fx  :

 32 32 22 2 2346 2436 2232 232 fxxxx xxx xxx xx 

  2 2 0232 3 023 or 2 02 (no real solution) xx xx x

The only real zero is 3 2

5.  3 236fxxx

17 f

and

24 f  . Since

1 f and

2 f have opposite signs, by the Intermediate Value Theorem, f has a real zero between 1 and 2.

as yx and as . yx

Now find the zeros of the function:  





4242 22 22 054054 41 04 or 01 2or 1 xxxx xx xx xx

 

There are four zeros, each of multiplicity 1, so the graph crosses the x-axis at each zero.

Next, find the y-intercept:  42 0544fxx

Determine symmetry:

42 42 54 54   fxxx xxfx



6.   2 2 1350 10 or 30 or 50 1 or 3 or 5 fxxxx xxx xxx

  fx has three distinct zeros.

7. 



 2 2 1350 10 or 30 or 50 1 (multiplicity 2) or 3 (multiplicity 1) or 5 (multiplicity 1)

8.  4232fxxx has at most three turning points. Using a graphing calculator, we see that there are indeed, three turning points.

9.  4254fxxx

Since the degree, 4, is even and the leading coefficient is 1, the end behavior is as shown:

So f is an even function and its graph is symmetric with respect to the y-axis. Now find the intervals on which the graph lies above or below the x-axis. The four zeros divide the x-axis into five intervals,   ,2,

 2,1,1,1,1,2, and 2,. Determine the sign of a test value in each interval

Interval

Test point Value of   fx Above/below x-axis

 ,2 3 40 below

2,1 1.5 2.1875 above

1,1 0 4 below

 1,2 1.5 2.1875 above   2,  3 40 below

Plot the zeros, y-intercepts, and test points, and then join the points with a smooth curve.

First, find the zeros of the function.

32 210 02010 21 xxx xxx xx 

The function has three distinct zeros, –2, 1, and 0. The zero x = –2 has multiplicity 3, so the graph crosses the x-axis at –2, and near –2 the function looks like

323 2221182.fxxx 

The zero x = 0 has multiplicity 1, so the graph crosses the x-axis at 0, and near 0 the function looks like

32 02018. fxxx 

The zero x = 1 has multiplicity 2, so the graph touches the x-axis at 1, and near 1 the function looks like  322 1121271.fxxx 

We can determine the end behavior of the polynomial by finding its leading term as a product of the leading terms of each factor.

 32 326 21 fxxxxxxxx 

Therefore, the end behavior of   fx is Case 1:

Now find the intervals on which the graph lies above or below the x-axis. The three zeros divide the x-axis into four intervals,   ,2,

 2,0,0,1,and 1,. Determine the sign of a test value in each interval

Interval

Test point

Value of

fx Above/below x-axis

  ,2 3 48 above

  2,0 1 –4 below

above

1,  2 128 above

Plot the zeros, y-intercepts, and test points, and then join the points with a smooth curve.

11. 333 7207.378 dm 3333 207.378 L

2.2 Exercises

Concepts and Vocabulary

1. Consider the polynomial 54 236. xxx

The degree of this polynomial is 5, its leading term is 5 2 x , its leading coefficient is 2, and its constant term is 6.

2. A number c for which   0 fc  is called a zero of the polynomial function f.

3. If c is a zero of even multiplicity for a polynomial function f, then the graph of f touches the x-axis at c.

4. If c is a zero of odd multiplicity for a polynomial function f, then the graph of f crosses the x-axis at c

5. False. The graph of a polynomial function is a smooth curve. That means it has no corners or cusps.

6. True

7. False. The polynomial function of degree n has, at most, n zeros.

8. True

Building Skills

9. Polynomial function; degree: 5; leading term: 5 2; x leading coefficient: 2

10. Polynomial function; degree: 4; leading term: 4 7; x leading coefficient: –7

11. Polynomial function; degree: 3; leading term: 3 2 ; 3 x leading coefficient: 2 3

12. Polynomial function; degree: 3; leading term: 3 2; x leading coefficient: 2

13. Polynomial function; degree: 4; leading term: 4 ; x  leading coefficient: 

14. Polynomial function; degree: 0; leading term: 5; leading coefficient: 5

15. Not a polynomial function (the graph has sharp corners; not a smooth curve; presence of x )

16. Not a polynomial function (the domain is restricted)

17. Not a polynomial function because the domain is not

18. Not a polynomial function (the graph is not continuous)

19. Not a polynomial function (presence of x )

20. Not a polynomial function (noninteger exponent)

21. Not a polynomial function because the domain is not

22. Not a polynomial function (negative exponent)

23. Not a polynomial function (graph is not continuous)

24. Not a polynomial function (the domain is restricted)

25. Not a polynomial function (the graph is not continuous)

26. Not a polynomial function (the graph has sharp corners; not a smooth curve)

27. Not a polynomial function (not the graph of a function)

28. Not a polynomial function (the graph has sharp corners; not a smooth curve)

29. c 30. f 31. a

32. e 33. d 34. b

 3 fxxx 

The

32 221fxxx

The leading term is 3 2, x so   fx  as x  and   fx  as . x 

37.  43 421fxxx

The leading term is 4 4, x so   fx  as

x  and   fx  as x 

38.  43 3 fxxxx 

The leading term is 4 , x so   fx  as x  and   fx  as x 

39.  2 221fxxx

The leading term is  23 22,xxx  so

  fx  as x  and   fx  as x 

40.  3 221fxxx

The leading term is  34 22,xxx  so

  fx  as x  and   fx  as . x 

41.  2 24 fxxx 

The leading term is  23 , xxx  so

  fx  as x  and   fx  as . x 

42.

3 32 fxxx 

The leading term is  34 , xxx  so

  fx  as x  and   fx  as x 

43.        3123fxxxx 

Zeros: –2, 1, 3

x = –2: multiplicity: 1, crosses the x-axis

x = 1: multiplicity: 1, crosses the x-axis

x= 3: multiplicity: 1, crosses the x-axis

44.        5123fxxxx 

Zeros: –2, –1, 3

x = –2: multiplicity: 1, crosses the x-axis

x = –1: multiplicity: 1, crosses the x-axis

x = 0: multiplicity: 1, crosses the x-axis

45.  2 221fxxx

Zeros: –2, 1 2

x = –2: multiplicity 2, touches but does not cross the x-axis

1 2 : x  multiplicity 1, crosses the x-axis

46.

3 221fxxx

Zeros: 1 2 ,2

1 2 : x  multiplicity 1, crosses the x-axis

x = 2: multiplicity 3, crosses the x-axis

47. 

  3 22 932fxxxx

Zeros: 2 3 3,,0,3

x = –3: multiplicity 1, crosses the x-axis

2 3 : x  multiplicity 3, crosses the x-axis

x = 0: multiplicity 2, touches but does not cross the x-axis

x = 3: multiplicity 1, crosses the x-axis

48. 

 2 32 432fxxxx

Zeros: 2 3 2,0,,2

x = –2: multiplicity 1, crosses the x-axis

x = 0: multiplicity 3, crosses the x-axis

2 3 : x  multiplicity 2, touches but does not cross the x-axis

x = 2: multiplicity 1, crosses the x-axis

49.

 2 2 132fxxx

Zero: 2 3 , multiplicity 2, touches but does not cross the x-axis

50. 

2 112fxxxx 

Zeros: –1, 2

x = –1: multiplicity 1, crosses the x-axis

x = 2; multiplicity 1, crosses the x-axis

51. 43 (2)22102; f  43 (3)331044 f 

Because the sign changes, there is a real zero between 2 and 3. The zero is approximately 2.09.

52. 42 (1)112(1)57; f  42 (2)222(2)53. f 

Because the sign changes, there is a real zero between 1 and 2. The zero is approximately 1.87.

53. 52 (2)29(2)1519; f  52 (3)39(3)15147. f 

Because the sign changes, there is a real zero between 2 and 3. The zero is approximately 2.28.

54. 5432 (0)05(0)8(0)4(0)055; f  5432 (1)15(1)8(1)4(1)1512. f 

Because the sign changes, there is a real zero between 0 and 1. The zero is approximately 0.68.

55. The least possible degree is 3. The zeros are –2, 1, and 3, each with multiplicity 1. The end behavior is   fx  as x  and   fx  as , x  so the leading coefficient is 1. The polynomial with smallest possible degree is 

213.xxx

56. The least possible degree is 3.The zeros are –1, 1, and 3, each with multiplicity 1. The end behavior is   fx  as x  and   fx  as , x  so the leading coefficient is –1. The polynomial with smallest possible degree is 

113.xxx 

57. The least possible degree is 3.The zeros are –3 (multiplicity 1) and 2 (multiplicity 2). The end behavior is   fx  as x  and   fx  as , x  so the leading coefficient is 1. The polynomial with smallest possible degree is  2 32.xx

58. The least possible degree is 3.The zeros are –1 (multiplicity 2) and 2 (multiplicity 1). The end behavior is   fx  as x  and   fx  as , x  so the leading coefficient is –1. The polynomial with smallest possible degree is  2 12.xx 

59. The least possible degree is 4. The zeros are –1 (multiplicity 2) and 2 (multiplicity 2). The end behavior is   fx  as x  and   fx  as , x  so the leading coefficient is 1. The polynomial with smallest possible degree is 22 12.xx

60. The least possible degree is 4. The zeros are –2 (multiplicity 2) and 2 (multiplicity 2). The end behavior is   fx  as x  and   fx  as , x  so the leading coefficient is –1. The polynomial with smallest possible degree is 22 22.xx 

61. The zeros are –2 (multiplicity 2), 2 (multiplicity 1), and 3 (multiplicity 1), so the least possible degree is 4. The end behavior is   fx  as

x  and   fx  as , x  so the leading coefficient is 1. The polynomial with smallest possible degree is  2 223.xxx

62. The zeros are –3 (multiplicity 1), 0 (multiplicity 2), and 2 (multiplicity 1), so the least possible degree is 4. The end behavior is   fx  as

2 12fxxxx

2 32.xxx

x  and   fx  as , x  so the leading coefficient is –1. The polynomial with smallest possible degree is

For exercises 63–74, use the procedure shown on pages 163–164 of the text to graph the function. 63.

2124fxxxx

2322fxxx

23 123fxxxx 

32 113fxxxx

22 11fxxxx

22 42fxxxx

22 12

2 2 92fxxxx

Applying the Concepts

75. a.  3 0.04825 404.14 hp 746 P 

b.  3 0.04825 6517.76 hp 746 P 

c.  3 0.04825 8033.12 hp 746 P  d. 

76. a.  4 128001.75120,050J

It reduced

77. a. 2 3(4)00 or 4. xxxx  x = 0, multiplicity 2; x = 4, multiplicity 1.

c. There are 2 turning points.

d. Domain: [0, 4]. The portion between the x-intercepts is the graph of R(x).

78.

R(x) is a maximum at x ≈ 2.67.

79. a. Domain [0, 1]

80.

v is a maximum when r ≈ 0.67.

81. a. 2 3 ()272730090,000 xx Rxxx

b. The domain of R(x) is the same as the domain of p 0 p  when 9003. p 

The domain is 0,9003.  

82. a.

About 900 pairs of slacks were sold.

b. $16,200

83. a.    2 ()124002 Nxxx 

b. The low end of the domain is 0. (There cannot be fewer than 0 workers.) The upper end of the domain is the value where productivity is 0, so solve N(x) = 0 to find the upper end of the domain. 2 22 (12)(4002)012 (reject this) or 40020200 xxx xx

102 (reject the negative solution). x  The domain is [0,102].

84. a.

The maximum number of oranges that can be picked per hour is about 5950.

b. The number of employees = 12 + 5 = 17.

85. a. ()(82)(152) Vxxxx 

86.

The largest possible value of the volume of the box is 90.74 cubic units.

87. a. 2 ()(1084) Vxxx  b.

88.

The largest possible volume of the box is 11,664 cubic inches.

89. a. 2 ()(622) Vxxx  b.

90.

The volume is greatest when x ≈ 20.67, so the dimensions of the suitcase with the largest volume are approximately 20.67 in. × 20.67 in. × 20.67 in.

91. a. 2 ()2(453) Vxxx 

The volume is greatest when x = 10, so the width of the bag is 10 inches, the length is 2(10) = 20 inches, and the height is 45 – 20 – 10 = 15 inches.

A worker is most efficient approximately five hours after 6:00 a.m. or 11:00 a.m. 95.  0.9630.880.1920.011723 fxxxx 

a. The year 2015 is represented by x = 7.

71.73 trillion f

[–2, 12] by [–2, 2.5]

It appears that there is one zero and two turning points.

[0, 8] by [–0.5, 2.5]

The maximum occurs at 3.27, x  which is during 2011. This fits the original data.

96.  2 3 0.230.00810.0002 0.000002 fxxx x  

a. The year 2025 is represented by x = 120.

 1201.32 f 

[0, 120] by [–0.5, 1.5]

It appears that there is one zero and no turning points.

[0, 8] by [–0.5, 2.5]

The zero occurs at 65, x  which is the year 1970. This fits the original data.

Beyond the Basics

97. The graph of 4 ()(1)fxx is the graph of 4 yx  shifted one unit right.

4 (1)01 xx . The zero is x = 1 with multiplicity 4.

98. The graph of 4 ()(1)fxx is the graph of 4 yx  shifted one unit left and then reflected across the x-axis. 4 (1)01 xx  . The zero is x = –1 with multiplicity 4.

99. The graph of 4 ()2fxx is the graph of 4 yx  shifted two units up.

4 4 202.xx There are no zeros.

100. The graph of 4 ()81fxx is the graph of 4 yx  shifted 81 units up.

4 4 81081.xx There are no zeros.

101. The graph of 4 ()(1)fxx is the graph of 4 yx  shifted one unit right and then reflected across the x-axis. 4 (1)01. xx  The zero is x = 1 with multiplicity 4.

102. The graph of 4 1 ()(1)8 2 fxx is the graph of 4 yx  shifted one unit right, compressed vertically by a factor of 2, and then shifted eight units down. 4 1 (1)80 2 x 

4 (1)16123 or 1. xxxx  Both zeros have multiplicity 1.

103. The graph of 5 ()1fxx is the graph of 5 yx  shifted one unit up. 5 10 x  5 11.xx The zero is –1 with multiplicity 1.

104. The graph of 5 ()(1)fxx is the graph of 5 yx  shifted one unit right. 5 (1)0 x  1. x  The zero is 1 with multiplicity 5.

105. The graph of 5 (1) ()8 4 fxx   is the graph of 5 yx  shifted one unit left, compressed vertically by one-fourth, reflected in the x-axis, and then shifted up eight units.

55 (1)(1) 808 44 xx  

5 (1)32121. xxx  The zero is 1 with multiplicity 1.

106. The graph of 5 (1) ()81 3 fxx   is the graph of 5 yx  shifted one unit left, compressed vertically by one-third, and then shifted up 81 units. 55 (1)(1) 81081 33 xx  

5 (1)243134. xxx  The zero is –4 with multiplicity 1.

107. If 0 < x < 1, then 2 1. x  Multiplying both sides of the inequality by 2 x , we obtain 42 xx  Thus, 42 0. xx

108. If x > 1, then 2 1. x  Multiplying both sides of the inequality by 2 x , we obtain 42 xx  Thus, 42 1. xx

109.

The graph of 24 yxx  represents the difference between the two functions 2 yx  and 4 , yx  so the maximum distance between the graphs occurs at the local maximum of 24 yxx  . The maximum vertical distance is 0.25. It occurs at x ≈ 0.71.

110. The graph of 35 yxx  represents the difference between the two functions 3 yx  and 5 , yx  so the maximum distance between the graphs occurs at the local maximum of 35 yxx  . The maximum vertical distance is 0.19. It occurs at x ≈ 0.77.

Answers will vary in exercises 111–114. Sample answers are given.

111. 2 ()(1)fxxx

112. ()(1)(1) fxxxx

113. 4 ()1 fxx 

114. 2 ()(1)(1) fxxxx

115. The smallest possible degree is 5, because the graph has five x-intercepts and four turning points.

116. The smallest possible degree is 5, because the graph has four turns.

117. The smallest possible degree is 6, because the graph has five turning points.

118. The smallest possible degree is 6, because the graph has five turning points.

119.  4 101fxxx

For large values of x,  4 10.fxx  For small values of x,   1. fxx

120.  3 1 fxxx

For large values of x,  3 fxx  For small values of x,   1. fxx

121.  42 1 fxxx

For large values of x,  4 fxx  For small values of x,  2 1. fxx

122.  32 1 fxxx

For large values of x,  3 . fxx  For small values of x,  2 1. fxx

Critical Thinking/Discussion/Writing

123. If a human is scaled up by a factor of 20, then the total weight will increase 3 208000  times but the bone strength will increase only 2 20400  times. The bones would probably break under the increased weight.

124. If a human is scaled down by a factor of 70, then the body surface area will decrease 2 704900  times while the volume will decrease 3 70343,000  times. The body will rapidly lose heat and will have a very hard time maintaining body temperature unless its metabolic rate increases drastically.

125. It is not possible for a polynomial function to have no y-intercepts because the domain of any polynomial function is (,),  which includes the point x = 0.

126. It is possible for a polynomial function to have no x-intercepts because the function can be shifted above the x-axis. An example is the function 2 1. yx

127. It is not possible for the graph of a polynomial function of degree 3 to have exactly one local maximum and no local minimum because the graph of a function of degree 3 rises in one direction and falls in the other. This requires an even number of turning points. Since the degree is 3, there can be only zero or two turning points. Therefore, if there is a local maximum, there must also be another turning point, which will be a local minimum.

128. It is not possible for the graph of a polynomial function of degree 4 to have exactly one local maximum and exactly one local minimum because the graph rises in both directions or falls in both directions. This requires an odd number of turning points. Since the degree is 4, there can be only one or three turning points. If there were exactly one local maximum and exactly one local minimum, there would be two turning points.

129. If  1 110 mm mm Pxaxaxaxa  and  1 110 , nn nn Qxbxbxbxb

PQxabxbxbxbabxbxbxb abxbxbxba 

1 11 1101110 1 11100 , mm nnnn mnnmnn nn nn

 which is a polynomial of degree .mn

130. If  1 110 mm mm Pxaxaxaxa   and

1 11 1101110 1 11100 , mm nnnn mnnmnn

Getting Ready for the Next Section

2.3 Dividing Polynomials and the Rational Zeros Test

2.3 Practice Problems 1. 232 32 2 2 34 01347 ()303 427 ()404 23 x xxxxx xxx xx xx x 

Quotient: 3x + 4; remainder: 2x + 3

2. 32705 639 2134

The quotient is 2 23, xx remainder 4 or 2 4 23. 3 xx x

3. 321187 6219 2732

The quotient is 2 273, xx remainder 2.

4.  11057 112151254 F

, so the remainder when  1105725Fxxx is divided by x 1 is 4.

5. 21010220 242852 12142632

The remainder is 32, so (2)32. f 

6. Since 2 is a zero of the function 32 32012, xxxx + 2 is a factor. Use synthetic division to find the depressed equation. 2312012 61412 3760

Thus,

322 320122376 xxxxxx

Now solve 2 3760 xx to find the two remaining zeros:

2 37603320 2 30 or 3203 or 3 xxxx xxxx

The solution set is 2 2,,3. 3

32 0.234.2550.34541.05 310310 103 Cxxxx CCxxQx CxxQx 

So, 3 is zero of C(x) 10 = 32 0.234.2550.34531.05 xxx 

We need to find another positive zero of C(x) 10. Use synthetic division to find the depressed equation.

30.234.2550.34531.05

0.6910.69531.05

0.233.56510.350

Solve the depressed equation 2 0.233.56510.350 xx using the quadratic formula: 2 3.565(3.565)4(0.23)(10.35) 2(0.23) 18 or2.5 x   

The positive zero is 18. Check by verifying that C(18) = 10.

8. The factors of the constant term, 8, are  1,2,4,8 , and the factors of the leading coefficient, 2, are  1,2 . The possible rational zeros are 1 ,1,2,4,8. 2     Use synthetic division to find one rational root: 22368 428 2140

The remainder is 0, so 2 is a zero of the function.    322 2368224 xxxxxx 

Now find the zeros of 2 24 xx using the quadratic formula: 2 1(1)4(2)(4) 133 2(1)2 x    , which are not rational roots. The only rational zero is { 2}.

From the similar right triangles EFC  and , ABC  we have 180180 , 2 y xx    or 2 180180. x y x  

Using the Pythagorean theorem in right triangle ABC gives    2 2 2 218021.xy Now substitute.

 2 2 2 2 218021. x x x     

Multiply both sides by 2 x to eliminate the fraction.

    2 2 222 2180221 xxxx  Now expand and simplify.      2222 43222 432 4418044441 44180720720441 42577207200 xxxxxx xxxxxx xxxxPx



 (continuedonnextpage)

(continued)

Use the factorization of 42 720235  to check for all possible rational zeros of the polynomial.



  

1,2,3,4,5,6,8,9,10, 12,15,18,20,24,30,36, 40,48,60,72,80,90,120, 144,180,240,360,720

We know that x must be a positive number less than 21 Using synthetic division, we see that x = 4 is a zero of   . Px 414257720720 432900720 182251800

So, the remainder is

328225180.Qxxxx 

Using synthetic division again, we see that x = 12 is another zero. 1218225180 12240180 120150

There are no other real zeros.

432 2 4257720720 4122015 Pxxxxx xxxx  

Since the height = x + 2, the top of the ladder touches the wall at either 6 feet above the ground or 14 feet above the ground.

2.3 Exercises Concepts and Vocabulary

1. Consider the equation   3235145. xxxxx  If both sides are divided by 2 1, x  then the dividend is 3 35,xx the divisor is 2 1, x  the quotient is ,x and the remainder is 45. x 

2. Consider the division 2 4 12 226.xx xx x   The dividend is 2 4,xx  the divisor is 2, x the quotient is 6, x  and the remainder is 12.

3. The Remainder Theorem states that if a polynomial ()Fx is divided by () xa , then the remainder R = F(a).

4. The Factor Theorem states that () xa is a factor of a polynomial ()Fx if and only if F(a) = 0.

5. True

6. True

7. False. The possible rational zeros of  32 991Pxxxx are 11 93,,1.

8. False. The remainder is a constant.

Building Skills

9. 2 2 32 2162 (63) 42 (42) 0 x xxx xx x x  

10. 2 32 32 2 2 7 22 2 23423 (46) 4 (46) 73 21 7 2 15 2 xx xxxx xx xx xx x x  

In exercises 9–14, insert zero coefficients for missing terms.

11. 32 432 43 32 32 2 2 3336 130637 (33) 36 (33) 33 (33) 67 (66) 13 xxx xxxxx xx xx xx xx xx x x     

12.

13.

14.

15.

16. First, arrange the terms in order of descending powers. 2 2432 432 32 32 2 2 232 356111310 (6210) 9913 (9315) 6210 (6210) 0 xx xxxxxx xxx

17. 11172 107 1075

The quotient is 2 7 x and the remainder is 5. 18. 22312 41426 271324

The quotient is 2 2713 xx and the remainder is 24.

19. 214710 2422 121112

The quotient is 2 211xx and the remainder is 12.

20. 311132 3123 1411

The quotient is 2 41xx and the remainder is 1.

21. 213245 2208 110413

The quotient is 32 4 xx and the remainder is 13.

22. 1153010 1477 14773

The quotient is 32477 xxx and the remainder is 3.

23. 51 24 13 24 1 2431 2 1 25

The quotient is 2 1 25 2 xx and the remainder is 3 4

24. 4 3 7 3 1 3811 3 13 394

The quotient is 2 394 xx and the remainder is 7 3 .

25. 1 2532 2 133 2661

The quotient is 2 266 xx and the remainder is 1.

26. 1 3282 3 113 3391

The quotient is 2 339 xx and the remainder is 1.

27. a. 11301 144 1445

The remainder is 5, so (1)5. f  b. 11301 122 1223

The remainder is 3, so (1)3. f 

c. 1 1301 2 177 248 7715 1 248

The remainder is 15 8 , so 115 28 f

d. 101301 101301300 1131301301

The remainder is 1301, so (10)1301. f 

28. a. 22301 41428 271427

The remainder is –27, so (2)27. f 

b. 12301 255 2554

The remainder is –4, so (1)4. f 

c. 1 2301 2 121 2420

The remainder is 0, so 1 0. 2 f

d. 72301 1477539 21177540

The remainder is 540, so (7)540. f 

29. a. 1153020 1633 163317

The remainder is –17, so (1)17. f 

b. 1153020 1477 147727

The remainder is –27, so (1)27. f 

c. 2153020 261836 1391856

The remainder is –56, so (2)56. f 

d. 2153020 2142244 17112224

The remainder is 24, so (2)24. f 

30. a. 0.110.50.3020 0.10.060.0240.0024 10.60.240.02420.0024

The remainder is –20.0024, so (0.1)20.0024. f 

b. 0.510.50.3020

0.50.50.10.05

11.00.20.119.95

The remainder is –19.95, so (0.5)19.95. f 

c. 1.710.50.3020

1.73.745.8489.9416

12.23.445.84810.0584

The remainder is 10.0584 , so (1.7)10.0584. f 

d. 2.310.50.3020

2.34.148.83220.3136

11.83.848.8320.3136

The remainder is 0.3136 , so (2.3)0.3136. f 

31. 32 (1)2(1)3(1)6(1)10 f  1 x is a factor of 32 2361. xxx Check as follows: 12361 251 2510

32. 32 (3)3(3)9(3)4(3)1203 fx  is a factor of 32 39412. xxx Check as follows: 339412 9012 3040

33. 432 (1)5(1)8(1)(1) f  2(1)40 

1 x  is a factor of 432 5824. xxxx Check as follows: 158124 5324 53240

34. 432 (3)3(3)9(3)4(3) f  9(3)90 

3 x  is a factor of 432 39499. xxxx Check as follows: 339499 90129 30430

35. 432 (2)22221802 fx  is a factor of 432 18. xxxx Check as follows: 2111118 261018 13590

36. 542 (3)(3)3(3)(3) f  8(3)150 

3 x  is a factor of 5423815. xxxx Check as follows: 31301815 300315 100150

37. 6543 (2)(2)(2)7(2)(2) f  2 8(2)5(2)20 

2 x  is a factor of 654327852. xxxxxx  Check as follows: 21171852 262642 1313210

38. 6543 (2)2(2)5(2)4(2)(2) f  2 7(2)7(2)20 

2 x is a factor of 65432 254772. xxxxxx  Check as follows: 22541772 4241062 2125310

39. 32 (1)0(1)3(1)(1) 011 fk kk

 

40. 32 (1)014(1)(1)210 1 fkk k  

41. 32 (2)02(2)(2)22 14207 fkk kk  

42. 22 2 (1)03(1)2(1)6 560(3)(2)0 3 or 2 fkkk kkkk kk

  

In exercises 43–46, use synthetic division to find the remainder.

43. 22449 408 2041

The remainder is 1, so 2 x is not a factor of 32 2449. xxx 

44. 339512 9015 3053

The remainder is –3, so 3 x  is not a factor of 32 39512. xxx 

45. 249314 8222 41116

The remainder is 6, so 2 x  is not a factor of 432 4934. xxxx

46. 338573 932493 3183190

The remainder is 90, so 3 x is not a factor of 432 38573. xxxx

47. The factors of the constant term, 5, are  1,5 , and the factors of the leading coefficient, 3, are  1,3 . The possible rational zeros are 15 33,1,,5.    

48. The factors of the constant term, 1, are  1 , and the factors of the leading coefficient, 2, are  1,2 . The possible rational zeros are 1 ,1. 2    

49. The factors of the constant term, 6, are  1,2,3,6 , and the factors of the leading coefficient, 4, are  1,2,4 .The possible rational zeros are 1133 ,,,1,, 4242      2,3,6.

50. The factors of the constant term, –35, are  1,5,7 , and the factors of the leading coefficient, 6, are  1,2,3,6 . The possible rational zeros are 11553515 ,,1,,,5,7,,,, 6232236     7773535 ,,,,,35. 63263    

51. The factors of the constant term, 4, are  1,2,4 , and the factors of the leading coefficient, 1, are  1 . The possible rational zeros are  1,2,4 . Use synthetic division to find one rational root: 11144

104 1040

So, 32244(1)(4) (1)(2)(2) xxxxx xxx   the rational zeros are {–2, 1, 2}.

52. The factors of the constant term, 2, are  1,2 , and the factors of the leading coefficient, 1, are  1 . The possible rational zeros are  1,2 . Use synthetic division to find one rational root: 11122 102 1020

So, 32222(1)(2) xxxxx the rational zero is {–1}.

53. The factors of the constant term, 6, are  1,2,3,6 , and the factors of the leading coefficient, 1, are  1 . The possible rational zeros are  1,2,3,6 . Use synthetic division to find one rational root: 11416 156 1560



So, 32246(1)(56) (1)(2)(3) xxxxxx xxx

 the rational zeros are {–1, 2, 3}.

54. The factors of the constant term, 6, are  1,2,3,6 , and the factors of the leading coefficient, 1, are  1 . The possible rational zeros are  1,2,3,6 . Use synthetic division to find one rational root: 31326 306 1020

So, 322326(3)(2) xxxxx the rational zero is {–3}.

55. The factors of the constant term, 6, are  1,2,3,6 , and the factors of the leading coefficient, 2, are  1,2 . The possible rational zeros are 13 22,1,,2,3,6    

Use synthetic division to find one rational root: 221136 4106 2530

322 2136(2)(253) (2)(3)(21) xxxxxx xxx   the rational zeros are 1 3,,2 2 

56. The factors of the constant term, –2, are  1,2 , and the factors of the leading coefficient, 3, are  1,3 . The possible rational zeros are 12 33,,1,2     . Use synthetic division to find one rational root: 2 3232 3 202 3030

322 2 3232(33) 3 xxxxx

the rational zero is 2 3 

57. The factors of the constant term, –2, are  1,2 , and the factors of the leading coefficient, 6, are  1,2,3,6 . The possible rational zeros are 1112 6323,,,,1,2     . Use synthetic division to find one rational root: 261312 1222 6110

322 6132(2)(61) (2)(21)(31) xxxxxx xxx

the rational zeros are 11 2,,23

58. The factors of the constant term, 6, are  1,2,3,6 , and the factors of the leading coefficient, 2, are  1,2 . The possible rational zeros are 13 22,1,,2,3,6   

Use synthetic division to find one rational root: 3 2346 2 306 2040

322 3 2346(24) 2 xxxxx

the rational zero is 3 2

59. The factors of the constant term, 2, are  1,2 , and the factors of the leading coefficient, 3, are  1,3 . The possible rational zeros are 12 33,,1,2  

Use synthetic division to find one rational root: 1 3782 3 122 3660

322 1 3782(366) 3 xxxxxx 

the rational zero is 1 3

60. The factors of the constant term, 4, are  1,2,4 , and the factors of the leading coefficient, 2, are  1,2 . The possible rational zeros are 1 ,1,2,4 2     . Use synthetic division to find one rational root. 1 2184 2 104 2080

322 1 284(28) 2 xxxxx

   the rational zero is 1 2  

61. The factors of the constant term, –2, are  1,2 , and the factors of the leading coefficient, 1, are  1 . The possible rational zeros are  1,2 . Use synthetic division to find one rational root: 111112 1212 12120

So, –1 is a rational zero. Use synthetic division again to find another rational zero:.

432322(1)(22) xxxxxxxx  21212 202 1010 43222(1)(2)(1) xxxxxxx

the rational zeros are {–1, 2}.

62. The factors of the constant term, –4, are  1,2,4 , and the factors of the leading coefficient, 2, are  1,2 . The possible rational zeros are 1 ,1,2,4

. Use synthetic division to find one rational root: 123834 2174 21740

So, –1 is a rational zero. Use synthetic division again to find another rational zero:

63. The factors of the constant term, 12, are  1,2,3,4,6,12  , and the factors of the leading coefficient, 1, are  1 . The possible rational zeros are  1,2,3,4,6,12  . Use synthetic division to find one rational root: 31113112 312312 14140

So, –3 is a rational zero. Use synthetic division again to find another rational zero: 432 32 1312 (3)(44) xxxx xxxx   11414 154 1540

So, –1 is also a rational zero. 432 2 1312 (3)(1)(54) (3)(1)(4)(1) xxxx xxxx xxxx 

 the rational zeros are {–3, –1, 1, 4}.

64. The factors of the constant term, –2, are  1,2 , and the factors of the leading coefficient, 3, are 1,3. The possible rational zeros are 12 33,,1,2     . Use synthetic division to find one rational root: 235152 6262 31310

So, –2 is a rational zero. Use synthetic division again to find another rational zero: 432 32 3552 (2)(331) xxxx xxxx

1 3131 3 101 3030 432 2 2 3552 1 (2)(33) 3 1 3(2)(1) 3 xxxx xxx xxx

the rational zeros are 1 2, 3

65. The factors of the constant term, 1, are  1 , and the factors of the leading coefficient, 1, are  1 . The possible rational zeros are  1 . Use synthetic division to find one rational root: 1121011 1198 11989

The remainder is 9 so, 1 is not a rational zero. Try 1: 1121011 131314 13131415

The remainder is 15 so, 1 is not a rational zero. Therefore, there are no rational zeros.

66. The factors of the constant term, 2, are  1,2 , and the factors of the leading coefficient, 1, are  1 .The possible rational zeros are  1,2 . Testing each value  (2), f  (1),(1),(2)fff shows that none cause the value of the function to equal zero, so there are no rational zeros.

67.  32582fxxxx

The possible rational zeros are 1,2. 11582 162 1620

So, 1 is a zero. Now solve the depressed equation 2 620.xx

636412 644

The solution set is  1,311. 

68.  32753fxxxx

The possible rational zeros are 1,3. 11753 183 1830

So, 1 is a zero. Now solve the depressed equation 2 830.xx

864413 852 212 413 x    

The solution set is  1,413.

69.  43331fxxxx

The possible rational zeros are 1. 113031 1221 12210

So, 1 is a zero. Now find a zero of the depressed equation  32221.gxxxx 11221 131 1310

So, 1 is a zero. Now solve the depressed equation 2 310xx

 39411 35 212 x   

The solution set is 35 1,. 2    

70.  432675418fxxxxx 

The possible rational zeros are  1,2,3,6,9,18.  31675418 394818 131660

So, 3 is a zero. Now find a zero of the depressed equation  323166.gxxxx 313166 3186 1620

So, 3 is a zero. Now solve the depressed equation 2 620xx

 636412 628 37 212 x   

The solution set is  3,37.

71.  32 2961fxxxx

The function has degree 3, so there are three zeros, one or three possible positive zeros; no possible negative zeros.

The possible rational zeros are 1 1,. 2

1 2961 2 141 2820

So, 12 is a zero. Now solve the depressed equation 2 2820 xx

2 88422 848 224 23 x

Solution set: 1 23,,23. 2

72.  32 2341fxxxx

The function has degree 3, so there are three zeros, one possible positive zero; zero or two possible negative zeros.

The possible rational zeros are 1

So, 12 is a zero. Now solve the depressed equation 2 2420 xx

Solution set: 1 12,,12. 2

73.  432536fxxxxx

The function has degree 4, so there are four zeros, zero or two possible positive zeros; zero or two possible negative zeros. The possible rational zeros are

 111536 1236 12360

So, 1 is a zero. Now find a zero of the depressed equation 322360. xxx 21236 206 1030

So, 2 is a zero. Now solve the depressed equation 2 30. x 

223033 xxx

Solution set:  2,3,1,3

74.  4322546fxxxxx 

The function has degree 4, so there are four zeros, zero or two possible positive real zeros; zero or two possible negative real zeros. The possible rational zeros are  1,2,3,6.  312546 3366 11220

So, 3 is a zero. Now find a zero of the depressed equation 32 220.xxx 11122 102 1020

So, 1 is a zero. Now solve the depressed equation 2 20. x  222022 xxx

Solution set:  2,1,2,3

75.  32 252fxxxx

The possible rational zeros are 1 2 ,1,2.

Using synthetic division, we find that x = 1 is a zero. 12512 232 2320

Now factor  2 232.Qxxx  2 232212 xxxx 

So, the complete factorization of   Px is    

 1212.xxx

76.  32 243fxxxx

The possible rational zeros are 3 1 22,,1,3.

Using synthetic division, we find that x = –1 is a zero.

12143 213 2130

Now factor  2 23.Qxxx

2 23231 xxxx 

So, the complete factorization of   Px is  2 123.xx

77.  32 231730fxxxx 

The possible rational zeros are 35 1 222 1,2,3,5,6,,,. 

Using synthetic division, we find that x = –2 is a zero.

2231730 4230 21150

Now factor  2 215.Qxxx

2 215253 xxxx 

So, the complete factorization of   Px is

 2253.xxx

78.  32 232312fxxxx 

The possible rational zeros are 3 1 22 1,2,3,4,6,12,,. 

Using synthetic division, we find that x = 4 is a zero.

4232312 82012 2530

Now factor  2 253.Qxxx  2 253213 xxxx 

So, the complete factorization of   Px is

 4213.xxx

79.  4329116fxxxxx 

The possible rational zeros are 1,2,3,6. 

Using synthetic division, we find that x = 2 is a zero.

2119116 22146 11730

So, 2 is a zero. Now find a zero of the depressed function  32 73.Qxxxx 31173 363 1210

So, 3 is a zero. Now find the zeros of the depressed function 2 21xx using the quadratic formula.   2 22411 28 212 222 12 2 x     

So, the complete factorization of   Px is 

 231212,xxxx  or 

231212.xxxx 

80.  4326308fxxxxx 

The possible rational zeros are 1,2,4,8. 

Using synthetic division, we find that x = –2 is a zero.

2161308 216348 181740

So, –2 is a zero. Now find a zero of the depressed function  328174.Qxxxx 418174 4164 1410

So, 4 is a zero. Now find the zeros of the depressed function 2 41xx using the quadratic formula.

 2 44411 412 212 423 23 2 x     

So, the complete factorization of   Px is

  242323,xxxx  or

    242323.xxxx 

For exercises 81–88, use synthetic division to find the zeros of each function and then use the methods shown in Section 2.2 to complete the graph.

81.  3221920fxxxx

Zeros: –5, –1, 4

82.  32 1424fxxxx

Zeros: –3, –2, 4

83.  32 1620fxxxx

Zeros: –2 (multiplicity 2), 5

84.  3221536fxxxx

Zeros: –3 (multiplicity 2), 4

85.  42222445fxxxx 

Zeros: –3 (multiplicity 2), 1, 5

86.  4323135136fxxxxx 

Zeros: –3 (multiplicity 2), –1, 4

87.  32 6132fxxxx

Zeros: 11 2,,23 

88.  32 61710fxxxx

Zeros: 52 ,1, 23  

Applying the Concepts

91. a.

 4030004030000RR and

 6030006030000.RR Thus, 40 and 60 are zeros of

 3000. Rx Therefore,

2 30004060 1002400 Rxaxx axx   Since (30, 2400) lies on R(x), we have  

  2400300030403060 60010302 a aa   Thus       2 300024060 22004800 Rxxx xx

b.   2 2 300022004800 22001800 Rxxx Rxxx 

c. The maximum weekly revenue occurs at the vertex of the function, 22,. bb R aa

The maximum revenue is $3200 if the phone is priced at $50.

92. a. 

  84725847250.RR and  

1247251247250.RR Thus, 8 and 12 are zeros of   4725. Rx Therefore,

2 4725812 2096 Rxaxx axx   Since (6, 4125) lies on R(x), we have

 4125472568612 6002650 a aa   Thus,

2 472550812 5010004800 Rxxx xx 

b.   2 2 47255010004800 50100075 Rxxx Rxxx  

c. The maximum weekly revenue occurs at the vertex of the function, 22,. bb R aa 

2 1000 10 2 250 0501010001075 4925 b a R    The maximum revenue is $4925 if tickets are priced at $10 each.

93. Since t = 11 represents 2002, we have (11)97.6()(11)()97.6 CCxxQx ()97.6(11)(). CttQt  So

0.00060.06132.082982.904973232.60.00060.06132.082914.6960 (11)(). tttttt xQx 

Use synthetic division to find Q(x):

110.00060.06132.082914.6960

0.00660.746914.6960

0.00060.06791.33600

Now solve the depressed equation to find another zero:  2 2 0.06790.06794(0.0006)(1.3360) 0.00060.06791.33600 2(0.0006)

0.06790.007816810.06790.0884 130.26 or 17.0939. 0.00120.0012

Since we must find t greater than 0, t = 17, and the year is 1991 + 17 = 2008.

94. Since t = 1 represents 1999, we have (1)737.7()(1)()737.7 ssttQt ()737.7(1)(). sttQt

So 3232 0.17792.829242.0240698.6831737.70.17792.829242.024039.0169 (1)() tttttt tQt

Use synthetic division to find Q(t): 10.17792.829242.024039.0169

0.17793.007139.0169

0.17793.007139.01690

Now solve the depressed equation to find another zero: 2 2 3.0071(3.0071)4(0.1779)(39.0169) 0.17793.007139.01690 2(0.1779) 3.007136.8071 25.5030 or 8.5997 0.3558

Since t must be positive, t ≈ 9, and the year is 1998 + 9 = 2007.

95.  32 0.00270.36815.8645195.2782Mtttt  Since   2185.0, M 

We must find two other zeros of  32 32 1850.00270.36815.8645195.2782185

ttt

0.00270.36815.864510.2782

Because 2 is a zero of F(t), use synthetic division to find Q(t). 20.00270.36815.864510.2782

0.00540.725410.2782

0.00270.36275.13910

Now use the quadratic formula to solve the depressed equation.

2 0.36270.362740.00275.1391 0.36270.0760 16.1 or 118 20.00270.0054 ttt

Thus, the model shows that the Marine Corps had about 186,000 when t ≈ 16, or in the year 1990 + 16 = 2006.

96.

We must find two other zeros of

FtUtttt ttt

0.03740.59342.05530.9522

Because 6 is a zero of F(t), use synthetic division to find Q(t).

60.03740.59342.05530.9522

0.22442.21400.9522

0.03740.36900.15870

Now use the quadratic formula to solve the depressed equation.

2 0.36900.369040.03740.1587 0.36900.1599 0.41 or 10.3 0.0748 20.0374 ttt

Reject the negative solution.

According to the model, there was a 7.7% unemployment level in 2002 + 10 = 2012.

97. Use synthetic division to solve the equation 32 62836108100 xxx  32 361085280. xxx The factors of the constant term, 528, are {1,2,3,4,6,8,11,48,66,88,  132,176,264,528}. 

The factors of the leading coefficient, 3, are {1,3}.  Only the positive, whole number possibilities make sense for the problem, so the possible rational zeros are {1, 2, 3, 4, 6, 8, 11, 16, 22, 44, 48, 66, 88, 132, 176, 264, 528}.

436108528 1224528 361320

Thus, x = 4.

98. The demand function is 2 ()33010 pxxx the revenue function is 23 33010. xxx 

Profit = revenue – cost = 2332 910(33010)(36108100) xxxxxx  32 910416222100 xxx  32 04162221010. xxx  The factors of the constant term, –1010, are {1,2,5,10,101,202,505,1010}.  The factors of the leading coefficient, –4, are  1,2,4 .

Find the zero using synthetic division: 54162221010 20201010 442020

Thus, x = 5.

99. The cost function gives the result as a number of thousands, so set it equal to 125: 3215550125 xxx 32155750. xxx The factors of the constant term, –75, are {1,3,5,15,  25,75}. The factors of the leading coefficient, 1, are  1 . Only the positive solutions make sense for the problem, so the possible rational zeros are {1, 3, 5, 15, 25, 75}. Use synthetic division to find the zero: 15115575 15075 1050

The total monthly cost is $125,000 when 15 hundred (1500) units are produced.

100. The demand function is 74 ()33pxx x  , so the revenue function is 2 ()3374.xpxxx

The break-even point occurs when revenue – cost = 0: 232 32 32 22 2 (3374)(15550)0 122240 (2)(1224)0 (2)12(2)0 (12)(2)0 12 xxxxx xxx xxx xxx xx x

So 1200 printers must be sold to break even.

101. Let x represent the edge of the taller box. Then x – 3 represents the edge of the shorter box. The volume of the taller box is 3 , x and the volume of the shorter box is  3 3. x We must solve  3 3 31843. xx

There are many possible rational zeros, so we start with an educated guess. The edge must be greater than 3 and less than 12. (Note that 3 121728,  which is close to the total volume of the two boxes.) The possible rational zeros between 3 and 12 are 5, 10, 11, and 11 2 Using synthetic division, we find that 11 is a zero. 1129271870 221431870 2131700

Using the quadratic formula to find the zeros of 2 213170 xx , we have

2 131342170 131191 , 224 x

which is not a real number. Thus, there are no other real zeros. Therefore, the taller box has edge 11 inches and the shorter box has edge 11 – 3 = 8 inches.

102. Let x represent the width of the pool. Then the volume of the deep water section is 3 , x and the volume of the shallow water section is 3. x The total volume is 536 cubic meters, so we have 33353635360. xxxx

There are many possible rational zeros, so we start with an educated guess. The edge must be greater than 1 and less than 9. (Note that 3 9729,  which is more the total volume of the pool.) The possible rational zeros between 1 and 8 are 2, 4, 8. Using synthetic division, we find that 8 is a zero.

8103536 864536 18670

Using the quadratic formula to find the zeros of 2 867xx , we have

 2 884167 8204 , 212 x    which is not a real number. Thus, there are no other real zeros.

Therefore, the dimensions of the deep water section are 8 m8 m8 m. 

103. The length and width of the box are 18 – 2x, and the height is x, so the volume is 2 (182).xx

2 432(182) xx  32 432472324 xxx  32 4723244320. xxx

The factors of the constant term, 432, are {1,2,3,4,6,8,9,12,  16,18,24,27,36,48,54, 72,108,144,216,432}.

 

The factors of the leading coefficient, 4, are  1,2,4 . There are many possible rational zeros, but the only ones that make sense for the problem are 1133 4242,,,1,,2,3,4,6,8.

Use synthetic division to find the zero: 3472324432 12180432 4601440

The corners should be 3 inches by 3 inches.

From the similar right triangles EFC  and , ABC  we have 2727 , 1 y xx    or 1 2727. x y x  

Using the Pythagorean theorem in right triangle

ABC gives    2 2 2 1278.xy

Now substitute.

2 2 2 1 1278. x x x   

Multiply both sides by 2 x to eliminate the fraction.

  2 2 222 12718 xxxx 

Now expand and simplify.



2222 43222 432 21272164 227542764 23654270 xxxxxx xxxxxx xxxxPx





The possible rational zeros of the polynomial are 1,3,9,27. 

We know that x must be a positive number less than 8 Using synthetic division, we see that x = 3 is a zero of   . Px 312365427 3156327 152190

So, the remainder is  325219.Qxxxx

There are no other real zeros.

432 32 2365427 35219 Pxxxxx xxxx  

The possible rational zeros of   Qx are 1,3,9. Using synthetic division, we see that x = 3 is a zero of   . Qx 315219 3249 1830

Thus,    432 2 2 2365427 383 Pxxxxx xxx  

Since the height = x + 1, the top of the ladder touches the wall at 3 + 1 = 4 feet above the ground.

Beyond the Basics

105. Divide 32 48113 xxx by 1 : 2 x 

1248113 253 41060

Now divide 2 4106 xx by 1 : 2 x 

124106 26 4120

Since 1 2 x     does not divide 412 x  , 1 2 x 

  is a root of multiplicity 2 of 32 48113 xxx

106. Divide 32 9384 xxx by 2 : 3 x   

239384 624 9360

Now divide 2 936 xx by 2 : 3 x  

23936 66 990

Since 2 3 x      does not divide 9x – 9, 2 3 x  

is a root of multiplicity 2 of 32 9384 xxx

107. Let x represent b. Then we can use synthetic division to divide 33 xa by x – a. 3 23 2 100 10 aa aaa aa

Then    3322 xaxaxaxa  or    3322 . babababa 

108. Let x represent b. Then we can use synthetic division to divide 44 xa by x – a. 4

109. If 2 1 x is a factor of 10 4, xax then so are x – 1 and x + 1. Use synthetic division to show that one or the other is not a factor.

Now see if x + 1 is a factor of

x + 1 is not a factor, so there is no real number a such that 2 1 x is a factor of 10 4. xax

110. If 22ba divides 22 , nnba then b – a and b + a, must divide 22nnba Let x represent b. Then we can use synthetic division to divide 22nn xa by x – a and by x + a

Thus, 22ba divides 22nnba

The

113. 43213 2220 22 xxxx

432 44430. xxxx The factors of the constant term, –3, are 1,3. The factors of the leading coefficient are  1,2,4 . So, the possible rational zeros are 1133 ,,,1,,3. 4242

Use synthetic division to find one zero:

So, 1 2 is a rational zero and

3 2 x  is another rational zero.

114. i. Simplifying the fraction if necessary, we can assume that p q is in lowest terms. Since p q is a zero of F, we have 0. Fp q



ii. Substitute p q for x in the equation ()0.Fx 

iii. Multiply the equation in (ii) by . n q

iv. Subtract 0 n aq from both sides of the equation.

v. The left side of the equation in (iv) is

11 11 121 11 nnn nn nnn nn apapqapq papapqaq    Therefore p is a factor.

vi. ab ab pp 

vii. Since p and q have no common prime factors, p must be a factor of 0 . a

viii. Rearrange the terms of the equation in (iii).

ix. The left side of the equation in (viii) is

11 110 121 110 nnn n nnn n apqapqaq qapapqaq    

Therefore q is a factor.

x. . ab ab qq 

xi. Since p and q have no common prime factors, q must be a factor of n a

Critical Thinking/Discussion/Writing

115.  3 2 fxxcx

Possible rational zeros are 1,2.

Use synthetic division to see which zeros lead to an integer value of c. 1102 111 1110 c c c 

  2101 cc  1102 111 1110 c c c  

  2103 cc  2102 2428 1240 c c c  

  22805 cc  2102 2428 1240 c c c 

  22803 cc 

The values of c for which  3 2 fxxcx has at least one rational zero are –1, 3, and 5.

116.  3 6 fxxcx

Possible rational zeros are 1,2,3,6. 

Use synthetic division to see which zeros lead to an integer value of c. 1106 111 1110 c c c 

  6107 cc  (continuedonnextpage)

(continued) 1106 111 1110 c

6106 6366216 16360 c

 66216035 cc  6106 6366216 16360 c c c 

 66216037 cc 

The values of c for which  3 6 fxxcx has at least one rational zero are –5, 1, 7, 11, 35, and 37.

117. Let 22 3330.xxx  The only possible rational zeros of this equation are 1 and 3 . Because 3 is neither of these, it must be irrational.

118.   3 3333 4440.xxx 

The only possible rational zeros of this equation are  1,2,4 . Since 3 4 is not in the set, it must be irrational.

119. 2 3232962 xxxx  2 670.xx The possible rational zeros of this equation are  1,7 . Since 32 is not in the set, it must be irrational.

120.   3 2/332/33 99810.xxx  The possible rational zeros of this equation are {1,3,9,27,  81}. Since 23 9 is not in the set, it must be irrational.

121. a. xa  is a factor of nn xa  if n is an odd integer. The possible rational zeros of nn xa  are 23 {,,,,}. n aaaa   Since xa  is a factor means that –a is a root, then ()0()nnnn aaaa  only for odd values of n.

b. xa  is a factor of nn xa if n is an even integer. The possible rational zeros of nn xa are 23 {,,,,}. n aaaa  Since xa  is a factor means that –a is a root, then ()0()nnnn aaaa  only for even values of n.

c. There is no value of n for which xa is a factor of nn xa  . The possible rational zeros of nn xa  are 23 {,,,,}. n aaaa   If xa is a factor, then a is a root, and 0 nnnn aaaa  , which is not possible.

d. xa is a factor of nn xa for all positive integers n. The possible rational zeros of nn xa are 23 {,,,,}. n aaaa   If xa is a factor, then a is a root, and 0 nn aa , which is true for all values of n. However, if n is negative, then 11 nn nn xa xa  and xa is not a factor.

122. a. According to exercise 121d, xa is a factor of nn xa for any positive integer n. Using this result with x = 7, a = 2, and n = 11, we find that 7 – 2 = 5 is a factor of 1111 72. Therefore, 721111 is divisible by 5.

b.     202010101010 191019101910. 

According to exercises 121b and 121d, xa  is a factor of nn xa if n is an odd integer, and xa is a factor of nn xa for any positive integer n. Using these results with x = 19, a = 10, and n = 10, we find that 19 – 10 = 9 and 19 + 10 = 29 are both factors of 1010 1910, and therefore, 19102020 is divisible by 9 and 29. Since 9 and 29 have no common factors, 19102020 must be divisible by their product, 9(29) = 261.

Getting Ready for the Next Section

Rational Functions

Practice Problems 1. 2 3 () 45 fxx xx 

The domain of f consists of all real numbers for which 2 450xx 2 450(5)(1)0 5 or 1 xxxx xx  

Thus, the domain is (,1)(1,5)(5,)  

2. a.  3 2 gx x  Let  1 fx x  Then   31 332. 22 gxfx xx

The graph of   2 yfx is the graph of   yfx  shifted two units to the right. This moves the vertical asymptote two units to the right. The graph of   32yfx is the graph of   2 yfx stretched vertically three units. The domain of g is     ,22,.  The range of g is     ,00,.  (continuedonnextpage)

(continued)

The vertical asymptote is x = 2. The horizontal asymptote is y = 0. b.  25 1 hxx x    2 125 22 3 xx x

253 2 11 hxx xx 

1 fx x  Then

253 2231. 11 x hxfx xx  

The graph of   yhx  is the graph of

  yfx  shifted one units to the left and then stretched vertically three units. The graph is then shifted two units up. This moves the vertical asymptote one unit to the left. The horizontal asymptote is shifted two units up. The domain of h is

,11,.  The range of h is  

 ,22,.  The vertical asymptote is x = 1. The horizontal asymptote is y = 2.

3. 2 1 () 310 fxx xx   

The vertical asymptotes are located at the zeros of the denominator.

2 3100(5)(2)0 5 or 2 xxxx xx 



The vertical asymptotes are x = 5 and x = 2.

4. 2 3(3)1 () (3)(3)3 9 303 fxxx xxxx xx  



The vertical asymptote is x = 3.

5. a. 25 () 34fxx x  

Since the numerator and denominator both have degree 1, the horizontal asymptote is 2 3 y 

b. 2 3 () 1 x gx x  

The degree of the numerator is greater than the degree of the denominator, so there are no horizontal asymptotes.

6. 2 2 () 1 fxx x 

First find the intercepts: 2 2 00 1 x x x 

The graph passes through the origin. Find the vertical asymptotes: 2 10(1)(1)0 1 or 1 xxx xx  

Find the horizontal asymptote: The degree of the numerator is less than the degree of the denominator, so the horizontal asymptote is the x-axis.

Test for symmetry: 22 2()2 ()() ()11 fxxxfx xx 

f is an odd function and is symmetric with respect to the origin.

Use test numbers to determine where the graph of f is above and below the x-axis. (continuedonnextpage)

Interval

The zero of   2 Rxx  is 2 and the zeros of  2 23231Dxxxxx  are

2 x  and x = 1.

Use test numbers to determine where the graph of f is above and below the horizontal asymptote y = 1.

Interval

7. 2 2 21 () 23 fxx xx  

First find the intercepts: 2 2 212 0 2 23 x x xx   ; 1 (0) 3 f 

Find the vertical asymptotes: 2 230(23)(1)0 3 or 1 2 xxxx xx

Find the horizontal asymptote: The degree of the numerator is the same as the degree of the denominator, so the horizontal asymptote is 2 1 2 y 

Test for symmetry: 22 2()121 () 2()()323 () and () fxxx xxxx fxfx

f is neither even nor odd, so there is no symmetry with respect to the y-axis or the origin.

Test point Value of 

fx Above/below x-axis 3 , 2

Notice that the graph crosses the horizontal asymptote at (2, 1).

First find the intercepts:

x xi

there is no x-intercept.

Find the vertical asymptotes: 2 202xxi there is no vertical asymptote. Find the horizontal asymptote: The degree of the numerator is the same as the degree of the denominator, so the horizontal asymptote is 1 1 1 y

(continuedonnextpage)

(continued)

Test for symmetry: 22 22 ()11 ()() ()22 fxxxfx xx

 



f is an even function and is symmetric with respect to the y-axis.

 2 22 11 1 22 fxx xx   

Neither R(x) = 1 nor  2 2 Dxx have real zeros. Since 2 1 2 x  is negative for all values of x, the graph of   fx is always below the line y = 1.

9. 2 2 () 1 fxx x  

First find the intercepts: 2 2 02 1   x xi x there is no x-intercept. (0)2 f 

Find the vertical asymptotes: 101xx

Find the horizontal asymptote: The degree of the numerator is greater than the degree of the denominator, so there is no horizontal asymptote. However, there is an oblique asymptote: 2 23 11 is the oblique 11 x xyx xx   asymptote.

The graph is above the line y = x + 1 on   1,  and below the line on   ,1.

Test for symmetry: 22 ()22 ()() 11 and () fxxxfx xx fx    f is neither even nor odd, so there is no symmetry with respect to the y-axis or the origin.

The intervals determined by the zeros of the numerator and of the denominator of

2 23 11 11 x fxxx xx 

divide the x-axis into two intervals,  ,1 and   1,. 

Use test numbers to determine where the graph of f is above and below the x-axis.

Interval Test point Value of   fx Above/below y = x + 1

2 6 above 10. (100) () 20 xx Rx x   a. 10(10010) (10)30 1020 R   billion dollars.

This means that if income is taxed at a rate of 10%, then the total revenue for the government will be 30 billion dollars. Similarly, R(20) = $40 billion, R(30) = $42 billion, R(40) = $40 billion, R(50) ≈ $35.7 billion, R(60) = $30 billion.

From the graphing calculator screen, we see that a tax rate of about 29% generates the maximum tax revenue of about $42.02 billion.

2.4 Exercises

Concept and Vocabulary

1. A function f is rational if it can be expressed in the form   Nx Dx where N(x) and D(x) are polynomials.

2. A line x = a is called a vertical asymptote of the rational function f if ()fx  as xa   or as xa 

3. A line y = k is called a horizontal asymptote of the rational function f if () fxk  as x  or as x  .

4. If a is a zero of denominator   Dx of the rational function

Nx fxDx

expressed in lowest terms, then x = a is a vertical asymptote of f

5. True

6. False. The graph of a rational function can cross its horizontal asymptote.

7. True

8. False Building Skills

9. (,4)(4,)

13. 2 60(3)(2)02,3xxxxx 

The domain of the function is (,2)(2,3)(3,).  

14. 2 670(7)(1)01,7xxxxx 

The domain of the function is (,1)(1,7)(7,).  

15. 2 680(4)(2)02,4xxxxx 

The domain of the function is (,2)(2,4)(4,).  

16. 2 320(2)(1)01,2xxxxx 

The domain of the function is (,1)(1,2)(2,).  

17. As 1,().xfx  

18. As 1,().xfx

19. As 2,().xfx  

20. As 2,().xfx

21. As ,()1.xfx

22. As ,()1.xfx

23. The domain of f is (,2)(2,1)(1,).  

24. There are two vertical asymptotes.

25. The equations of the vertical asymptotes of the graph are x = –2 and x = 1.

26. The equation of the horizontal asymptote of the graph is y = 1.

27.

Domain: (,4)(4,)  

Range: (,0)(0,)  

Vertical asymptote: x = 4

Horizontal asymptote: y = 0

Domain: (,3)(3,)  

Range: (,0)(0,)  

Vertical asymptote: x = 3

Horizontal asymptote: y = 0

Domain:

Range:

Horizontal

Domain:

Range:

Domain: (,2)(2,)  

Range: (,3)(3,)  

Vertical asymptote: x = 2

Horizontal asymptote: y = 3

Domain: (,3)(3,)  

Range: (,1)(1,)  

Vertical asymptote: x = 3

Horizontal asymptote: y = 1

Domain: (,4)(4,)  

Range: (,5)(5,)  

Vertical asymptote: x = 4

Horizontal asymptote: y = 5

Domain: (,5)(5,)  

Range: (,2)(2,)  

Vertical asymptote: x = 5

Horizontal asymptote: y = 2

In exercises 35 44, to find the vertical asymptotes, first eliminate any common factors in the numerator and denominator, and then set the denominator equal to zero and solve for x

35. 1 x  36. 2 x 

37. 4,3xx 38. 34 , 23 xx

39. 2 2 1(1)(1) (). (2)(3) 6 hxxxx xxxx

The equations of the vertical asymptotes are 3 and 2.

40. 2 2 4(2)(2) (). (34)(1) 34 hxxxx xxxx

The equations of the vertical asymptotes are 4 and 1. 3

41. 2 2 68(4)(2) (). (4)(3) 12 fxxxxx xxxx

Disregard the common factor. The vertical asymptote is 3. x 

42. 2 3 9(3)(3) (). (2)(2) 4 fxxxx xxxxx

The equations of the vertical asymptotes are 0,2, and 2. xxx

43. There is no vertical asymptote.

44. There is no vertical asymptote.

For exercises 45 52, locate the horizontal asymptote as follows:

• If the degree of the numerator of a rational function is less than the degree of the denominator, then the x-axis (y = 0) if the horizontal asymptote.

• If the degree of the numerator of a rational function equals the degree of the denominator, the horizontal asymptote is the line with the equation n m a yb  , where n a is the coefficient of the leading term of the numerator and mb is the coefficient of the leading term of the denominator.

• If the degree of the numerator of a rational function is greater than the degree of the denominator, then there is no horizontal asymptote.

45. 0 y  46. 0 y  47. 2 3 y  48. 3 4 y 

49. There is no horizontal asymptote.

50. 0 y  51. 0 y 

52. There is no horizontal asymptote.

53. d 54. f 55. e

56. b 57. a 58. c

59. 2 00 3 x x x  is the x-intercept. 2(0) 00 03 y  is the y-intercept. The vertical asymptote is x = 3. The horizontal asymptote is y = 2. 2()2 ()() ()33 fxxxfx xx   and   ()fxfx there are no symmetries. The intervals to be tested are (,3) and (3,). The graph is above the horizontal asymptote on (3,)  and below the horizontal asymptote on (,3). 

60. 00 1 x x x  is the x-intercept.

0 00 01 y  is the y-intercept. The vertical asymptote is x = 1. The horizontal asymptote is y = –1. () () ()3fxx x  () and() 3 xfxfx x  there are no symmetries. The intervals to be tested are (,1), and (1,). The graph is above the horizontal asymptote on (,1)  and below the horizontal asymptote on (1,). 

61. 2 00 4 x x x  is the x-intercept.

2 0 00 04 y  is the y-intercept. The vertical asymptotes are x = –2 and x = 2. The horizontal asymptote is the x-axis.

22 ()() ()44 xx fxfx xx  the function is odd, and the graph is symmetric with respect to the origin. The intervals to be tested are (,2),(2,0),(0,2) and (2,). The graph is above the x-axis on (2,0)(2,)   and below the x-axis on (,2)(0,2).  

62. 2 00 1 x x x  is the x-intercept.

2 0 00 10 y  is the y-intercept. The vertical asymptotes are x = –1 and x = 1. The horizontal asymptote is the x-axis.

22 ()() 1()1 xx fxfx xx  the function is odd, and the graph is symmetric with respect to the origin. The intervals to be tested are (,1),(1,0),(0,1) and (1,). The graph is above the x-axis on (,1)(0,1)   and below the x-axis on (1,0)(1,).  

63. 2 2 2 00 9 x x x  is the x-intercept.

2 2 0 00 04 y  is the y-intercept. The vertical asymptotes are x = –3 and x = 3. The horizontal asymptote is y = –2.

22 () ()() ()99 xx fxfx xx  the function is even, and the graph is symmetric with respect to the y-axis. The intervals to be tested are (,3),(3,3),  and (3,).  The graph is above the horizontal asymptote on (3,3) and below the horizontal asymptote on (,3)(3,)  

64. 2 2 4 02 x x x  is the x-intercept.

2 2 40 0  there is no y-intercept. The vertical asymptote is the y-axis. The horizontal asymptote is y = –1. 22

22 4()4 ()() () xx fxfx xx  the

function is even, and the graph is symmetric with respect to the y-axis. The intervals to be tested are (,0),and (0,). The graph is above the horizontal asymptote on (,0)(0,).  

65. 2 2 0 2 x  there is no x-intercept.

2 2

01 02 y  is the y-intercept. The

vertical asymptotes are 2 x  . The horizontal asymptote is the xaxis.

22 22 ()() ()22 fxfx xx  the function is even, and the graph is symmetric with respect to the y-axis. The intervals to be tested are (,2),(2,2) and (2,).

The graph is above the x-axis on   ,2   2,   and below the x-axis on   2,2 .

66. 2 2 0 3 x  there is no x-intercept.

2 22 0 3 03 y  is the y-intercept. The vertical asymptotes are 3 x  . The horizontal asymptote is the x-axis.

22 22 ()() ()32 fxfx xx  the function is even, and the graph is symmetric with respect to the y-axis. The intervals to be tested are (,3),(3,3) and (3,).

The graph is above the x-axis on   3,3 and below the x-axis on  

 ,33,  

67. 1 01 (2)(3) x x xx    is the x-intercept. 0111 (02)(03)66 y    is the y-intercept. The vertical asymptotes are x = –3 and x = 2. The horizontal asymptote is the x-axis. ()1 ()() (()2)(()3)fxxfx xx    and ()fx  there are no symmetries. The intervals to be tested are (,3),(3,1),  (1,2), and (2,). The graph is above the x-axis on (3,1)(2,)   and below the x-axis on (,3)(1,2).  

68. 1 01 (1)(2) x x xx   is the x-intercept.

0111

(01)(02)22 y   is the y-intercept.

The vertical asymptotes are x = –1 and x = 2. The horizontal asymptote is the x-axis. ()1 ()() and (()1)(()2) x fxfx xx   ()fx  there are no symmetries. The intervals to be tested are (,1),(1,1),  (1,2), and (2,).  The graph is above the x-axis on (1,1)(2,)   and below the x-axis on (,1)(1,2).  

69.  2 2 32 fxx xx   2 2 0200 32 x xx xx   is the x-intercept.    2 20 00 0302 y   is the y-intercept.  2 32121,2xxxxxx  are the vertical asymptotes. The horizontal asymptote is the x-axis.      2 2 2 32 2 or 32 x fx xx xfxfx xx    

There are no symmetries. The intervals to be tested are (,2),(2,1),  (1,0)and (0,).

The graph is below the x-axis on (,2)(1,0)   and above the x-axis on (2,1)(0,).  

70.  2 2 68 fxx xx   2 2 0200 68 x xx xx   is the x-intercept.

2 20 00 0608 y   is the y-intercept.

 2 68242,4xxxxxx  are the vertical asymptotes. The horizontal asymptote is the x-axis.

    2 2 2 68 2 or 8 x fx xx xfxfx xx    

There are no symmetries. The intervals to be tested are (,0),(0,2),  (2,4)and (4,).  The graph is below the x-axis on (,0)(2,4)   and above the x-axis on (0,2)(4,).  

71. 

2 1 1 fxx xx    2 1

0101 1 x xx xx   is the x-intercept.

 2 01 001  is undefined, so there is no y-intercept.  2 100,1xxxx are the vertical asymptotes. The horizontal asymptote is the x-axis.

2 2 1 1 1 or 1 fxx xx xfxfx xx

There are no symmetries. The intervals to be tested are (,1),(1,0),  (0,1)and (1,).  The graph is above the x-axis on (,1)(1,)   and below the x-axis on (1,0)(0,1). 

72.   2 1 1 fxx xx

2 1 0101 1 x xx xx

 is the x-intercept.

 2 01 001  is undefined, so there is no y-intercept.

 2 100,1xxxx are the vertical asymptotes. The horizontal asymptote is the x-axis.

   2 1 or 1 fxxfxfx xx  

There are no symmetries. The intervals to be tested are (,1),(1,0),  (0,1)and (1,).  The graph is below the x-axis on (0,1) and above the x-axis on (,1)(1,0)(1,).  

73.  2 3 712 fxx xx  

2 331 0 344 712 xx xxxxx 

there is no x-intercept.  2 031 4 07012   is the y-intercept.

There is a hole at (3, –1) and x = 4 is the vertical asymptote. The horizontal asymptote is the x-axis.

  2 2 3 712 3 or 712 fxx xx xfxfx xx  

 

There are no symmetries. The intervals to be tested are (,3),(3,4),  and (4,).  The graph is below the x-axis on (,1)(3,4)   and above the x-axis on (4,). 

74.  2 2 6 fxx xx  

2 221 0 233 6 xx xxxxx    there is no x-intercept. There is a hole at   1 5 2,.

2 021 3

006   is the y-intercept. x = 3 is the vertical asymptote. The horizontal asymptote is the x-axis.

2 2 2 6 2 or 6 fxx xx xfxfx xx     

There are no symmetries. The intervals to be tested are (,2),(2,3),  and (3,).  The graph is below the x-axis on (,2)(2,3)   and above the x-axis on (3,). 

75. 2 2 00 1 x x x   is the x-intercept.

2 2 0 00 01 y   is the y-intercept. There is no vertical asymptote. The horizontal asymptote is y = 1. 22 22 () ()() ()11 xx fxfx xx   the function is even, and the graph is symmetric with respect to the y-axis. The intervals to be tested are (,0) and (0,). The graph is above the x-axis on (,0)(0,)   and below the horizontal asymptote on   ,. 

76. 2 2 2 00 4 x x x   is the x-intercept.

2 2 2(0) 00 04 y   is the y-intercept. There is no vertical asymptote. The horizontal asymptote is y = 2. 22 22 2() ()() ()44 xx fxfx xx   the function is even, and the graph is symmetric with respect to the y-axis. The intervals to be tested are (,0) and (0,). The graph is above the x-axis on (,0)(0,)   and below the horizontal asymptote on   ,.  77.

  02fxx are the x-intercepts. 

 2 322 322 2 4 44044 9 9909 9 xx xxx xxx xx  4 9 is the y-intercept. However, there is a hole at 4 0, 9

since

3 0900 

2 90330xxxx x = –3 and x = 3 are the vertical asymptotes.

(continuedonnextpage)

(continued)

The degree of the numerator is the same as the degree of the denominator, so the horizontal asymptote is y = 1.

3 33 333 4 44 () 99 9 () the function is even, and the fxxxxxxx xxxxxx fx

 graph is symmetric with respect to the y-axis. The intervals to be tested are (,3),(3,0),  (0,3), and (3,).  The graph is above the horizontal asymptote on (,3)(3,)   and below the horizontal asymptote on

3,00,3.

2 2 2 32 032 8 x x x    there are no x-intercepts.

2 2 03232 4 8 08 y    is the y-intercept. However, there is a hole at (0, 4) since  3 0800.  Since there is no real solution for 2 80, x  there are no vertical asymptotes. The horizontal asymptote is y = 1 since the degrees of the numerator and the denominator are equal and the leading coefficients are the same.

     even, and the graph is symmetric with respect to the y-axis. The intervals to be tested are (,0) and (0,). The graph is above the horizontal asymptote on (,0)(0,)  

() the function is

79. 2 (2) 0 2 x x  there is no x-intercept. (There is a hole at x = 2.)

2 (02) 22 02 y  is the y-intercept. There are no vertical asymptotes. There are no horizontal asymptotes.

2 (2)

()() and() 2 x fxfxfx x 

there are no symmetries. The intervals to be tested are (,2)  and (2,).  The graph is above the x-axis on (2,)  and below the x-axis on (,2)  .

80. 2 (1) 0 1 x x  there is no x-intercept. (There is a hole at x = 1.)

2 (01) 11 01 y  is the y-intercept. There are no vertical asymptotes. There are no horizontal asymptotes.

2 (1) ()() and() 1 x fxfxfx x  there are no symmetries. The intervals to be tested are (,1)  and (1,).  The graph is above the x-axis on (1,)  and below the x-axis on (,1) 

(continuedonnextpage)

(continued)

81. The x-intercept is 1 and the vertical asymptote is x = 2. The horizontal asymptote is y = 2, so the degree of the numerator equals the degree of the denominator, and the ratio of the leading terms of the numerator and the denominator is 2.

Thus, the equation is of the form 1 2 x ya

The y-intercept is 1, so we have 01 12.

Thus, the equation is 

82. The x-intercept is 2 and the vertical asymptote is x = 1. The horizontal asymptote is 1 2 y  , so the degree of the numerator equals the degree of the denominator, and the ratio of the leading terms of the numerator and the denominator is 1 2 . Thus, the equation is of the form 2 1 x ya x

The y-intercept is 1, so we have 021 1. 012

Thus, the equation is 

2 21fxx x

83. The x-intercepts are 1 and 3, and the vertical asymptotes are x = 0 and x = 2. The horizontal asymptote is y = 1, so the degree of the numerator equals the degree of the denominator, and the ratio of the leading terms of the numerator and the denominator is 1. Thus, the equation is of the form

There is no y-intercept, so the equation is

84. The x-intercepts are 2 and 3. The vertical asymptotes are x = 1 and x = 2. The horizontal asymptote is y = 1, so the degree of the numerator equals the degree of the denominator, and the ratio of the leading terms of the numerator and the denominator is 1. Thus, the equation is of the form

The y-intercept is 3, so we have

0203 31. 0102 aa    Thus, the equation is 

85. 2 211 2. x x xx  

The oblique asymptote is 2.yx 

86. 2 11 . x x xx 

The oblique asymptote is yx 

87. 3 22 11 . x x xx 

The oblique asymptote is yx 

88. 32 22 211 21. xx x xx

The oblique asymptote is 21.yx 89. 2 2 2 11 21 22 3 x xxx xx x

The oblique asymptote is 2. yx

90. 2 2 21 1232 22 2 1 1 x xxx xx x x

The oblique asymptote is 21.yx 91. 232 3 2 2 2 1201 21 22 1 x xxxx xx xx x x

The oblique asymptote is y = x 2. Note that there is a hole in the graph at x = 1. 92. 232 3 4001 4 41 x xxxx xx x

The oblique asymptote is .yx 

Applying the Concepts

93. a. ()0.52000Cxx b. ()0.5()0.520002000 Cxx Cx xxx

c. 2000 (100)0.520.5 100 2000 (500)0.54.5 500 2000 (1000)0.52.5 1000

These show the average cost of producing 100, 500, and 1000 trinkets, respectively.

d. The horizontal asymptote of ()Cx is 0.5 y  . It means that the average cost approaches the fixed daily cost of producing each trinket as the number of trinkets approaches .

94. a. 2 0.00267000 () 7000 0.0026 Cxxx x x x  

b. 7000 (100)0.002(100)675.8 100 7000 (500)0.002(500)619 500 7000 (1000)0.002(1000)611 1000

These show the average cost of producing 100, 500, and 1000 CD players, respectively.

c. The oblique asymptote is 0.0026.yx For large values of x, this is a good approximation of the average cost of producing x CD players.

95. a.

b. 4(50)1 (50)4 min 10050 f  

4(75)1 (75)12 min 10075 f   4(95)1 (95)76 min 10095 f  

4(99)1 (99)397 min 10099 f  

c. (i) As 100,().xfx

(ii) The statement is not applicable because the domain is x < 100.

d. No, the bird doesn’t ever collect all the seed from the field.

96. a. 1000 (50)20 10050 1000 (75)40 10075 1000 (90)100 10090 1000 (99)1000 10099

These show the estimated cost (in millions of dollars) of catching and convicting 50%, 75%, 90%, and 99% of the criminals, respectively.

c. As 100,().xCx

d. 1000 303000301000 100 20003066.67% x x xx  

97. a. 2 3(50)50 (50)$3.02 billion 50(10050) C  

c. 22 2 22 2 350350 30 (100) 100 300030350 333000500 xx xxxx xxx xx

2 300030004(33)(50) 2(33) 30008,993,40030002998.9 6666 90.89 or 0.017 x  

Reject the negative solution. Approximately 90.89% of the impurities can be removed at a cost of $30 billion.

98. a. The horizontal asymptote is y = aa is called the saturation level because as the concentration of the nutrient is increased, the growth rate is pushed close to a

b. () 22 kka gkaakkk  

99. a. 8(0)16 (0)16 thousand16,000 2(0)1 P   

b. The horizontal asymptote is 4. y  This means that the population will stabilize at 4000.

100. a.

b. The horizontal asymptote is 0. y  This means that, as time passes, the concentration of the drug approaches 0.

The concentration of the drug in the bloodstream is maximal at 1 hour after the injection.

d. 2 2 5 22520 1 1 (21)(2)0 or 2 2 t tt t txtt

The concentration level equal 2 ml/l at 1 2 hr and 2 hr after the injection.

101. a. 10200,000 () 2500 fxx x   b. 10(10,000)200,000 (10,000)$40 10,0002500 C  

c. 10200,000 20 2500 2050,00010200,000 10250,00025,000 x x xx xx  

More than 25,000 books must be sold to bring the average cost under $20.

d. The vertical asymptote is x = 2500. This represents the number of free samples. The horizontal asymptote is y = 10. This represents the cost of printing and binding one book.

102. a. (1)(1)111 (1) 1 qppqpq p qppq p  

c. 0.25 0.3333.33% 10.25 q 

Beyond the Basics

103. Stretch the graph of 1 y x  vertically by a factor of 2, and then reflect the graph about the x-axis.

104. Shift the graph of 1 y x  two units right, and then reflect the graph about the x-axis.

105. Shift the graph of 2 1 y x  two units right.

106. Shift the graph of 2 1 y x  one unit left.

107. Shift the graph of 2 1 y x  one unit right and two units down.

108. Shift the graph of 2 1 y x  two units left and three units up.

109. Shift the graph of 2 1 y x  six units left.

110. 22 2 22 22 2 222 3 6931828 31827 1 318283(69)1

Shift the graph of 2 1 y x  three units left and three units

111. 22 2 1 2122 21 1 xxxx xx

22 22 2 222 22(21)1 2121 2111 1 2121(1) xxxx xxxx xx xxxxx

Shift the graph of 2 1 y x  one unit right and one unit up.

112. 22 2 22 22 2 222 2 21243 242 5 2432(21)5 2121 2(21)55 2 2121(1) xxxx xx xxxx xxxx

Shift the graph of 2 1 y x  one unit left, stretch the graph vertically by a factor of 5, reflect it in the x-axis, and shift it two units up.

113. a. If c is a zero of f(x), then x = c is a vertical asymptote because c is not a factor of the numerator.

b. Because the numerators are positive, the signs of f(x) and g(x) are the same.

c. If the graphs intersect for some value x, then f(x) = g(x).

2 1

d. If f(x) increases (decreases, remains constant), then the denominator of g(x) increases (decreases, remains constant). Therefore, g(x) decreases (increases, remains constant).

114.

115.  1 1 ()23() and 23 23 becomes 23 fxxfx x yxxy

Thus, the functions are different.

116.  1 1 12 ()() and 21 11 becomes 22 2121 (1)21 2121 (). 11 xxfxfx xx xy yx xy xyxyxyyx yxx xx yyfx xx

117. g(x) has the oblique asymptote 23yx . () as , and () as . gxxgx x 

118. f(x) has no oblique asymptote. ()fx  as , and ()as . xfxx  For large |x|, the graph behaves like the graph of 2 . yx 

119. The horizontal asymptote is y = 1. 2 22 2 2 1223 23 xx xxxx xx

1 31. 3 xx The point of intersection is 1 ,1. 3

120. The horizontal asymptote is 1 . 2 y 

1. x  The point of intersection is 1 1,. 2 

121. Vertical asymptote at 3 x  the denominator is 3 x . Horizontal asymptote at 1 y  the ratio of the leading coefficients of the numerator and denominator = 1. The x–intercept = 2  the numerator is x 2. So the equation is of the form  22 or . 33 fxxx xx 

122. Vertical asymptotes at x = 1 and x = 1  the denominator is    11xx . Horizontal asymptote at 1 y  the degrees of the numerator and denominator are the same. The x–intercept = 0  the numerator is 2 x . So the equation is of the form

 22 2 . 11 1 fxxx xxx  

123. Vertical asymptote at 0 x  the denominator is x. The slant asymptote y = x  the degree of the numerator is one more than the degree of the denominator and the quotient of the numerator and the denominator is . The x-intercepts 1 and 1  the numerator is

11.xx So the equation is of the form

2 2 1 1 x xx x

However, the slant asymptote is y = x, so a = 1 and the equation is

124. Vertical asymptote at 3 x  the denominator is x 3. The slant asymptote y = x + 4  the degree of the numerator is one more than the degree of the denominator, and the quotient of the numerator and the denominator is x + 4. Thus, (x 3)(x + 4) + a is the numerator.

Therefore, the equation is

125. Vertical asymptote at 2 x  the denominator is x 2. Horizontal asymptote at 1 y  the degrees of the numerator and denominator are the same and the leading coefficients of the numerator and denominator are the same. So the numerator is xa  , where a is chosen so that 4 (0)2(). 2 ffxx x  

126. Vertical asymptotes at 2 and 1 xx the denominator is (2)(1) xx . Horizontal asymptote at 0 y  the degree of the numerator < the degree of the denominator. So the numerator is xa  , where a is chosen so that  4 (0)2. (2)(1)ffxx xx   Verify that the x-intercept is 4:

 44 40. 4241 f  

127. () as 1 and () as 1 the denominator is zero if 1. fxxfx xx    Since 1 x  from both directions, the denominator is 2 (1). x ()4 as fx 

x  the horizontal asymptote is 4. So the leading coefficient of the numerator is 4 and the degree of the numerator is the same as the degree of the denominator. The numerator is 2 4 xa  , where a is chosen so that   2 2 412 1 01. 2 121 a fa    

So, 2 2 41 (). (1) fxx x 

128. No vertical asymptotes  the denominator has no real zeros. ()fx symmetric about the yaxis ()fx  is an even function. So the degree of the numerator and the degree of the denominator can equal 2. Then a possible denominator is 2 1. x  ()2 as fx 

x  the horizontal asymptote is 2. So the numerator is 2 2 xa  where a is chosen so that (0)00. fa 2 2 2 (). 1 fxx x  

129. Vertical asymptote at 1 x  the denominator could be 1. x Because the oblique asymptote is 32,yx the numerator is (32)(1), xxa  where a can be any number (no x-intercepts or function values are given). Let a = 1. So. (32)(1)1 () 1fxxx x   2 31 1 xx x  

130. A rational function cannot have both a horizontal asymptote and an oblique asymptote because there would values of x which are mapped to two different function values. In that case, it wouldn’t be a function.

Critical Thinking/Discussion/Writing

131. 2 12 111 ,5 T R RRR  a. 11 111 55 1111 555 T TT RR R RRRRR 

b. 1 1 1 5 0 5 R R R   is the x-intercept.

1 1 5 5(0) 00 505 R Ry  is the y-intercept. The vertical asymptote is x = –5. The horizontal asymptote is y = 5.

 11 1 11 55 () 55 T RR RRfx RR   and   ()fxfx there are no symmetries. The intervals to be tested are (,5),(5,0) and (0,). The graph is above the horizontal asymptote on (,5)  and below the horizontal asymptote on (5,0)(0,).   The graph is below the x-axis on (–5, 0) and above the x-axis on (,5)(0,).  

c. 1R must be greater than 0, so 05. TR  The horizontal asymptote is 5, TR  so as

1 ,5. TRR 132. 2 ,2 1 Kcc   

a. 2 2 0 1 K     is the x-intercept.

2 20 00 10 K  is the y-intercept. The vertical asymptote is 1.   The degree of the numerator is greater than the degree of the denominator, so there is an oblique asymptote.

2 22 22 11    

The oblique asymptote is 22. K  

2 2 2 2 () 11 fx      and   ()fxfx there are no symmetries.

Use the intervals determined by the zeros of the numerator and of the denominator of   2 22 2222 11 K    

to create a sign graph. The numerator 2 has no zero and the denominator has one zero, 1. The intervals to be tested are (,1),and (1,). The graph is above the oblique asymptote on (,1)  and below the oblique asymptote on (1,). 

b. As 0    (meaning as  approaches 0 from the right), 0. K  In other words,  is a positive number that is becoming smaller and smaller. As it approaches 0, 2 2. K  

c. As 1   (meaning as  approaches 1 from the left), . K  In other words,  is a positive number > 0 that is becoming closer and closer to 1. As it approaches 1, the denominator approaches 0, so Ostwald’s law cannot hold for strong elctrolytes.

133. Answers may vary. Sample answers are given:

134. Answers may vary. Sample answers are given:

135. Since y = ax + b is the oblique asymptote, we know that 

. rx Rxaxb Dx  We are told that the graph of   Rx intersects the asymptote at the points

1122 ,,,,,, nncRccRccRc  so   Rxaxb  when

0. rx Dx  Thus,

Recall that the numerator of a rational function must have degree greater than that of the denominator in order for there to be an oblique asymptote, so the denominator,

, Dx can be any polynomial whose degree is greater than or equal to n + 2. Thus, a possible rational function is

where

Dx is a polynomial of degree n + 1 and none of its zeros are at 12,,,. n ccc

Getting Ready for the Next Section

The

For exercises 141–144, instead of graphing the function to determine the solution set as shown in Section 2.1, we use test points.

141. 

120xx

The zeros occur at x = –2 and x = 1, so the intervals to be tested are

 ,2,2,1, and   1,. 

Interval Test point

Value of   fx Result

 ,2 –3 4 +

2,1 0 –2 –

 1,  2 4 +

 120xx for

,21,.

142.     240xx

The zeros occur at x = –2 and x = –4, so the intervals to be tested are 

  ,4,4,2, and

 1,.

Interval Test point Value of

,4 –5 3 +

 4,2 –3 –1 –

2,  0 8 +

143.

2 3100520xxxx

The zeros occur at x = –5 and x = 2, so the intervals to be tested are   ,5,5,2, and

2,. 

Interval Test point

,5

Value of

fx Result

2.5 Polynomial and Rational Inequalities

2.5 Practice Problems 1. 322 32 25214215 49360 xxxxx xxx  

Now solve 3249360. xxx

The possible rational zeros are 1,2, 3,4,6,9,12,18,36. 

Using synthetic division, we see that x = 3 is a factor. 314936 32136 17120

32249363712 3340 xxxxxx xxx  

So, the zeros are x = –4, x = –3, and x= 3, and the intervals to be tested are   ,4,  4,3,3,3, and   3,. 

Interval Test point



144.

2 670710xxxx

The zeros occur at x = –1 and x = 7, so the intervals to be tested are   ,1,1,7, and

 7,. 

Interval Test point

 7,  10 33 + 2 670xx for 1,7.

Value of   Px Result

  ,4 –5 –16 –

 4,3 3.5 1.625 +  3,3 0 –36 –  3,  5 144 +

Solution set:   4,33,.  

 2 210xxx

The boundary points are x = 0, x = –2, and x = 1, so the intervals are     ,2,2,0,

  0,1, and   1,.  The leading term of  22 4 21, Pxxxxxxxx  so the end behavior of the graph is   fx  as x  and   fx  as . x  Because 0 and 1 are zeros of odd multiplicity, the graph of   Px crosses the x-axis at these points. Because –2 is a zeros of even multiplicity, the graph touches but does not cross the x-axis at x = –2.

(continuedonnextpage)

(continued)

To solve the inequality, examine the graph to find the intervals on which   Px is positive. The solution set is

,22,01,.

2 2220xxxx

Note that the quadratic term 2 22xx has no real zeros and is always positive. The real zeros of 

2 222xxxx are x = 0 and x = 2.

So, the intervals are   ,0,0,2, and

 2,  3 51 + [–6, 6] by [–6, 6]

We can verify the solution by examining the graph. Solution set:    ,02,.  4.

numerator: 21206 xx

The boundary points are 0, 4, and 6. The lines x = 0 and x = 4 are the vertical asymptotes of

. Rx The intervals are

,0,0,4,

4,6, and   6,.  Note that 0 and 4 are not included in the intervals to be tested because they cause the denominator of   Rx to equal 0.

Interval Test point

[–10, 10] by [–10, 10]

We can verify the solution by examining the graph. Solution set: 

,04,6.

5151 330 33 5533 24 0 33 xx xx xxx Rx xx       numerator: 2402 xx denominator : 303xx

The boundary points are 2 and 3. The line x = 3 is the vertical asymptote of   Rx The intervals are     ,3,3,2, and   2,. 

Note that –3 is not included in the intervals to be tested because it causes the denominator of   Rx to equal 0. To determine the sign of   , Rx form the corresponding polynomial

      243.Pxxx The signe rule says that   Px has the same sign as   Rx on the intervals listed above.   Px is a quadratic polynomial with leading coefficient 2 20. a 

(continuedonnextpage)

(continued)

From the graph, it is clear that   0 Rx  when 32. x  Thus, the solution set is   3,2.

 2 2 0 31 xx Rx xx   

numerator:  2 200,2xxxx

denominator:     3103,1xxxx 

The lines x = –3 and x = 1 are the vertical asymptotes of   Rx The intervals to be tested are

,3,3,2,2,0,0,1, and

 1,.  Note that –3 and 1 are not included in the intervals to be tested because they cause the denominator of   Rx to equal 0.

Interval Test point

 ,3

Value of   Rx Result

2.5 Exercises

Concepts and Vocabulary

1. For the polynomial inequalities   0 Px  or   0, Px  the zeros of the polynomial function   Px are always included in the solution set.

2. For the polynomial inequalities   0 Px  or   0, Px  the zeros of the polynomial function

  Px are always excluded from the solution set.

3. To solve a rational inequality, we begin by locating all zeros and vertical asymptotes (if they exist).

4. For the rational inequalities   0, Rx  

  0,0,RxRx or   0, Rx  the zeros of the denominator of the rational function   Rx (vertical asymptotes or holes) are always excluded from the solution set.

5. True

6. False. A polynomial function does not change sign at its zero if its graph touches, but does not cross the x-axis. A polynomial function changes sign at its zero if its graph crosses the x-axis.

7. False

8. True

Building Skills

9. From the graph,   0 Px  on   2,13,. 

10. From the graph   0 Px  on   2,03,.  

11. From the graph,   0 Px  on     3,22,.  

12. From the graph,   0 Px  on   2,. 

13. From the graph,   0 Px  on       ,22,23,.  

14. From the graph,   0 Px  on     2,11,3. 

[–10, 10] by [–10, 10]

We can verify the solution by examining the graph. Solution set:     3,22,01, 

or 

  3,01,. 

15. From the graph,   0 Px  on 3,2.

16. From the graph   0 Px  on 

 ,31,. 

17. 

 340xx

The zeros are x = 3 and x = 4. So, the intervals to be tested are 

,3,3,4, and

4,. 

Interval Test point

18.

250xx

The zeros are x = –2 and x = 5. So, the intervals to be tested are

,2,2,5,

Interval Test point

and

Value of

 Px Result

,2 –5 30 +

2,5 0 –10 –

5,  6 8 + Solution set: 

2,5

19.

  180xx

The zeros are x = –1 and x = 8. So, the intervals to be tested are   ,1,1,8, and

Interval Test point

8,. 

Value of   Px Result

 ,1 –2 10 +

1,8 0 –8 –

8,  10 22 +

Solution set: 

,18,  

20.

460xx

The zeros are x = –4 and x = –6. So, the intervals to be tested are

,6,6,4, and   4,.

Interval Test point

Value of   Px Result   ,6 –10 24 +

6,4 –5 –1 –  4,  0 24 +

Solution set:  6,4

21. 

    21530 xxx

The zeros are x = 1, x = –5, and x = 3. So, the intervals to be tested are 

  ,5,5,1,

 1,3, and   3,. 

Interval Test point

Value of   Px Result  ,5 –6 –126 –

5,1 0 30 +

1,3 2 –14 –

 3,  6 330 +

Solution set:    ,51,3 

22.

31240 xxx

The zeros are x = –1, x = 2, and x = 4. So, the intervals to be tested are     ,1,1,2,   2,4, and   4,. 

Interval Test point

Value of   Px Result

 ,1 –2 –72 –  1,2 0 24 +

  2,4 3 –12 –

  4,  5 54 +

Solution set:     1,24,  

23.       1470 xxx 

The zeros are x = 1, x = 4, and x = –7. So, the intervals to be tested are   ,7,7,1,

1,4, and   4,. 

Interval Test point Value of   Px Result

  ,7 –10 462 +

 7,1 0 –28 –

 1,4 2 18 +

  4,  5 –48 –

Solution set:   7,14,  

24.

2450xxx

The zeros are x = –2, x = 4, and x = –5. So, the intervals to be tested are   ,5,5,2,

2,4, and   4,. 

Interval Test point

Value of   Px Result  ,5 –6 40 +

 5,2 –4 –16 –

 2,4 0 40 +

  4,  5 –70 –

Solution set:   ,52,4 

25.  2 210xxx

The zeros are x = 0, x = –2, and x = 1. So, the intervals to be tested are     ,2,2,0,   0,1, and   1,. 

Interval Test point

Value of   Px Result

  ,2 –3 36 +   2,0 –1 –2 –

 0,1 0.5 –0.3125 –

  1,  2 16 +

Solution set:    2,00,1 

26.  2 2160xxx

The zeros are x = –2, x = 1, and x = 6. So, the intervals to be tested are     ,2,2,1,   1,6, and   6,. 

Interval Test point Value of   Px Result

 ,2 –3 144 +

2,1 0 –12 –

 1,6 2 –16 –  6,  7 324 +

Solution set:     ,26,  

27.  2 130xxx

The zeros are x = 0, x = 1, and x = 3. So, the intervals to be tested are   ,0,0,1, 1,3, and   3,. 

Interval Test point

Value of   Px Result

  ,0 –1 8 +

0,1

1,3 2 –4 –  3,  5 200 +

Solution set:

     ,00,13,,13,  

28.  2 4310xxx

The zeros are x = –4, x = 3, and x = 1. So, the intervals to be tested are   ,4,4,1,

1,3, and   3,. 

Interval Test point

Value of   Px Result

  ,4 –5 384 +

 4,1 0 –36 –

 1,3 2 6 +

  3,  5 144 + (continuedonnextpage)

(continued)

Note that  2 4310xxx when x = 3, so include that value in the solution set.

Solution set:  4,13 

29.    2 2110 xxx

The real zeros are x = 0 and x = 1. So, the intervals to be tested are     ,0,0,1, and

 1,. 

Interval Test point

Value of   Px Result

  ,0 –1 8 +

0,1 0.5 –0.625 –

 1,  2 20 +

Solution set:     ,01,  

30.    2 3230 xxx 

The real zeros are x = 0 and x = –2. So, the intervals to be tested are     ,2,2,0, and   0,. 

Interval Test point

Value of   Px Result

 ,2 –3 –108 –

2,0 –1 12 +

0,  1 –36 –

Solution set:   2,0

31.  22220 210 xxxx xx  

The zeros are x = 2 and x = –1. So, the intervals to be tested are     ,1,1,2, and   2,. 

Interval Test point

Value of   Px Result

 ,1 –3 10 +   1,2 0 –2 –

 2,  3 4 +



The zeros are x = –5 and x = 4. So, the intervals to be tested are     ,5,5,4, and   4,. 

Interval Test point Value of 

Px Result

,5 –6 –10 –

 5,4 0 20 +

4,  5 –10 –

Solution set:   5,4

33. 221033100 xxxx  2 310Pxxx cannot be factored, so there are no rational zeros. We must examine the graph. This is a parabola that opens upward and with vertex

331 2224,,. bb aa P 

[–10, 10] by [0, 20]

The graph is always greater than 0, so the solution set is   ,. 

Solution set:     ,12,   32.   22 2 20200 200540 xxxx xxxx 

34. 2242240 xxxx  2 34Pxxx cannot be factored, so there are no rational zeros. We must examine the graph. This is a parabola that opens upward and with vertex

   22,1,1. bb aa P 

[–10, 10] by [0, 20]

The graph is never less than or equal to 0, so the solution set is .

35. 32460xxx

The possible rational zeros are 1,2,3,6. 

Using synthetic division, we find that x = 1 is a zero.

11416 156 1560

So,

32246156 123 xxxxxx xxx

Interval Test point Value of   Px Result

  1,2 1.5 –1.375 –  2,  3 14 +

Solution set:     ,41,2 

37. 33121612160 xxxx

The possible rational zeros are 1,2,4,8,16. 

Using synthetic division, we find that x = –2 is a zero.

,3,3,2,

The zeros are x = –2, x = –3, and x = 1. So, the intervals to be tested are

2,1, and

Interval

3,2 –2.5 0.875 +

2,1 0 –6 –

 1,  2 20 +

Solution set: 

36. 32 1080xxx

The possible rational zeros are 1,2,4,8. 

Using synthetic division, we find that x = 1 is a zero.

111108 128 1280 So,

322108128 124 xxxxxx xxx

The zeros are x = –4, x = 1, and x = 2. So, the intervals to be tested are 

  ,4,4,1,

 1,2, and   2,. 

Interval Test point Value of   Px Result   ,4 –5 –42 –

 4,1 0 8 +

2101216 2416 1280 So,

321216228 224 xxxxx xxx  

The zeros are x = –2 and x = 4. So, the intervals to be tested are   ,2,2,4, and   4,. 

Interval

Test point

Value of   Px Result

  ,2 –3 –7 – 2,4 0 –16 –  4,  5 49 +

Note that 3 1216xx when x = –2, so include that value in the solution set.

Solution set:    24,  

38.  33 3 32320 320 xxxx xx  

The possible rational zeros are 1,2.

Using synthetic division, we find that x = –2 is a zero. 21032 242 1210

So,     32 2 32221 21 xxxxx xx  

The zeros are x = –2 and x = 1. So, the intervals to be tested are   ,2,2,1, and   1,. 

(continuedonnextpage)

(continued)

Interval Test point Value of   Px Result   ,2 –3 16 +  2,1 0 –2 –  1,  2 –4 –

Solution

39. 322 32 33523 5280 xxxxx xxx

The possible rational zeros are 1,2,4,8. 

Using synthetic division, we find that x = 2 is a zero. 21528 268 1340

So,

322528234 241 xxxxxx xxx

The zeros are x = –1, x = 2, and x = 4. So, the intervals to be tested are   ,1,1,2, 2,4, and   4,. 

Interval Test point Value of   Px Result

 ,1 –2 –24 –

 1,2 0 8 +

 2,4 3 –4 –

  4,  5 18 +

Solution set:   ,12,4

The possible rational zeros are 1,2,3,4,6,12.

Using synthetic division, we find that x = 1 is a zero. 1121112 1112 11120

32221112112 143 xxxxxx xxx

The zeros are x = –3, x = 1, and x = 4. So, the intervals to be tested are   ,3,3,1, 1,4, and   4,. 

Interval Test point Value of   Px Result

,3 –4 –40 – 3,1 0 12 +  1,4 3 –12 –

 4,  5 32 +

Solution set:   3,14,  

424222 2 33030 330 xxxxxx xxx  

The zeros are 3,0. 

So, the intervals to be tested are

Note that 42 3 xx  when x = 0, so include that value in the solution set.

Solution set:  

42.

The zeros are 2,0. 

So, the intervals to be tested are

,2,2,0, 

0,2,

and

43. From the graph, 

44. From the graph, 

45. From the graph, 

46. From the graph,   0 Rx  on     1,00,1. 

47. From the graph,   0 Rx  on    1,02,. 

48. From the graph,   0 Rx  on     ,01,2. 

49. From the graph,   0 Rx  on       ,11,01,.  

50. From the graph,   0 Rx  on       1,00,11,.  

For exercises 51–76, verify your answers by graphing the function.

51. 1 0 2 x x   numerator: 101xx denominator: 202xx

The line x = –2 is the vertical asymptote of   , Rx so –2 is not included in the intervals to be tested The boundary points are –2 and 1. The intervals are     ,2,2,1, and   1,. 

Interval Test point

Solution set:     ,21,  

52. 3 0 4 x x  

numerator: 303xx denominator: 404xx

The line x = 4 is the vertical asymptote of 

, Rx so 4 is not included in the intervals to be tested. The boundary points are –3 and 4. The intervals are     ,3,3,4, and   4,. 

Interval Test point

Value of   Rx Result

,3 –4 1 8 +

3,4 0 3 4 –

 4,  5 8 +

Solution set:   3,4

53. 2 0 8 x x  

numerator: 202xx

denominator : 808xx

The line x = 8 is the vertical asymptote of   , Rx so 8 is not included in the intervals to be tested. The boundary points are –2 and 8. The intervals are    ,2,2,8, and   8,. 

Interval Test point

Value of   Rx Result   ,2 –4 1 6 +   2,8 0 1 4 –

 8,  10 6 +

Solution set:     ,28,  

54. 5 0 1 x x  

numerator: 505xx

denominator : 101xx

The line x = 1 is the vertical asymptote of   , Rx so 1 is not included in the intervals to be tested. The boundary points are –5 and 1. The intervals are  

,5,5,1, and   1,. 

Interval Test point Value of   Rx Result

55. 32 0 43 x x  numerator: 2 3 320xx

The line 4 3 x  is the vertical asymptote of

56. 12 0 32 x x   numerator: 1 2 120xx  denominator : 3 2 320xx 

The line 3 2 x  is the vertical asymptote of   , Rx so 3 2 is not included in the intervals to be tested. The boundary points are 1 2 and 3 2

The intervals are

Solution set:  3 1 22 ,  

57.     12 0 1 xx x   

numerator:     1202,1xxxx  denominator : 101xx

The line x = –1 is the vertical asymptote of   , Rx so –1 is not included in the intervals to be tested. The boundary points are –2, –1, and 1. The intervals are

Interval Test point Value of

58.  3 0 2 xx x  

numerator:   300,3xxxx 

denominator : 202xx

The line x = –2 is the vertical asymptote of   , Rx so –2 is not included in the intervals to be tested. The boundary points are –2, 0, and 3. The intervals are 

 ,2,2,0,0,3, and 

3,. 

Interval Test point

Solution set:

59.

Value of

 Rx Result

60.  1 0 4 xx x   numerator:   100,1xxxx  denominator : 404xx The line x = –4 is the vertical asymptote of   , Rx so –4 is not included in the intervals to be tested. The boundary points are –4, 0, and 1. The intervals are     ,4,4,0,  0,1 and   1,. 

Interval Test point

 52 0 3 xx x    numerator:     5205,2xxxx  denominator : 303xx

The line x = –3 is the vertical asymptote of   , Rx so –3 is not included in the intervals to be tested. The boundary points are –5, –3, and 2. The intervals are

 3,2 and   2,. 

,5,5,3,

,5 –8 –6 –

Value of   Rx Result

  ,4 –5 –30 –  4,0 –2 3 +

 0,1 1 2 1 18 –

  1,  4 3 2 +

Solution set:   ,40,1 

2 11 0 43 xx xx   

numerator:  2 1101,1xxxx  denominator :     4304,3xxxx 

The lines x = –4 and x = 3 are the vertical asymptotes of   , Rx so –4 and 3 are not included in the intervals to be tested. The boundary points are –4, –1, 1, and 3. The intervals are     ,4,4,1, 1,1,   1,3, and   3,. 

Interval Test point

Value of   Rx Result

  ,4 –5 –12 –

 4,1 –2 3 10 +

 1,1 0 1 12 +

 1,3 2 3 2 –

  3,  5 8 +

Solution set:         4,11,13,4,13,  

62.



2 42 0 21 xx xx  

numerator:     4204,2xxxx 

denominator :  2 2101,2xxxx 

The lines x = 1 and x = 2 are the vertical asymptotes of   , Rx so 1 and 2 are not included in the intervals to be tested. The boundary points are –4, –2, 1, and 2. The intervals are   ,4,4,2,

2,1,

1,2, and

Interval Test point Value of   Rx Result

 1,2 3 2 154 +

Solution set: 

 4,21,22,  

63.  3 23 0 1 xx x  

numerator:  3 2302,3 xxxx 

denominator : 101xx

The line x = –1 is the vertical asymptote of   , Rx so –1 is not included in the intervals to be tested. The boundary points are –1, 2, and 3. The intervals are     ,1,1,2, 2,3, and   3,. 

Interval Test point Value of   Rx Result 

1,2 0 –24 –

2,3 5 2 1 56 +

Solution set:

64.

3 41 0 2 xx x 

numerator:  3 4101,4xxxx 

denominator : 202xx

The line x = 2 is the vertical asymptote of

, Rx so 2 is not included in the intervals to be tested. The boundary points are 1, 2, and 4. The intervals are    ,1,1,2,   2,4, and   4,. 

Interval Test point Value of   Rx Result

,1 0 2 +  

65.

Solution set:     1,24,  

5959 440 44 5944 7 00 44 xx xx xxx xx

numerator: 707xx

denominator : 404xx

The line x = 4 is the vertical asymptote of   , Rx so 4 is not included in the intervals to be tested. The boundary points are –7 and 4. The intervals are     ,7,7,4 and   4,. 

Interval Test point

Value of   Rx Result

  ,7 –8 1 12 +   7,4 0 7 4 –

  4,  5 12 +

Solution set:     ,74,  

66.

217217 330 55 21735 2 00 55 xx xx xxx xx

numerator: 202xx 

denominator : 505xx

The line x = 5 is the vertical asymptote of   , Rx so 5 is not included in the intervals to be tested. The boundary points are –2 and 5. The intervals are     ,2,2,5 and   5,. 

Interval Test point

Value of   Rx Result

 ,2 –3 1 8 –

2,5 0 2 5 +

Solution set:     ,25,   67.

88 220 33 823 2 00 33 xx xx xxx xx

numerator: 202xx  denominator : 303xx

The line x = –3 is the vertical asymptote of

 , Rx so –3 is not included in the intervals to be tested. The boundary points are –3 and 2. The intervals are     ,3,3,2, and

 2,. 

Interval Test point Value of   Rx Result

,3 –4 –6 –





numerator: 0 x 

denominator : 303xx

The line x = 3 is the vertical asymptote of   , Rx so 3 is not included in the intervals to be tested. The boundary points are 0 and 3. The intervals are   ,0,0,3 and   3,. 

,0 –3 1 2 +

0,3 1 1 2 –

3,  4 4 +

Solution set:  0,3 69.

2121 0 3232 223 7 00 3232 xxxx xxx xxxx

numerator: 707xx denominator :     3202,3xxxx 

The lines x = –2 and x = 3 are the vertical asymptotes of   , Rx so –2 and 3 are not included in the intervals to be tested. The boundary points are –7, –2, and 3. The intervals are 

 ,7,7,2,

Interval

Solution

 2,3, and   3,. 

numerator: 909xx 

denominator :

The lines x = –2 and x = 5 are the vertical asymptotes of   , Rx so –2 and 5 are not included in the intervals to be tested. The boundary points are –9, –2, and 5. The intervals are

,9,9,2,

2,5, and

 ,9 –10 1 120 +

The lines x = –2 and x = 5 are the vertical asymptotes of   , Rx so –2 and 5 are not included in the intervals to be tested. The boundary points are –2, 1 4 , and 5. The intervals are

numerator: 1 3 310xx denominator :

 133,1xxxx 

The lines x = –3 and x = –1 are the vertical asymptotes of   , Rx so –3 and –1 are not included in the intervals to be tested. The boundary points are –3, –1, and 1 3 The intervals are 

1 3 ,.

  ,3,3,1,  1 3 1,,   and

Solution set: 

73. 2 8868 6600 xx xx xxx 

numerator:

denominator : 0 x 

The line x = 0 (the y-axis) is the vertical asymptote of   , Rx so 0 is not included in the intervals to be tested. The boundary points are 0, 2, and 4. The intervals are

,0,0,2,

and

Interval Test point Value of

Rx Result

75. 2 3 0 4 x x  numerator: 300 xx denominator : 2 402xx

The lines x = –2 and x = 2 are the vertical asymptotes of   , Rx so –2 and 2 are not included in the intervals to be tested. The boundary points are –2, 0, and 2. The intervals are     ,2,2,0,   0,2, and   2,. 

Interval Test point Value of   Rx Result   ,2 –4 –1 –

 2,0 –1 1 +

 0,2 1 –1 –

2,  4 1 +

Solution set:     2,02, 

Solution set: 

,02,4

74. 2 121212 1100 xx xx xxx 

numerator:

denominator : 0 x 

The line x = 0 (the y-axis) is the vertical asymptote of   , Rx so 0 is not included in the intervals to be tested. The boundary points are –3, 0, and 4. The intervals are   ,3,  

 3,0,0,4, and   4,. 

Interval Test point Value of   Rx Result

 ,3 –4 –2 –

  3,0 –1 10 +

  0,4 1 –12 –  4,  6 3 +

Solution set:     3,04,  

76. 2 2 1 0 4 x x  

numerator: 2 10 x  has no real solutions. denominator : 2 402xx

The lines x = –2 and x = 2 are the vertical asymptotes of   , Rx so –2 and 2 are not included in the intervals to be tested. The boundary points are –2 and 2. The intervals are

   ,2,2,2, and   2,. 

Interval Test point Value of   Rx Result

 ,2 –3 2 +

2,2 0 1 4 –

 2,  3 2 +

Solution set:   2,2

Applying the Concepts

77. 2 0.05 dvv   2 0.05252556.25 d 

In a 25-mile-per-hour zone, the stopping distance is about 56.25 feet, so the driver was going over the speed limit because it took over 75 feet to stop.

78. 2 2 163295110 1632150 tt tt

2 323241615 216 3264328 3232 4024 1.25 or 0.75 3232

The height of the stone will be at least 110 feet between 0.75 second and 1.25 seconds.

79. 3218111020QQQ 

The possible rational zeros are 1,2,3,6,17,34,51,102. 

Using synthetic division, we find that 3 is a factor. 311811102 345102 115340

The zeros are Q = –2, Q = 3, and Q = 17. Because we are looking for an amount, it is not necessary to test any negative values. So, the intervals to be tested are   0,3,   3,17, and   17,. 

Interval Test point Value of   Px Result



The company will make a profit if they sell between 3000 and 17,000 units of lip gloss. 80.

The length of the box is 9 –2x, the width is 6 – 2x, and the height is x. Then, the volume is given by     9262.Vxxx 

32 32 926220 4305420 43054200 xxx xxx xxx

The possible rational zeros are 1,2,4,5,10,20. 

Using synthetic division, we find that 2 is a factor.

24305420 84420 422100

32 2 2 4305420 242210 222115 22215 xxx xxx xxx xxx

The zeros are 1 2 ,2, and 5. Because we are looking for a length, it is not necessary to test any negative values. The maximum value of x cannot be 3 or greater because that would eliminate the entire width. So, the intervals to be tested are  11 22 0,,,2,   and   2,3.

If 0.52, x  the volume of the resulting open-top box will be at least 20.

81. We want to find the value of Q that makes . ABTCTC  322 32 6141052034 1134240 QQQQQ QQQ

The possible rational zeros are 1,2,3,4,6,8,12,24. 

Using synthetic division, we find that 1 is a factor. 11113424 11024 110240 (continuedonnextpage)

(continued)

32 2 113424 11024 146

The zeros are Q = 1, Q = 4, and Q = 6. Because we are looking for an amount, it is not necessary to test any negative values. So, the intervals to be tested are

0,1,1,4,4,6, and   6,. 

Note that 400 3 133.3. 

The likelihood that the next card dealt will be a jack, queen, king, or ace will be greater than 50% if between 134 and 208 cards are dealt.

83. The speed of the ferry against the current is v – 2, and the speed of the ferry with the current is v + 2. The time going against the current is 40 2 , v and the time going with the current is 40 2 v  So, we have

It is more cost effective to produce headphones at company A when between 0 and 1000 headphones are produced and when between 4000 and 6000 headphones are produced.

640.2640.2 0.50.50 208208 640.20.5208 0 208 640.21040.5400.3 00 208208 xx xx xx x xxx xx 

numerator: 400 3 400.30xx 

denominator : 2080208 xx 

The line x = 208 is the vertical asymptote of   , Rx so 208 is not included in the intervals to be tested. The boundary points are 400 3 and 208.

Because we are looking for an amount, it is not necessary to test any negative values. There are 208 cards in four decks, so it is not necessary to test values greater than 208. The intervals are

  400 3 0, and   400 3 ,208. Interval Test

numerator:

denominator :     222,2vvvv 

The lines v = –2 and v = 2 are the vertical asymptotes of   , Rv so –2 and 2 are not included in the intervals to be tested. Because we are looking for a speed, it is not necessary to test values less than or equal to 0. The boundary points are 0, 2, and 18. The intervals are   0,2,   2,18, and   18,. 

Interval Test point

Value of   Rv Result

  0,2 1 –31.17 –

 2,18 5 14.55 +

  18,  20 –0.46 –

The ferry should travel at least 18 miles per hour to limit the round trip to at most 4.5 hours.

 4 3 12340,12tttt 

denominator : 2 160 t  has no real zeros. Because we are looking for an amount, it is not necessary to test any negative values. The boundary points are 4 3 and 12. The intervals to be tested are

44 33 0,,,12,

and

12,. 

Interval Test point Value of

4 3 0,

Rx Result

2 1 +

12,  13 –0.19 –

The drug concentration is least 3 mg/dl for 4 3 hr12 hr, t  or between 1 hr 20 min and 12 hr, inclusive.

85. We are looking for the values of I that make . ABPP 

 numerator:  2 4400,4IIIIII  denominator :     5205,2 IIII 

The lines I = –5 and I = –2 are the vertical asymptotes of   , RI so –5 and –2 are not included in the intervals to be tested. Because we are looking for an amount, it is not necessary to test any negative values. The boundary points are 0 and 4. The intervals to be tested are   0,4 and   4,. 

Interval

Test point

Value of   Rx

Result

  0,4 1 1 6 –

  4,  5 1 14 +

ABPP  for 04. I  86.

00 11 0.4252 0 1 0.4212 0 1 xx

 1 2 0.42120,2 xxxx 

denominator : 2 1 x  has no real zeros

Because we are looking for an amount, it is not necessary to test any negative values. The boundary points are 1 2 and 2. The intervals to be tested are  11 22 0,,,2,   and   2,. 

Interval Test point Value of 

The efficiency of this rocket is at least 80% for 1 2 2. x 

We want to know when   1.4. fx  Graph   fx and y = 1.4 in the window [0, 20] by [–1, 7], and then find the intersection.

Vinyl sales exceeded 1.4 million between 13.5 years after 1993, or during 2007, and 2013.

88.  25.2691.24050.0640.00103623 fxxxx 

We want to know when   17.65. fx  Graph   fx and y = 17.65. Use the window [0, 40] by [17, 19] to be able to see the intersections clearly.

91.

The intervals to be tested are

 ,5,5,3, and   3,. 

Interval Test point

The intersections occur at 13.3, x  which is during 1988, 18.3, x  which is during 1993, and at 30.2, x  which is during 2005. The price of a music album was less than $17.65 between the years 1988 and 1993, and between 2005 and 2014.

Beyond the Basics

89.  2 412fxxx

The domain consists of those values of x for which 2 4120.xx

2 4120620 6,2 xxxx xx  

The intervals to be tested are   ,2,2,6, and   6,. 

Interval Test point Value of   Px Result

–3 9 +

2,6 0 –12 –

6,  7 9 +

The domain is    ,26,.

90.  2 152 fxxx 

The domain consists of those values of x for which 2 1520. xx 

2 1520530 5,3 xxxx xx 



Value of   Px Result

,5 –6 –9 – 5,3 0 15 +

 3,  4 –9 –

The domain is 5,3.

 3 4 fxxx 

The domain consists of those values of x for which 3 40. xx  3 40220 0,2,2 xxxxx xxx 

The intervals to be tested are

 ,2,2,0,0,2, and   2,. 

Interval Test point Value of   Px Result   ,2 –3 15 +  2,0 –1 –3 –

 0,2 1 3 +

  2,  3 –15 –

The domain is   ,20,2. 

92.  4 8 fxxx 

The domain consists of those values of x for which 4 80. xx     43 2 8080 22400,2 xxxx xxxxxx

 

The intervals to be tested are

  ,2,2,0, and   0,. 

Interval Test point

Value of   Px Result

  ,2 –3 57 +

 2,0 –1 –7 –

  0,  1 9 +

The domain is    ,20,. 

93.  2 5 fxx x  

The domain consists of those values of x for which 2 0. 5 x x  

numerator: 202 xx 

denominator : 505xx

The line x = –5 is the vertical asymptote of   , Rx so –5 is not included in the intervals to be tested. The boundary points are –5 and 2. The intervals to be tested are     ,5,5,2, and   2,. 

Interval Test point

Value of   Px Result

 ,5 –6 –8 –

 5,2 0 2 5 +

 2,  3 1 8 –

The domain is   5,2.

94.  3 2 fxx x  

The domain consists of those values of x for which 3 0. 2 x x   numerator: 303xx denominator : 202xx

The line x = 2 is the vertical asymptote of   , Rx so 2 is not included in the intervals to be tested. The boundary points are –3 and 2. The intervals to be tested are

  ,3,3,2, and   2,. 

Interval Test point

Value of   Px Result

 ,3 –6 3 8 +

The domain is 

,32,.

95.  12 1 33fxx xx 

The domain consists of those values of x for which 12 10. 33 x xx 

numerator: 202xx denominator : 303xx

The line x = 3 is the vertical asymptote of   , Rx so 3 is not included in the intervals to be tested. The boundary points are 2 and 3. The intervals to be tested are    ,2,2,3, and   3,. 

Interval Test point

Value of   Px Result

  ,2 0 2 3 +

 3,  4 2 +

The domain is     ,23,. 

25311 311 2 55 1 5 fxxxx xx x x

The domain consists of those values of x for which 3111 20. 55 xx xx    numerator: 101xx  denominator : 505xx

The line x = –5 is the vertical asymptote of   , Rx so –5 is not included in the intervals to be tested. The boundary points are –5 and –1. The intervals to be tested are

    ,5,5,1, and   1,.

Interval Test point

Value of   Px Result

  ,5 –6 –5 –

 5,1 –3 1 +   1,  0 1 5 –

The domain is   5,1.

97.  2 22 31 1 44 fxx xx 

The domain consists of those values of x for which 2 22 31 10. 44 x xx 

numerator: 2 101xx

denominator : 2 402xx

The lines x = –2 and x = 2 are the vertical asymptotes of   , Rx so –2 and 2 are not included in the intervals to be tested. The boundary points are –2, –1, 1 and 2. The intervals to be tested are     ,2,2,1,   1,1,1,2, and   2,. 

Interval Test point Value of

Px Result

,2 –4 5 4 +

2,1 –1.5

The domain is    

,21,12,.  

98.  2 15252 22 fxxxx xx  

The domain consists of those values of x for which 2 15252 0. 22 xx x xx   numerator:  2 1 2 25202120 ,2 xxxx xx   denominator : 200 xx The line x = 0 (the y-axis) is the vertical asymptotes of   , Rx so 0 is not included in the intervals to be tested. The boundary points are 0, 1 2 , and 2. The intervals to be tested are   ,0,  11 22

The domain is

3221112,Pxxxx range [–10, 18]

We must solve 

10 Px  and   18. Px 

First inequality: 32 32 2111210 211220 xxx xxx  

Possible rational zeros: 1,2,11,22 

Using synthetic division, we find that 2 is a factor.

2121122 2022 10110

  32221122211 21111 xxxxx xxx  

The zeros are x = 2, 11, x  and 11. x 

The intervals to be tested are  ,11,    11,2,  

2,11,

and

11,.



Note that 113.32. 

Interval Test point Value of   Px Result

,11    –4 –30 –11,2

–2 28 + 2,11

4 10 +

The solution set for the first inequality is

(continuedonnextpage)

(continued)

Second inequality: 32 32 2111218 21160 xxx xxx

Possible rational zeros: 1,2,3,6  Using synthetic division, we find that –2 is a factor. 212116 286 1430

Use the quadratic formula to find the zeros of 2 43.xx

Verify graphically by graphing   1 , yPx  2 10, y  and 3 18, y  and then finding the intersections.

[–10, 10] by [–20, 20] 100.  323420,Pxxxx range [8, 14]

We must solve   8 Px  and   14. Px 

The zeros are x = –2, 27, x  and 27. x  The intervals to be tested are

,2, 2,27,   27,27,    and  27,.   

Interval Test point Value of   Px Result

  ,2 –3 –18 –

2,27   –1 2 + 27,27,    0 –6 –

27,    5 14 +

The solution set for the second inequality is

 ,227,27,   

 ,20.65,4.65. 

Now we must see where the two solution sets intersect. To make the work easier, use the decimal equivalents.

3.32,23.32, ,20.65,4.65 3.32,20.65,23.32,4.65 11,227,211,27

The number line below shows the overlap.

First inequality: 32 32 34208 34120 xxx xxx  

Possible rational zeros: 1,2,3,4,6,12  Using synthetic division, we find that 2 is a factor. 213412 2212 1160



 322341226 232 xxxxxx xxx  

The zeros are x = –2, x = 2 and x = 3 The intervals to be tested are   ,2,

2,2, 2,3, and   3,. 

Interval Test point Value of   Px Result

  ,2 –3 –30 –

 2,2 0 12 +

 2,3 2.5 –1.125 –  3,  4 12 +

The solution set for the first inequality is   2,23,.  

Second inequality: 32 32 342014 3460 xxx xxx  

Possible rational zeros: 1,2,3,6  (continuedonnextpage)

(continued)

Using synthetic division, we find that 1 is a factor.

Use the quadratic formula to find the zeros of 2 26.

The zeros are 17,1,xx and 17. x

The intervals to be tested are

Interval

Test point

Value of

Px Result  ,17    –2 –6 –17,1   0 6 + 1,17

The solution set for the second inequality is

 ,1.651,3.65.

Now we must see where the two solution sets intersect. To make the work easier, use the decimal equivalents.

2,171,23,17 

2,23,,1.651,3.65

2,1.651,23,3.65

The number line below shows the intersection.

Verify graphically by graphing   1 , yPx  2 10, y  and 3 18, y  and then finding the intersections.

[–10, 10] by [–2, 22]

101.  2 , 3 x Rx x   range [2, 6]

We must solve   2 Rx  and   6. Rx 

First inequality:  22 220 33 223 8 00 33 xx xx xxx xx

 

numerator: 808xx  denominator : 303xx

The line x = 3 is the vertical asymptote of   , Rx so 3 is not included in the intervals to be tested. The boundary points are 3 and 8. The intervals to be tested are     ,3,3,8, and   8,.  Interval

The solution set for the first inequality is   3,8.

Second inequality:

22 660 33 263 520 00 33 xx xx xxx xx  

 

numerator: 52004 xx  denominator : 303xx

The line x = 3 is the vertical asymptote of   , Rx so 3 is not included in the intervals to be tested. The boundary points are 3 and 4. The intervals to be tested are     ,3,3,4, and   4,. 

(continuedonnextpage)

(continued)

Interval Test point

Value of   Px Result

 ,3 2 –10 –

  3,4 3.5 5 +

  4,  8 –4 –

The solution set for the second inequality is

    ,34,.  Now we must see where the two solution sets intersect.

  3,8,34,4,8 

The number line below shows the intersection.

Verify graphically by graphing   1 , yRx 

2 10, y  and 3 18, y  and then finding the intersections.

[–4, 16] by [–10, 10]

102.  3 , 7 x Rx x  range [1, 3]

We must solve   1 Rx  and   3. Rx 

First inequality:

numerator: 21005 xx  denominator : 707xx

The line x = 7 is the vertical asymptote of   , Rx so 7 is not included in the intervals to be tested. The boundary points are 5 and 7. The intervals to be tested are

,5,5,7,

and   7,. 

Interval Test point

Value of   Px Result

 7,  8

The solution set for the first inequality is   5,7. Second inequality:

numerator: 42406 xx  denominator : 707xx

The line x = 7 is the vertical asymptote of   , Rx so 7 is not included in the intervals to be tested. The boundary points are 6 and 7. The intervals to be tested are    ,6,6,7, and

The solution set for the second inequality is     ,67,.  Now we must see where the two solution sets intersect.        5,7,67,5,6 

The number line below shows the intersection.

Verify graphically by graphing   1 , yRx  2 1, y  and 3 3, y  and then finding the intersections.

[–4, 16] by [–10, 10]

103.  2 5 , 1 x Rx x  range [0, 1]

We must solve   0 Rx  and   1. Rx 

First inequality: 2 5 0 1 x x 

numerator: 505 xx 

denominator : 2 101xx

The lines x = –1 and x = 1 are the vertical asymptotes of   , Rx so –1 and 1 are not included in the intervals to be tested. The boundary points are –1, 1, and 5. The intervals to be tested are

,1,1,1,1,5, and

5,. 

Interval Test point Value

Px Result

The solution set for the first inequality is

Second inequality:

numerator:

denominator : 2 101xx

The lines x = –1 and x = 1 are the vertical asymptotes of   , Rx so –1 and 1 are not included in the intervals to be tested.

The boundary points are –3, –1, 1, and 2. The intervals to be tested are    ,3,3,1,

 1,1,1,2, and   2,. 

The solution set for the second inequality is

,31,12,.   Now we must see where the two solution sets intersect.

,11,5 ,31,12, ,32,5

The number line below shows the intersection.

Verify graphically by graphing   1 , yRx 

2 1, y  and 3 3, y  and then finding the intersections.

[–10, 10] by [–10, 10]

104.  2 21 , 4 x Rx x  range [0, 1]

We must solve   0 Rx  and   1. Rx 

First inequality:

2 21 0 4 x x 

numerator: 1 2 210xx

denominator : 2 402xx

The lines x = –2 and x = 2 are the vertical asymptotes of   , Rx so –2 and 2 are not included in the intervals to be tested.

(continuedonnextpage)

(continued)

The boundary points are –2, 1 2 , and 2. The intervals to be tested are

Interval Test point

The solution set for the first inequality is

denominator : 2 402xx

The lines x = –2 and x = 2 are the vertical asymptotes of   , Rx so –2 and 2 are not included in the intervals to be tested. The boundary points are –2, –1, 2, and 3. The intervals to be tested are     ,2,2,1,

  1,2,2,3, and   3,. 

Interval Test point Value of   Px Result

The solution set for the second inequality is       ,21,23,.   Now we must see where the two solution sets intersect.  



1 2 1 2 2,2, ,21,23, 1,3,

The number line below shows the intersection.

Verify graphically by graphing   1 , yRx  2 1, y  and 3 3, y  and then finding the intersections.

[–6, 6] by [–6, 6]

105. 2 1121 2200 xx xx xxx  

numerator:  2 2 210101xxxx 

denominator : 0 x 

The boundary points are 0 and 1, so the intervals to be tested are   ,0,   0,1, and   1,. 

Interval Test point

Value of   Px Result   ,0  –1 –4 – 0,1 0.5 1 2 +   1,  2 1 2 +

Solution set:   0, 

106.

4444 432 22220 282432140 xxxx xxxx

From the hint, we know that  2 1 x is a factor. After using long division or synthetic division twice, we have

there are no real zeros. The intervals to be tested are 

,1 and

The solution set is

Therefore, the inequality is true for all real numbers. 107.

22 114114 0 4 0 44 0 xaxaxaxa aaxaxxax axax aaxx axax





denominator :   00, axaxxxa 

The boundary points are 0, 2 , a and a. The intervals to be tested are   2 ,0,0,, a

and

,.a

Substitute the test values for x in

22 44 aaxx axax 

Interval Test point

The solution set is

Therefore, x is a positive number less than a

332323 2 23320 20 xaxaxxaa axax 

The zeros of   Px are x = a and x = –2a

Because x is positive, it is not necessary to test negative values. Substitute the test values for x in  2 2.axax 

Interval Test point Value of   Px Result

0, a 2 a 3 5 8 a +

 , a  a + 1 3a + 1 +

The solution set is     0,,0,. aa  

Therefore, x is a positive number.

Critical Thinking/Discussion/Writing

For exercises 109–114, answers will vary. Verify your answer by graphing.

109.    4,5 450xx

110.     ,24, 240xx   

111.      1,23, 1230xxx   

112.    2 3,14 3140xxx 

113.      1,04,5 1450xxxx 

114.

,20,14, 2140xxxx

For exercises 115–120, answers will vary. Verify your answer by graphing.

115.

2,6 2 0 6 x x

119.

1,23, 3 0 12 x xx

121. If

0, Px  then

Px could be

2 213.xxx

Solving  2 2130xxx leads to the solution set

,21,33,,21,.  

Thus,  2 213Pxxxx  satisfies both conditions.

122. Use the graphing transformation rules to find the solution sets.

a.     ,64,2 Px 

b.     21,23,Px  

c.     1 2 4,812, Px  

123.     10 xax

For 1, a  the solution set is

   ,1,. a   For a = 1, the solution set is

  ,.  For a > 1, the solution set is

   ,1,. a  

124. 0 1 xa x 

For 1, a  the solution set is

    ,1,. a 

 For a = 1, the solution set is

    ,11,.  For a > 1, the solution set is 

  ,1,. a 

125.       110xaxx

For 1, a  the solution set is   ,11,. a   For a = –1, the solution set is    11,.  For –1 < a < 1, the solution set is   1,1,. a   For a = 1, the solution set is

  1,. For a > 1, the solution set is   1,1,. a  

126.  0 11 xa xx  

For a < –1, the solution set is     ,11,. a  

For a = –1, the solution set is   1,. 

For –1 < a < 1, the solution set is

 1,1,. a   For a = 1, the solution set is

 1,11,.  For a > 1, the solution set is

1,1,. a  

127. P has to be a polynomial of odd degree, and its degree has to be less than or equal to 21. k  An example is       1,12,46,   is the solution set of

 112460.Pxxxxxx

The solution set has k = 2 open intervals of the form   , ab and one interval of the form

  ,.c    Px has degree 5 = 2k + 1.

128. P has to be a polynomial of even degree, and its degree has to be greater than or equal to 2. k An example is       3,21,12,4  is the solution set of

3211240. Px xxxxxx  

The solution set has 3 k  open intervals of the form   ,.ab   Px has degree 6 = 2k.

Getting Ready for the Next Section

129.

130.

 222211xixixixx 

222 2 52522542541 254 xixixix x  

2.6 Zeros of a Polynomial Function 2.6 Practice Problems

1. 52 ()2351 fxxxx

There is one sign change in ()fx , so there is one positive zero. 52 52 ()2()3()5()1 2351 fxxxx xxx 



There are two sign changes in ()fx , so there are either 2 or 0 negative zeros.

2. 32 ()44 fxxxx

The possible rational zeros are  1,2,4.

There are two sign changes in ()fx , so there are either 2 or 0 positive zeros.

32 32 ()()()4()4 44 fxxxx xxx 



There is one sign change in ()fx , so there is 1 negative zero. Testing each positive possibility shows that 2 is an upper bound. Testing the negative values shows that –2 is a lower bound. 21144 224 1120 21144 264 1320

3. a.  ()3(2)(1)(1)(1) 3(2)(1)(1)(1) Pxxxxixi xxxixi  

b.     2 32 432 432 ()3(2)(1)(1)(1) 3(2)(1)22 3(2)342 3264 3361812 Pxxxxixi xxxx xxxx xxxx xxxx  





4. Since 2 3i is a zero of multiplicity 2, so is 2 + 3i. Since i is a zero, so is i. The eight zeros are 3, 3, 2 3i, 2 3i, 2 + 3i, 2 + 3i, i, and i.

5. The function has degree four, so there are four zeros. Since one zero is 2i, another zero is 2i

So 2 (2)(2)4 xixix  is a factor of ()Px .

Now divide to find the other factor:

So, 432 2 ()36128 (2)(2)(32) (2)(2)(2)(1) Pxxxxx xixixx xixixx

The zeros of () are 1,2,2, and 2.Pxii

6. 432 ()8222816  fxxxxx

The function has degree 4, so there are four zeros. There are three sign changes, so there are either 3 or 1 positive zeros.

Now find the zeros of 2 22xx using the quadratic formula: 2 2(2)4(1)(2) 24 1 2(1)2 

The zeros are 2, 4, 1 + i, and 1 i.

2.6 Exercises

Concepts and Vocabulary

1. If the nonzero terms of a polynomial are written in descending order, then a variation of sign occurs when the sign of two consecutive terms differ.

2. The number of positive zeros of a polynomial function P(x), is either equal to the number changes of sign of P(x) or less than that number by an even integer.

3. The Fundamental Theorem of Algebra states that a polynomial function of degree 1 n  has at least one complex zero.

4. If P is a polynomial function with real coefficients and if z = a + bi is a zero of P, then zabi  is also a zero of P(x).

5. True.

6. False. If 2 + 3i is a zero of P(x) with real coefficients, then so is 2 3i

7. True

432 43 ()()8()22()28()16 8222816 fxxxxx xxxx

 

There are no sign changes in ()fx , so there are no negative zeros. The possible rational zeros are  1,2,4,8,16.  Using synthetic division to test the positive values, we find that one zero is 2: 218222816 2122016 161080

The zeros of the depressed function 326108 xxx are also zeros of P

The possible rational zeros of the depressed function are  1,2,4,8. Examine only the positive possibilities and find that 4 is a zero: 416108 488 1220

So, 4328222816 xxxx  2 (2)(4)(22) xxxx  .

8. False. 164i 

Building Skills

9. 32 32 32 ()5234; ()5()2()3()4 5234 fxxxx fxxxx xxx







There are two sign changes in ()fx , so there are either 2 or 0 positive zeros. There is one sign change in ()fx , so there is one negative zero.

 

10. 32 32 32 ()393; ()3()()9()3 393 gxxxx gxxxx xxx



There is one sign change in ()gx , so there is one positive zero. There are two sign changes in ()gx , so there are either 2 or 0 negative zeros.



11. 32 32 32 ()252; ()2()5()()2 252 fxxxx fxxxx xxx

There are two sign changes in ()fx , so there are either 2 or 0 positive zeros. There is one sign change in ()fx , so there is one negative zero.

12. 432 432 432 ()38523; ()3()8()5()2()3 38523 gxxxxx gxxxxx xxxx 

There are three sign changes in ()gx , so there are either 3 or 1 positive zeros. There is one sign change in ()gx , so there is 1 negative zero.

13. 532 532 532 ()25321 ()2()5()3()2()1 25322 hxxxxx hxxxxx xxxx

There are three sign changes in ()hx , so there are either 1 or 3 positive zeros. There are two sign changes in ()hx , so there are either 2 or 0 negative zeros.

14. 643 643 643 ()5721; ()5()7()2()1 5721 Fxxxx Fxxxx xxx

There are three sign changes in ()Fx , so there are either 3 or 1 positive zeros. There is one sign change in ()Fx , so there is 1 negative zero.



15. 432 432 432 ()34537; ()3()4()5() 3()7 34537 Gxxxxx Gxxxx x xxxx

There are three sign changes in ()Gx , so there are either 3 or 1 positive zeros. There is one sign change in ()Gx , so there is 1 negative zero.

16. 532 532 532 ()53274; ()5()3()2() 7()4 53274 Hxxxxx Hxxxx x xxxx

There are three sign changes in ()Hx , so there are either 3 or 1 positive zeros. There are two sign changes in ()Hx , so there are 0 or 2 negative zeros.

17. 

42 42 42 24 2424 fxxx fxxxxx 



There are no sign changes in f(x), nor are there sign changes in f( x). Therefore, there are no positive zeros and no negative zeros.

18. 

42 42 42 356 356356 fxxx fxxxxx  

There are no sign changes in f(x), nor are there sign changes in f( x). Therefore, there are no positive zeros and no negative zeros.

19.

53 53 53 23 23 23 gxxxx gxxxx xxx

There are no sign changes in g(x), nor are there sign changes in g( x). Therefore, there are no positive zeros and no negative zeros.

20.

53 53 53 245 245 245 gxxxx gxxxx xxx

There are no sign changes in g(x), nor are there sign changes in g( x). Therefore, there are no positive zeros and no negative zeros.

21.

53 53 53 24 24 24 hxxx hxxx xx



There is one sign change in h(x), and no sign changes in g( x). Therefore, there is one positive zero and there are no negative zeros.

22.

53 53 53 231 231 231 hxxx hxxx xx

There are no sign changes in h(x), and one sign change in g( x). Therefore, there are no positive zeros and there is one negative zero.

23. The possible rational zeros are 1 ,1,3. 3    

There are three sign changes in ()fx , so there are either 3 or 1 positive zeros. There are no sign changes in ()fx , so there are no negative zeros. Testing each positive possibility shows that 13 is an upper bound. Testing the negative values shows that 13 is a lower bound.

24. The possible rational zeros are 13721 ,1,,3,,7,21.

There are two sign changes in ()gx , so there are either 2 or 0 positive zeros. There is one sign change in ()gx , so there is one negative zero. Testing each positive possibility shows that 72 is an upper bound. Testing the negative values shows that –3 is a lower bound.

25. The possible rational zeros are 17

There are no sign changes in ()Fx , so there are no positive zeros. There are three sign changes in ()Fx , so there are either 3 or 1 negative zeros. Because there are no positive zeros, the smallest positive possible zeros, 13 , is an upper bound. Verify this with using synthetic division. Testing the negative values shows that 73 is a lower bound.

26. The possible rational zeros are  1,2,4.

There is one sign change in ()Gx , so there is one positive zero. There are two sign changes in ()Gx , so there are either 2 or 0 negative zeros. Testing each positive possibility shows that 1 is an upper bound. Testing the negative values shows that –4 is a lower bound.

27. The possible rational zeros are  1,31.

There are two sign changes in ()hx , so there are either 2 or 0 positive zeros. There are two sign changes in ()hx , so there are either 2 or 0 negative zeros. Testing each positive possibility shows that 31 is an upper bound. Testing the negative values shows that –31 is a lower bound.

311315931

31105432,209998,200 134103932,200998,231

311315931

3186826,443820,012 12885326,452820,043

28. The possible rational zeros are 113 33,1,,13.

There are three sign changes in ()Hx , so there are either 3 or 1 positive zeros. There is one sign change in ()Hx , so there is one negative zero. Testing each positive possibility shows that 13 is an upper bound. Testing the negative values shows that –1 is a lower bound.

13320281913

39247357546,722 319275359446,709 1320281913 3235132 323513219

29. The possible rational zeros are 111777 ,,,1,,,,7. 632632     There are two sign changes in ()fx , so there are either 2 or 0 positive zeros. There are two sign changes in ()fx , so there are either 2 or 0 negative zeros. Testing each positive possibility shows that 72 is an upper bound. Testing the negative values shows that 72 is a lower bound.

7 614377 2 2177119392 62234112399 (continuedonnextpage)

(continued) 7 614377 2 1891421 2170 24 2031449 62027 24

30. The possible rational zeros are 111555 ,,,,1,,,5. 632632     There is one sign change in ()gx , so there is one positive zero. There are three sign changes in ()gx , so there are either 3 or 1 negative zeros. Testing each positive possibility shows that 12 is an upper bound. Testing the negative values shows that 5 is a lower bound. 1 6232595 2 313195 62638100 56232595 30353001545 67603091540

31. 22250255 xxxi 

32. 22 164 9160 93 xxxi

33. 22 (2)90(2)9 2323 xx xixi  

34. 22 (1)270(1)27 133133 xx xixi

 

35. 22449(2)9 2323 xxx xixi

 

36. 222116(1)16 1414 xxx xixi

 

37. 32 2 80(2)(24)0 224(1)(4) 2 or 2(1) 2416212 13 22 xxxx xx i   

 

The solution set is  2,13 i  .

38. 422 2 10(1)(1)0 (1)(1)(1)01 or 1 or xxx xxxxx xi   

39. (2)(3)(3)02 or 3 xxixixxi 

40. (1)(2)(26)(26)0 1 or 2 or 3 xxixixi xxixi  

41. The remaining zero is 3. i

42. The remaining zero is 2. i 

43. The remaining zero is 2. i 

44. The remaining zeros are i and 1. i 

45. The remaining zeros are i and 3. i

46. The remaining zeros are 2,4, ii and 1. i

47. 432 ()2((5))((5))(3)(3) 22070180468 Pxxixixixi xxxx  



48. 432 ()3((23))((23)) ((14))((14)) 318114282663 Pxxixi xixi xxxx

49. 2 5432 ()7(5)(1)((3))((3) 71197772415 35001750 Pxxxxixi xxxx x   

50. 3 6543 ()4(3)((23))((23)) 428100156 Pxxxxixi xxxx 



51. The function has degree four, so there are four zeros. Since one zero is 3, i another zero is 3. i So 2 (3)(3)9 xixix  is a factor of ()Px Now divide to find the other factor: 2 2432 42 3 3 999 9 9 9 0 xx xxxxx xx xx xx  



432 2 ()99 (3)(3)() (3)(3)(1) the zeros of Pxxxxx xixixx xixixx

   () are1,0,3,3. Pxii

52. The function has degree four, so there are four zeros. Since one zero is 1, i another zero is 1. i  So 2 ((1))((1))22 xixixx  is a factor of ()Px . Now divide to find the other factor:

((1))((1))(1) ((1))((1))(1)(1)

53. The function has degree five, so there are five zeros. Since one zero is 3, i another zero is 3. i  So 2 ((3))((3))610 xixixx  is a factor of ()Px . Now divide to find the other factor: 32 25432 543 432 432





5432



2 ()522220 ((3))((3))(2) ((3))((3))(2) ((3))((3))(2)(1) Pxxxxxx xixixxx xixixxx xixixxx



 the zeros of () are2,0,1,3,3. Pxii 

54. The function has degree five, so there are five zeros. Since one zero is i, another zero is –i. So 2 ()()1 xixix is a factor of ()Px . Now divide to find the other factor: 32 25432 53 432 42 32 3 2 2 211176 12111917176 22 111717 1111 17617 1717 66 66 0 xxx xxxxxx xx xxx xx xxx xx x x       5432 32 ()2111917176 ()()(211176) Pxxxxxx xixixxx  

Now find the zeros of 32 211176. xxx

There are either three or one positive, rational zeros. The possible zeros are 13 22,1,,2,3,6.

Using synthetic division, we find that x = 2 is one of the zeros: 2211176 4146 2730 5432 32 2 ()2111917176 ()()(211176) ()()(2)(273) ()()(2)(21)(3) Pxxxxxx xixixxx xixixxx xixixxx

    the zeros of 1 () are ,2,3,,. 2 Pxii

55. The function has degree 3, so there are three zeros. There are three sign changes, so there are 1 or 3 positive zeros. The possible rational zeros are  1,17. Using synthetic division we find that one zero is 1: 1192517 1817 18170 32292517(1)(817). xxxxxx 

Now solve the depressed equation 2 8170xx 2 8(8)4(1)(17) 84 2(1)2 xx

  

4.xi  The zeros are 1,4. i 

56. The function has degree 3, so there are three zeros. There are two sign changes, so there are either 2 or 0 positive zeros. There is one sign change in ()fx , so there is one negative zero.

The possible rational zeros are 1,13. Using synthetic division, we find that one zero is 1: 115713 1613 16130 3225713(1)(613). xxxxxx

Now solve the depressed equation 2 2 6130 6(6)4(1)(13) 616 2(1)2

32. xi  The zeros are 1,32. i 

57. The function has degree 3, so there are three zeros. There are two sign changes, so there are either 2 or 0 positive zeros. There is one sign change in ()fx , so there is one negative zero.

The possible rational zeros are 12

Using synthetic division to test the negative values we find that one zero is

58. The function has degree 3, so there are three zeros. There are three sign changes, so there are either 3 or 1 positive zeros. There are no sign changes in ()fx , so there are no negative zeros. The possible rational zeros are

4 ,2,4. 3 

  Using synthetic division to test the positive values we find that one zero is 1 3 :

322 1 3124(312) 3 xxxxx

2 11 3(4)3(2)(2). 33 xxxxixi

The zeros are 1 ,2. 3 i 

59. The function has degree 4, so there are four zeros. There are two sign changes, so there are 4, 2, or 0 positive zeros. There are no sign changes in ()fx , so there are no negative zeros. The possible rational zeros are 139 ,1,,3,,9. 222   

Using synthetic division to test the positive values we find that one zero is 1: 121023249 28159 281590

The zeros of the depressed function 32 28159 xxx are also zeros of P. The possible rational zeros are 139 ,1,,3,,9. 222    

Using synthetic division to test the positive values we find that one zero is 1: 128159 269 2690

Thus, 432 21023249 xxxx     2 2 1269.xxx  Now solve the depressed equation 2 2690. xx (continuedonnextpage)

(continued) 2 6(6)4(2)(9) 636 2(2)4

6633 422 ii x   The zeros of P are 33 1,. 22 i 

60. The function has degree 4, so there are four zeros. There are two sign changes, so there are either 2 or 0 positive zeros. There are two sign change in ()fx , so there are either 2 or 0 negative zeros. The possible rational zeros are 12142848 ,,,,,,1,,2,, 99393933

 4,8. Using synthetic division to test the negative values we find that one zero is –2: 293014168 1824208 9121040

The zeros of the depressed function 32 912104 xxx are also zeros of P. The possible rational zeros are 121424 ,,,,,1,,2,4. 993933  

Using synthetic division to test the negative values we find that one zero is –2: 2912104 18124 9620

Now solve the depressed equation 2 2 9620 6(6)4(9)(2) 636 2(9)18 xx xx

6611 1833   i xi The zeros are 11 2,. 33  i

61. The function has degree 4, so there are four zeros. There are three sign changes, so there are either 3 or 1 positive zeros. There is one sign change in ()fx , so there is one negative zero. The possible rational zeros are  1,2,3,5,6,10,15,30.  Using synthetic division to test the negative values, we find that one zero is –3: 31453830 3214830 1716100

The zeros of the depressed function 3271610 xxx are also zeros of P. The possible rational zeros of the depressed function are  1,2,5,10. Since we have already found the negative zero, we examine only the positive possibilities and find that 1 is a zero: 1171610 1610 16100 432453830 xxxx  2 (3)(1)(610) xxxx  . Now solve the depressed equation: 2 2 6(6)4(1)(10) 6100 2(1) xxx 

The zeros are –3, 1, 3. i 

62. The function has degree 4, so there are four zeros. There is one sign change, so there is one positive zero. There are three sign changes in ()fx , so there are either 3 or 1 negative zeros. The possible rational zeros are  1,2,3,6,9,18.  Using synthetic division to test the positive values we find that one zero is 1: 1117918

12918

129180

The zeros of the depressed function 322918 xxx are also zeros of P. We can factor by grouping to find the next zero: 3222918(2)9(2) xxxxxx  2 (2)(9). xx  So –2 is a zero, as are 3. i 

The zeros are –2, 1, 3. i 

63. The function has degree 5, so there are five zeros. There are five sign changes, so there are either 5, 3, or 1 positive zeros. There are no sign changes in ()fx , so there are no negative zeros. The possible rational zeros are 13 22,1,,2,3,6.    Using synthetic division to test the positive values, we find that one zero is 1 2 : (continuedonnextpage)

(continued)

1 2111917176 2 15756 2101410120

The zeros of the depressed function 432 210141012 xxxx  are also zeros of P. Use synthetic division again to find the next zero, 2: 2210141012 412412 26260 5432 2111917176 xxxxx 

32 1 (2)(2626) 2 xxxxx

32 1 2(2)(33). 2 xxxxx

Use factoring by grouping to factor 3233xxx : 32233(3)1(3) xxxxxx  2 (1)(3) xx  . So the remaining zeros are 3 and i The zeros are 1 ,2,3,. 2 i

64. The function has degree 5, so there are five zeros. There are four sign changes, so there are either 4, 2, or 0 positive zeros. There is one sign change in ()fx , so there is one negative zero. The possible rational zeros are  1,2,4.

Using synthetic division to test the negative values, we find that one zero is –2: 21218104 2814124 147620

The zeros of the depressed function 4324762 xxxx are also zeros of P Use synthetic division again to find the next zero, 1: 114762 1342 13420

543228104 xxxxx  32 (2)(1)(342) xxxxx 

The zeros of the depressed function 32342 xxx are also zeros of P. Use synthetic division again to find the next zero, 1: 11342 122 1220

543228104 xxxxx  32 (2)(1)(342) xxxxx  2 (2)(1)(1)(22). xxxxx 

Now solve the depressed equation: 2 220xx 2 2(2)4(1)(2) 24 1. 2(1)2 xi    The zeros are 2,1 (multiplicity 2),1. i 

65. Since one zero is 3i, 3i is another zero. There are two zeros, so the degree of the equation is at least 2. Thus, the equation is of the form

   2 339.fxaxixiax The graph passes through (0, 3), so we have   2 1 309. 3 aa  Thus, the equation of the function is    2 1 9. 3 fxx

66. Since one zero is i, another zero is i. From the graph, we see that 3 is another zero. There are three zeros, so the degree of the equation is at least 3. Thus, the equation is of the form

   2 331fxaxxixiaxx 

The graph passes through (0, 3), so we have    2 303011. aa  Thus, the equation of the function is    2 31.fxxx

67. Since i and 2i are zeros, so are i, and 2i. From the graph, we see that 2 is also a zero. There are five zeros, so the degree of the equation is at least 5. Thus, the equation is of the form

 22 222 142 fxaxixixixix axxx  

The graph passes through (0, 4), so we have

  22 1 4010402. 2 aa 

Thus, the equation is

  22 1 142 2 fxxxx  or

 22 1 142. 2 fxxxx 

68. Since 2i is a zero, so is 2i. From the graph, we see that 1, 0, and 1 are also zeros. There are five zeros, so the degree of the equation is at least 5. Thus, the equation is of the form

The y-intercept is (0, 0), so we have

22 000401 a  which is true for all values of a. Thus, the equation is

Beyond the Basics

69. There are three cube roots. We know that one root is 1. Using synthetic division, we find 32 01(1)(1) xxxx  :

Solve the depressed equation 2 2 114(1)(1) 10 2(1)

So, the cube roots of 1 are 1 and 13 . 22 i 

70. The solutions of the equation and the zeros of the polynomial are the same. There are n roots.

71. The polynomial has degree 6, so there are six zeros. There are no sign changes in ()Px , nor are there sign changes in ()Px , so according to Descartes’s Rule of Signs, there are neither positive nor negative zeros. Therefore, all six zeros must be complex.

72. The polynomial has degree 5, so there are five zeros. There are no sign changes in ()Px , while there is one sign change in ()Px . According to Descartes’s Rule of Signs, there is one negative zero, and the remaining four zeros must be complex.

73. The polynomial has degree 7, so there are seven zeros. There are no sign changes in ()Px , while there is one sign change in ()Px . According to Descartes’s Rule of Signs, there is one negative zero, and the remaining six zeros must be complex.

74. The polynomial has degree 3, so there are three zeros. There is one sign change in ()Px , while there are no sign changes in ()Px . According to Descartes’s Rule of Signs, there is one positive zero, and the two remaining zeros must be complex.

75. Since 1 + 2i is a zero, so is 1 2i. The equation is of the form

21212 225 fxaxxixi axxx  

The y-intercept is 40, so we have    2 400202054 aa 

Thus, the equation is

2 4225.fxxxx  Because a < 0, as yx and as . yx

76. Since 2 3i is a zero, so is 2 + 3i. The equation is of the form

 2 12323 1413 fxaxxixi axxx  

The y-intercept is 26, so we have 

 2 2601040132 aa 

Thus, the equation is    2 21413.fxxxx  Because a > 0, as yx and as . yx

77. Since 3 + i is a zero, so is 3 i. The equation is of the form

22 1133 1610 fxaxxxixi axxx  

The y-intercept is 20, so we have      22 2001060102 aa 

Thus, the equation is



  22 21610.fxxxx  Because a < 0, as yx and as . yx

78. Since 1 2i and 3 2i are zeros, so are 1 + 2i and 3 + 2i. The equation is of the form 





22 1212 3232 25613 fxaxixi xixi axxxx

 (continuedonnextpage)

(continued)

The y-intercept is 130, so we have

Thus, the equation is  

 22 1300205060132 aa 

 22 225613.fxxxxx 

Because a > 0, as yx and as . yx

79.  2 22, Pxxixixi  122 2 120 iii ii  2 222xixixix 

80.  2 32,2 Pxxixxi  2132 22 10 ii i i

2 322 xixxixi 

81. 323434, Pxxixixixi 

 13434 34 1340 iiii iii 

32234343441 xixixixxxixxxi 

82. 32427814,2 Pxxixixixi 

 21427814 2814 1470 iiii iii   

322427814472 xixixixxxi 

Now solve 2 470xx to find linear factors.

Thus,

32242781447223232 23232 xixixixxxixxxi xixixi

Critical Thinking/Discussion/Writing

83. Factoring the polynomial, we have 1 110 nn nn axaxaxa  12 ()()(). nn axrxrxr   Expanding the right side, we find that the coefficient of 1 n x is 12 nnnn ararar and the constant term is 12 (1) n nn arrr . Comparing the coefficients with those on the left side, we obtain 112(). nnn aarrr 

Because 0, n a  1 12 n n n a rrr a   and 0 12 (1). n n n a rrr a  

84. Because 1r and 2r are roots of the equation, we have

85. a. 336206200. xxxx There is one sign change, so there is one positive root. There are no sign changes in ()fx , so there are no negative roots. Therefore, there is only one real solution.

b. Substituting vu for x and remembering that 33 20 and 2, vuuv we have 332233322 3333 ()6()336()()(33)6() ()3()6()()(63)() 20(63(2))()20 vuvuvvuvuuvuvuvuvuvu vuvuvuvuvuuvvu vu 

Therefore, xvu  is the solution.

c. To solve the system 33 20,2vuvu , solve the second equation for v, and substitute that value into the first equation, keeping in mind that u cannot be zero, so division by u is permitted: 3 3333 3 3 3 28 2020 8 200. vuuu uu u u

3 2 Let, so 8 2002080 ua aaa a

d. If q is positive, then, according to Descartes’s Rule of Signs, the polynomial 3 xpxq  has one positive zero and no negative zeros. If q is negative, then it has no positive zeros and one negative zero. Either way, there is exactly one real solution. From (b), we have 33 vuq  , , 3 p vu  and .xvu 

Substituting, we have 33 3223 3322 33 33 ()() 33() ()(33)() ()3()() ()(3)() 3 3 xpxvupvu vvuvuupvu vuvuvupvu vuvuvupvu vupuvvu p qpxq

    

is the xvu  solution. Solving the system 33 vuq  , , 3 p vu  we obtain 3 3333 33 3 3 pp vuquuq uu

Let, so we have 00 33 ua pp aqaqa a   (continuedonnextpage)

86. 32610806,10,8 xxxabc 

Substituting 6 2 3 xyy as in (1e), we have 32 (2)6(2)10(2)80 yyy  322 (6128)(62424) 102080 yyyyy y   3324024. yyyy Then, using the results of (1d), we have 23 3 23 3 442 223 442 223

Using a calculator, we find that y = –2. So 224. x  Now use synthetic division to find the depressed equation: 416108 488 1220 3226108(4)(22)0. xxxxxx  2 2 224(1)(2) 220 2(1) 2422 1 22 xxx i i 

The solution set is  4,1. i 

Getting Ready for the Next Section

87. a.     22 3535 33 2221 53 3333 yxyx yxyx

b.  21 11 33 y  88. a. The slope of the line 456 xy 46 546 55yxyx  is 4 . 5 The slope of the line perpendicular to this line is 5 4 The equation we are seeking is

 5 32 4 yx

 5 32 4 5551 23. 4442 yx yxyx 

2.7 Variation

2.7 Practice Problems

A battery of 275 volts is needed to produce 60 amperes of current.

3. 22 2 48(2)12 12(5)300 ykxkk y

4. 1260 5 60 20 3 kk Ak B A 

6. Mars 12 22 Mars Mars 2 Mars mmmM FGmgG rR M gG R

Note that the radius of Mars is given in kilometers, which must be converted to meters.

2.7 Exercises Concepts and Vocabulary

1. y varies directly as x if y = kx

2. y varies inversely as x if . k yx 

3. y varies directly as the nth power of x if n ykx 

4. z varies jointly as x and y if z = kxy

5. True

6. False. A varies directly as the second power of r

7. False.

8. True Building Skills

9. 11 ;1530;(28)14 22xkykkx 

10. ;32;(7)3321 222ykxkky 

11. 222 ;64216;5(16)400sktkks 

12. 33 3 ;270310; 80102 ykxkk xx  

13. 33 ;333;99 1113 kk rkr u 

14. 4 ;244;4 161 kk yky z 

15. 333 81 ;18; 8 24 kk BkB A 

16. 3 3 3 3 3 3 ;10102; 2 10220001 4040 32 kk yk x x xx  

17. ;42(2)(3)7;562(7) 4 zkxykky y  

18. 1137(1)1 ;1; 226142 kqk mkm p 

19. 222 ;3242;2(5)50zkxkkz 

20. 333 721 ;972; 3 26 kk uku t 

21. 22 2 1 ;36(17)(6); 17 1324 (4)(9) 1717 PkTQkk P  

22. 1 ;91381; 13 115 (5)9 1313 akbckk a  

23. 22 2 16 ;2454 3 54 2724 2 kxk zk y x xx  

24. 33 22 3 22 4(3)1 ;9; 3 2 1 (27)(2) 72 3 883 kuvk zk w w ww  

25. 168 6 12 y y 

26. 17 85 22110 z z 

27. 00 100 200 2 y y xx 

28. 23981 24 9 yy y 

Applying the Concepts

29. yHx  , where y is the speed of the galaxies and x is the distance between them.

30. RkP  , where k is a constant.

31. a. 30.5 yx  , where y is the length in centimeters and x is the length in feet.

b. (i) 30.5(8)244 cm y 

(ii) 1 30.55162.67 cm 3 y   

c. (i) 5730.51.87 ft xx  (ii) 12430.54.07 ft xx 

32. a. 2.2 yx  , where y is the weight in pounds and x is the weight in kilograms.

b. (i) 2.2(0.125)0.275 lb y  (ii) 2.2(4)8.8 lb y  (iii) 2.2(2.4)5.28 lb y 

c. (i) 272.212.27 kg xx  (ii) 1602.272.73 kg xx 

33. 77 ;720;(100)35 g 2020PkQkkP 

34. ;6004015;Wkhkk  15(25)$375. W  The constant of proportionality is the hourly wage.

35. 222 ;64216;916 0.75 sec dktkkt t  

36. 55 ;104;(6)15 lb 22 FkxkkF 

37. 2 ;206000; 300 6000 60 lb/in. 100 kk Pk V P  

38. 9 2609 ;36;1355 KK kTT k PkP VV 

a. 3 9(300) 4013.5 in. 5 V V  b. 2 9(280) 12.92 lb/in. 5(39) P 

39. a. The astronaut is 6000 + 3960 miles from the Earth’s center.

2 22 2 2 ;120120(3960) 3960 120(3960) 18.97 lb (60003960) kk Wk d W   

b. 2 2 2 2 200200(3960) 3960 200(3960) 201.01 lb 3950 k k W  

40. You should buy at the higher altitude because an object weighs less the further it is from the Earth’s center.

41. 1740 km = 1,740,000 m = 6 1.74010 m 

(6.67)(7.4)(10) 1.63 m/sec 1.74010 M

42. 696,000 km = 696,000,000 m = 6.96108 m 

(6.67)(2)(10) 2.7510 m/sec 6.9610 M gG R

43. a. 22 2 ;32032,000 10 32,000 1280 candla 5 kk Ik d I

b. 2 2 32,000 400808.94 ft dd d  from the source.

44. a. ;489626 26 skdkk sd  

b. (i) 26(60)37.95 mph s  (ii) 26(150)60 mph s  (iii) 26(200)69.28 mph s 

c. 2 7026635 6351225204.17 ft dd dd  

45. 22 24 pklkppp lll  the length is quadrupled if the period is doubled.

46. 3 ;4001004 4(120)480 cm VkAkk V  

47. a. 2 HkRN 

b. 22 (2)44 kRNkRNH the horsepower is multiplied by 4 if the radius is doubled.

c. 22 (2)22 kRNkRNH the horsepower is doubled if the number of pistons is doubled.

d. 2 22 2 (2) 2422 RkRNkRNH kN  

the horsepower is halved if the radius is halved and the number of pistons is doubled.

48. Convert the dimensions given in inches to feet. 2 2 2 11 32 ;576172,800 25 15 172,800 26 3000 lb 20 k kwd sk l s



Beyond the Basics

49. a. 23Eklv 

b. 23 3 1920(10)(8)0.0375 80 kk 

c. 23 0.0375(8)(25)37,500 watts E 

d. 2323 (2)88 klvklvE 

e. 2323 (2)44 klvklvE  f. 2323 (2)(2)4(8)32 klvklvE 

50. a. 22 1002 ;20.08 25 2 d kIk Ik d 

b. The distance from the bulb to the point on the floor is 22 2313 m.   2 2 0.08(200) 1.23 watts/m 13 d I 

c. If the bulb is raised by 1 meter, then the distance from the bulb to the point on the floor is 22 3332 m. 

2 2 0.08(200) 0.89 watts/m 32 d I 

51. a. 3434 4 ;75(75) 352.94 mkwk k  

b. 34 2.94(450)287.25 watts m 

c. 34343434 (4)42.832.83m kwkwkw

52.

The gravitational attraction between the sun and the Earth is approximately 168.92 times as strong as the gravitational pull between the Earth and the moon.

53. a. 23 2 12 4 () r T

b. Because gravity is in terms of cubic meters per kilogram per second squared, convert the distance from kilometers to meters: 811 1.510km1.510m

4(1.510) (3.1510) 6.6710 4(3.375)10 9.922510 6.6710 M M

24 sun 1114 30 133.2410 6.67109.922510 2.0110

54. Because gravity is in terms of cubic meters per kilogram per second squared, convert the distance from kilometers to meters: 8 384,000km3.8410m  . 27.3 days 27.3(24)(60)(60) sec 2,358,720 sec   283 2 11 Earth 283 Earth 112 24 Earth 4(3.8410) 2,358,720 6.6710 4(3.8410) 6.67102,358,720 6.0210 kg M M M

55. a. (),RkNPN  where k is the constant of proportionality.

b. 6 451000(9000)510 kk 

c. 6 510(5000)(5000)125 R  people per day.

d. 6 2 100510(10,000) 20,000,00010,000 NN NN

2 27 10,00020,000,0000 10,00010,0004(1)(210) 2(1) 10,00020,000,000 2764 or 7236 2 NN N

Critical Thinking/Discussion/Writing

56. 1.31.56 the voltage 1.2 kV IIRkV R kV IIRkV R   must increase by 56%.

57. a. 2 vkw  . The diamond is cut into two pieces whose weights are 2 5 w and 3 5 w The value of the first piece is 4 (1000)$160 25   

, and the value of the second piece is 9 (1000)$360 25     . The two pieces together are valued at $520, a loss of $480.

b. The stone is broken into three pieces whose weights are 59 25525,, www  and 11 . 25 w So, the values of the three pieces are 2 1 (25,000)$1000, 5



2 9 (25,000)$3240, 25



and 2 11 (25,000)$4840, 25



respectively. The total value is $9,080, a loss of $15,920.

c. The weights of the pieces are 23 ,,, 1515155 wwww  45 , and , 15153 www  respectively. If the original value is x, then 222 22 121 85,000 15155 41 153 11 85,000$112,500 the 45 xxxx xx xxx

original value of the diamond. A diamond whose weight is twice that of the original diamond is worth 4 times the value of the original diamond = $450,000.

58. ()pkdnf  80(40)(30) and kf

180(60)(35). kf

Solving the first equation for k we have

Substitute that value into the second equation and solve for f: 2 180(60)(35) 30 4200120 180 30 54001804200120 12006020 ff f f ff ff

59.

60. 24;202442 0242144. The greatest number skwkk ww

Getting Ready for the Next Section 61. 3 28

2 2 11 3

3  63. 3 3 3 111 28 2

64. 4 4 1 216 2

65.

13 132 22224 xxxx

 66.

233 233 5555 xxxxx  

67.  32 325 23 5 2 22 2 x xxx x  

68. 

 2 1 1 222 xxxxxx 

Chapter 2 Review Exercises

Building Skills

1.(i) opens up (ii) vertex: (1, 2)

(iii) axis: x = 1

(iv) 22 0(1)22(1) there xx  are no x-intercepts.

(v) 2 (01)23 is the yy  y-intercept.

(vi) The function is decreasing on (,1)  and increasing on (1,). 

2.(i) opens up (ii) vertex: (–2, –3)

(iii) axis: x = –2

(iv) 22 0(2)33(2)

23 are the -intercepts. xx xx 

(v) 2 (02)31 is the yy  y-intercept.

(vi) The function is decreasing on (,2)  and increasing on (2,). 

3.(i) opens down

(ii) vertex: (3, 4)

(iii) axis: x = 3

(iv) 22 02(3)42(3)

32 are the -intercepts.

(v) 2 2(03)414 is the yy  y-intercept.

(vi) The function is increasing on (,3)  and decreasing on (3,). 

4.(i) opens down

(ii) vertex: (–1, 2) (iii) axis: x = –1

(iv) 22 1 0(1)24(1) 2 3 and 1 are the -intercepts. xx xxx

(v) 2 13 (01)2 is the 22yy  y-intercept.

(vi) The function is increasing on (,1)  and decreasing on (1,). 

5.(i) opens down (ii) vertex: (0, 3)

(iii) axis: x = 0 (y-axis)

(iv) 22 3 023 2 6 are the -intercepts. 2 xx xx

(v) 2 2(0)33 is the yy  y-intercept.

(vi) The function is increasing on (,0)  and decreasing on (0,). 

6.(i) opens up

(ii) To find the vertex, write the equation in standard form by completing the square: 2 2 2 241 122(21) 2(1)3 the vertex is (1,3).

(iii) axis: x = –1

(iv) 2 2 0241 444(2)(1) 424 2(2)4 6 1 are the -intercepts. 2 xxx xx 

(v) 2 2(0)4(0)11 is the yy  y-intercept.

(vi) The function is decreasing on (,1)  and increasing on (1,). 

7.(i) opens up

(ii) To find the vertex, write the equation in standard form by completing the square: 2 2 2 243 322(21) 2(1)1 the vertex is (1,1). yxx yxx yx 

(iii) axis: x = 1

(iv) 2 2 0243 4(4)4(2)(3) 48 2(2)4 xx x

there are no x-intercepts.

(v) 2 2(0)4(0)33 is the yy  y-intercept.

(vi) The function is decreasing on (,1)  and increasing on (1,). 

8.(i) opens down

(ii) To find the vertex, write the equation in standard form by completing the square: 2 2 2 23 111 32 8216 125 2 48 125 the vertex is 48,. yxx yxx yx

(iii) axis: 1 4 x 

(iv) 2 023 3 0(23)(1) and 2 1 are the -intercepts. xx xxx xx

  

(v) 2 2(0)033 is the yy  y-intercept.

(vi) The function is increasing on 1 , 4  

and decreasing on 1 ,. 4

9.(i) opens up

(ii) To find the vertex, write the equation in standard form by completing the square.

2 2 2 321 121 13 339 1212 3 the vertex is 3333,. yxx yxx yx

(iii) axis: 1 3 x 

(iv) 2 2 0321 2(2)4(3)(1) 28 2(3)6

there are no x-intercepts.

(v) 2 3(0)3(0)11 is the yy  y-intercept.

(vi) The function is decreasing on 1 , 3

and increasing on 1 ,. 3

10.(i) opens up

(ii) To find the vertex, write the equation in standard form by completing the square: 2 2 2 354 25525 43 12336 523 3 612 523 the vertex is 612,. yxx yxx yx

(iii) axis: 5 6 x 

(iv) 2 2 0354 5(5)4(3)(4) 523 2(3)6 xx x

there are no x-intercepts.

(v) 2 3(0)5(0)44 is the yy  y-intercept.

(vi) The function is decreasing on 5 , 6

and increasing on 5 ,. 6

In exercises 11–14, find the vertex using the formula 22,. bb f aa

11. 0 a  the graph opens up, so f has a minimum value at its vertex. The vertex is 44 2(1)2(1),(2,1).

12. 0 a  the graph opens down, so f has a maximum value at its vertex. The vertex is 88 2(4)2(4),(1,1).

13. 0 a  the graph opens down, so f has a maximum value at its vertex. The vertex is

14. 0 a  the graph opens up, so f has a minimum value at its vertex. The vertex is

15. Shift the graph of 3 yx  one unit left and two units down.

16. Shift the graph of 4 yx  one unit left and two units up.

17. Shift the graph of 3 yx  one unit right, reflect the resulting graph about the y-axis, and shift it one unit up.

18. Shift the graph of 4 yx  three units up.

19.(i) () as () as fxx fxx  

(ii) Zeros: x = –2, multiplicity 1, crosses the x-axis; x = 0, multiplicity 1, crosses the x-axis; x = 1, multiplicity 1, crosses the x-axis.

(iii) x-intercepts: –2, 0, 1; 0(01)(02)0 is the yy  y-intercept.

(iv) The intervals to be tested are (2,0) , (0,1),(1,2), and (2,).  The graph is above the x-axis on (2,0)(1,)  and below the x-axis on (,2)(0,1). 

(v) ()(1)(2)() fxxxxfx  f is not even.

()((1)(2))() fxxxxfx  f is not odd. There are no symmetries.

(vi)

20.(i) () as () as fxx fxx

 

(ii) 32 ()(1) fxxxxx (1)(1).xxx  Zeros: x = –1, multiplicity 1, crosses the x-axis; x = 0, multiplicity 1, crosses the x-axis; x = 1, multiplicity 1, crosses the x-axis.

(iii) x-intercepts: –1, 0, 1; 3 000yy is the y-intercept.

(iv) The intervals to be tested are (,1)  , (1,0),(0,1), and (1,). The graph is above the x-axis on (1,0)(1,)  and below the x-axis on (,1)(0,1). 

(v) ()()()33 fxxxxx 

()fx  f is odd. f is symmetric with respect to the origin.

(vi)

21.(i) () as () as fxx fxx  

(ii) Zeros: x = 0, multiplicity 2, touches but does not cross the x- axis; x = 1, multiplicity 2, touches but does not cross the x- axis.

(iii) x-intercepts: 0, 1; 22 0(01)0 is the yy  y-intercept.

(iv) The intervals to be tested are (,0)  , (0,1), and (1,).  The graph is below the x-axis on (,0)(0,1)(1,). 

(v) 2 2 ()()(1) (1)() or () fxxx xxfxfx

 ()fx  f is neither even nor odd. There are no symmetries. (vi)

22.(i) () as () as fxx fxx  

(ii) Zeros: x = 0, multiplicity 3, crosses the x-axis; x = 2, multiplicity 2, touches but does not cross the x-axis.

(iii) x-intercepts: 0, 2; 32 0(02)0 is the yy  y-intercept.

(iv) The intervals to be tested are (,0)  , (0,2), and (2,).  The graph is above the x-axis on (,0)  and below the x-axis on (0,2)(2,). 

(v) 3 3 ()()(2) (2)() or () fxxx xxfxfx

 f is neither even nor odd. There are no symmetries.

(vi)

23.(i) () as () as fxx fxx  

(ii) Zeros: x = –1, multiplicity 1, crosses the x-axis; x = 0, multiplicity 2, touches but does not cross; x = 1, multiplicity 1, crosses the x-axis.

(iii) x-intercepts: –1, 0, 1; 22 0(01)0 is the yy  y-intercept.

(iv) The intervals to be tested are (,1)  , (1,0),(0,1), and (1,). The graph is above the x-axis on (1,0)(0,1)  and below the x-axis on (,1)(1,). 

(v) 22 22 ()()(()1) (1)() fxxx xxfx

f is even, and f is symmetric with respect to the y-axis.

(vi)

24.(i) () as () as fxx fxx  

(ii) Zero: x = 1, multiplicity 2, touches but does not cross.

(iii) x-intercept: 1; 22 (01)(01)1 is the yy  y-intercept.

(iv) The intervals to be tested are (,1)  and (1,).  The graph is below the x-axis on (,1)(1,). 

(v) 22 ()(1)(()1)() or fxxxfx  ()fx  f is neither even nor odd. There are no symmetries.

(vi)

xxxxx xx xx xx xx xx x x   

28. 232 32 2 2 1 26347 26 27 26 413 x xxxxx xxx xx xx x 

29. 310123 399 1336

Quotient: 2 33xx remainder –6.

30. 64350 24126786 421131786

Quotient: 2 421131 xx remainder –786.

31. 12357165 251017 251017182

Quotient: 32 251017 xxx remainder 182.

32. 2320116132 616326292 38163146224

Quotient: 432 38163146 xxxx  remainder –224.

33.(i) 32 (2)23(2)11(2)2911 f  (ii) 2131129 2218 11911

The remainder is –11, so f(2) = –11.

34.(i) 32 (2)2(2)(2)15(2)216 f  (ii) 221152 4618 23916

The remainder is 16, so f(–2) = 16.

35.(i) 42 (2)(3)2(3)5(3)1088 f  (ii) 3102510 392178 1372688

The remainder is 88, so f(–3) = 88.

36.(i) 5 (1)123 f  (ii) 1100002 11111 111113

The remainder is 3, so f(1) = 3.

37. 217148 2108 1540

The remainder is 0, so 2 is a zero. Now find the zeros of the depressed function

2 54xx (4)(1) xx  4 and 1 are also zeros. So the zeros of 327148 xxx are 1, 2, and 4.

38. 223124 4144 2720

The remainder is 0, so –2 is a zero. Now find the zeros of the depressed function 2 272 xx : 2 7(7)4(2)(2) 2(2) x   733 4  

So the zeros of 32 23124 xxx are –2 and 733 4 

39. 1 314136 3 156 315180

The remainder is 0, so 1 3 is a zero. Now find the zeros of the depressed function 2 315183(2)(3) xxxx –2 and –3 are also zeros. So the zeros of 32 314136 xxx are –3, –2, and 1 3 .

40. 1 419132 4 152 42080

The remainder is 0, so 14 is a zero. Now find the zeros of the depressed function 2 2 20204(4)(8) 4208 2(4) 20433533 . 82 xx    

So the zeros of 32 419132 xxx are 533 2  and 1 4

41. The factors of the constant term are  1,2,3,6 , and the factors of the leading coefficient are 1. The possible rational zeros are  1,2,3,6

42. The factors of the constant term are  1,2,4,8,16  , and the factors of the leading coefficient are  1,3,9. The possible rational zeros are 121 993,,,     4284168 ,,,1,,,2,,4, 939393  16 ,8,16. 3    

43. 32 32 32 ()5112; ()5()11()2() 5112 fxxxx fxxxx xxx   

There are no sign changes in ()fx , so there are no positive zeros. There are two sign changes in (),fx so there are 2 or 0 negative zeros.

322 5112(5112) (51)(2) xxxxxx xxx   1 the zeros are 2,,0. 5



44. 32 32 32 ()256 ()()2()5()6 256 fxxxx fxxxx xxx





There is one sign change in ()fx , so there is one positive zero. There are two sign changes in (),fx so there are 2 or 0 negative zeros. The possible rational zeros are  1,2,3,6 . Use synthetic division to find one zero: 31256 336 1120

The remainder is 0, so –3 is a zero. Now find the zeros of the depressed function.

2 2(2)(1)xxxx –1 and 2 are also zeros. So the zeros of 32256 xxx are –3, –1, and 2.

45. 32 32 32 ()3412 ()()3()4()12 3412 fxxxx fxxxx xxx







There is one sign change in ()fx , so there is one positive zero. There are two sign changes in (),fx so there are 2 or 0 negative zeros. The possible rational zeros are  1,2,3,4,  6,12. Use synthetic division to find one zero: 313412 3012 1040

The remainder is 0, so –3 is a zero. Now find the zeros of the depressed function

2 4(2)(2)xxx –2 and 2 are also zeros. So the zeros of 323412 xxx are –3, –2, and 2.

 

46. 32 32 32 ()29125; ()2()9()12()5 29125 fxxxx fxxxx xxx



There are three sign changes in (),fx so there are 3 or 1 positive zeros. There are two sign changes in (),fx so there are 2 or 0 negative zeros. The possible rational zeros are 15 22,1,,5.    

Use synthetic division to find one zero: 129125 275 2750

The remainder is 0, so 1 is a zero. Now find the zeros of the depressed function 2 275(25)(1)52 xxxx and 1 are also zeros. The zeros of 32 29125 xxx are 52 and 1 (multiplicity 2).

47. The function has degree three, so there are three zeros. Use synthetic division to find the depressed function: 21076 246 1230

Now find the zeros of the depressed function 2 23(3)(1)3 and 1 are xxxx zeros. The zeros of the original function are –3, 1, 2.

48. The function has degree four, so there are four zeros. Since 1 is a zero of multiplicity 2, 2 (1) x is a factor of 43232. xxxx

Use synthetic division twice to find the depressed function: 111312 1212 12120 11212 132 1320

Alternatively, divide 43232 xxxx by 22 (1)21 xxx  . Now find the zeros of the depressed function 2 32(2)(1)xxxx 2 and 1 are zeros. The zeros of the original function are –2, –1, and 1.

49. The function has degree four, so there are four zeros. Use synthetic division twice to find the depressed function: 11261827 13927 139270 313927 3027 1090

Alternatively, divide 432261827 xxxx  by 2 (1)(3)23. xxxx  Now find the zeros of the depressed function 2 93 xi  are zeros. The zeros of the original function are –1, 3, and 3i 

50. The function has degree three, so there are three zeros. Since one zero is 2 – i, another zero is 2 + i. Divide 32 4193215 xxx by 2 ((2))((2))45 xixixx  to find the depressed function:

Now find the zeros of the depressed function: 3 430. 4 xx Alternatively, find the possible rational zeros, and then use synthetic division to find the zero. The zeros of the original function are 3 ,2, and 2. 4 ii

51. The function has degree four, so there are four zeros. Since one zero is –1 + 2i, another zero is –1 – 2i. Divide 43229820 xxxx by 2 ((12))((12))25 xixixx  to find the depressed function: 2 2432 432 2 2 4 2529820 25 4820 4820 0

Now find the zeros of the depressed function: 2 402. xxi  The zeros of the original function are 2 and 12. ii

52. The function has degree five, so there are five zeros. Since two zeros are 2 + 2i, 2 – 2i are also zeros. Divide 54327243264 xxxx by  2 22 ((22))((22))(48) xixixx  4328326464 xxxx  to find the depressed function. (Note it may be easier to divide by 2 48xx and then to divide the quotient by 2 48xx .) The quotient is 1, x  so –1 is a zero. The zeros of the original function are 1 and 22. i 

53. The function has degree three, so there are three zeros. The possible rational zeros are  1,2,4. Using synthetic division, we find that one zero is 1: 11144 104 1040

Now find the zeros of the depressed function: 2 402.xx

The solution set is {–2, 1, 2}.

54. The function has degree three, so there are three zeros. The possible rational zeros are 13 22,1,,2,3,6. 

  

Using synthetic division, we find that one zero is 12 : 1221126 106 20120

Now find the zeros of the depressed function: 2 21206. xx

The solution set is  12,6.

55. The function has degree three, so there are three zeros. The possible rational zeros are 1133 ,,,1,,3. 4242     Using synthetic division, we find that one zero is –1: 14073 443 4430

Now find the zeros of the depressed function: 2 4430(23)(21)0 xxxx 32 or 12. xx

The solution set is  12,1,32

56. The function has degree three, so there are three zeros. The possible rational zeros are  1,2,3,6. Using synthetic division, we find that one zero is –3: 31186 366 1220

xx x    

Now find the zeros of the depressed function: 2 2 220 2(2)4(1)(2) 212 13. 2(1)2

The solution set is  3,13

57. The function has degree four, so there are four zeros. The possible rational zeros are 1,2. Using synthetic division, we find that one zero is 2: 211112 2222 11110

Now find the zeros of the depressed function:

32 10. xxx Again using synthetic division, we find that one zero is –1: 11111

101 1010

The zeros of the depressed function

2 10 x  are .i The solution set is  1,2, i 

58. The function has degree four, so there are four zeros. The possible rational zeros are  1,2,  3,4,6,12. Using synthetic division, we find that one zero is 2: 11113112 101312 1013120

Now find the zeros of the depressed function:

3 13120.xx Again using synthetic division, we find that one zero is –1: 1101312 1112 11120

2 120(4)(3)04xxxxx  or x = 3. The solution set is  3,1,1,4 .

59. The only possible rational roots are 1,2. None of these satisfies the equation.

60. The only possible rational roots are  5 1 33,1,,5. None of these satisfies the equation.

61. 32 (1)16(1)2821 f 

32 (2)26(2)284 f 

Because the sign changes, there is a real zero between 1 and 2. The zero is approximately 1.88.

62. 32 (1)13(1)3(1)76 f  32 (2)23(2)3(2)77 f 

Because the sign changes, there is a real zero between 1 and 2. The zero is approximately 1.60.

63. 1 101 x x  is the x-intercept. There is no y-intercept. The vertical asymptote is the y-axis (x = 0). The horizontal asymptote is y = 1. 1 ()1() and(),fxfxfx x  so there are no symmetries. Testing the intervals (,0),  and (0,),  we find that the graph is below the horizontal asymptote on (,0)  and above the horizontal asymptote on (0,). 

64. 2 02 x x x  is the x-intercept. There is no y-intercept. The vertical asymptote is the y-axis (x = 0). The horizontal asymptote is y = –1.

2 ()() and(), fxxfxfx x   so there are no symmetries. Testing the intervals (,0), and (0,), we find that the graph is above the horizontal asymptote on (0,)  and below the horizontal asymptote on (,0).  (continuedonnextpage)

65. 2 00 1 x x x  is the x-intercept.

2 0 00 01 y  is the y-intercept. The vertical asymptotes are x = 1 and x = –1. The horizontal asymptote is the x-axis.

2 ()() and(), ()1 fxxfxfx x  so there is symmetry with respect to the origin. Testing the intervals (,1),(1,0), (0,1),  and (1,),  we find that the graph is above the x-axis on (,1)(0,)   and below the x-axis on (1,0)(0,1). 

Exercise 65

66.

Exercise 66

2 2 9 03 4 x x x  are the x-intercepts.

2 2 0999 44 04 y  is the y-intercept. The vertical asymptotes are x = 2 and x = –2. The horizontal asymptote is y = 1.

2 ()9 ()() ()1 x fxfx x  the function is even and the graph is symmetric with respect to the y-axis. Testing the intervals (,2),  (2,2), and (3,), we find that the graph is above the horizontal asymptote on (2,2) and below the al asymptote on (,2)(2,).  

67. 3 2 00 9 x x x  is the x-intercept.

3 2 0 00 09 y  is the y-intercept. The vertical asymptotes are x = 3 and x = –3. There is no horizontal asymptote.

2 () ()() but()(), ()9 fxxfxfxfx x  so there is symmetry with respect to the origin.

3 2 9, 9 x xx x  so the oblique asymptote is y = x.

Testing the intervals (,3),(3,0),(0,3),  and (3,),  we find that the graph is above the oblique asymptote on (3,0) (3,)   and below the oblique asymptote on (,3)(0,3).  

68.

2 1 01 28 x x xx   is the x-intercept.

2 0111 88 02(0)8 y   is the y-intercept. The vertical asymptotes are x = 4 and x = –2. The horizontal asymptote is the x-axis.

2 ()1 ()() and(), ()2()8 fxxfxfx xx   so there are no symmetries. Testing the intervals (,2),(2,1),(1,4),  and (4,),  we find that the graph is above the x-axis on (2,1)(4,))   and below the x-axis on (,2)(1,4).   (continuedonnextpage)

69. 4 2 00 4 x x x  is the x-intercept.

4 2 0 00 04 y  is the y-intercept. The vertical asymptotes are x = 2 and x = –2. There is no horizontal asymptote.

4 2 () ()() ()4 fxxfx x  , so the function is even, and the graph is symmetric with respect to the y-axis.

Testing the intervals (,2),  (2,0),(0,2), and (2,), we find that the graph is above the x-axis on (,2)(2,)   and below the x-axis on (2,0)(0,2). 

70. 2 2 6 02 12 xx x xx   is the x-intercept.

2 2 00611 22 0012 y   is the y-intercept. The vertical asymptote is x = 4. The horizontal asymptote is y = 1.

2 2 ()()6 ()() and ()()12 xx fxfx xx   (),fx  so there are no symmetries.

Testing the intervals (,3),(3,4),  and (4,),  we find that the graph is above the horizontal asymptote on (4,)  and below the horizontal asymptote on (,3)(3,4).  

71.       2320xxx

The real zeros are x = –2, x = 2, and x = 3. So, the intervals to be tested are   ,2,  2,2,2,3, and   3,. 

Interval Test point

Value of   Px Result

  ,2 –3 –30 –

 2,2 0 12 +

 2,3 2.5 –1.125 –

  3,  4 12 +

Solution set:   ,22,3 

72.  2 1120xxx

The real zeros are x = –2, x = –1, and x = 1. So, the intervals to be tested are   ,2,

 2,1,1,1, and   1,. 

Interval Test point

Value of   Px Result

  ,2 –3 16 +

 2,1 –1.5 –0.3125 –

 1,1 0 –2 –

  1,  2 36 +

Solution set:      ,211,  

73. 322 32 4774 5730 xxxx xxx

The possible rational zeros are 1,3

Using synthetic division, we find that x = 3 is a zero.

31573 363 1210

322 2 573321 31 xxxxxx xx

The zeros are x = 1 and x = 3. So, the intervals to be tested are

,1,1,3, and

Interval Test point Value of

3,. 

3,  4 9 +

Solution set: 

74. 322 32 24832 5280 xxxxx xxx 

The possible rational zeros are 1,2,4,8. 

Using synthetic division, we find that x = 2 is a zero. 21528 268 1340

322528234 241 xxxxxx xxx 

The zeros are x = –1, x = 2 and x = 4. So, the intervals to be tested are 

,1,1,2,

 2,4 and

4,. 

Interval Test point Value of

Px Result

1,2 0 8 +

2,4 3 –4 –

Solution set:

13 0 2 xx x   

numerator:     1301,3xxxx  denominator : 202xx

The line x = –2 is the vertical asymptote of   , Rx so –2 is not included in the intervals to be tested. The boundary points are –3, –2, and 1. The intervals are   ,3,   3,2,   2,1, and   1,. 

Interval Test point Value of   Rx Result

,3 –5 –4 –

1,  2 5 4 +

Solution set:     3,21,  

76.  2 14 0 2 xx x  

numerator:  2 1401,4xxxx 

denominator : 202xx

The line x = 2 is the vertical asymptote of   , Rx so 2 is not included in the intervals to be tested. The boundary points are –1, 2, and 4. The intervals are   ,1,   1,2,   2,4, and   4,. 

Interval Test point

Value of   Rx Result  ,1 –2 3 2 +   1,2 0 2 +   2,4 3 –16 –

  4,  5 12 +

Solution set:    12,4  (Remember to include –1 in the solution set because it is a zero.)

77.

2323 330 44 2334 15 00 44 xx xx xxx xx

numerator: 15015xx 

denominator : 404xx

The line x = –4 is the vertical asymptote of   , Rx so –4 is not included in the intervals to be tested. The boundary points are –15 and –4. The intervals are

,15,15,4

and

Interval Test point Value of

Rx Result

–12 –

Solution set: 

,154, 

2323 0

numerator: 12012xx 

denominator :   400,4xxxx

The lines x = –4 and x = 0 are the vertical asymptotes of   , Rx so –4 and 0 are not included in the intervals to be tested. The boundary points are –12, –4, and 0. The intervals are     ,12,12,4,   4,0, and   0,. 

  ,12 –13 1 117 +

 12,4 –6 1 2 –

 4,0 –3 3 +

 0,  1 13 5 –

Solution set:     ,124,0 

Applying the Concepts

79. ;1243;3(5)15ykxkky 

80. 12 ;412;3 34 kk pkp q 

81. 222 ;2025;3(5)45sktkks 

82. 222 192 ;3192;124 8 kk ykx xx 

83. The maximum height occurs at the vertex, 2020 (100,1000). 2(110)2(110) f

The maximum height is 1000. To find where the missile hits the ground, solve 2 1 2000 or 200. 10 xxxx 

The missile hits the ground at x = 200.

84. Let x = the length of one piece, and the area of the square formed by that piece is 2 x . Then 20 x  the length of the other piece, 2 (20) x  the area of the square formed by that piece. The total area is 222(20)240400. xxxx  The minimum is at the x-coordinate of the vertex, 40 10. 2(2)  Each piece must be 10 cm long.

85. The area of each section is 400 square feet. Since the width is x, 400 y x  The total amount of fencing needed is 2400 4. x x  Using a graphing calculator, we find that this is a minimum at 24.5 feet. x  So 400 16.3 feet. 24.5 y  The dimensions of each pen should be approximately 24.5 ft by 16.3 ft. (continuedonnextpage)

(continued)

86. Since x = the width, 400 y x  Then 2400

6 y x  the length of the heavy fencing.

The farmer needs 2400 2 x x  feet of heavy fencing, so it will cost 240012,000 5210. xx xx 

The low fencing will cost 3(2)6. xx  The total cost of the fencing is 12,000 16. x x 

Using a graphing calculator, we find this is a minimum when 27.4 x  .

87. a.

So 400 14.6 27.4 y  . The cost will be minimized when the dimensions of each pen are 27.4 ft by 14.6 ft.

2 2 2 425800105,000 0 50 271503350 0 50 xx x xx x

   

numerator:     2715033500 xx  150350 73 , xx

denominator:  2 50050 xx 

The line x = –50 is the vertical asymptote of   , Rx so –50 is not included in the intervals to be tested. In the context of this problem, x > 0, so it is not necessary to test negative values. The boundary points are 0, 150 7 , and 350 3 The intervals are

150150350 773 0,,,, and   350 3 ,. 

Interval Test point

Value of   Rx Result

150 7 0, 10 128 9 –  150350 73 , 50 8 +   350 3 ,  200 –10 –

The power absorbed by the circuit is greater than 42 for 150350 73 x 

The maximum occurs at x = 50.

88. a. The total area of the paper is 100 square inches. If the width of the paper is x, then the length is 100 x . The width of the printed area is x –2(1) = x – 2, and the length of the printed area is   1003100 2 23. xx  The area of the printed region is    100 23. x x (continuedonnextpage)

We want to solve    100 2356. x x 

2 100 2356 100 23560 200 3106560 200 3500 350200 0 x x x x x x x x xx x

numerator: 2 3502000 xx 

20 3 10320010,xxx 

denominator: x = 0

Test intervals: 

 2020 33 0,,,10,10,  

Interval Test point Value of   Rx Result

 20 3 0,   5 –5 –20 3 ,10   8 1 +   10,  12 8 3 –The width of the paper should be between 20 3 6.67  in. and 10 in.

b. The area of the printed region is given by    100 23. x Ax [0, 30] by [0, 100]

The maximum value of x, the width, is about 8.16 in., so the maximum length is 100 8.16 12.25 in.  The maximum area of the printed region is about 57.01 in2

89. a. The revenue is 24x. Profit = revenue – cost, so 2 3 ()241503.9 1000 Pxxxx 

2 3 20.1150. 1000 xx 

b. The maximum occurs at the x-coordinate of the vertex:  3 1000 20.1 3350 2  c. 2 3 1000 3.9150 () () 3150 3.9 1000 Cxxx Cx xx x x   

90. a. 19,340 19,340 feet3.66 miles 5280 69.1 11.59 inches 3.662.3 p   

b. 69.1 0069.1, 2.3 a   which is impossible. So there is no altitude at which the pressure is 0.

91. 280407;7(35)$245 kks 

92. 22 2 1 ;2530 36 1 (66)121 feet 36 dkskk s  

93. 2 , ki I d  where I = illumination, i = intensity, and d = distance from the source. The illumination 6 inches from the source is 6 2 300 , 6 k I  while the illumination x inches from the source is 2 300 x k I x  . 6 2 IIx

Chapter 2 Practice Test A

1. 2 2 6(6)4(1)(2) 620 2(1) xxx   628 37 2 x   are the x-intercepts

94. 1.2 ;1.2295 295 1.2 (310)1.26 cubic feet. 295 VkTkk V

95. ;309000 300 kk Ik R

a. 9000 36 amp 250 I 

b. 9000 60150 ohms R R

96. 22 2 4 ;201440 12 14403 24.94 tons 10

97. ()Rkipi 

a. 6 255(0.15)(20,000)(0.85)(20,000) 1 510 200,000 k  

b. 1 (10,000)(10,000) 200,000 500 people per day R  

c. 2 1 95(20,000) 200,000 20,00019,000,0000 (1000)(19,000)01000 or 19,000 xx xx xxx x

98. 2 12 () Fkqqd  . If the distance is quadrupled, then the force is 1212 new 22 . (4)16 Fkqqkqq dd  So the original force is divided by 16: 96 6 16  units.

3. The vertex is at , 22 bb f aa

1414 2(7)2(7),(1,10).

4. The denominator is 0 when x = –4 or x = 1. The domain is (,4)(4,1)(1,). 

5. Using either long division or synthetic division, we find that the quotient is 2 43xx and the remainder is 0. 21256 286 1430

7. The function has degree three, so there are three zeros. Since 2 is a zero, use synthetic division to find the depressed function: 22288 448 2240

Now find the zeros of 2 224: xx 22 2242(2)2(2)(1) xxxxxx  the zeros are x = –2 and x = 1. The zeros of the original function are –2, 1, 2.

9. Using synthetic division to find the remainder, we have (2)53. P  2157917 262670 13133553

10. The function has degree three, so there are three zeros. There are two sign changes in ()fx , so there are either 2 or 0 positive zeros. There is one sign change in ()fx , so there is one negative zero. The possible rational zeros are  1,2,4,5,10,20.  Using synthetic division, we find that –2 is a zero: 215420 21420 17100

Now find the zeros of the depressed function 22710:710(5)(2) xxxxxx  the zeros are x = 2 and x = 5. The zeros of the original function are –2, 2, 5.

11. The function has degree four, so there are four zeros. Factoring, we have 432 15 xxx  22(15)xxx  0 is a zero of multiplicity

2. Now find the zeros of 2 15: xx 2 114(1)(15) 161 2(1)2 x   

The zeros of the original function are 0 and 161 2 

12. The factors of the constant term, 9, are  1,3,9 , while the factors of the leading coefficient are 1,2. The possible rational zeros are 139 ,1,,3,,9. 222   

13. () as ;() as . fxxfxx

14. 22 2 ()(4)(2) (2)(2)(2) fxxx xxx   the zeros are –2 (multiplicity 3) and 2 (multiplicity 1).

15. There are two sign changes in ()fx , so there are either two or zero positive zeros. There is one sign change in ()fx , so there is one negative zero.

16. The zeros of the denominator are x = 5 and x = –4, so those are the vertical asymptotes. The degree of the numerator equals the degree of the denominator, so the horizontal asymptote is y = 2.

17.

numerator: 21407 xx  denominator:

4204,2xxxx 

The intervals to be tested are   ,7,

 7,4,4,2, and   2,. 

Interval Test point Value of   Rx Result

Solution set:

7,42,   18. 2 2 8 63 2 kx y t k k     2 312 4 3 y 

19. The minimum occurs at the x-coordinate of the vertex: 30 15 2(1)  thousand units.

20. (172)(82) Vxxx 

Chapter 2 Practice Test B

1. B 2 2 530 554(1)(3) 513 2(1)2 xx x 

2. D. The graph of 2 ()4(2)fxx is the graph of 2 () fxx  shifted two units right, reflected across the x-axis, and then shifted 4 units up

3. A 1212 ,, 222(6)2(6) (1,11) bb ff aa

4. B. The denominator is 0 when x = –3 or x = 2.

5. D 31086 393 1313

6. C 433 ()2(2). Pxxxxx  So the zeros are 0 (multiplicity 3) and –2 (multiplicity 1).

7. C 33266130 95130 317100 The zeros of the depressed function 2 31710 xx are 2 and 5. 3 xx

8. B 2 32 32 2 2 534 2310211712 1015 617 69 812 812 0 xx xxxx xx xx xx

9. C 31471015 33126 114221 (3)21. P 

10. A. The polynomial has degree three, so there are three zeros. The possible rational zeros are  1,2,3,4,6,12.  Using synthetic division, we find that one zero is 3: 311812 31212 1440

The zeros of the depressed function 2 44xx  are x = 2 (multiplicity 2).

11. A. The polynomial has degree three, so there are three zeros. Factoring, we have 32230(30)(6)(5) xxxxxxxxx 

12. D. The factors of the constant term, 60, are  1,2,3,4,5,6,10,12,15,20,   30,60. Since the leading coefficient is 1, these are also the possible rational zeros.

13. C 14. B 22 2 (1)(1)(1)(1)(1). xxxxx 

15. C. There are three sign changes in ()Px , so there are 3 or 1 positive zeros. There are two sign changes in ()Px , so there are 2 or 0 negative zeros.

16. C. The zeros of the denominator are x = 3 and x = –4, so those are the vertical asymptotes. The degree of the numerator equals the degree of the denominator, so the horizontal asymptote is y = 1.

17. C

numerator: 707

18. B

19. B. The minimum occurs at the x-coordinate of the vertex: 24 12. 2(1) 

20. D. (102)(122) Vxxx 

Cumulative Review Exercises Chapters 1–2

1.   2 2121 22 123513

2.  1212 , 22 2(8)5(3),3,4 22 Mxxyy        

3.   2 2 028042 404 or 202 002088 xxxx xxxx f 

The x-intercepts are ( 2, 0) and (4, 0), and the y-intercept is (0, 8).

4. Write the equation in slope-intercept form: 1 360362 3 xyyxyx 

The slope is 1 3 , and the y-intercept is (0, 2). 3(0)60606,xxx  so the x-intercept is (6, 0).

5. Write the equation in slope-intercept form: 1 26623 2 xyxyxy 

The slope is 1 2 , and the y-intercept is (0, 3).

The x-intercept is x = 2(0) 6 = 6. (continuedonnextpage)

6. 22 (2)(3)16 xy



7. 22 22 22 2440 (2)(4)4 (21)(44)414 xyxy xxyy xxyy

  22 (1)(2)9 xy

The center is (–1, 2). The radius is 3.

8. 23(1)35yxyx 

9. 25 235 33xyyx  the slope is 2 3 2211 3(1)333yxyx 

10. 3 230. 2 xx

The domain is 33 ,, 22

11. 4202. xx  The domain is (,2). 

12. 2 2 2 22 22 (2)(2)2(2)311; (3)32(3)36; ()()2()3 2223 2(1)23 f f fxhxhxh xxhhxh xhxhh  

  

   222 2 22 ()() (2(1)23)(23) 2(1)22 222222 22 fxhfx h xhxhhxx h hxxhh h hxxxhhhhxh hh hx

13. a.   2 1 fgxx

b.     2 11gfxxx

c.   4 ffxxx 

   2 2421122ggxxxx

14. a. (1)3(1)25;(3)4(3)111; (4)6 ff f  

15. 23.yx Interchange x and y, and then solve for y 1 313 23(). 222 x xyyxfx  

16. a. Shift the graph of yx  two units left.

b. Shift the graph of yx  one unit left, stretch vertically by a factor of two, reflect about the x-axis, and then shift up three units.

17. The factors of the constant term, –6, are  1,2,3,6 , and the factors of the leading coefficient, 2, are 1,2. The possible rational zeros are 13 22,1,,2,3,6.

18. a.

b. 2 2410 xx

Use the quadratic formula to find the zeros.

20. a.

The intervals to be tested are   ,1, 1,3, and   3,.  From the graph in part (a), we see that the solution set is 1,3.

b.   2 2 120 1201,2 xx xxx  

The intervals to be tested are   ,2, 2,1, and   1,.  From the graph in part (a), we see that the solution set is     2,11,2,.  

Interval Test point

21. a.

b. 2 2 1 0 4 x x 

numerator: 2 101xx denominator: 2 402xx

The intervals to be tested are   ,2,    2,1,1,1,1,2, and   2,.  From the graph in part (a), we see that the solution set is      ,21,12,.  

22. Since one zero is 1 + i, another zero is 1 i So 2 ((1))((1))22 xixixx  is a factor of ()fx . Now divide to find the other factor:

432 2 ()3224 ((1))((1))(2) ((1))((1))(2)(1) fxxxxx xixixx xixixx

the zeros of () are1,2,1,1. fxii

23. ;643;399ykxkky 

24. Profit = revenue – cost 2 2 150(0.021003000) 0.02503000. xxx xx   The maximum occurs at the vertex: 5050 , 2(0.02)2(0.02) (1250,$28,250).

25. 100.0248.8299025,000 0.249025,000 0.024925000 Now solve the depressed equation 2 0.024925000. xx 2 49494(0.02)(2500) 2(0.02) 4926014951 50 or 0.040.04 x x     2500 x  . There cannot be a negative amount of units sold, so another break-even point is 50.