Solutions for College Algebra 4th Us Edition by Ratti by 6alsm

Chapter 2 Graphs and Functions

2.1 The Coordinate Plane

2.1 Practice Problems

22 12121 22 22 2602 8268

22 23131 22 22 1652 5334

22 33232 2 2 22 1250 3534

Yes, the triangle is an isosceles triangle.

We are asked to find the distance between the points A(60, 0) and B(0, 60).

22 222 (,)600060 6060260 60284.85 ft

6.  21 56113 ,, 2222 M

2.1 Concepts and Vocabulary

1. A point with a negative first coordinate and a positive second coordinate lies in the second quadrant.

2. Any point on the x-axis has second coordinate 0.

3. The distance between the points   11 , Pxy and   22 , Qxy is given by the formula

 22 2121 ,.dPQxxyy

4. The coordinates of the midpoint M(x, y) of the line segment joining   11 , Pxy and   22 , Qxy are given by  1212 ,,. 22 xxyy xy

5. True

6. False. The point (7, 4) is 4 units to the right and 6 units below the point (3, 2).

7. False. Every point in quadrant II has a negative x-coordinate.

8. True.

2.1 Building Skills

9. (2, 2): Q1; (3, 1): Q4; ( 1, 0): x-axis ( 2, 5): Q3; (0, 0): origin; ( 7, 4): Q2 (0, 3): y-axis; ( 4, 2): Q2

10. a. Answers will vary. Sample answer: (2,0), (1,0),(0,0),(1,0),(2,0) The y-coordinate is 0.

The set of all points of the form (x, 1) is a horizontal line that intersects the y-axis at 1.

11. a. If the x-coordinate of a point is 0, the point lies on the y-axis.

The set of all points of the form (–1, y) is a vertical line that intersects the x-axis at –1.

12. a. A vertical line that intersects the x-axis at –3.

b. A horizontal line that intersects the y-axis at 4.

13. a. 0 y  b. 0 y 

c. 0 x  d. 0 x 

14. a. Quadrant III b. Quadrant I c. Quadrant IV d. Quadrant II In Exercises 15 24, use the distance formula,  22 2121 dxxyy  and the midpoint formula, (,),1212 22 xxyy xy       .

15. a. 222 (22)(51)44 d 

b.  2215 22,2,3 M     

16. a. 222 (23)(55)(5)5 d 

b. 3(2)55,0.5,5 22 M   

17. a. 22 22 (2(1))(3(5)) 3213 d  

b. 125(3),0.5,4 22 M   

18. a. 22 22 (7(4))(91) (3)(10)109 d  

b.  4(7)1(9),5.5,4 22 M   

19. a. 22 22 (3(1))(6.51.5) 4(8)8045

d  

b. 131.5(6.5),1,2.5 22 M    

d 



20. a. 22 22 (10.5)(10.5) 510 (0.5)(1.5)2.5 22

b.  0.510.5(1),0.75,0.25 22 M

21. a.  2 22 22(54)11 d 

b.  2245 22,2,4.5 M     



22. a. 22 2 (()())() (2)2 dvwvwtt ww



b.  ()() 22,, vwvwtt Mvt   

23. a.



22 2222 2222 2 ()() 22 24222 2()2 dkttk ktkttktk ttkkttkk tktk

 



b. , 22 tkkt M 

24. a.

 22 2222 22 22 2 ()() 22 242 22 2()2 dnmmn nmnmmmnn mmnn mmnn mnmn  





b. ()() , 22 , 22 mnnm M mnnm     

25. P = (–1, –2), Q = (0, 0), R = (1, 2) 

dPQ

dQR

dPR





 22 22 22 22 (,)0(1)0(2)5 (,)(10)(20)5 (,)1(1)2(2) 242025





Because d(P, Q) + d(Q, R) = d(P, R), the points are collinear.

26. P = (–3, –4), Q = (0, 0), R = (3, 4)

dPQ

dQR

dPR

27. P = (4, –2), Q = (1, 3), R = (–2, 8)  22 22 (,)143(2)34 (,)(21)(83)34 dPQ dQR

dPR





 22 22 (,)248(2) (6)10136234





Because d(P, Q) + d(Q, R) = d(P, R), the points are collinear.

28. It is not possible to arrange the points in such a way so that d(P, Q) + d(Q, R) = d(P, R), so the points are not collinear.

29. P = (–1, 4), Q = (3, 0), R = (11, –8) 

dPQ

dQR

dPR





 22 22 22 22 (,)3(1)0442 (,)(113)((8)0)82 (,)11(1)84 (12)(12)288122





Because d(P, Q) + d(Q, R) = d(P, R), the points are collinear.

30. It is not possible to arrange the points in such a way so that d(P, Q) + d(Q, R) = d(P, R), so the points are not collinear.

31. It is not possible to arrange the points in such a way so that d(P, Q) + d(Q, R) = d(P, R), so the points are not collinear.

32. P = (1, 7), Q = (–3, 7.5), R = (–7, 8)   22 22 22 22 (,)317.5716.25 (,)(7(3))(87.5) 16.25 (,)7187 (8)165216.25 dPQ dQR

dPR







Because d(P, Q) + d(Q, R) = d(P, R), the points are collinear.



 22 22 22 22 (,)0(3)0(4)255 (,)(30)(40)255 (,)3(3)4(4) 6810010









Because d(P, Q) + d(Q, R) = d(P, R), the points are collinear.

33. First, find the midpoint M of PQ  4008 22,2,4 M   

Now find the midpoint R of PM.    42 04 22,3,2 R     

Finally, find the midpoint S of MQ

 2048 22,1,6 S     

Thus, the three points are ( 3, 2), ( 2, 4), and ( 1, 6).

34. First, find the midpoint M of PQ

412 816 22,4,4 M

Now find the midpoint R of PM

44 84 22,2,0 R

Finally, find the midpoint S of MQ.

412 416 22,10,8 S

Thus, the three points are ( 2, 0), (4, 4), and (10, 8).

35.

dPQ

dQR

dPR

22 22 22 (,)1(5)4517 (,)4(1)1432 (,)4(5)1517

The triangle is isosceles.

36.

dPQ

dQR

dPR



22 22 22 (,)63625 (,)165652 (,)13525



The triangle is an isosceles triangle.

37.

38.

dPQ dQR

dPR



22 22 22 (,)0(4)7817 (,)305713 (,)3(4)5810





The triangle is scalene.

dPQ

dQR

dPR



22 22 22 (,)161672 (,)5(1)3(1)42 (,)5636130





The triangle is scalene.

dPQ dQR dPR



22 22 22 (,)909(1)145 (,)591(9)229 (,)501(1)29



The triangle is scalene.

40.



22 22 22 (,)4(4)5465 (,)042565 (,)0(4)24213 dPQ dQR dPR





The triangle is isosceles. 41.



 

 22 22 (,)111(1)22 (,)3(1)31 32313231 822 dPQ dQR

 22 (,)313(1) 32313231 822

The triangle is equilateral.

(continued)

All the sides are equal, so the quadrilateral is either a square or a rhombus. Now find the length of the diagonals:

22 22 (,)14711(12)172 (,)22(1)43172

The diagonals are equal, so the quadrilateral is a square.

44. First find the lengths of the sides:

22 22 (,)9811(10)2 (,)8912(11)2

dRS dSP

22 22 (,)7811(12)2 (,)8710(11)2

All the sides are equal, so the quadrilateral is either a square or a rhombus. Now find the length of the diagonals.

dPR dQS

22 22 (,)8812(10)2 (,)7911(11)2

The diagonals are equal, so the quadrilateral is a square.





45.  22 2 22 2 522(1) 449 541325413 04120(6)(2) 2 or 6 x xx xxxx xxxx xx











46.  22 2 2 2 2 132(10)(3) 14469 6153 1696153 06160(8)(2) 8 or 2 y yy yy yy yyyy yy









47. P = (–5, 2), Q = (2, 3), R = (x, 0) (R is on the x-axis, so the y-coordinate is 0).

 22 22 (,)(5)02 (,)(2)(03) dPRx dQRx

 

   

 22 22 2222 22 (5)02(2)(03) (5)(02)(2)(03) 10254449 1029413 1416 8 7 xx xx xxxx xx x x 

The coordinates of R are 8 ,0 7   

48. P = (7, –4), Q = (8, 3), R = (0, y) (R is on the y-axis, so the x-coordinate is 0).  22 22 (,)07(4) (,)(08)(3) dPRy dQRy

22 22 07(4) (08)(3) y y 

 2 2 22 49(4)64(3) 498166469 865673 148 4 7 yy yyyy yy y y

The coordinates of R are 4 0, 7 

2.1 Applying the Concepts 49.

50.  20102016308324 , 22 2013,316 M

The population is 2013 was about 316 million.

55. 2014 is the midpoint of the initial range, so

20122016326425 , 22 2014,375.5

Americans spent about $376 billion on prescription drugs in 2014.

56. 2014 is the midpoint of the initial range, so

2012201624973696 , 22 2014,3096.5

There were about 3097 million Internet users in 2014.

57. Percentage of Android sales in June 2013: 51.5%

58. Percentage of iPhone sales in December 2012: 49.7%

59. Android sales were at a maximum in June 2014.

60. iPhone sales were at a maximum in December 2012.

61. Denote the diagonal connecting the endpoints of the edges a and b by d. Then a, b, and d form a right triangle. By the Pythagorean theorem, 222abd  . The edge c and the diagonals d and h also form a right triangle, so 222 cdh  . Substituting 2 d from the first equation, we obtain 2222 abch  .

62. a.

b. 22 22 (,)(800200)(1200400) 1000 (,)(2000800)(3001200) 1500

The distance traveled by the pilot = 1000 + 1500 = 2500 miles.

c. 22 (,)(2000200)(300400) 3,250,00032510000 10032510051350013 1802.78 miles

63. First, find the initial length of the rope using the Pythagorean theorem: 22 241026 c  . After t seconds, the length of the rope is 26 – 3t. Now find the distance from the boat to the dock, x, using the Pythagorean theorem again and solving for x: 222 22 22 2 (263)10

So we have 5 2.50 2 x x   and 4 5.5 2 y   7 y 

The coordinates of D are (0, 7).

6761569100

5761569

5761569 tx ttx ttx ttx 





2.1 Beyond the Basics

64. The midpoint of the diagonal connecting (0, 0) and (a + b, c) is 22,. abc     The midpoint of the diagonal connecting (a, 0) and (b, c) is also 22,. abc     Because the midpoints of the two diagonals are the same, the diagonals bisect each other.

65. a. If AB is one of the diagonals, then DC is the other diagonal, and both diagonals have the same midpoint. The midpoint of AB is  2534 22,3.5,3.5

     . The midpoint of DC = 38 (3.5,3.5),. 22 xy     

So we have 3 3.54 2 x x   and 8 3.5 2 y   1 y 

The coordinates of D are (4, –1).

b. If AC is one of the diagonals, then DB is the other diagonal, and both diagonals have the same midpoint. The midpoint of AC is

2338 22,2.5,5.5 

   . The midpoint of DB = 54 (2.5,5.5),. 22 xy   

c. If BC is one of the diagonals, then DA is the other diagonal, and both diagonals have the same midpoint. The midpoint of BC is  5348 22,4,6      . The midpoint of DA is 23 (4,6),. 22 xy      So we have 2 46 2 x x   and 3 69 2 y y  

The coordinates of D are (6, 9).

66. The midpoint of the diagonal connecting (0, 0) and (x, y) is , 22 xy     . The midpoint of the diagonal connecting (a, 0) and (b, c) is 22,. abc      Because the diagonals bisect each other, the midpoints coincide. So 22 xab   xab  , and 22 yc yc Therefore, the quadrilateral is a parallelogram.

67. a. The midpoint of the diagonal connecting (1, 2) and (5, 8) is  1528 22,3,5.     

The midpoint of the diagonal connecting (–2, 6) and (8, 4) is 2864 22,(3,5).      Because the midpoints are the same, the figure is a parallelogram.

b. The midpoint of the diagonal connecting (3, 2) and (x, y) is 32 22,. xy    The midpoint of the diagonal connecting (6, 3) and (6, 5) is (6, 4). So 3 69 2 x x   and 2 46 2 y y  

68. Let P(0, 0), Q(a, 0), R(a + b, c), and S(b, c) be the vertices of the parallelogram.

22 (0)(00). PQRSaa   2 2

22 ()(0) . QRPSabac bc 



The sum of the squares of the lengths of the sides = 222 2(). abc  22 (,)(). dPRabc   2 2 (,)(0). dQSabc 

The sum of the squares of the lengths of the diagonals is     ()()2222 abcabc  222222 22 aabbcaabbc  222222 2222(). abcabc 

69. Let P(0, 0), Q(a, 0), and R(0, b) be the vertices of the right triangle. The midpoint M of the hypotenuse is 22,. ab

70. Let P(0, 0), Q(a, 0), and R(0, a) be the vertices of the triangle.

Using the Pythagorean theorem, we have 2222222 12 2 2 caacaca acc

71. Since ABC is an equilateral triangle and O is the midpoint of AB, then the coordinates of A are ( a, 0).

AB = AC = AB = 2a. Using triangle BOC and the Pythagorean theorem, we have 2 22222 22222 2 433 BCOBOCaaOC aaOCaOCOCa

Thus, the coordinates of C are   0,3a and the coordinates of D are   0,3. a

72.

To show that M is the midpoint of the line segment PQ, we need to show that the distance between M and Q is the same as the distance between M and P and that this distance is half the distance from P to Q

(continuedonnextpage)

(continued)

Similarly,

2.1 Critical Thinking/Discussion/Writing

73. a. y-axis

b. x-axis

74. a. The union of the x- and y-axes

b. The plane without the x- and y-axes

75. a. Quadrants I and III

b. Quadrants II and IV

76. a. The origin

b. The plane without the origin

77. a. Right half-plane

b. Upper half-plane

78. Let (x, y) be the point. The point lies in if

Quadrant I x > 0 and y > 0

Quadrant II x < 0 and y > 0

Quadrant III x < 0 and y < 0

Quadrant IV x > 0 and y < 0

2.1 Getting Ready for the Next Section

79. a. 22 22 111121 224442 xy

22 22 2222 1 2244

2 222 2 2 121112 234913

2222 22 124122 3491625

81. a. 3 223 110 2323 xy

b. 3 443 110 4343 xy

82. a.

12 12 1212 112 xy xy

32 32 112 3232 xy xy

83. 2 222 2 6 663 2 69 xxxx xx

2 2 22 2 8 884 2 816 xxxx xx

2 22239 333 24yyyyyy

2 22525 55 24yyyy

2 2 22 24 aa xaxxax

2 2 22 24 yy xxyxxy

2.2 Graphs of Equations

2.2 Practice Problems

1. 2 1 yx

x y = x2 + 1 (x, y)

2 y = ( 2)2 + 1 ( 2, 3)

1 y = ( 1)2 + 1 ( 1, 0)

0 y = (0)2 + 1 (0, 1)

1 y = (1)2 + 1 (1, 0)

2 y = (2)2 + 1 (2, 3)

2. To find the x-intercept, let y = 0, and solve the equation for x: 2 0232 xx 

 1 0212 or 2 2 xxxx  . To find the y-intercept, let x = 0, and solve the equation for y:  2 203022.yy 

The x-intercepts are 1 2 and 2; the y-intercept is 2.

3. To test for symmetry about the y-axis, replace x with –x to determine if (–x, y) satisfies the equation.

 2 22211xyxy  , which is the same as the original equation. So the graph is symmetric about the y-axis.

4. x-axis:  3 223 , xyxy  which is not the same as the original equation, so the equation is not symmetric with respect to the x-axis.

y-axis:  2 323 , xyxy  which is the same as the original equation, so the equation is symmetric with respect to the y-axis.

origin: 23 23 , xyxy  which is not the same as the original equation, so the equation is not symmetric with respect to the origin.

5. 4277324ytt

a. First, find the intercepts. If t = 0, then y = 324, so the y-intercept is (0, 324). If y = 0, then we have

  

  42 42 22 2 077324 773240 8140 99409,9,2 tt tt tt tttti



So, the t-intercepts are ( 9, 0) and (9, 0). Next, check for symmetry.

t-axis: 4277324ytt  is not the same as the original equation, so the equation is not symmetric with respect to the t-axis.

y-axis: 4277324ytt  4277324,ytt which is the same as the original equation. So the graph is symmetric with respect to the y-axis.

origin: 4277324ytt  4277324,ytt  which is not the same as the original equation. So the graph is not symmetric with respect to the origin. Now, make a table of values. Since the graph is symmetric with respect to the y-axis, if (t, y) is on the graph, then so is ( t, y). However, the graph pertaining to the physical aspects of the problem consists only of those values for t ≥ 0.

t y = t4 + 77t2 + 324 (t, y)

0 324 (0, 324)

1 400 (1, 400)

2 616 (2, 616)

3 936 (3, 936)

4 1300 (4, 1300)

5 1624 (5, 1624)

6 1800 (6, 1800)

7 1696 (7, 1696)

8 1156 (8, 1156)

9 0 (9, 0) (continuedonnextpage)

(continued)

2.2 Concepts and Vocabulary

1. The graph of an equation in two variables, such as x and y, is the set of all ordered pairs (a, b) that satisfy the equation.

2. If (–2, 4) is a point on a graph that is symmetric with respect to the y-axis, then the point (2, 4) is also on the graph.

3. If (0, 5) is a point of a graph, then 5 is a y- intercept of the graph.

c. The population becomes extinct after 9 years.

6. The standard form of the equation of a circle is ()()222 xhykr  (h, k) = (3, 6) and r = 10

The equation of the circle is 22 (3)(6)100. xy

7.  22 2136,2,1,6 xyhkr 

This is the equation of a circle with center (2, 1) and radius 6.

8. 22 46120xyxy 224612 xxyy Now complete the square: 2244691249 xxyy

 22 2325xy

This is a circle with center ( 2, 3) and radius 5.

4. An equation in standard form of a circle with center (1, 0) and radius 2 is  2 2 14xy .

5. False. The equation of a circle has both an 2 x -term and a 2 y -term. The given equation does not have a 2 y -term.

6. False. The graph below is an example of a graph that is symmetric about the x-axis, but does not have an x-intercept.

7. False. The center of the circle with equation  22 349xy is   3,4.

8. True

2.2 Building Skills

In exercises 9 14, to determine if a point lies on the graph of the equation, substitute the point’s coordinates into the equation to see if the resulting statement is true.

9. on the graph: (–3, –4), (1, 0), (4, 3); not on the graph: (2, 3)

10. on the graph: (–1, 1), (1, 4), 5 ,0 3    ; not on the graph: (0, 2)

11. on the graph: (3, 2), (0, 1), (8, 3); not on the graph: (8, –3)

12. on the graph: (1, 1), 1 2, 2    ; not on the graph: (0, 0), 1 3, 3    

13. on the graph: (1, 0),

2,3,2,3 ; not on the graph: (0, –1)

14. Each point is on the graph.

36.

For exercises 37–46, read the answers directly from the given graphs.

37. x-intercepts: –1, 1 y-intercepts: none symmetries: y-axis

38. x-intercepts: none y-intercepts: –1, 1 symmetries: x-axis

39. x-intercepts: ,0,  y-intercepts: 0 symmetries: origin

40. x-intercepts: , 22  y-intercepts: 2 symmetries: y-axis

41. x-intercepts: –3, 3 y-intercepts: –2, 2 symmetries: x-axis, y-axis, origin

42. x-intercepts: –2, 2 y-intercepts: –3, 3 symmetries: x-axis, y-axis, origin

43. x-intercepts: –2, 0, 2 y-intercepts: 0 symmetries: origin

44. x-intercepts: –2, 0, 2 y-intercepts: 0 symmetries: origin

45. x-intercepts: –2, 0, 2 y-intercepts: 0, 3 symmetries: y-axis

46. x-intercepts: 0, 3 y-intercepts: –2, 0, 2 symmetries: x-axis

51. To find the x-intercept, let y = 0, and solve the equation for x: 34(0)124. xx  To find the y-intercept, let x = 0, and solve the equation for y: 3(0)4123. yy  The x-intercept is 4; the y-intercept is 3.

52. To find the x-intercept, let y = 0, and solve the equation for x: 5 23(0)5. 2 xx To find the y-intercept, let x = 0, and solve the equation for y: 5 2(0)35. 3 yy  The x-intercept is 52 ; the y-intercept is 53

53. To find the x-intercept, let y = 0, and solve the equation for x: 0 15. 53 x x  To find the y-intercept, let x = 0, and solve the equation for y: 0 13. 53 y y  The x-intercept is 5; the y-intercept is 3.

54. To find the x-intercept, let y = 0, and solve the equation for x: 0 12. 23 x x  To find the y-intercept, let x = 0, and solve the equation for y: 0 13. 23 y y  The x-intercept is 2; the y-intercept is 3 .

55. To find the x-intercept, let y = 0, and solve the equation for x: 2 02. 1 x x x   To find the y-intercept, let x = 0, and solve the equation for y: 02 2. 01 y   The xintercept is –2; the y-intercept is –2.

56. To find the x-intercept, let y = 0, and solve the equation for x: 02 2. 01 xx  To find the y-intercept, let x = 0, and solve the equation for y: 2 02. 1 y y y   The xintercept is –2; the y-intercept is 2.

57. To find the x-intercept, let y = 0, and solve the equation for x: 2 0684 or xxx  2. x  To find the y-intercept, let x = 0, and solve the equation for y: 2 06(0)8 y  8. y  The x-intercepts are 2 and 4; the y-intercept is 8.

58. To find the x-intercept, let y = 0, and solve the equation for x: 2 05(0)66.xx  To find the y-intercept, let x = 0, and solve the equation for y: 2 0562 or yyy 

3. y  The x-intercept is 6; the y-intercepts are 2 and 3.

59. To find the x-intercept, let y = 0, and solve the equation for x: 22042. xx To find the y-intercept, let x = 0, and solve the equation for y: 22 042. yy  The x-intercepts are –2 and 2; the y-intercepts are –2 and 2.

60. To find the x-intercept, let y = 0, and solve the equation for x:  2 2 10913 2 or 4 xx xx  

To find the y-intercept, let x = 0, and solve the equation for y:  2 222 019198 822 yyy y  

The x-intercepts are –2 and 4; the y-intercepts are 22.

61. To find the x-intercept, let y = 0, and solve the equation for x: 2 093. xx  To find the y-intercept, let x = 0, and solve the equation for y: 2 903.yy The x-intercepts are –3 and 3; the y-intercept is 3.

62. To find the x-intercept, let y = 0, and solve the equation for x: 2 011. xx  To find the y-intercept, let x = 0, and solve the equation for y: 2 01 y  no solution. The x-intercepts are –1 and 1; there is no y-intercept.

63. To find the x-intercept, let y = 0, and solve the equation for x: (0)1 x  no solution. To find the y-intercept, let x = 0, and solve the equation for y: (0)1 y  no solution. There is no x-intercept; there is no y-intercept.

64. To find the x-intercept, let y = 0, and solve the equation for x: 22 011 xx  there is no real solution. To find the y-intercept, let x = 0, and solve the equation for y: 2 011.yy There is no x-intercept; the y-intercept is 1.

In exercises 65–74, to test for symmetry with respect to the x-axis, replace y with –y to determine if (x, –y) satisfies the equation. To test for symmetry with respect to the y-axis, replace x with –x to determine if (–x, y) satisfies the equation. To test for symmetry with respect to the origin, replace x with –x and y with –y to determine if (–x, –y) satisfies the equation.

65. 2 1 yx  is not the same as the original equation, so the equation is not symmetric with respect to the x-axis.

2 ()1yx 2 1 yx , so the equation is symmetric with respect to the y-axis.

2 ()1yx  2 1 yx  , is not the same as the original equation, so the equation is not symmetric with respect to the origin.

66. 22 ()11xyxy  , so the equation is symmetric with respect to the x-axis.

2 1 xy  is not the same as the original equation, so the equation is not symmetric with respect to the y-axis.

2 ()1xy  2 1 xy  is not the same as the original equation, so the equation is not symmetric with respect to the origin.

67. 3 yxx is not the same as the original equation, so the equation is not symmetric with respect to the x-axis.

3 () yxx 3 yxx

3 ()yxx  is not the same as the original equation, so the equation is not symmetric with respect to the y-axis.

33 () yxxyxx 

33 (), yxxyxx  so the equation is symmetric with respect to the origin.

68. 3 2 yxx is not the same as the original equation, so the equation is not symmetric with respect to the x-axis.

3 2()() yxx 3 2 yxx

3 2() yxx  is not the same as the original equation, so the equation is not symmetric with respect to the y-axis.

2()()233 yxxyxx 

2()233 yxxyxx  , so the equation is symmetric with respect to the origin.

69. 5242yxx is not the same as the original equation, so the equation is not symmetric with respect to the x-axis.

5()2()42yxx 5242yxx  , so the equation is symmetric with respect to the y-axis.

442 5()2()52 yxxyxx  is not the same as the original equation, so the equation is not symmetric with respect to the origin.

70. 32642 yxxx  is not the same as the original equation, so the equation is not symmetric with respect to the x-axis.

642 3()2()() yxxx 

32642 yxxx  , so the equation is symmetric with respect to the y-axis.

642 3()2()() yxxx  32642 yxxx  is not the same as the original equation, so the equation is not symmetric with respect to the origin.

71. 3253yxx is not the same as the original equation, so the equation is not symmetric with respect to the x-axis.

3()2()53yxx  3253yxx  is not the same as the original equation, so the equation is not symmetric with respect to the y-axis.

5353 3()2()32 yxxyxx  5353 (32)32 yxxyxx  , so the equation is symmetric with respect to the origin.

72. 2 2 yxx is not the same as the original equation, so the equation is not symmetric with respect to the x-axis.

2 2() yxx 2 2 yxx  , so the equation is symmetric with respect to the y-axis.

2 2() yxx  2 2 yxx is not the same as the original equation, so the equation is not symmetric with respect to the origin.

73. 22 22 ()2()121 xyxyxyxy  is not the same as the original equation, so the equation is not symmetric with respect to the x-axis.

22 22 ()2()121 xyxyxyxy  is not the same as the original equation, so the equation is not symmetric with respect to the y-axis.

22 ()()2()()1 xyxy 

22 21xyxy , so the equation is symmetric with respect to the origin.

74. 2222 ()16 16 xyxy , so the equation is symmetric with respect to the x-axis.

()162222 16 xyxy  , so the equation is symmetric with respect to the y-axis.

2222 ()()1616 xyxy  , so the equation is symmetric with respect to the origin.

For exercises 75–78, use the standard form of the equation of a circle, ()()222 xhykr  .

75. Center (2, 3); radius = 6

76. Center (–1, 3); radius = 4

77. Center (–2, –3); radius = 11

78. Center 13 , 22    ; radius = 3 2

79. 22(1)4xy

80. 22 (1)1 xy

81. 22 (1)(2)2 xy

82. 22 (2)(3)7 xy

83. Find the radius by using the distance formula: 22 (13)(5(4))97 d  The equation of the circle is 22 (3)(4)97. xy

84. Find the radius by using the distance formula: 22 (12)(15)255 d  . The equation of the circle is 22 (1)(1)25. xy

85. The circle touches the x-axis, so the radius is 2. The equation of the circle is 22 (1)(2)4. xy

86. The circle touches the y-axis, so the radius is 1. The equation of the circle is 22 (1)(2)1 xy

87. Find the diameter by using the distance formula: 22 (37)(64)104226 d 

So the radius is 26 . Use the midpoint formula to find the center: 7(3)46,2,5 22 M 

   . The equation of the circle is 22 (2)(5)26 xy

88. Find the center by finding the midpoint of the diameter: 2835,5,1 22 C    

Find the length of the radius by finding the length of the diameter and dividing that by 2. 22 (28)(35)10010 d 

Thus, the length of the radius is 5, and the equation of the circle is 22 (5)(1)25. xy

89. a. 22 2240xyxy 22224 xxyy

Now complete the square: 222121411 xxyy 22 (1)(1)6 xy . This is a circle with center (1, 1) and radius 6

b. To find the x-intercepts, let y = 0 and solve for x:

  2 22 2 (1)(01)6116 151515 xx xxx  

Thus, the x-intercepts are   15,0  and   15,0.

To find the y-intercepts, let x = 0 and solve for y:

  2 22 2 (01)(1)6116 151515 yy yyy  

Thus, the y-intercepts are   0,15  and   0,15.

90. a. 22 42150xyxy

224215 xxyy

Now complete the square: 2244211541 xxyy

22 (2)(1)20. xy This is a circle with center (2, 1) and radius 25

b. To find the x-intercepts, let y = 0 and solve for x: 22 (2)(01)20 x   22 2120219 219219 xx xx  

Thus, the x-intercepts are   219,0  and   219,0.

To find the y-intercepts, let x = 0 and solve for y: 22 (02)(1)20 y   2 2 4(1)20116 143,5 yy yy  

Thus, the y-intercepts are (0, 3) and (0, 5).

91. a. 22 2240 xyy

22 2(2)0 xyy 22 20.xyy

Now complete the square: 22222101(1)1. xyyxy 

This is a circle with center (0, –1) and radius 1.

b. To find the x-intercepts, let y = 0 and solve for x: 222(01)100 xxx 

Thus, the x-intercept is (0, 0).

To find the y-intercepts, let x = 0 and solve for y:

22 0(1)1110,2 yyy 

Thus, the y-intercepts are (0, 0) and (0, 2).

92. a. 22 3360 xyx

22 3(2)0 xyx 2220.xxy

Now complete the square: 2222 2101(1)1.xxyxy 

This is a circle with center (–1, 0) and radius 1.

b. To find the x-intercepts, let y = 0 and solve for x:

22 (1)01110,2 xxx  Thus, the x-intercepts are (0, 0) and ( 2, 0).

To find the y-intercepts, let x = 0 and solve for y: 222 (01)100 yyy  Thus, the y-intercept is (0, 0).

93. a. 222200. xyxxxy

Now complete the square: 2211 0 44xxy 2 2 11 . 24xy

This is a circle with center 1 ,0 2     and radius 1 2

b. To find the x-intercepts, let y = 0 and solve for x: 2 2 1111 0 2422 xx

0,1. x  Thus, the x-intercepts are (0, 0) and (1, 0). To find the y-intercepts, let x = 0 and solve for y: 2 22 1111 0 2444 yy

2 00.yy Thus, the y-intercept is (0, 0).

94. a. 2222101. xyxy The radius cannot be negative, so there is no graph.

b. There are no intercepts.

2.2 Applying the Concepts

95. The distance from P(x, y) to the x-axis is x while the distance from P to the y-axis is y So the equation of the graph is . xy 

96. The distance from P(x, y) to (1, 2) is 22 (1)(2) xy while the distance from P to (3, –4) is 22 (3)(4) xy .

So the equation of the graph is 2222 (1)(2)(3)(4) xyxy  2222 (1)(2)(3)(4) xyxy  (continuedonnextpage)

(continued)

97. If you save $100 each month, it will take 24 months (or two years) to save $2400. So, the graph starts at (0, 0) and increases to (2, 2400). It will take another 30 months (or 2.5) years to withdraw $80 per month until the $2400 is gone. Thus, the graph passes through (4.5, 0).

98. If you jog at 6 mph for 10 minutes, then you have traveled

 1 6 61  mile. So the graph starts at (0, 0) and increases to (10, 1). Resting for 10 minutes takes the graph to (20, 1). It will take 20 minutes to walk one mile at 3 mph back to the starting point. Thus, the graph passes through (40, 0).

99. a. July 2018 is represented by t = 0, so March 2018 is represented by t = –4. The monthly profit for March is determined by 2 0.5(4)3(4)8$12 P

million.

b. July 2018 is represented by t = 0, so October 2018 is represented by t = 3. So the monthly profit for October is determined by 2 0.5(3)3(3)8$5.5 P

million. This is a loss.

Because t = 0 represents July 2018, t = –6 represents January 2018, and t = 5 represents December 2018.

d. To find the t-intercept, set P = 0 and solve for t: 2 00.538 tt  2 3(3)4(0.5)(8) 325 2(0.5)1 2 or8 t   

The t-intercepts represent the months with no profit and no loss. In this case, t = 8 makes no sense in terms of the problem, so we disregard this solution. t = 2 represents Sept 2018.

e. To find the P-intercept, set t = 0 and solve for P: 2 0.5(0)3(0)88.PP 

The P-intercept represents the profit in July 2018.

100. a.

b. To find the P-intercept, set t = 0 and solve for P: 2 0.002(0)0.51(0)17.5 P  17.5. P  The P-intercept represents the number of female college students (in millions) in 2005.

101. a. t Height = 2 16128320 tt 

feet

c. 010 t 

d. To find the t-intercept, set y = 0 and solve for t: 2 016128320 tt  2 016(820) tt  0(10)(2) tt  10 or 2. tt

The graph does not apply if t < 0, so the t-intercept is 10. This represents the time when the object hits the ground. To find the y-intercept, set t = 0 and solve for y: 2 16(0)128(0)320320.yy 

This represents the height of the building.

102. a.

b. 060 t 

c. The total time of the experiment is 60 minutes or 1 hour.

2.2 Beyond the Basics





103. 22 22 22 22 42200 4220 44212041 (2)(1)25 xyxy xxyy xxyy xy





So this is the graph of a circle with center (2, –1) and radius 5. The area of this circle is 25.  22 42310xyxy 224231 xxyy 22 22 44213141 (2)(1)36 xxyy xy 



2.2 Graphs of Equations

So, this is the graph of a circle with center (2, –1) and radius 6. The area of this circle is 36.  Both circles have the same center, so the area of the region bounded by the two circles equals 362511.  

104. Using the hint, we know that the center of the circle will have coordinates (0, k).

Use the Pythagorean theorem to find k. 222224516259 3 kkk k  

The equations of the circles are  2 22 35. xy

2.2 Critical Thinking/Discussion/Writing

105. The graph of 2 2 yx  is the union of the graphs of 2  yx and 2.  yx

106. Let (x, y) be a point on the graph. The graph is symmetric with regard to the x-axis, so the point (x, –y) is also on the graph. Because the graph is symmetric with regard to the y-axis, the point (–x, y) is also on the graph. Therefore the point (–x, –y) is on the graph, and the graph is symmetric with respect to the origin. The graph of 3 yx  is an example of a graph that is symmetric with respect to the origin but is not symmetric with respect to the x- and y-axes.

107. a. First find the radius of the circle: 22 (,)(60)(81)85dAB

The center of the circle is 6018 , 22

9 3,. 2

So the equation of the circle is 2 2 985 (3). 24 xy

To find the x-intercepts, set y = 0, and solve for x: 2 2 22 985 (3)0 24 8185 (3)691 44 x xxx

2 680xx

The x-intercepts are the roots of this equation.

b. First find the radius of the circle: 22 22 (,)(0)(1) (1) dABab ab   22 (1) . 2 ab r  

The center of the circle is 011 ,, 2222 abab 

So the equation of the circle is 221(1)22 224 abab xy

To find the x-intercepts, set y = 0 and solve for x: 22 22 2 222 2 22222 2 2 1(1) 0 224 (1)(1) 444 442121 4440 0 abab x abab xax xaxabbabb xaxb xaxb



The x-intercepts are the roots of this equation.

c. a = 3 and b = 1. Approximate the roots of the equation by drawing a circle whose diameter has endpoints A(0, 1) and B(3, 1).

The center of the circle is 3 ,1 2    and the radius is 3 2 . The roots are approximately (0.4, 0) and (2.6, 0).

108. a. The coordinates of the center of each circle are (r, r) and (3r, r).

b. To find the area of the shaded region, first find the area of the rectangle shown in the figure below, and then subtract the sum of the areas of the two sectors, A and B.

53 19 19 64 12 215 5 3412

2 2363262 3 xyyxyx 

22 21021 33 2228 2 3333

124. 0.10.200.20.10.5 xyyxyx 

2.3 Lines

2.3 Practice Problems

1.  358 6713 m  A slope of 8 13 means that the value of y decreases 8 units for every 13 units increase in x.

2.  2 2,3, 3 Pm

2 32 3 2 32 3 24213 3 3333 yx yx yxyx

3.  4610 5 312 m 

Use either point to determine the equation of the line. Using ( 3, 4), we have

Using ( 1, 6), we have

5. The slope is 2 3 and the y-intercept is 4. The line goes through (0, 4), so locate a second point by moving two units down and three units right. Thus, the line goes through (3, 2).

6. x = 3. The slope is undefined, and there is no y-intercept. The x-intercept is 3. y = 7. The slope is 0, and the y-intercept is 7.

7. First, solve for y to write the equation in slope-intercept form: 34244324 xyyx  3 6 4 yx

. The slope is 3 4 , and the y-intercept is 6. Find the x-intercept by setting y = 0 and solving the equation for x: 33 0668 44  xxx . Thus, the graph passes through the points (0, 6) and (8, 0).

8. Use the equation 2.665Hx  

1 2 2.64365176.8 2.64465179.4 H H

The person is between 176.8 cm and 179.4 cm tall, or 1.768 m and 1.794 m.

9. a. Parallel lines have the same slope, so the slope of the line is 3744 2533 m 

Using the point-slope form, we have  4 5231542 3 3154843230 yxyx yxxy

b. The slopes of perpendicular lines are negative reciprocals. Write the equation 4510 xy in slope-intercept form to find its slope: 4510 xy 41 541 55yxyx  . The slope of a line perpendicular to this lines is 5 4 .

Using the point-slope form, we have

5 434453 4 41651554310 yxyx yxxy



10. Because 2016 is 10 years after 2006, set x = 10. Then y = 0.44(10) + 6.70 = 11.1 There were 11.1 million registered motorcycles in the U.S. in 2016.

2.3 Concepts and Vocabulary

1. The slope of a horizontal line is 0 ; the slope of a vertical line is undefined.

2. The slope of the line passing through the points   11 , Pxy  and   22 , Qxy  is given by the formula 21 21 yy m xx 

3. Every line parallel to the line 32yx has slope, m, equal to 3 .

4. Every line perpendicular to the line 32yx has slope, m, equal to 1 3 .

5. False. The slope of the line 1 4 5 yx is equal to 1 4 .

6. False. The y-intercept of the line 23yx is equal to –3.

7. True

8. True

2.3 Building Skills

9. 734 ; 413 m  the graph is rising.

10. 044 2; 202 m  the graph is falling.

11. 2(2)0 0; 268 m  the graph is horizontal.

12. 7(4)11 slope is undefined; 3(3)0 m  the graph is vertical.

13. 3.525.5 2.2; 30.52.5 m  the graph is falling.

14. 3(2)1 1 231 m  ; the graph is rising.

15.  514 4 1 122 m   ; the graph is rising.

16.   330333 2 23 1313 m   ; the graph is rising.

17. 3  18. 2 

19. 4  20. 1 

21. 1  passes through the points (2, 3) and ( 5, 4). 1 437 1. 527 m  

22. 2  is a horizontal line, so it has slope 0.

23. 3  passes through the points (2, 3) and (0, 1). 3 13 2. 02 m  

24. 4  passes through the points ( 3, 3) and (0, 1).

4 134 033 m  

25.   0,5;3 m  35yx

26.   0,9;2 m  29yx

27. 1 4 2 yx

28. 1 4 2 yx

29. 33 1(2)13 22 3 4 2 yxyx yx 



30. 222 (1) 555yxyx 

31. 40(5)404yxyy 

32. Because the slope is undefined, the graph is vertical. The equation is 5. x 

33. 01 1 10 m  . The y-intercept is (0, 1), so the equation is 1. yx

34. 31 2 10 m  . The y-intercept is (0, 1), so the equation is 21.yx

35. 33 0 3(1) m  Because the slope = 0, the line is horizontal. Its equation is 3. y 

36. 716 2(5)7 m  . Now write the equation in point-slope form, and then solve for y to write the equation in slope-intercept form. 6 1(5) 7 yx 630 1 77 yx 637 77 yx

37. 1(1)2 1(2)3 m  . Now write the equation in point-slope form, and then solve for y to write the equation in slope-intercept form. 2 1(2) 3

38. 9(3)6 6(1)7 m  . Now write the equation in point-slope form, and then solve for y to write the equation in slope-intercept form. 6 3(1) 7

39. 17 2 7 44 11 2 0 22 m

. Now write the equation in point-slope form, and then solve for y to write the equation in slope-intercept form. 77 22 22  yxyx

40. 3(7)10 440 m  the slope is undefined. So the graph is a vertical line. The equation is 4. x 

41. 5 x  42. 1.5 y 

43. 0 y

45. 14 y

47. 2 4 3 yx

44. 0 x 

46. 25yx

48. 63yx

49. 404 ; 0(3)3 m  4 4 3 yx

50. 202 ; 0(5)5 m  2 2 5 yx

51. 7 y  52. 4 x 

53. 5 y  54. 3 x 

55. 32yx

The slope is 3 and the y-intercept is (0, –2). 2 03232 3 xxx 

The x-intercept is   2 3 ,0.

56. 23yx

The slope is –2 and the y-intercept is (0, 3). 3 02323 2 xxx 

The x-intercept is   3 2 ,0.

57. 1 240242. 2 xyyxyx 

The slope is 12 , and the y-intercept is (0, 2).

To find the x-intercept, set y = 0 and solve for x: 2(0)404xx 

58. 1 39933 3 xyxyyx 

The slope is 13 , and the y-intercept is (0, 3).

To find the x-intercept, set y = 0 and solve for x: 3(0)99xx .

59. 3 32603623 2 xyxyxy  .

The slope is 32 , and the y-intercept is (0, 3).

To find the x-intercept, set y = 0 and solve for x: 32(0)60362 xxx 

60. 24152154 xyxy  115 24xy . The slope is 12 , and the y-intercept is 154 . To find the x-intercept, set y = 0 and solve for x: 24(0)15 x  152 x 

61. 505xx . The slope is undefined, and there is no y-intercept. The x-intercept is 5.

62. 5 250 2 yy . The slope is 0, and the y-intercept is 52 . This is a horizontal line, so there is no x-intercept.

63. 0 x  . The slope is undefined, and the y-intercepts are the y-axis. This is a vertical line whose x-intercept is 0.

64. 0 y  . The slope is 0, and the x-intercepts are the x-axis. This is a horizontal line whose y-intercept is 0.

For exercises 65–68, the two-interecepts form of the equation of a line is 1. xy ab 

65. 1 43 xy 66. 1 32 xy 

67. 236 2361; 66632 xyxy xy

x-intercept = 3; y-intercept = 2

68. 341203412 3412 1; 12121243 xyxy xyxy 

x-intercept = –4; y-intercept = 3

69. 23 231211 121264 xyxy xy

The x-intercept is 6 and the y-intercept is –4.

70. 34 341211 121243 xyxy xy

The x-intercept is 4 and the y-intercept is –3.

71. 52 521011 101025 xyxy xy 

The x-intercept is –2 and the y-intercept is 5.

72. 45 452011 202054 xyxy xy 

The x-intercept is –5 and the y-intercept is 4.

73. 945 1. 725 m  The equation of the line through (2, 4) and (7, 9) is 41(2)yx y = x + 2. Check to see if (–1, 1) satisfies the equation by substituting x = –1 and y = 1: 11211.  So (–1, 1) lies on the line.

74. 325 1. 275 m  The equation of the line through (7, 2) and (2, –3) is 21(7)yx 5. yx Check to see if (5, 1) satisfies the equation by substituting x = 5 and y = 1: 15510.  So (5, 1) does not lie on the line.

75. The given line passes through the points (0, 3) and (4, 0), so its slope is 3 4 Any line parallel to this line will have the same slope. The line that passes through the origin and is parallel to the given line has equation

3 4 .yx 

76. From exercise 75, the slope of the given line is 3 4 Any line perpendicular to this line will have slope 4 3 . The line that passes through the origin and is perpendicular to the given line has equation 4 3 .yx 

77. The red line passes through the points (–2, 0) and (0, 3), so its slope is 3 2 . The blue line passes through (4, 2) and has the same slope, so its equation is

78. The red line passes through the points (–2, 0) and (0, 3), so its slope is 3 2 The green line passes through (4, 2) and has slope 2 3 , so its equation is

3214

79. The slope of 31yx is 3. The slope of 32yx is also 3. The lines are parallel.

80. The slope of 22yx is 2. The slope of 22yx is –2. The lines are neither parallel nor perpendicular.

81. The slope of 24yx is 2. The slope of 1 2 4 yx is 1 2 . The lines are perpendicular.

82. The slope of 31yx is 3. The slope of 1 3 1 yx is 1 3 . The lines are neither parallel nor perpendicular.

83. The slope of 387 xy is 38 , while the slope of 570 xy is 57 . The lines are neither parallel nor perpendicular.

84. The slope of 1023 xy is –5. The slope of 51 xy is also –5, so the lines are parallel.

85. The slope of 48xy is 14 . The slope of 41yx is 4 , so the lines are perpendicular.

86. The slope of 31yx is 3. The slope of 620 yx is 13 . The lines are perpendicular.

87. Both lines are vertical lines. The lines are parallel.

88. The slope of 237 xy is 23 , while 2 y  is a horizontal line. The lines are neither parallel nor perpendicular.

89. The equation of the line through (2, –3) with slope 3 is   332336yxyx  39.yx

90. The equation of the line through (–1, 3) with slope –2 is

321321 yxyx  32221.yxyx 

91. A line perpendicular to a line with slope 1 2 has slope 2. The equation of the line through (–1, 2) with slope 2 is 

221yx   221222 24. yxyx yx 



92. A line perpendicular to a line with slope 1 3 has slope –3. The equation of the line through (2, –1) with slope –3 is

132yx

 132136 35. yxyx yx  

93. The slope of the line joining (1, –2) and (–3, 2) is   22 1. 31  The equation of the line through (–2, –5) with slope –1 is

  5252 527. yxyx yxyx

 

94. The slope of the line joining (–2, 1) and (3, 5) is  514 325  The equation of the line through (1, 2) with slope 4 5 is  4 215241 5 51044546 46 55 yxyx yxyx yx







95. The slope of the line joining (–3, 2) and (–4, –1) is

A line perpendicular to this line has slope 1 3 .

The equation of the line through (1, –2) with slope 1 3 is

96. The slope of the line joining (2, 1) and (4, –1) is 11 1. 42 

A line perpendicular to this line has slope 1 The equation of the line through (–1, 2) with slope 1 is

21213.

97. The slope of the line 65yx is 6. The lines are parallel, so the slope of the new line is also 6. The equation of the line with slope 6 and y-intercept 4 is 64.yx

98. The slope of the line 1 2 5 yx is 1 2 . The lines are parallel, so the slope of the new line is also 1 2 The equation of the line with slope 1 2 and y-intercept 2 is 1 2 2. yx

99. The slope of the line 65yx is 6. The lines are perpendicular, so the slope of the new line is 1 6 The equation of the line with slope 1 6 and y-intercept 4 is 1 6 4. yx

100. The slope of the line 1 2 5 yx is 1 2 The lines are perpendicular, so the slope of the new line is 2. The equation of the line with slope 2 and y-intercept –4 is 24.yx

101. The slope of x + y = 1 is 1. The lines are parallel, so they have the same slope. The equation of the line through (1, 1) with slope 1 is 1(1)yx 11yx 2 yx .

102. The slope of 237 xy  is 23 . The lines are parallel, so they have the same slope. The equation of the line through (1, 0) with slope 23 is 222 0(1)333yxyx 

103. The slope of 3918 xy is 13 . The lines are perpendicular, so the slope of the new line is 3 . The equation of the line through (–2, 4) with slope 3 is 43((2))yx 43632.yxyx 

104. The slope of 214 xy  is 2. The lines are perpendicular, so the slope of the new line is 12. The equation of the line through (0, 2) with slope 12 is 1 2 2. yx

2.3 Applying the Concepts

105. a. The y-intercept represents the initial expenses.

b. The x-intercept represents the point at which the teacher breaks even, i.e., the expenses equal the income.

c. The teacher’s profit if there are 16 students in the class is $640.

d. The slope of the line is   640750 1390695 160168 

The equation of the line is 695 750. 8 Pn

106. a. The y-intercept represents the initial prepaid amount.

b. The x-intercept represents the total number of minutes the cellphone can be used.

c. The slope of the line is 015151 . 750755 

The equation of the line is 1 15. 5 Pt

d. The cost per minute is 1 $20 5  ¢.

107. slope = rise41 run4010 

108. 4 miles = 21,120 feet. |slope| = rise run  200025 21,120264 

109. 8 in. in two weeks  the plant grows 4 in. per week. John wants to trim the hedge when it grows 6 in., so he should trim it every 6 4 1.5 weeks10 days. 

110. 2 min. min.231 12.4 min. 5 in.31 in.5 x x 

The water will overflow in about 12 min.

111. a. x = the number of weeks; y = the amount of money in the account after x weeks; 7130yx

b. The slope is the amount of money deposited each week; the y-intercept is the initial deposit.

112. a. x = the number of sessions of golf; y = the yearly payment to the club; 351000yx

b. The slope is the cost per golf session; the y-intercept is the yearly membership fee.

113. a. x = the number of months owed to pay off the refrigerator; y = the amount owed; 15600yx

b. The slope is the amount that the balance due changes per month; the y-intercept is the initial amount owed.

114. a. x = the number of rupees; y = the number of dollars equal to x rupees. 1 0.019802 50.5 yxx 

b. The slope is the number of dollars per rupee. The y-intercept is the number of dollars for 0 rupees.

115. a. x = the number of years after 2010; y = the life expectancy of a female born in the year 2010 + x; 0.1780.8yx

b. The slope is the rate of increase in life expectancy; the y-intercept is the current life expectancy.

116. a. 1400(2)14,000$11,200 v 

b. 1400(6)14,000$5600 v 

To find when the tractor will have no value, set v = 0 and solve the equation for t: 0140014,00010 years tt 

117. There are 30 days in June. For the first 13 days, you used data at a rate of 435 33.5 13 

MB/day. At the same rate, you will use   33.517569.5  MB for the rest of the month.

435 + 569.5 = 1004.5

So, you don’t need to buy extra data. You will have about 20 MB left.

118. For the first three hours, you traveled at 195 65 mph.

3 

5201956532565 5 drttt t  

You will arrive at your destination five hours after 12 pm or 5 pm.

119. 540,000yx

120. a. 0.2530Cx

c. 0.25(60)30$45 y 

d. 47.750.253071 miles xx 

121. a. The two points are (100, 212) and (0, 32). So the slope is 212321809 . 10001005 

The equation is 99 32(0)32 55 FCFC 

b. One degree Celsius change in the temperature equals 95 degrees change in degrees Fahrenheit.

c. C 9 32 5 FC 40ºC 104°F

25º C 77°F –5ºC 23°F –10ºC 14°F

d. 9 100F3237.78C 5 CC 

9 90F3232.22C 5 CC 

9 75F3223.89C 5 CC 

9 10F3223.33C 5 CC 

9 20F3228.89C 5 CC 

e. 9 97.6F3236.44C 5 CC  ; 9 99.6F3237.56C 5 CC 

f. Let x = F = C. Then 9 32 5 xx 4 3240 5 xx  . At –40°, °F = °C.

122. a. The two points are (4, 210.20) and (10, 348.80). So the slope is 348.80210.20138.6 23.1 1046  . The equation is 348.823.1(10)yx  23.1117.8yx

b. The slope represents the cost of producing one modem. The y-intercept represents the fixed cost.

c. 23.1(12)117.8$395yy 

123. a. The year 2005 is represented by t = 0, and the year 2011 is represented by t = 6. The points are (0, 2425) and (6, 4026). So the slope is 40262425 266.8 6  The equation is 2425266.8(0)yt  266.82425yt

c. The year 2008 is represented by t = 3. So 266.8(3)24253225.4yy  . Note that there cannot be a fraction of a person, so. there were 3225 women prisoners in 2008.

d. The year 2017 is represented by t = 12. So 266.8(12)24255626.6yy  . There will be 5627 women prisoners in 2017.

124. a. The two points are (5, 5.73) and (8, 6.27).

The slope is 6.275.730.54 0.18. 853 

The equation is 5.730.18(5)Vx 0.184.83.Vx

b. The slope represents the monthly change in the number of viewers. The V-intercept represents the number of viewers when the show first started.

c. 0.18(11)4.836.81VV  million

125. The independent variable t represents the number of years after 2000, with t = 0 representing 2000. The two points are (0, 11.7) and (5, 12.7). So the slope is 12.711.7 0.2 5  . The equation is 11.70.2(0)pt 0.211.7.pt The year 2010 is represented by t = 10. 0.2(10)11.713.7%.pp 

126. The year 2004 is represented by t = 0, so the year 2009 is represented by t = 5. The two points are (0, 82.7) and (5, 84.2). So the slope is 84.282.71.5 0.3 55  . The equation is 0.382.7.yt

127. a.

212.4yx

b. [0, 6, 1] by [0, 12, 1]

c. The price in the table is given as the number of nickels. 35¢ = 7 nickels, so let x = 7.   2712.41.6 y 

Thus, no newspapers will be sold if the price per copy is 35¢. Note that this is also clear from the graph, which appears to cross the x-axis at approximately x = 6.

128. a.

0.0910.3yx

[0, 700, 100] by [0, 80, 10]

c. The advertising expenses in the table are given as thousands of dollars, so let x = 700.   0.0970010.373.3 y 

Sales are given in thousands, so approximately 73.3100073,300  computers will be sold.

2.3 Beyond the Basics

129. 3 39312 1(2)

130. The y-intercept is –4, so its coordinates are (0, –4). Substitute x = 0, y = –4 into the equation and solve for c

131. a. Let A = (0, 1), B = (1, 3), C = (–1, –1). 31134 101122;2ABBCmm 11 2 10 AC m 

The slopes of the three segments are the same, so the points are collinear.

b. 22 22 (,)(10)(31)5 (,)(11)(13)25   dAB dBC 22 (,)(10)(11)5  dAC

Because d(B, C) = d(A, B) + d(A, C), the three points are collinear.

132. a. Let A = (1, 0.5), B = (2, 0), C = (0.5, 0.75). 00.50.750 210.520.5;0.5ABBC mm

0.750.5 0.5 0.51 AC m 

The slopes of the three segments are the same, so the points are collinear.

b. 2 2 (,)(12)0155 242 dAB     1322 (,)20 24 4535 164

dBC

dAC

 13122 (,)1242 55 164



Because d(B, C) = d(A, B) + d(A, C), the three points are collinear.

133. a. 413842 ; 1125(1)3ABBC mm .

The product of the slopes = –1, so ABBC  b. 22 22 22 (,)(11)(41)13 (,)(5(1))(84)52 (,)(51)(81)65 dAB dBC dAC

 

  222 (,)(,)(,), dABdBCdAC  so the triangle is a right triangle.

134. 2(1)3121 ; 1(4)5312 2132(1)1 ; 2352(4)2

ABBC CDAD mm mm 

So, and ABCDBCAD  , and ABCD is a parallelogram.

For exercises 135 and 136, refer to the figures accompanying the exercises in your text.

135. AD and BC are parallel because they lie on parallel lines 1l and 2 . lAB and CD are parallel because they are parallel to the x-axis. Therefore, ABCD is a parallelogram.

ABCD  and ADCB  because opposite sides of a parallelogram are congruent.

ABDCDB   by SSS. Then

1 rise run BD m CD  and 2 rise . run BD m AB  Since

AB = CD, 12 . BDBD mm CDAB 

136. OKABLO because OLAKd  and .BLOKc  Then, 1 rise run d m c  and

2 rise . run cc m dd  12 1. dc mm cd

137. Let the quadrilateral ABCD be such that and .  ABCDABCD  Locate the points as shown in the figure.

Because ABCD  , the y-coordinates of C and D are equal. Because ABCD  , the x-coordinates of the points are as shown in the figure. The slope of AD is dc The slope of BC is 0 . dd bcbc   So . ADBC  22 (,). dADdc   2 222 (,)(). dBCdbcbdc  So ADBC 

138. Let 112233 (,),(,),(,) and AxyBxyCxy 44 (,)Dxy be the vertices of the quadrilateral.

Then the midpoint 1 M of AB is 1212 , 22 xxyy 

; the midpoint 2 M of BC is 2323 , 22 xxyy 

; the midpoint 3 M of

CD is 3434 , 22 xxyy  

; and the midpoint

4 M of AD is 1414 22,. xxyy  

The slope of 12MM is 23 12 13 23 12 13 22 22 yyyy yy xxxxxx     

The slope of 23MM is 2334 24 2334 24 22 . 22 yyyy yy xxxxxx   

The slope of 34MM is 34 14 3113 34 14 3113 22 22 yyyy yyyy xxxxxxxx

The slope of 14MM is 1214 24 1214 24 22 22 yyyy yy xxxxxx   

So 1234 MMMM  and 2314 MMMM  , and 1234MMMM is a parallelogram.

139. Let (x, y) be the coordinates of point B. Then 22 22 (,)12.5(2)(2) (2)(2)156.25 and 42 4(2)3(2) 32 42 . Substitute this into the first 33 AB dABxy xy

Solve this equation using the quadratic formula: 2 1001004(25)(1306.25) 2(25) 10010,000130,625 50 100140,625100375 5050 9.5 or 5.5

Now find y by substituting the x-values into the slope formula: 42 12 39.52 y y  or 42 8. 35.52 y y  So the coordinates of B are (9.5, 12) or (–5.5, –8).

140. Let (x, y) be a point on the circle with 11 (,) xy and 22 (,) xy as the endpoints of a diameter. Then the line that passes through (x, y) and 11 (,) xy is perpendicular to the line that passes through (x, y) and 22 (,) xy , and their slopes are negative reciprocals. So 12 12 yyxx xxyy 

1212 ()()()() yyyyxxxx

1212

141. 12012 505 OPm 

Since the tangent line ℓ is perpendicular to OP , the slope of ℓ is the negative reciprocal of 12 5 or 5 12 . Using the point-slope form, we have

55 125125 1212

5255169 12 12121212yxyx

142.

11 11 0 0 OP yy m xx

Since the tangent line ℓ is perpendicular to OP , the slope of ℓ is the negative reciprocal of 1 1 y x or 1 1 x y

Using the point-slope form, we have

2222

Since the equation of the circle is 222 xya  , we substitute 2 a for 22 11xy  to obtain 2 11 xxyya  .

143.

The family of lines has slope 2. The lines have different y-intercepts.

144.

The family of lines has y-intercept 2. The lines have different slopes.

145.

The lines pass through (1, 0). The lines have different slopes.

146.

The lines pass through ( 1, 2). The lines have different slopes.

2.3 Critical Thinking/Discussion/Writing

147. a.

This is a family of lines parallel to the line 2.yx  They all have slope –2.

Section 2.3 Lines 199

This is a family of lines that passes through the point (0, –4). Their y-intercept is –4.

148.  11 1122 22 12211221 21 12 ymxbmxbmxb ymxb mxmxbbxmmbb bb x mm

a. If 12 0 mm and 12 , bb  then 2112 1212 bbbb x mmmm 

b. If 12 0 mm and 12 , bb  then 21 12 . bb x mm 

c. If 12 0 mm and 12 , bb  then 2112 1221 bbbb x mmmm 

d. If 12 0 mm and 12 , bb  then 2121 1221 bbbb x mmmm 

2.3 Getting Ready for the Next Section

149.

 2 40220 202 or 202 xxx xx xx    Solution: {–2, 2}

150.

 2 10110 101 or 101 xxx xx xx    Solution: {–1, 1}

151.

 2 20 120 xx xx   101 or 202xxxx  Solution: {–1, 2}

152.

153.

157.

130xx

Solve the associated equation:

 1301 or 3. xxxx 

So, the intervals are

158. 2 230xx

1,3.

2 230310xxxx

Now solve the associated equation:

 3103 or 1. xxxx 

So, the intervals are    ,1,1,3,and 3,.

Interval Test point Value of 2 23xx Result

 ,1 –2 5 +  1,3 0 –3 –

  3,  5 12 +

The solution set is    ,13,. 

2.4 Functions

2.4 Practice Problems

1. a. The domain of R is {2, 2, 3} and its range is {1, 2}. The relation R is a function because no two ordered pairs in R have the same first component.

b. The domain of S is {2, 3} and its range is {5, 2}. The relation S is not a function because the ordered paired (3, 2) and (3, 5) have the same first component.

2. Solve each equation for y. a. 21212222 xyxy 2 21; xy not a function

b.  1 25525; 2 xyxyxy  a function

3.  2 25 gxxx 

a.  2 020500 g 

b.  2 121517 g 

c.   2 22 22 25 2255 24525 gxhxhxh xxhhxh xhxxhh   

4.          lengthheight3122 22244 sq. units TLMSA  

5. a.  1 1 fx x  is not defined when 101 xx  or when 101. xx . Thus, the domain of f is  ,1

b.  2 23gxxx is not defined when 2 230.xx Use the test point method to see that 2 230xx on the interval (–3, 1). Thus, the domain of g is 

 ,31,.

6.  2 , fxx  domain X = [ 3, 3]

a.  2 1010103.16fxxx

Since 103  and 103, neither solution is in the interval X = [ 3, 3]. Therefore, 10 is not in the range of f.

b.  2 442fxxx

Since 3 < 2 < 2 < 3, 4 is in the range of f.

c. The range of f is the interval [0, 9] because for each number y in this interval, the number xy  is in the interval [ 3, 3].

The graph is not a function because a vertical line can be drawn through three points, as shown.

Domain:   3,; range:   2,23 

9.  2 45yfxxx 

a. Check whether the ordered pair (2, 7) satisfies the equation:  ? 2 72425 77   

The point (2, 7) is on the graph.

b. Let y = 8, then solve for x:  22 845043 0313 or 1 xxxx xxxx  

The points ( 3, 8) and ( 1, 8) lie on the graph.

c. Let x = 0, then solve for y:  2 04055 y 

The y-intercept is 5.

d. Let y = 0, then solve for x:  2 045051 5 or 1 xxxx xx  

The x-intercepts are 5 and 1.

10. The range of C(t) is [6, 12).       11 22 1110611.989611.995CC

11. From Example 11, we have 22 500 APx  and 1200 PDx  feet. If c = the cost on land, the total cost C is given by 

 22 22 1.3 1.35001200 1.35001200 CcPDcAP cxcx cxx 

12. a.   1200100,000Cxx

b.   2500 Rxx 

25001200100,000 1300100,000 PxRxCx xx x  

d. The break-even point occurs when

 CxRx . 1200100,0002500 100,000130077 xx xx 

Metro needs 77 shows to break even.

2.4 Basic Concepts and Skills

1. In the functional notation   yfx  , x is the independent variable.

2. If (2)7 f  , then 2 is in the domain of the function f, and 7 is in the range of f

3. If the point (9, 14) is on the graph of a function f, then (9)14. f 

4. If (3, 7) and (3, 0) ar both points on a graph, then the graph cannot be the graph of a function.

5. False.

6. False. For example, if  1 , fx x  then a = 1 and b = 1 are both in the domain of f. However, a + b = 0 is not in the domain of f

7. True. x = 7 and the square root function is defined for all positive numbers.

8. False. The domain of f is all real x for x > 2. Values of x ≤ –2 make the fraction undefined.

2.4 Building Skills

9. Domain: {a, b, c}; range: {d, e}; function

10. Domain: {a, b, c}; range: {d, e, f}; function

11. Domain: {a, b, c}; range: {1, 2}; function

12. Domain: {1, 2, 3}; range: {a, b, c, d}; not a function

13. Domain: {0, 3, 8}; range:{ –3, –2, –1, 1, 2}; not a function

14. Domain: {–3, –1, 0, 1, 2, 3}; range: {–8, –3, 0, 1}; function

15. 22xyyx  ; a function

16. 11xyyx ; a function

17. 1 y x  ; a function

18. 1 1 xyy x  ; a function

19. 222 yxyxyx  ; not a function

20. or xyyxyx  ; not a function

21. 1 25 y x  ; a function

22. 2 1 1 y x  ; a function

23. 2323 yxyx ; a function

24. 3 35153 5 xyyx  ; a function

25. 2 88 xyyx  ; not a function

26. 2 or xyyxyx  ; not a function

27. 3 232 55 xyyx  ; a function

28. 3 3 88 xyyx  ; a function

In exercises 29 32, 2 ()31, fxxx 2 (),gx x  and ()2. hxx 

29. 2 (0)03(0)11 2 (0)(0) is undefined 0 f gg   (0)202 h  2 22 ()31 ()()3()131 faaa fxxxxx  

30. 2 2 (1)13(1)11;(1)2; 1 fg 2 (1)211;(); hga a  2 2 22 ()gx xx 

31. 2 (1)(1)3(1)15; f  2 (1)(1) is undefined; 1 gg (1)2(1)3;()2; hhcc  ()2()2 hxxx 

32. 2 2 (4)43(4)15;(4)1; 4 fg (4)242(4) is undefined; hh  2 (2); 2 gk k   2 22 ()()3()1 2331 fakakak aakkak  

33. a. 2 2(0) (0)0 40 f 

b. 2 2(1)223 (1) 3 3 41 f 

c. 2 2(2)4 (2)(2) is undefined 0 42 ff 

d. 2 2(2)4 (2)(2) is 0 4(2) ff  undefined

e. 22 2()2 () 4()4 fxxx xx 

34. a. 2 (0)2(0)04(0) is undefined gg 

b. 2 (1)2(1)14(1) is undefined gg 

c. 2 (2)2(2)244 g 

d. 2 (3)2(3)(3)465 g 

e. 2 2 ()2()()4 24 gxxx xx 

35. The width of each rectangle is 1. The height of the left rectangle is  2 1123. f  The height of the right rectangle is  2 2226. f 

Aff  

1112 13169 sq. units

36. The width of each rectangle is 1. The height of the left rectangle is  2 0022. f  The height of the right rectangle is  2 1123. f 

1011 12135 sq. units

39. The denominator is not defined for x = 9. The domain is (,9)(9,)  

40. The denominator is not defined for x = –9. The domain is (,9)(9,)  

41. The denominator is not defined for x = –1 or x = 1. The domain is (,1)(1,1)(1,). 

42. The denominator is not defined for x = –2 or x = 2. The domain is (,2)(2,2)(2,). 

43. The numerator is not defined for x < 3, and the denominator is not defined for x = –2. The domain is [3,) 

44. The denominator is not defined for x ≥ 4. The domain is (,4) 

45. The denominator equals 0 if x = –1 or x = –2. The domain is (,2)(2,1)(1,)  

46. The denominator equals 0 if x = –2 or x = –3. The domain is (,3)(3,2)(2,)  

47. The denominator is not defined for x = 0. The domain is (,0)(0,)  

48. The denominator is defined for all values of x The domain is (,)  .

49. a function

50. not a function

59. ()7hx  , so solve the equation 2 71 xx  2 60(3)(2)02xxxxx  or 3. x 

51. a function

60. ()7Hx  , so solve the equation 2 78 xx  2 114(1)(1) 10 2(1) xxx   13 2 x   there is no real solution.

61. a. 2 12(11)711,  which is false. Therefore, (1, 1) does not lie on the graph of f

52. not a function

b. 22 2 12(1)72(1)6 (1)31313 xx xxx  

The points  13,1 and  13,1 lie on the graph of f c.  2 20175yy 

The y-intercept is (0, 5).

53. not a function

d.   22 2 0217721 7714 11 222 xx xx   14 1 2 x 

54. a function

55. (4)2;(1)1;(3)5;(5)7 ffff 

56. (2)5;(1)4;(3)0;(4)5 gggg 

57. (2)5;(1)4;(0)3;(1)4 hhhh 

58. (1)4;(0)0;(1)4 fff

The x-intercepts are 14 1,0 2    and 14 1,0. 2     

62. a.  2 10321221012,  which is false. Therefore, ( 2, 10) does not lie on the graph of f

b. ()12fx  , so solve the equation 2 31212 xx  22 22 3121244 440(2)0 202 xxxx xxx xx   

c.  2 301200yy 

The y-intercept is (0, 0).

d.  2 0312034 0 or 4 xxxx xx   The x-intercepts are (0, 0) and ( 4, 0).

63. Domain: [ 3, 2]; range: [ 3, 3]

64. Domain: [ 1, 3]; range: [ 2, 4]

65. Domain:   4,  ; range: [ 2, 3]

66. Domain:   ,4 ; range: [ 1, 3]

67. Domain:   3,  ; range:  1,43 

68. Domain:     ,11,4  Range: ( 2, 4]

69. Domain:    ,42,24,   Range:  2,23 

70. Domain:     ,21,   Range:   , 

71.   9,  72. [–1, 7]

73. –3, 4, 7, 9 74. 6

75.   

 74,15,52fff

76.   

  44,17,33fff

77.    3.75,2.25,312,  

78. 

79.   9,  80.

81.       41,13,34ggg

82. 

55268gg

83.  9,5 84.   5, 

2.4 Applying the Concepts

85. A function because there is only one high temperature per day.

86. A function because there is only one cost of a first-class stamp on January 1 each year.

87. Not a function because there are several states that begin with N (i.e., New York, New Jersey, New Mexico, Nevada, North Carolina, North Dakota); there are also several states that begin with T and S.

88. Not a function because people with a different name may have the same birthday.

89. 2 ();(4)16;AxxA A(4) represents the area of a tile with side 4.

90. ();(3)2733in.;VxxV V(3) represents the volume of a cube with edge 3.

91. It is a function. 2 ()6;(3)54SxxS

92. ();(59)1.5 39.37 x fxf meters

93. a. The domain is [0, 8]. b. 2 (2)128(2)16(2)192 h  2 (4)128(4)16(4)256 h  2 (6)128(6)16(6)192 h 

c. 2 012816016(8) tttt  0 or 8 tt . It will take 8 seconds for the stone to hit the ground. d.

94. After 4 hours, there are (0.75)(16) = 12 ml of the drug.

After 8 hours, there are (0.75)(12 + 16) = 21 ml. After 12 hours, there are (0.75)(21 + 16) = 27.75 ml.

After 16 hours, there are (0.75)(27.75 + 16) = 32.81 ml.

After 20 hours, there are (0.75)(32.81 + 16) = 36.61 ml.

95.

96.

97. Note that the length of the base = the width of the base = x

100. The volume of the tank is 2 64, Vrh   so 2 64 h r  The top is open, so the surface area is given by 22 2 2 64 22 128 . rrhrr r r r  

101. The volume of the pool is 2 2 288 288.Vxhh x 

98.

The length of the rectangle is 2x and its height is 22

b. 22 2 Alwxrx 

99. The piece with length x is formed into a circle, so 2. 2 Cxrrx  

Thus, the area of the circle is 2 2 2 . 24 Arxx 

The piece with length 20 x is formed into a square, so  1 20420. 4 Pxssx 

Thus, the area of the square is

2 2 2 11 2020. 416 sxx

The sum of the areas is  2 2 1 20 416 Axx  

102.

The total area to be tiled is 2 2881152 44xhx xx

The cost of the tile is 11526912 6. xx

The area of the bottom of the pool is 2 , x so the cost of the cement is 2 2. x Therefore, the total cost is 2 6912 2.Cx x 

Using the Pythagorean theorem, we have 

22 2 12 22 150030210030 150030210030 dtt dtt

103.

Using the distance formula we have

b. c. 6501275256252525 xxx  25,000 TVs can be sold at $650 per TV.

106. a. 2 ()(127525)127525 Rxxxxx  domain [1, 30]

104.

The distance from A to P is 222525xx mi. At 4 mi/hr, it will take Julio 2 25 4 x  hr to row that distance. The distance from P to C is (8 x) mi, so it will take Julio 8 5 x hr to walk that distance. The total time it will take him to travel is 2 258 . 45 xx T  

105. a. (5)127525(5)1150. p  If 5000 TVs can be sold, the price per TV is $1150. (15)127525(15)900. p  If 15,000 TVs can be sold, the price per TV is $900. (30)127525(30)525. p  If 30,000 TVs can be sold, the price per TV is $525.

b. 2 2 2 (1)1275(1)25(1)1250 (5)1275(5)25(5)5750 (10)1275(10)25(10)10,250

R R R    2 2 2 2 (15)1275(15)25(15)13,500 (20)1275(20)25(20)15,500 (25)1275(25)25(25)16,250 (30)1275(30)25(30)15,750 R R R R

This is the amount of revenue (in thousands of dollars) for the given number of TVs sold (in thousands).

d. 2 2 2 4700127525 511880 51514(1)(188) 2(1) 4 or 47 xx xx x xx    

 47 is not in the domain, so 4000 TVs must be sold in order to generate revenue of 4.7 million dollars.

107. a. ()5.575,000Cxx

b. ()0.6(15)9 Rxxx 

c. ()()()9(5.575,000) 3.575,000 PxRxCxxx x  

d. The break-even point is when the profit is zero: 3.575,000021,429 xx 

e. (46,000)3.5(46,000)75,000 $86,000 P  

The company’s profit is $86,000 when 46,000 copies are sold.

108. a. ()0.5500,000;()5 CxxRxx  . The break-even point is when the profit is zero (when the revenue equals the cost): 50.5500,0004.5500,000 xxx  111,111.11 x  . Because a fraction of a CD cannot be sold, 111,111 CD’s must be sold.

b. ()()() 750,0005(0.5500,000) 1,250,0004.5277,778

The company must sell 277,778 CDs in order to make a profit of $750,000.

2.4 Beyond the Basics

109. 2 4224 4 xxyxxyx

110. 3223xyyyxy 33 (2)3()22 yxyfx xx  3

Domain: (,2)(2,).(4) 2 f  

111.   2 2 2 12 1 x xyxy x   2 22 (); Domain: (,);(4) 17 1 x fxf x  

112. 2222 yxxyyxyx   2 2()2222 xxyxxyfx xx  

Domain: 1 [0,);(4) 9 f 

113. ()() fxgx  because they have different domains.

114. ()() fxgx  because they have different domains.

115. ()() fxgx  because they have different domains. ()gx is not defined for x = –1, while ()fx is defined for all real numbers.

116. ()() fxgx  because they have different domains. ()gx is not defined for x = 3, while ()fx is not defined for x = 3 or x = –2.

117. ()() fxgx  because    3103fg  and     5265.fg 

118. ()() fxgx  because   216 f  while   213. g 

119. 2 (2)15(2)231563 faaa 3. a 

120. 222 (6)2866680 gbbbb (2)(4)02 or 4. bbbb

121. 3(6)2 (6)001829 2(6) a haa b   3(3)2(9) (3) is undefined has a zero 2(3) h b   in the denominator. So 606. bb 

122.   22 ()2323 fxxfxx  2 22 ()(23)4129 fxxxx 

123. 22 222 2 11111 () 1 gxxgx x xxx x

22 111 ()0 gxgxx xxx

124. 1 1 11 1 () 1 11 1 1 (1)(1) 21 1 (1)(1) 2 1 x xxxfxf xxx x xx x xxxx x

125. 3 () 45 35(35)3(41) 3 4141 () (1220)(5(41)) 35 45 41 41 (35)(123)17 (1220)(205)17

2.4 Critical Thinking/Discussion/Writing

126. Answers may vary. Sample answers are given

a. 2 yx b. 1 2 y x 

c. 2 yx  d. 1 2 y x 

127. a. 2 0 axbxc

b. yc 

c. The equation will have no x-intercepts if 2 40bac .

d. It is not possible for the equation to have no y-intercepts because ().yfx 

128.a. () fxx  b. ()0fx 

c. () fxx 

d. 2 () fxx  (Note: the point is the origin.)

e. ()1fx 

f. A vertical line is not a function.

129.a. {(a, 1), (b, 1)}

{(a, 1), (b, 2)}

{(a, 1), (b, 3)}

{(a, 3), (b, 1)}

{(a, 3), (b, 2)}

{(a, 3), (b, 3)}

{(a, 2), (b, 1)} {(a, 2), (b, 2)}

{(a, 2), (b, 3)}

There are nine functions from X to Y.

b. {(1, a)}, {(2, a)}, {(3, a)}

{(1, a)}, {(2, a)}, {(3, b)}

{(1, a)}, {(2, b}, {(3, a)}

{(1, b)}, {(2, a)}, {(3, a)}

{(1, b)}, {(2, b}, {(3, a)}

{(1, b)}, {(2, a)}, {(3, b)}

{(1, b)}, {(2, b)}, {(3, b)}

There are eight functions from Y to X

130. If a set X has m elements and a set of Y has n elements, there are m n functions that can be defined from X to Y. This is true since a function assigns each element of X to an element of Y. There are m possibilities for each element of X, so there are m m nnnnn



possible functions.

2.4 Getting Ready for the Next Section

131. 24122168 xxx 

The solution set is   ,8.

132.   59715977

221 or 1 xxxx xxx  

The solution set is   1,. 

133. 2 0 x 

Solve the associated equation: 2 00.xx

So, the intervals are     ,0 and 0,.

Interval Test point Value of 2 x Result

  ,0 

The solution set is     ,00,  

134.     350 xx 

Solve the associated equation:     3503 or 5. xxxx 

So, the intervals are    ,5,5,3,and 3,.

Interval Test point Value of     35 xx 

Result

  ,5 –10 –65 – 5,3 0 15 +   3,  5 –20 –

The solution set is 5,3.

2.5 Properties of Functions

2.5 Practice Problems

1. The function is decreasing on     0,3,12,13, and   15,24; increasing on   3,12 and

2. Relative maxima of 3640 at x = 12 and 4070 at x = 15; relative minima of 40 at x = 3 and 3490 at x = 13.

Relative minimum of 0 at x = 0

Relative maximum of 1 at x = 1

4.  2 11 vrr 

[0, 13, 1] by [0, 250, 25]

Mrs. Osborn’s windpipe should be contracted to a radius of 7.33 mm for maximizing the airflow velocity.

5.  2 fxx 

Replace x with x:

Thus, the function is even.

6.  3 fxx 

Replace x with x:

3 3 fxxxfx

Thus, the function is odd.

7. a. 42 42 ()3()5() 35() () is even. gxxx xxfx gx

b.  5353 53 ()4()2()42 42() () is odd. fxxxxx xxfx fx

The average rate of change is –500, so the number of bacteria per cubic centimeter decreases by 500 each day after adding the chlorine.

2 22 1;2,4 2123;41415 153 12 6 422 fxxab ff fbfa ba

2.5 Concepts and Vocabulary

1. A function f is decreasing if 12xx  implies that 

 12fxfx 

2.   fa is a relative maximum of f if there is an interval   12 , xx containing a such that

fafx  for every x in the interval   12,.xx

3. A function f is even if f( x) = f(x) for all x in the domain of f

4. The average rate of change of f as x changes from x = a to x = b is ()() ,. fbfaab ba 

5. True

6. False. A relative maximum or minimum could occur at an endpoint of the domain of the function.

7. True

8. False. The graph of an odd function is symmetric with respect to the origin.

2.5 Building Skills

9. Increasing on   , 

10. Decreasing on   , 

11. Increasing on   ,2, decreasing on   2, 

12. Decreasing on   ,3, increasing on   3, 

13. Increasing on   ,2, constant on   2,2, increasing on   2, 

14. Decreasing on   ,1, constant on   1,4, decreasing on

4, 

15. Increasing on  ,3 and   1 2 ,2 , decreasing on   1 2 3, and   2, 

16. Increasing on     3,1,0,1, and   2,.  Decreasing on

,3,1,0, and   1,2.

17. No relative extrema

18. No relative extrema

19. (2, 10) is a relative maximum point and a turning point.

20. (3, 2) is a relative minimum point and a turning point.

21. Any point on (x, 2) is a relative maximum and a relative minimum point on the interval (–2, 2). Relative maximum at ( 2, 2); relative minimum at (2, 2). None of these points are turning points.

22. Any point on (x, 3) is a relative maximum and a relative minimum point on the interval (–1, 4). Relative maximum at (4, 3); relative minimum at ( 1, 3). None of these points are turning points.

23. (–3, 4) and (2, 5) are relative maxima points and turning points.   1 2 ,2 is a relative minimum and a turning point.

24. (–3, –2), (0, 0), and (2, –3) are relative minimum points and turning points. (–1, 1) and (1, 2) are relative maximum points and turning points.

For exercises 25 34, recall that the graph of an even function is symmetric about the y-axis, and the graph of an odd function is symmetric about the origin.

25. The graph is symmetric with respect to the origin. The function is odd.

26. The graph is symmetric with respect to the origin. The function is odd.

27. The graph has no symmetries, so the function is neither odd nor even.

28. The graph has no symmetries, so the function is neither odd nor even.

29. The graph is symmetric with respect to the origin. The function is odd.

30. The graph is symmetric with respect to the origin. The function is odd.

31. The graph is symmetric with respect to the y-axis. The function is even.

32. The graph is symmetric with respect to the y-axis. The function is even.

33. The graph is symmetric with respect to the origin. The function is odd.

34. The graph is symmetric with respect to the origin. The function is odd.

For exercises 35 48, ()()() is even fxfxfx  and ()()() is odd. fxfxfx 

35. 44 ()2()424() () is even. fxxxfx fx 

36. 44 ()3()535() () is even. gxxxgx gx 

 

37. 33 3 ()5()3()53 (53)() () is odd. fxxxxx xxfx fx

38. 3 3 ()2()4() 24() () is odd. gxxx xxgx gx

 

 

39.  ()2()424 () () is neither even nor odd. fxxx fxfx fx

 

40.  ()3()737 () () is neither even nor odd. gxxx gxgx gx

41. 22 11 ()() ()44 () is even. fxfx xx fx

 

42.      2 2 44 2 2 1 1 is even. xx gxgx xx gx     

43. 33 22 () ()() ()11 () is odd. xx fxfx xx fx

 

44. 44 33 4 3 ()33 () 2()3()23 3 ()() is odd. 23 xx gx xxxx xfxfx xx

45.

46.

5353 53 53 3 3 1 3 13 fxxx xxxx xxfx xxxx

Thus, f(x) is even.

33 55 3 3 5 5 ()2()2 () 2()3()23 12 2 23 123 xxxx gx xxxx xxxx fx xxxx

Thus, f(x) is even.

48.

Thus, f(x) is neither even nor odd.

22 3()737 () ()33 xx gxgxgx xx 

Thus, g(x) is neither even nor odd.

49. a. domain: (,);  range: (,3] 

b. x-intercepts: ( 3, 0), (3, 0) y-intercept: (0, 3)

c. increasing on (,0),  decreasing on (0,) 

d. relative maximum at (0, 3)

e. even

50. a. domain: (,);  range: (,) 

b. x-intercepts: ( 4, 0), (0, 0), (4, 0) y-intercept: (0, 0)

c. decreasing on   ,2 and   2,  , increasing on   2,2

d. relative maximum at (2,3); relative minimum at ( 2, 3)

e. odd

51. a. domain: ( 3, 4); range: [ 2, 2]

b. x-intercept: (1, 0); y-intercept: (0, 1)

c. constant on ( 3, 1) and (3, 4) increasing on ( 1, 3)

d. Since the function is constant on ( 3, 1), any point (x, 2) is both a relative maximum and a relative minimum on that interval. Since the function is constant on (3, 4), any point (x, 2) is both a relative maximum and a relative minimum on that interval.

e. neither even not odd

52. a. domain: ( 3, 3); range: { 2, 0, 2}

b. x-intercept: (0, 0); y-intercept: (0, 0)

c. constant on ( 3, 0) and (0, 3)

d. Since the function is constant on ( 3, 0), any point (x, 2) is both a relative maximum and a relative minimum on that interval. Since the function is constant on (0, 3), any point (x, 2) is both a relative maximum and a relative minimum on that interval.

e. odd

53. a. domain: ( 2, 4); range: [ 2, 3]

b. x-intercept: (0, 0); y-intercept: (0, 0)

c. decreasing on ( 2, 1) and (3, 4) increasing on ( 1, 3)

d. relative maximum: (3, 3) relative minimum: ( 1, 2)

e. neither even nor odd

54. a. domain:   , 

range:   , 

b. x-intercepts: (2, 0), (3, 0) y-intercept: (0, 3)

c. decreasing on       ,00,33,  

d. no relative minimum relative maximum: (0, 3)

e. neither even nor odd

55. a. domain: (,);  range:   0, 

b. no x-intercept; y-intercept: (0, 1)

c. increasing on   , 

d. no relative minimum or relative maximum

e. neither even nor odd

56. a. domain:     ,00,   range:   , 

b. x-intercepts: ( 1.5, 0), (1.5, 0) no y-intercept

c. decreasing on   ,0 increasing on   0, 

d. no relative minimum or relative maximum

e. even



57. ()27;1,3 (3)2(3)71;(1)2(1)79 fxxab ff

 (3)(1) average rate of change 3(1) 19 2 4 ff

58. ()49;2,2 (2)4(2)91;(2)4(2)917 fxxab ff 

59. ()3;1,5 (5)3515;(1)313 fxxcab fccfcc 

60. ();1,7 (7)7;(1) fxmxcab fmcfmc

(7)(1) average rate of change 7(1) 7 8 8 8 ff

61. 2 22 ()1;2,0 (0)011;(2)(2)13 hxxab hh

(0)(2) average rate of change 0(2) 13 2 2 hh

62. 2 22 ()2;3,4 (4)2414;(3)237 hxxab hh

(4)(3) average rate of change 43 14(7) 7 1 hh

63. 2 22 ()(3);1,3 (4)(33)0;(1)(31)4 fxxab ff

(3)(1) average rate of change 31 04 2 2 ff 

64. 2 22 ()(2);1,5 (5)(52)9;(1)(12)9 fxxab ff 

(5)(1) average rate of change 5(1) 99 0 6 ff

65. 3 33 ();1,3 (3)327;(1)(1)1 gxxab gg

  (3)(1) average rate of change 3(1) 27(1) 7 4 gg  

66. 3 33 ();1,3 (3)327;(1)(1)1 gxxab gg



 (3)(1) average rate of change 3(1) 271 7 4 gg

67. 1 ();2,6 11 (2);(6)26 hxab x hh



 11 62 (6)(2) average rate of change 62 1 412 hh 

68. 4 ();2,4 3 444 (4);(2)4 43723 hxab x hh

69. () fxhxh  ()() fxhfxxhxh  ()() 1 fxhfxh hh

70. ()3()2332 fxhxhxh  ()()332(32)3 fxhfxxhxh

71.

72.



56556 55656 5 ()()5 5 fxhxhxh fxhfxxhx h fxhfxh hh

73.

    

()() fxhmxhbmxmhb fxhfxmxmhbmxb mh fxhfxmh m hh





 222 222 2 ()()2 2 fxhaxhcaxahc fxhfxaxahcaxc ah fxhfxah a hh

75. ()()2222 fxhxhxxhh  222 2 ()()2 2 fxhfxxxhhx xhh 

 2 ()()2 2 fxhfxxhh xh hh  



76. 2 22 22 ()()() 2 2 fxhxhxh xxhhxh xxhxhh

  222 2 ()() 2() 2 fxhfx xxhxhhxx xhhh    2 ()()2 21 fxhfxxhhh xh hh  



77. 2 22 22 ()2()3() 24233 24323 fxhxhxh xxhhxh xxhxhh



 222 2 ()() 24323(23) 423 fxhfx xxhxhhxx xhhh

   2 ()()423 423 fxhfxxhhh hh xh   

78. 2 22 22 ()3()2()5 363225 362325 fxhxhxh xxhhxh xxhxhh



  22 2 2 ()()3623 25(325) 632 fxhfxxxhxh hxx xhhh

   2 ()()632 632 fxhfxxhhh hh xh

  

79. ()4fxh ()()440 fxhfx ()()0 0 fxhfx hh  

80. ()3fxh ()()3(3)0 fxhfx ()()0 0 fxhfx hh  

81. 1 ()fxh

82. 1 ()

2.5 Applying the Concepts

83. a. Increasing: (2006, 2009), (2011, 2012), (2013, 2014)

Decreasing: (2009, 2011), (2012, 2013)

b. Relative maxima: 251.1 at x = 2009, 293.2 at x = 2012

Relative minima: 21.5 at x = 2011, 187.0 at x = 2013

84. a. Increasing: (Jan., June), (July, Sept.) Decreasing: (June, July), (Sept., Dec.)

b. Relative maxima: 185 in June, 185 in Sept. Relative minima: 132 in July

85. domain:   0, 

The particle’s motion is tracked indefinitely from time t = 0.

86. range:  7,5

The particle takes on all velocities between –7 an 5. Note that a negative velocity indicates that the particle is moving backward.

87. The graph is above the t-axis on the intervals (0, 9) and (21, 24). This means that the particle was moving forward between 0 and 9 seconds and between 21 and 24 seconds.

88. The graph is below the t-axis on the interval (11, 19). This means that the particle is moving backward between 11 and 19 seconds.

89. The function is increasing on (0, 3), (5, 6), (16, 19), and (21, 23). However, the speed |v| of the particle is increasing on (0, 3), (5, 6), (11, 15), and (21, 23). Note that the particle is moving forward on (0, 3), (5, 6), and (21, 23), and moving backward on (11, 15).

90. The function is decreasing on (6, 9), (11, 15), and (23, 24). However, the speed |v| of the particle is decreasing on (6, 9), (16, 19), and (23, 24). Note that the particle is moving forward on (6, 9) and (23, 24), and moving backward on (16, 19).

91. The maximum speed is between times t = 15 and t = 16.

92. The minimum speed is 0 on the intervals (9, 11), (19,21), and (24, ∞).

93. The particle is moving forward with increasing velocity.

94. The particle is moving backward with decreasing speed.

95.

b. The length of the squares in the corners must be greater than 0 and less than 6, so the domain of V is (0, 6).

c. [0, 6, 1] by [ 25, 150, 25] range: [0, 128]

d. V is at its maximum when x = 2.

96. a. Let x = one of the numbers. Then 32 x  the other number. The sum of the numbers is 32 . Sx x 

b. [0, 50, 10] by [ 10, 70, 10]

The minimum value of approximately 11.31 occurs at x ≈ 5.66.

97. a. ()21010,500Cxx

b. (50)210(50)10,500$21,000 C 

It costs $21,000 to produce 50 notebooks per day.

c. average cost = $21,000 $420 50 

d. 21010,500 315 21010,500315 10,500105100 x x xx xx    

The average cost per notebook will be $315 when 100 notebooks are produced.

98. 

2 2 2 234 1213145 3233345 fxxx f f  

The secant passes through the points (1, 5) and (3, 5).

The equation of the secant is

551555 510 yxyx

99.

The average rate of increase was 0.35 million, or 350,000, students per year.

100.

  60 5 6060 557;1555 51 ft t ff 

(5)(1) average rate of decrease 51 755 12 4 ff

The average rate of decrease is 12 gallons per minute.

101. a.  2 003044 f 

The particle is 4 ft to the right from the origin.

b.  2 4434432 f 

The particle started 4 ft from the origin, so it traveled 32 4 = 28 ft in four seconds.

c.  2 3333422 f 

The particle started 4 ft from the origin, so it traveled 22 4 = 18 ft in three seconds. The average velocity is 1836ftsec 

d.   2 2 2232414 5535444 f f  

The particle traveled 44 14 = 30 ft between the second and fifth seconds. The average velocity is   305210ftsec 

102. a.   2 2 00.0100.205050 40.0140.245050.96

P P  

The population of Sardonia was 50 million in 2000 and 50.96 million in 2004.

b.  2 100.01100.2105053 P 

The average rate of growth from 2000 to 2010 was 5350 0.3 10  million per year.

2.5 Beyond the Basics

103.

104.

105. In order to find the relative maximum, first observe that the relative maximum of

Then

2 10 x

Thus, the x-coordinate of the relative maximum is 1.

The relative maximum is ( 1, 5). There is no relative minimum.

106.  10if 5 5if 55 if 5 xx fxx xx

The point (0, 5) is a relative maximum, but not a turning point. 107.

109.

1111 11 11 11 11 1 11 fxhfx h xhxxhx hxhx xhx hxhx h hxhx xhx

11 1

xxhxhx hhxxh xhxxhx hxxhxhx xhx hxxhxhx xxhxxh

2.5 Critical Thinking/Discussion/Writing

111. f has a relative maximum at x = a if there is an interval   1 , ax with 1 axb  such that 

 , fafx  or

, fxfa  for every x in the interval 

1 ,.xb

112. f has a relative minimum at x = b if there is 1x in  , ab such that  

fxfb  for every x in the interval   1 ,.xb

113. Answers will vary. Sample answers are given.

a.   fxx  on the interval [ 1, 1]

if 01 1if 0 xx fx

0if 0 or 1 1if 01 and is rational 1if 01 and is irrational

2.5 Getting Ready for the Next Section

119. If 3, x  then 12. x  120. 280280284 xxxx  121.

3 32 222822 f  (ii) 

3 323 44428

3 323 44428 fii

122.  23 fxx  (i)  2313 3 2244 f  (ii)    2 232 3 88824 f  (iii)     2 232 3 88824 f 

2.6 A Library of Functions

2.6 Practice Problems

1. Because g( 2) = 2 and g(1) = 8, the line passes through the points ( 2, 2) and (1, 8).  826 2 123 m 

Use the point-slope form:    821822 2626 yxyx yxgxx

2. Using the formula Shark length = (0.96)(tooth height) 0.22, gives: Shark length = (0.96)(16.4) 0.22 = 15.524 m

4. Domain:

For the first line segment:

253 312 3 5121031 2 210332313 313 22 m yxyx yxyx yx 

For the second line segment: 42 1 53 451 m yxyx

The piecewise function is  313 if 13 22 1if 35 xx gx xx

504555675 20057575 xx fx

b. The fine for driving 60 mph is

 5046055$70. 

c. The fine for driving 90 mph is

20059075$275. 

7. The graph of f is made up of two parts: a line segment passing through (1, 5) and (3, 2) on the interval [1, 3], and a line segment passing through (3, 2) and (5, 4) on the interval [3, 5].

, and graph

2.6 Concepts and Vocabulary

1. The graph of the linear function () fxb  is a horizontal line.

2. The absolute value function can be expressed as a piecewise function by writing

if 0 () if 0 fxxxx xx

3. The graph of the function 2 2if 1 () if 1 fxxx axx       will have a break at x = 1 unless a = 3.

4. The line that is the graph of   23fxx has slope –2.

5. True. The equation of the graph of a vertical line has the format x = a

6. False. The absolute value function,  if 0 () if 0 fxxxx xx    is an example of a piecewise function with domain   ,. 

7. True

8. False. The function is constant on [0, 1), [1, 2), and [2, 3).

2.6 Building Skills

In exercises 9 18, first find the slope of the line using the two points given. Then substitute the coordinates of one of the points into the slope-intercept form of the equation to solve for b.

9. The two points are (0, 1) and (–1, 0). 01 1. 10 m  11(0)1 bb 

()1fxx

10. The two points are (1, 0) and (2, 1). 10 1. 21 m  011 bb  .

()1fxx

11. The two points are (–1, 1) and (2, 7).

71 2. 2(1) m    1213bb

()23fxx

12. The two points are (–1, –5) and (2, 4). 4(5) 3. 2(1) m  43(2)2. bb 

()32.fxx

13. The two points are (1, 1) and (2, –2).

21 3. 21 m  13(1)4. bb 

()34.fxx

14. The two points are (1, –1) and (3, 5).

5(1) 3. 31 m  13(1)4. bb  ()34.fxx

15. The two points are (–2, 2) and (2, 4). 421 . 2(2)2 m

1 4(2)3. 2 bb  1 ()3. 2 fxx

16. The two points are (2, 2) and (4, 5).

523 422 m  3 2(2)1. 2 bb  3 ()1. 2 fxx

17. The two points are (0, –1) and (3, –3). 3(1)2 303

2 1(0)1. 3 bb

2 ()1. 3 fxx

18. The two points are (1, 1/4) and (4, –2).

3 2(4)1. 4 bb  3 ()1. 4 fxx

19.  if 2 () 2if 2 fxxx x   

 12;22;33fff b.

20. 2if 0 () if 0 xx gxxx    a.       12;00;11ggg b.

21.  1if 0 () 1if 0 x gxx   

a.     151;121ff

c. domain: 

 ,00,   range: { 1, 1}

22. 24if 1 () 2if 1 xx gxxx   

 31;13;310ggg

c. domain:

range:

23. 2 2if 0 () if 0 xx fx xx

Domain:   ,; range:   , 

24. 2 if 1 () if 1 xx fx xx     

Domain:   ,; range:   0, 

25. 1 if 0 () if 0 x gxx xx

Domain:   ,; range:   , 

26. 3 if 1 () if 1 hxxx xx      

Domain:   ,; range:   , 

27.  3 if 1 () if 1 xx fx xx

Domain:

28. 3 if 0 () if 0 xx gx xx

Domain:

29. 3 if 1 () if 1 xx gx xx

1 if 0 () if 0 fxx x xx

Domain:   ,; range:

31. if 21 () if 1 xx fx xx

Domain:   2,; range:   0, 

32. 3 if 81 () 2if 1 xx gx xx      

Domain:   8,; range: 

33.  if 1 () 2if 14fxxx xx

Domain:   ,4;  range:

34. if 0 () if 04 xx hx xx

Domain:   ,4;  range:   0, 

35.  23if 2 121 3if 1 xx fxxx xx

Domain:

36.  21if 1 2112 2if 2 xx fxxx

Domain:   ,; range:   1, 

37. The graph of f is made up of two parts.

For x < 2, the graph is made up of the half-line passing through the points ( 1, 0) and (2, 3).

 303 1 213 011 m yxyx

For x ≥ 2, the graph is a half-line passing through the points (2, 3) and (3, 0).

03 3 32 03339 m yxyx 

Combining the two parts, we have

 1if 2 39if 2 fxxx xx 

38. The graph of f is made up of two parts. For x < 2, the graph is made up of the half-line passing through the points (2, 1) and (0, 3).

31 4 2 022 23 m yx  

For x ≥ 2, the graph is a half-line passing through the points (2, 1) and (4, 0).

101 242 11 042 22 m yxyx

Combining the two parts, we have

23if 2 1 2if 2 2 xx fx xx

39. The graph of f is made up of three parts. For x < 2, the graph is the half-line passing through the points ( 2, 2) and ( 3, 0).     022 2 321 02323 26 m yxyx yx

(continuedonnextpage)

(continued)

For 2 ≤ x < 2, the graph is a horizontal line segment passing through the points ( 2, 4) and (2, 4), so the equation is y = 4.

For x ≥ 2, the graph is the half-line passing through the points (2, 1) and (3, 0).

2.6 Applying the Concepts

41. a. () 33.81fxx 

Domain: [0,);  range: [0,). 

b. 3 (3)0.0887 33.81  f

This means that 3 oz ≈ 0.0887 liter.

Combining the three parts, we have

40. The graph of f is made up of four parts. For x ≤ 2, the graph is the half-line passing through the points ( 2, 0) and ( 4, 3).

For 2 < x ≤ 0, the graph is a line segment passing through the points ( 2, 0) and (0, 3).

For 0 < x ≤ 2, the graph is a line segment passing through the points (0, 3) and (2, 0).

For x ≥ 2, the graph is the half-line passing through the points (2, 0) and (4, 3).

Combining the four parts, we have

c. 12 (12)0.3549 liter. 33.81 f 

42. a. (0)1.8(0)212212. B 

The y-intercept is 212. This means that water boils at 212°F at sea level.

01.8212117.8 hh 

The h-intercept is approximately 117.80. This means that water boils at 0°F at approximately 117,800 feet above sea level.

b. Domain: closed interval from 0 to the end of the atmosphere, in thousands of feet.

c. 98.61.821263. hh  Water boils at 98.6°F at 63,000 feet. It is dangerous because 98.6°F is the temperature of human blood.

43. a. 1 (0)(0)11. 33 P  The y-intercept is 1.

This means that the pressure at sea level (d = 0) is 1 atm. 1 0133. 33 dd 

d can’t be negative, so there is no d-intercept.

b. 1 (0)1 atm;(10)(10)11.3 atm; 33 1 (33)(33)12 atm; 33 1 (100)(100)14.03 atm. 33 PP P P 

 

c. 1 51132 feet 33 dd 

The pressure is 5 atm at 132 feet.

44. a. (90)10551.1(90)1154 ft/sec V 

The speed of sound at 90ºF is 1154 feet per second.

b. 110010551.140.91F  TT

The speed of sound is 1100 feet per second at approximately 40.91ºF.

45. a. ()506000Cxx

b. The y-intercept is the fixed overhead cost.

c. 11,500506000110 x  110 printers were manufactures on a day when the total cost was $11,500.

46. a. The rate of change (slope) is 100. Find the y-intercept by using the point (10, 750): 750100(10)250. bb  The equation is ()100250.fpp

b. (15)100(15)2501250 f 

When the price is $15 per unit, there are 1250 units.

c. 1750100250$20. pp  1750 units can be supplied at $20 per unit.

47. a. 90030 Rx 

b. (6)90030(6)720 R 

If you move in 6 days after the first of the month, the rent is $720.

c. 6009003010 xx  You moved in ten days after first of the month.

48. a. Let t = 0 represent the year 2009. The rate of change (slope) is 995976 9.5. 02  The y-intercept is 995, so the equation is ()9.5995.ftt

b.     49.54995957 f 

The average SAT score will be 957 in 2013.

c. 9.59959009.59510 ttt  2009 + 10 = 2019.

The average SAT score will be 900 in 2019.

49. The rate of change (slope) is 10040 1 2080 

Use the point (20, 100) to find the equation of the line: 10020120. bb  The equation of the line is 120. yx Now solve 5012070. xx 

Age 70 corresponds to 50% capacity.

50. a. 2 (5)(60)24 25 y 

The dosage for a five-year-old child is 24 mg.

b. 2 60(60)12.5 25 aa 

A child would have to be 12.5 years old to be prescribed an adult dosage.

51. a. The rate of change (slope) is 50302 . 42015027 

The equation of the line is 2 30(150) 27 2 (150)30. 27 yx yx  

b. 21210 (350150)3044.8 2727yy 

There can’t be a fractional number of deaths, so round up. There will be about 45 deaths when x = 350 milligrams per cubic meter.

c. 2 70(150)30690 27 xx 

If the number of deaths per month is 70, the concentration of sulfur dioxide in the air is 3 690 mg/m .

52. a. The rate of change is 1 3 The y-intercept is 47 12 , so the equation is 147 (). 312yLSS

b. 147 (4)(4)5.25 312 L 

A child’s size 4 shoe has insole length 5.25 inches.

c. 61147 6.556.5 10312  xx

A child whose insole length is 6.1 inches wears a size 6.5 shoe.

53. a.

c. (i) 6000.04$15,000 xx 

b.(i) (12,000)0.04(12,000)$480 T 

(ii) (20,000)8000.06(20,00020,000)

$800 T  

(iii) (50,000)8000.06(50,00020,000)

$2600 T  

(ii) 12000.04$30,000 xx  , which is outside of the domain. Try   12008000.0620,000$26,667 xx 

(iii)   23008000.0620,000$45,000 xx 

54. a.  0.1if 09225 922.500.15(9225)if 922537,450 5156.250.25(37,450)if 37,45090,750, 18,481.250.28(90,750)if 90,750189,300 46,075.250.33(189,300)if 189,300411,500 119,401.250.35( xx xx xx fxxx xx x

 411,500)if 411,500413,200 119,996.250.396(413,200)if 413,200 x xx

b. (i)   (35,000)922.500.1535,0009225$4788.75$4789 f 

(ii) (100,000)18,481.250.28(100,00090,750)$21,071.25$21,071 f  (iii) (500,000)119,996.250.396(500,000413,200)$154,369.05$154,369 f 

c. (i)

3500922.500.1592252577.50.15922517183.339225$26,408 xxxx  (ii) 12,7005156.250.25(37,450)7543.750.25(37,450)30,17537,450 $67,625 xxx x   (iii) 35,00018,481.250.28(90,750)16518.750.28(90,750) 58995.5490,750$149,746 xx xx

 

2.6 Beyond the Basics

55.     231335914 aaa 

56. 133323353 5 3 aaa a  

57. a. Domain: (,)  ; range: [0,1)

b. The function is increasing on (n, n + 1) for every integer n

c.  ()()(), fxxxfxfx  so the function is neither even nor odd.

58. a. Domain: (,0)[1,)   range: 1 :0, an integer nn n    

b. The function is constant on (n, n + 1) for every nonzero integer n

c.  1 ()()(), fxfxfx x  so the function is neither even nor odd.

59. a. (i) (2)40 WCI 

(ii)

(16)91.4(91.440) 0.0203(16)0.304160.474 21 WCI    (iii) (50)1.6(40)559 WCI 

2.6 Critical Thinking/Discussion/Writing

63. D 64. C

65. a. If f is even, then f is increasing on  ,1 and decreasing on   1,0.

b. (i) 5891.4(91.4) (0.0203(36)0.304360.474

5891.4(91.4)(1.5672)

5851.841.56724F T T T TT 

5891.4143.241.5672

   (ii) 101.65528F TT 

60. a. 

()20()145 20145 Cxfx x 

2.3202.3145 20(3)145 2024585 C 

It will cost 85¢ to mail a first-class letter weighing 2.3 oz.

61.  ()24Cxx

62. a.  150 if 100 () 0.2100150if 100 x Cx xx    

b. If f is odd, then f is decreasing on  ,1 and increasing on   1,0.

66. a. If f is even, then f has a relative maximum at x = –1 and a relative minimum at x = –3.

b. If f is odd, then f has a relative minimum at x = –1 and a relative maximum at x = –3.

2.6 Getting Ready for the Next Section

67. If we add 3 to each y-coordinate of the graph of f, we will obtain the graph of y = f(x) + 3.

68. If we subtract 2 from each x-coordinate of the graph of f, we will obtain the graph of y = f(x + 2).

69. If we replace each x-coordinate with its opposite in the graph of f, we will obtain the graph of y = f( x).

70. If we replace each y-coordinate with its opposite in the graph of f, we will obtain the graph of y = f(x).

74.

For exercises 75–78, refer to section 1.3 in your text for help on completing the square.

75.  2 2 816;4xxx

76.  2 2 69;3xxx

77. 2 2 211 ; 393 xxx

78. 2 2 442 ; 5255 xxx

2.7 Transformations of Functions

2.7 Practice Problems

The graph of g is the graph of f shifted one unit up. The graph of h is the graph of f shifted two units down.

The graph of g is the graph of f shifted one unit to the right. The graph of h is the graph of f shifted two units to the left.

3. The graph of   23 fxx is the graph of   gxx  shifted two units to the right and three units up.

4. The graph of  2 12 yx can be obtained from the graph of 2 yx  by first shifting the graph of 2 yx  one unit to the right. Reflect the resulting graph about the xaxis, and then shift the graph two units up

5. The graph of y = 2x 4 is obtained from the graph of y = 2x by shifting it down by four units. We know that  if 0 if 0. yy y yy   

This means that the portion of the graph on or above the x-axis (0) y  is unchanged while the portion of the graph below the x-axis (y < 0) is reflected above the x-axis. The graph of   24yfxx is given on the right.

The graph of g is the graph of f stretched vertically by multiplying each of its y-coordinates by 2.

8. Start with the graph of yx  . Shift the graph one unit to the left, then stretch the graph vertically by a factor of three. Shift the resulting graph down two units.

Shift the graph one unit right to graph

 1. yfx

Compress horizontally by a factor of 2.

Multiply each x-coordinate by 1 2 to graph

 21.yfx

Compress vertically by a factor of 1 2

Multiply each y-coordinate by 1 2 to graph

 1 2 21.yfx

Shift the graph up three units to graph

1 2 213.yfx

2.7 Concepts and Vocabulary

1. The graph of ()3yfx is found by vertically shifting the graph of ()yfx  three units down.

2. The graph of (5)yfx is found by horizontally shifting the graph of ()yfx  five units to the left.

3. The graph of ()yfbx  is a horizontal compression of the graph of ()yfx  is b is greater than 1.

4. The graph of ()yfx  is found by reflecting the graph of ()yfx  about the yaxis.

5. False. The graphs are the same if the function is an even function.

6. True

7. False. The graph on the left shows 2 yx  first shifted up two units and then reflected about the x-axis, while the graph on the right shows 2 yx  reflected about the x-axis and then shifted up two units.

8. True

2.7 Building Skills

9. a. The graph of g is the graph of f shifted two units up.

b. The graph of h is the graph of f shifted one unit down.

10. a. The graph of g is the graph of f shifted one unit up.

b. The graph of h is the graph of f shifted two units down.

11. a. The graph of g is the graph of f shifted one unit to the left.

b. The graph of h is the graph of f shifted two units to the right.

12. a. The graph of g is the graph of f shifted two units to the left.

b. The graph of h is the graph of f shifted three units to the right.

13. a. The graph of g is the graph of f shifted one unit left, then two units down.

b. The graph of h is the graph of f shifted one unit right, then three units up.

14. a. The graph of g is the graph of f reflected about the x-axis.

b. The graph of h is the graph of f reflected about the y-axis.

15. a. The graph of g is the graph of f reflected about the x-axis.

b. The graph of h is the graph of f reflected about the y-axis.

16. a. The graph of g is the graph of f stretched vertically by a factor of 2.

b. The graph of h is the graph of f compressed horizontally by a factor of 2.

17. a. The graph of g is the graph of f vertically stretched by a factor of 2.

b. The graph of h is the graph of f horizontally compressed by a factor of 2.

18. a. The graph of g is the graph of f shifted two units to the right, then one unit up.

b. The graph of h is the graph of f shifted one unit to the left, reflected about the x-axis, and then shifted two units up.

19. a. The graph of g is the graph of f reflected about the x-axis and then shifted one unit up.

b. The graph of h is the graph of f reflected about the y-axis and then shifted one unit up.

20. a. The graph of g is the graph of f shifted one unit to the right and then shifted two units up.

b. The graph of h is the graph of f stretched vertically by a factor of three and then shifted one unit down.

21. a. The graph of g is the graph of f shifted one unit up.

b. The graph of h is the graph of f shifted one unit to the left.

22. a. The graph of g is the graph of f shifted one unit left, vertically stretched by a factor of 2, reflected about the y-axis, and then shifted 4 units up.

b. The graph of h is the graph of f shifted one unit to the right, reflected about the x-axis, and then shifted three units up.

23. e 24. c 25. g 26. h

27. i 28. a 29. b 30. k

31. l 32. f 33. d 34. J

2 fxx 

62.  1 3 fxx 

63.  2 21 fxx

Start with the graph of  2 , fxx  then shift it two units right and one unit up.

64.  2 35 fxx

Start with the graph of  2 , fxx  then shift it three units right and five units down.

65.  2 53fxx

Start with the graph of  2 , fxx  then shift it three units right. Reflect the graph across the x-axis. Shift it five units up.

66.  2 21fxx

Start with the graph of  2 , fxx  then shift it one unit left. Reflect the graph across the x-axis. Shift it two units up.

67.   13 fxx

Start with the graph of   , fxx  then shift it one unit left and three units down.

68.   21 fxx

Start with the graph of   , fxx  then shift it two units right and one unit up

69.   12fxx

Start with the graph of   , fxx  then shift it one unit left. Reflect the graph across the yaxis, and then shift it two units up.

70.   23 fxx

Start with the graph of   , fxx  then shift it two units left. Reflect the graph across the xaxis, and then shift it three units down.

71.   12 fxx

Start with the graph of   , fxx  then shift it one unit right and two units down.

72.   31 fxx

Start with the graph of   , fxx  then shift it three units left. Reflect the graph across the x-axis, and then shift it one unit up.

73.  1 3 1 fx x 

Start with the graph of   1 , fxx  then shift it one unit right and three units up.

74.  1 2 2 fx x  

Start with the graph of   1 , fxx  then shift it two units left. Reflect the graph across the x-axis and then shift up two units up.

75.  2 211fxx

Start with the graph of  2 , fxx  then shift it one unit left. Stretch the graph vertically by a factor of 2, then shift it one unit down.

76.   2 1 12 3 fxx

Start with the graph of  2 , fxx  then shift it one unit left. Compress the graph vertically by a factor of 1/3, then shift it two units up.

77.   2 1 23 2 fxx

Start with the graph of  2 , fxx  then shift it three units right. Compress the graph vertically by a factor of 1/2, reflect it across the x-axis, then shift it two units up.

78.  2 133fxx

Start with the graph of  2 , fxx  then shift it three units right. Stretch the graph vertically by a factor of 3, reflect it across the x-axis, then shift it one unit up.

79.   213fxx

Start with the graph of   , fxx  then shift it one unit left. Stretch the graph vertically by a factor of 2, and then shift it three units down.

80.   221fxx

Start with the graph of   , fxx  then shift it two units right. Compress the graph horizontally by a factor of 1/2, and then shift it one unit up.

Start with the graph of   , fxx  then shift it one unit right. Stretch the graph vertically by a factor of 2, then reflect it across the x-axis. Shift the graph up two units.

Start with the graph of

, fxx  then shift it three units right. Compress the graph vertically by a factor of 1/2, then reflect it across the y-axis. Reflect the graph across the x-axis, and then shift the graph down one unit.

111. a. b.

112. a. b.

113. a. b.

114. a.

115. a. b.

116. a.

117. a. b.

118. a.

2.7 Applying the Concepts

119. ()()800gxfx

120. ()1.05() hxfx 

121.     ()1.02500 pxfx

122.

() 1.02if 30,000 fxfx gxfxfx

123. a. Shift one unit right, stretch vertically by a factor of 10, and shift 5000 units up.

b. (400)5000104001$5199.75 C 

124. a. For the center of the artery, R = 3 mm and r = 0.

22 1000309000 mmminute v 

b. For the inner linings of the artery, R = 3 mm and r = 3 mm

22 1000330 mmminute v 

c. Midway between the center and the inner linings, R = 3 mm and r = 1.5 mm

 22 100031.56750 mmminute v 

125. a. Shift one unit left, reflect across the x-axis, and shift up 109,561 units.

b. 2 2 69,160109,561(1) 40,401(1) 2011200$2.00 p p pp

c. 2 2 0109,561(1) 109,561(1) 3311330$3.30 p p pp

126. Write R(p) in the form 2 3()phk : 22 2 2 ()36003(200)

Complete the square 3(20010,000)30,000 3(100)30,000 Rppppp pp p

To graph this, shift R(p) 100 units to the right, stretch by a factor of 3, reflect about the xaxis, and shift by 30,000 units up.

127. The first coordinate gives the month; the second coordinate gives the hours of daylight. From March to September, there is daylight more than half of the day each day. From September to March, more than half of the day is dark each day.

128. The graph shows the number of hours of darkness.

2.7 Beyond the Basics

129. The graph is shifted one unit right, then reflected about the x-axis, and finally reflected about the y-axis. The equation is

130. The graph is shifted two units right and then reflected about the y-axis. The equation is 

2.gxfx 

131. Shift two units left, then 4 units down.

132.   22 2 ()6699 (3)9 fxxxxx x 

Shift three units right, then 9 units down.

133.

22 2 ()2211 (1)1 fxxxxx x

Shift one unit right, reflect about the x-axis, then shift one unit up.

134.   22 2 ()2211 (1)1 fxxxxx x

136.

Shift one unit right, stretch vertically by a factor of 2, then shift two units down.

2 2 2 2 ()263.5 233.5 232.253.522.25 21.51 fxxx xx xx x 



 Shift 1.5 units left, stretch vertically by a factor of 2, then shift one unit down.

137.

Shift one unit left, reflect about the x-axis, then shift one unit up. 135.



  22 2 2 ()283243 244324 2211 fxxxxx xx x 



Shift two units left, stretch vertically by a factor of 2, reflect across the x-axis, then shift eleven units up.

Shift 0.5 unit right, stretch vertically by a factor of 2, reflect across the x-axis, then shift 0.5 unit down.

2.7 Critical Thinking/Discussion/Writing

145. a.   2 yfx is the graph of   yfx  shifted two units left. So the x-intercepts are –1 – 2 = –3, 0 – 2 = –2, and 2 – 2 = 0.

b.   2 yfx is the graph of   yfx  shifted two units right. So the x-intercepts are –1 + 2 = 1, 0 + 2 = 2, and 2 + 2 = 4.

c.   yfx  is the graph of   yfx  reflected across the x-axis. The x-intercepts are the same, –1, 0, 2.

d.   yfx  is the graph of   yfx  reflected across the y-axis. The x-intercepts are the opposites, 1, 0, –2.

e.   2 yfx  is the graph of   yfx  compressed horizontally by a factor of 1/2. The x-intercepts are 1 2 ,0,1.

f.   1 2 yfx  is the graph of   yfx  stretched horizontally by a factor of 2. The x-intercepts are –2, 0, 4.

146. a.   2 yfx is the graph of   yfx  shifted two units up. The y-intercept is 2 + 2 = 4.

b.   2 yfx is the graph of   yfx  shifted two units down. The y-intercept is 2 – 2 = 0.

c.   yfx  is the graph of   yfx  reflected across the x-axis. The y-intercept is the opposite, –2.

d.   yfx  is the graph of   yfx  reflected across the y-axis. The y-intercept is the same, 2.

e.   2 yfx  is the graph of   yfx  stretched vertically by a factor of 2. The y-intercept is 4.

f.   1 2 yfx  is the graph of   yfx  compressed horizontally by a factor of 1/2. The y-intercept is 1.

147. a.   2 yfx is the graph of   yfx  shifted two units left. The domain is  12,323,1.  The range is the same, 2,1.

b.   2 yfx is the graph of   yfx  shifted two units down. The domain is the same, 1,3. The range is  22,124,1. 

c.   yfx  is the graph of   yfx  reflected across the x-axis. The domain is the same, 1,3. The range is the opposite, 1,2.

d.   yfx  is the graph of   yfx  reflected across the y-axis. The domain is the opposite, 3,1. The range is the same, 2,1.

e.   2 yfx  is the graph of   yfx  stretched vertically by a factor of 2. The domain is the same, 1,3. The range is

22,214,2.

f.   1 2 yfx  is the graph of   yfx  compressed horizontally by a factor of 1/2. The domain is the same, 1,3. The range is

111 2222,11,.

148. a.   2 yfx is the graph of   yfx  shifted two units left. So the relative maximum is at x = 1 – 2 = –1, and the relative minimum is at x = 2 – 2 = 0.

b.   2 yfx is the graph of   yfx  shifted two units down. The locations of the relative maximum and minimum do not change. Relative maximum at x = 1, relative minimum at x = 2.

c.   yfx  is the graph of   yfx  reflected across the x-axis. The relative maximum and relative minimum switch. The relative maximum occurs at x = 2, and the relative minimum occurs at x = 1.

d.   yfx  is the graph of   yfx  reflected across the y-axis. The relative maximum and relative minimum occur at their opposites. The relative maximum occurs at x = –1, and the relative minimum occurs at x = –2.

e.   2 yfx  is the graph of   yfx  stretched vertically by a factor of 2. The locations of the relative maximum and minimum do not change. Relative maximum at x = 1, relative minimum at x = 2.

f.   1 2 yfx  is the graph of   yfx  compressed horizontally by a factor of 1/2. The locations of the relative maximum and minimum do not change. Relative maximum at x = 1, relative minimum at x = 2.

2.7 Getting Ready for the Next Section

222 557946143

2332262565

222 5623912153

333225355

154.

155.  2 23 56 fxx xx  

The function is not defined when the denominator is zero.

156.  2 2 4 fxx x 

The function is not defined when the denominator is zero.

2 402202,2xxxx

The domain is 

,22,22,.  

157.   23fxx

The function is defined only if 2x 3 ≥ 0. 3 23023 2 xxx 

The domain is 3 ,. 2 

158.  1 52 fx x 

The function is defined only if 5 2x > 0. 5 52025 2 xxx 

The domain is 5 ,. 2  

159. 1 0 10 x x 

First, solve 101xx and 10010.xx So the intervals are (,1),(1,10), and (10,).

Interval Test point Value of 1 10 x x Result (,1)  0 1 10 + (1,10) 5 4 5 –(10,)  15 14 5 +

Note that the fraction is undefined if x = 10.

The solution set is (1,10).

160.  2 3 0 1 x 

Set the denominator equal to zero and solve for x  2 101 xx 

The intervals are  ,1 and   1,. 

Interval Test point Value of  2 3 1 x Result

 ,1 0 3 –  1,  2 3 –

There is no solution. The solution set is 

161. 2 28 0 1 x x   

Set the numerator and denominator equal to zero and solve for x. 280284 xxx  2 10 x  has no real solution.

The intervals are   ,4 and   4,. 

Interval Test point Value of 2 28 1 x x   Result   ,4 0 8 +   4,  5 1 13 –

The solution set is   4,. 

162.      31 0 51 xx xx  

Set the numerator and denominator equal to zero and solve for x      3103,1 5105,1 xxx xxx  

The intervals are     ,1,1,1,

    1,3, 3,5,and 5,.  (continuedonnextpage)

1,3 fxxgxx 

The domain of f is

and the domain of g is

,3.

The intersection of Df and Dg

The domain of fg is [1, 3].

The domain of f g is [1, 3).

The domain of g f is (1, 3]. 3.

005001gfgfgg

2 2,21fxxgxx

2 2 22 2 221 2441 8821289

2 22 21 22112

2 2 2 42 42 21 2211 24411 883 ggxggxgx x xx xx

2 1, 3 fxxgx x 

Let

 | is defined. Axgx 

gx is not defined if x = 3, so

,33,. A  

| is defined. Bxfgx 

2231 1 333fgxxx xxx

   fgx is not defined if x = 3 or if 1 0. 3 x x  101xx

Interval Test point Value of 1 3 x x Result

,1 0 1 3 +

1,3 2 1

3,  4 3 +

 fgx is not defined for   1,3, so

   ,13,. B   (continuedonnextpage)

(continued)

The domain of fg  is     ,13,. AB 

6.     ;3 fxxgxx    33 fgfxx   fg  is not defined if 303xx or 3. x  fg  is defined for x < 3. Thus, the domain of fg  is   ,3.

7.   2 1,4 fxxgxx 

a.    2 2 4 41 fgxfgxfx x   

The function  2 4 gxx  is defined for 22 40422. xxx  So, 2,2. A 

The function     fgx is defined for 22 222 41041 4133 33 xx xxx x    So, 3,3. B  

The domain of fg  is 3,3. AB   

b.  

2 41 415 gfxgfxx xx

The function   1 fxx is defined for 101.xx So,   1,. A 

The function     gfx is defined for 505, or 5. xxx  So,

,5. B

The domain of gf  is 1,5. AB   8.   2 2 1 ,21 21 Hxgxx x   If  1 fx x  , then

2 2 33 39 Afgtfgft

b.  2 2 996324At

The area covered by the oil slick is 3241018   square miles.

10. a.   4500 rxx

b.   0.060.94 dxxxx 

c. i.

 0.940.944500rdxrxx 

 4500 0.944500 0.944230 drxdx x x  

0.9442300.944500 270 drxrdx xx

2.8 Concepts and Vocabulary

1. 

fgxfxgx 

2. The domain of the function f + g consists of those values of x that are common to the domains of f and g

3. The composition of the function f with the function g is written as fg  and is defined by

fgxfgx

4. The domain of the composite function fg  consists of those values of x in the domain of g for which   gx is in the domain of f

5. False. For example, if   2 fxx  and

2 , gxx  then

2 24.gfxgfxgxx

while

6. True

7. False. The domain of fg  may include

0, gx  but the domain of f g cannot include

0. gx 

8. True

2.8 Building Skills 9.

(1)2(1) 12 (1)1 ff gg

fgfg

 2 1(1)(1) 1(1)110

2 0(0)(0) (10)(01)0 fgfg

2 2(2)(2) (12)(21)9 fgfg

 (1)112 10 (1)11 ff gg

1(1)(1) 1 2(1)10 12 fgfg

 0(0)(0) 12 2(0)11 2 02

 2(2)(2) 15 2(2)1 2 22

d.  1 (1)13 12 1 (1)2(1)19 33 ff gg

   2 1(1)(1) 1 3(1) (1)6(1)8 159 4 1515 fgfg 

 2 0(0)(0) 0 (30)3 06(0)8 fgfg

 2 2(2)(2) 22 (32)1 0 26(2)8 fgfg

the product does not exist. d.  2 1 1 (1)1 16(1)8 3 1 (1)3126 ff gg

25. a. 2 3; domain: (,) fgxx

b. 22 33; domain: (,) fgxxxx 

c. 232 (3)3; domain: (,) fgxxxx 

d. 2 3 ; domain: (,0)(0,) fx gx

e. 2 ; domain: (,3)(3,) 3 gx fx

26. a. 2 21; domain: (,) fgxx

b. 22 2121; domain: (,) fgxxxx

c. 232 (21)2; domain: (,) fgxxxx

d. 2 21 ; domain: (,0)(0,) fx gx

e. 2 11 ; domain: 2122,,gx fx

27. a. 

  323212524; domain: (,) fgxxxx 

323212526; domain: (,)

 32 532 125 2525; domain: (,) fgxx xxx

d. 3 2 1 ;domain: (,) 25 fx gx

e. 2 3 25 ;domain: (,1)(1,) 1 gx fx

28. a.

  22 2 468 264;domain: (,) fgxxx xx

22468612; domain: (,)

 22 432 468 642432; domain: (,) fgxxx xxxx



 2 2 22 42 244 68 xx fxx gxxx xx

  

The denominator of the original fraction = 0 if x = 2 or x = 4, so the domain is (,2)(2,4)(4,).  

e.      2 2 24 684 222 4 xx gxxx fxxx x   

The denominator of the original fraction = 0 if x = 2 or x = 4, so the domain is (,2)(2,2)(2,).  

29. a. 21; domain: [0,) fgxx

b. 21; domain: [0,) fgxx

c. (21)2; domain: [0,) fgxxxxx  

d. 21 ; domain: (0,) fx gx 

e. ; the numerator is defined only 21 for 0, while the denominator0 when 111 , so the domain is 2220,,. gx fx x x  

30. a. 1; domain: [0,) fgxx 

b. 1; domain: [0,) fgxx 

c.   1; domain: [0,) fgxx

d. 1 ; domain: (0,) fx gx 

e.   ; the numerator is defined only 1 for 0, while the denominator0 when 1, so the domain is 0,11,. gx fx x x    

31. a. 63; domain: [3,) fgxx 

b. 63; domain: [3,) fgxx 

c.   63; domain: [3,) fgxx 

d. 6 ; domain: (3,) 3 fx gx 

3 ; the numerator is defined only 6 for 3, while the denominator0 when 6, so the domain is 3,66,. gx fx

32. a.   21; domain: ,1 fgxx

b.   21; domain: ,1 fgxx

c.  

21; domain: ,1 fgxx 

d.  2 ; domain: ,1 1 fx gx

33. a.

1 ; the numerator is defined only 2 for 1, while the denominator0 when 2, so the domain is .22,1. gx fx x

22 21 1 1 121 = 1 211 11 fg xx xxxx xx xxxxx xxxx

domain: ,11,00,

The denominator equals zero when x = 0 or x = 1, so the domain is

,11,00,11,.

34. a. 11 11 fg xx

Neither f nor g is defined for x = 0, so the domain is (,0)(0,). 

b. 112 11; domain: (,0)(0,). fg xxx

c. 22 11111 1; domain: (,0)(0,). fgx xxxxx

d. 11 1 1 11 x fxx x g xx 

Neither f nor g is defined for x = 0, so the domain is (,0)(0,).  

e. 11 1 11 1 1 gxx fxx xx 

Neither f nor g is defined for x = 0, and g/f is not defined for x = 1, so the domain is (,0)(0,1)(1,). 

35. a. 22 111 fgxx xxx

Neither f nor g is defined for x = 1, so the domain is (,1)(1,).

b. 22 . 111 domain:(,1)(1,). fgxx xxx

c.  2 22 . 11 1 domain:(,1)(1,). fgxx xxx

d. 2 2 1 . Neither nor 1 fxfg x gx x    is defined for x = 1, and fg is not defined for x = 0, so the domain is   (,1)1,0(0,). 

e. 1 . Neither nor 2 2 1 x fxxfg g x    is defined for x = 1, so the domain is

(,1)1,.

a.  5141099 111 91 9 1 fgxxx xxx x x

Neither f nor g is defined for x = 1, so the domain is

(,1)1,. 

b. 5141011 111fgxxx xxx

Neither f nor g is defined for x = 1, so the domain is

(,1)1,.

c. 2 2 51410204610 11 21 fgxxxx xxxx

Neither f nor g is defined for x = 1, so the domain is

d. 51 51 1 410 410 1 x fx x x gx x 

Neither f nor g is defined for x = 1 and fg is not defined for 52, x  so the domain is

55 ,,11,. 22

e. 410 410 1 51 51 1 x gx x x fx x 

Neither f nor g is defined for x = 1 and gf is not defined for 15, x  so the domain is  11 ,11,,.

f is not defined for x = 1, g is not defined for x = ±1, and f + g is not defined for either 1 or 1, so the domain is

,11,11,.

2 2 222 2 32 22 2 1 22 1 111 2 2 11 21 2 1 11 xxx fgxx xxxx xxx xxx xx xxxxx xxx

f is not defined for x = 1, g is not defined for x = ±1, and fg is not defined for 1, so the domain is 

,11,11,.

c. 23 232 22 1 11 fgxxx xxxxx

f is not defined for x = 1, g is not defined for x = ±1, and fg is not defined for either 1 or 1, so the domain is

,11,11,. 

d.  2 22 2 1 1 1 2 122 1 x fxx

f is not defined for x = 1, g is not defined for x = ±1, and fg is not defined for either 1, 0, or 1, so the domain is

,11,00,11,.

Neither f nor g is defined for x = 1 and gf is not defined for x = 0 or x = 1, so the domain is

,11,00,11,.

2 2 3 25 3 920 54 34 35 55 x fx gx xx xx xx xxxx

f is not defined for x = 5 and x = 5, g is not defined for x = 4 and x = 5, and fg is not defined for x = 5, so the domain is

,55,44,33,55,. 

f is not defined for x = 5 and x = 5, g is not defined for x = 4 and x = 5, and f + g is not defined for 5, 5 or 4, so the domain is

,55,44,55,.

22 33 25920 33 5545 3435 554 fgxx xxx xx xxxx xxxx

22 32 32 12815 425100 927 425100 xxxx xxx x xxx

f is not defined for x = 5 and x = 5, g is not defined for x = 4 and x = 5, and f + g is not defined for 5, 5, or 4, so the domain is

,55,44,55,.

22 2 22 2 432 33 25920 3 25920 69 95225500 fgxx xxx x xxx xx xxxx

f is not defined for x = 5 and x = 5, g is not defined for x = 4 and x = 5, and fg is not defined for 5, 5, or 4, so the domain is

39. 

2 2 3 920 3 25 55 35 34 54 x gxx fx x xx xx xxxx

f is not defined for x = 5 and x = 5, g is not defined for x = 4 and x = 5, and gf is not defined for x = 4, so the domain is

,55,44,3 3,55,.

  

  1;5 fxxgxx 

a. 15 fgxx 

f is not defined for x < 1, g is not defined for x > 5. The domain is [1, 5].

b. 1 5 fx gx 

f is not defined for x < 1, g is not defined for x > 5. The denominator is zero when x = 5. The domain is [1, 5).

40.     2; 2 fxxgxx 

a. 2 224fgxxx 

f is not defined for x < 2, g is not defined for x < –2. The domain is   2,. 

b. 2 2 fx gx  

f is not defined for x < 2, g is not defined for x < –2. The denominator is zero when x = –2. The domain is   2,. 

41.   2 2;9 fxxgxx 

a. 2 29 fgxx 

f is not defined for x < –2, g is defined for [–3, 3]. The domain is 2,3.

b. 2 2 9 fx gx  

f is not defined for x < –2, g is defined for [–3, 3]. The denominator is zero when x = –3 or x = 3. The domain is   2,3.

42.   22 4;25 fxxgxx 

a. 22 425 fgxx 

f is defined for x ≤ –2 or x ≥ 2, g is defined for [–5, 5]. The domain is  5,22,5. 

b. 2 2 4 25 fx gx 

f is defined for x ≤ –2 or x ≥ 2, g is defined for [–5, 5]. The denominator is zero when x = –5 or x = 5. The domain is    5,22,5. 

43.

 22 2 2 21321; 222139; 32(3)1319 gfxxx gf gf

44.  

2 2 2 2 3113211; 2321126; 33(3)1111 gfxxxx gf gf   

45.    2 222(2)3111 fg 

46.  2 222(2)1347 gf

48.  2 522(5)13159 gf 

50.

2 11 22135 22 gf

51.    22 22()3145fgccc

52.

  22 223145fgccc  53. 

2 2 2 2213 24413 881 gfaa aa aa

 2 2 2 22()13 24413 881 gfaa aa aa    

 122(1)117 ff  

56.

 2 2 122(1)331 gg 

57. 

1 ;105 fxgxx x 

1 105 fgx x  

The domain of fg  is the set of all real numbers such that 1050, x  or 2. x  The domain of fg  is     ,22,. 

58.   1 ; fxgxx x   1 fgx x  

The domain of f is the set of all real numbers such that 0. x  The domain of   gx is   0,.  Therefore, the domain of fg  is

0,.  59.

;28 28 fxxgxx fgxx

The domain of fg  is the set of all real numbers such that 280, x  or 4. x  The domain of fg  is   4,. 

60.

; fxxgxx fgxx

The domain of fg  is the set of all real numbers such that 0, x  or 0. x  The domain of fg  is   ,0.

61.  222 11 1 1 fgxx xx xx

The domain of g is (,0)(0,).   Since –1 is not in the domain of f, we must exclude those values of x that make g(x) = –1. 1 11 x x 

Thus, the domain of fg  is

,11,00,. 

62.  11 22(3) 1 33 33 . 11 fgx x xx xx xx

The domain of g is (,3)(3,).   Since 1 is not in the domain of f, we must exclude those values of x that make g(x) = 1. 2 1231 3 xx x 

Thus, the domain of fg  is

,33,11,.  

63. 

 (23)313. fgxxx

The domain of g is (,).  Since f is not defined for   ,3, we must exclude those values of x that make g(x) < 3. 1 23331 3 xxx 

Thus, the domain of fg  is 1 ,. 3

64.  2525 (25)115 fgxxx xx

The domain of g is (,).  Since f is not defined for x = 1 we must exclude those values of x that make g(x) = 1. 1 25151 5 xxx

Thus, the domain of fg  is 11 ,,. 55

65.  2 1; domain: (,) fgxx 

66.

312; domain: (,) fgxx

67. a. 

 2(4)325; domain: (,) fgxxx 

 (23)421; domain: (,) gfxxx

c.  

 2(23)349; domain: (,) ffxxx 

 (4)48; domain: (,) ggxxx

68. a. 

 (35)338; domain: (,) fgxxx 

b. 

3(3)5314; domain: (,) gfxxx

c. 

(3)36; domain: (,) ffxxx

d.     3(35)5920; domain: (,) ggxxx 

69. a.  22 12(1)21; domain: (,) fgxxx

b.  22 1(12)442; domain: (,) gfxxxx 

c.     12(12)41; domain: (,) ffxxx 

d.  2242 1(1)22; domain: (,) ggxxxx

70. a.  22 2(2)343; domain: (,) fgxxx

 

b.  22 2(23)82418; domain: (,) gfxxxx 

c.     2(23)349; domain: (,) ffxxx 

d.    2 24 228; domain: (,) ggxxx 

71. a.  2 2 2 2(21)3(21) 2(441)63 821;domain: (,) fgxxx xxx xx  

b.  22 2(23)1461; domain: (,) gfxxxxx   

c. 

2 22 4322 432 223323 2412969 824249; domain: (,) ffxxxxx xxxxx xxxx

d. 

  2(21)143; domain: (,) ggxxx

72. a.  22 (2)3(2)46; domain: (,) fgxxxxx

b.  22 2(3)26; domain: (,) gfxxxxx 

c. 

2 22 4322 432 333 6939 6129; domain: (,) ffxxxxx xxxxx xxxx

d.  

2(2)4;domain: (,) ggxxx

73. a.    2 ;domain: [0,) fgxxx

b.  2 ;domain: (,) gfxxx

c. 

2 24 ;domain: (,) ffxxx

d.  4 ;domain: [0,) ggxxx 

74. a.    2 222 222;domain: [2,) fgxxx xx  

b.  2 22;domain: (,) gfxxx 

c.  222 4322 432 (2)2(2) 4424 464; domain: (,) ffxxxxx xxxxx xxxx

d.  22;domain: [2,) ggxx 

The domain of g is (,0)(0,).   Since 1 2 is not in the domain of f, we must find those values of x that make   1 2 . gx 

2 2 11 22 2

Thus, the domain of is ,22,00,22,.

b.   2 2 1 21 1 21 gfx x 

The domain of f is 11 ,,. 22

Since 0 is not in the domain of g, we must find those values of x that make   0. fx  However, there are no such values, so the domain of gf  is 11 ,,. 22

c.  11 221 1 21 21 21 2121 3223 ffx x xx xx xx 

The domain of f is 11 ,,. 22

21 23 x x is defined for 33 ,,, 22

so the domain of ff  is 1133 ,,,. 2222

d.  4 2 1 1 ggxx x 

The domain of g is (,0)(0,),   while 4 ggx   is defined for all real numbers. Thus, the domain of gg  is (,0)(0,).

76. a.  (1)1 1 111fgxxxx xxx

The domain of g is (,1)(1,).   Since f is defined for all real numbers, there are no values that must be excluded. Thus, the domain of fg  is (,1)(1,).

b.  11 (1)1 gfxxx xx   

The domain of f is all real numbers. Since g is not defined for x = –1, we must exclude those values of x that make f(x) = –1.

110xx Thus, the domain of gf  is (,0)(0,).  

c.     (1)12; domain: (,) ffxxx

d.  11 1 21 1 11 xx xxx ggx xxxx xx

The domain of g is 

 ,11,,

while 21 x x  is defined for

11 22 ,,.

The domain of gg  is

11 22 ,11.,.

77. a.    41;domain: ,3 fgxx

b.   41;domain: 1,17 gfxx 

c.    11; domain: 2, ffxx 

d.   44;domain: 12,4 ggxx 

78. a.    2 4422 fgxxx   domain: [ 2, 2]

b.   2 2 44gfxx  domain: 6,22,6   

c.     2 2 42 42 44 8164 812 ffxx xx xx

2 22 22 4444 44 xx xxx

domain: [ 2, 2]

The domain of g is (,4)(4,). 

The denominator of fg  is 0 when

 so the domain of fg

The domain of f is (,2)(2,).   The denominator of gf  is 0 when 7 5 , x  so, the domain of gf  is

The domain of f is (,2)(2,). 

The denominator of ff  is 0 when 5, x  so, the domain of ff  is

The domain of g is (,4)(4,).   The denominator of gg  is 0 when 19 3 , x  so the domain of gg  is

The domain of g is (,1)(1,). 

The denominator of fg  is 0 when x = 2, so the domain of fg

,11,22,.

The domain of f is (,3)(3,).   The denominator of gf  is never 0, so, the domain of gf  is (,3)(3,).  

c. 

The domain of f is (,3)(3,).   The denominator of ff  is 0 when

, x  so, the domain of ff

The domain of g is (,1)(1,).   The denominator of gg  is never 0 so the domain of gg  is (,1)(1,).   In exercises 81 90, sample answers are given. Other answers are possible.

81. ()2(),()2 Hxxfxxgxx

82. ()32(),()32

83.  10 2102 ()3(),()3 Hxxfxxgxx 

84. 22 ()35()5,()3 Hxxfxxgxx 

85. 11 ()(),()35 35 Hxfxgxx xx 

86. 55 ()(),()23 23 Hxfxgxx xx  

87. 3 22 3 ()7(),()7 Hxxfxxgxx 

88. 4 2 4 2 ()1(), ()1 Hxxxfxx gxxx  

89. 3 3 11 ()(),()1 1 Hxfxgxx xx 

90. 3 3 ()1(),()1 Hxxfxxgxx 

2.8 Applying the Concepts

91. a. ()fx is the cost function.

b. ()gx is the revenue function.

c. ()hx is the selling price of x shirts including sales tax.

d. ()Px is the profit function.

 

92. a. ()(50005) 4(50005)12,000 20,0002012,000 32,00020 CpCp p p p





b. 2 ()(50005)50005 Rppxpppp 

 

c. 2 2 ()()() 50005(32,00020) 5502032,000 PpRpCp ppp pp



93. a. ()()()25(3505) 20350 PxRxCxxx x 



b. (20)20(20)35050. P  This represents the profit when 20 radios are sold.

c. ()20350;5002035043 Pxxxx 

This function represents the revenue in terms of the cost C

94. a. ()0.04 gxx 

b. ()hx is the after tax selling price of merchandise worth x dollars.

c.   ()0.023 fxhx

d. ()Tx represents the total price of merchandise worth x dollars, including the shipping and handling fee.

95. a. ()0.7 fxx 

b. ()5gxx

96. a. ()0.8 fxx 

b. ()0.9 gxx 

0.7(5)(0.75) 0.73.50.75 $1.50

0.90.80.72 gfxxx

0.80.90.72 fgxxx 

e. They are the same.

97. a. ()1.1;()8 fxxgxx

b.       1.181.18.8fgxxx   . This represents a final test score computed by first adding 8 points to the original score and then increasing the total by 10%.

c.     1.18gfxx  This represents a final test score computed by first increasing the original score by 10% and then adding 8 points.

d.    



701.170885.8; 701.170885.0; fg gf  

fgxgfx

1.18.89073.82fgxxx 

 1.189074.55gfxxx  

98. a. ()fx is a function that models 3% of an amount x

b. ()gx represents the amount of money that qualifies for a 3% bonus.

c. Her bonus is represented by     . fgx 

d. 2000.03(17,5008000)$485 

e. 5212000.03(8000)$18,700 xx 

99. a. 2 () fxx  

b. 2 ()(30)gxx  

c. ()() gxfx represents the area between the fountain and the fence.

d. The circumference of the fence is 2(30) x     10.52(30)4200 (30)200 x x     3020020030.xx   22 22 ()()(30) (60900) 60900. Now substitute 20030 for to compute the estimate: 1.75[60(20030)900]

1.75(12,000900)$16,052. gxfxxx xxx x x           

100. a. 2 2 ()180(28)(4) 1440360(4) fxxx xx    

b. ()2(180)36022 gxxxxx 

c. ()() fxgx represents the area of the track.

d. (i) First find the radius of the inner track: 270 9002360xx    . Use this value to compute ()().fxgx (continuedonnextpage)

(continued)

101.

(ii) The outer perimeter

c. They are the same.

102.

c. They are the same.

2.8 Beyond the Basics

103. a. When you are looking for the domain of the sum of two functions that are given as sets, you are looking for the intersection of their domains. Since the x-values that f and g have in common are 2, 1, and 3, the domain of f + g is { 2, 1, 3}. Now add the y-values.

2303 12(2)0 3022 fg fg

Thus, f + g = {( 2, 3), (1, 0), (3, 2)}.

b. When you are looking for the domain of the product of two functions that are given as sets, you are looking for the intersection of their domains. Since the x-values that f and g have in common are 2, 1, and 3, the domain of f + g is { 2, 1, 3}.Now multiply the y-values.

Thus, fg = {( 2, 0), (1, 4), (3, 0)}.

c. When you are looking for the domain of the quotient of two functions that are given as sets, you are looking for the intersection of their domains and values of x that do not cause the denominator to equal zero. The x-values that f and g have in common are 2, 1, and 3; however, g( 2) = 0, so the domain is {1, 3}. Now divide the y-values.

Thus,   1,1,3,0. f g 

d. When you are looking for the domain of the composition of two functions that are given as sets, you are looking for values that come from the domain of the inside function and when you plug those values of x into the inside function, the output is in the domain of the outside function.

     20,fgf  which is undefined

021,fgf

123,fgf

321fgf

Thus,        0,1,1,3,3,1. fg  

104. When you are looking for the domain of the sum of two functions, you are looking for the intersection of their domains. The domain of f is [ 2, 3], while the domain of g is [ 3, 3].

The intersection of the two domains is [ 2, 3], so the domain of f + g is [ 2, 3].

For the interval [ 2, 1],

f + g = 2x + (x + 1) = 3x + 1.

For the interval (1, 2)

f + g = (x + 1) + (x + 1) = 2x + 2.

For the interval [2, 3],

f + g = (x + 1) + (2x 1) = 3x.

Thus,

31if 21 22if 12 3if 23. xx fgxxx xx

105. a. ()()(())()() ()() is an even function. fxhxhxhxhx fxfx

 

 

b.  ()()(())()() () is an odd function. gxhxhxhxhx gxgx

c.  ()()() ()()() ()()2() ()()()() () 222 fxhxhx gxhxhx fxgxhx hxfxgxfxgx

    

h(x) is the sum of an even function and an odd function.

108.   1if 20 1if 02fxx xx      1,22fxxx

 1if 20 11if 01 1if 12 x fxxxx xx       

 gxfxfx 

If 20, x  then   11. gxxxx 

If 01, x  then       110.gxxx

If 12, x  then

106. a. 22 2 ()23() (even), ()23 (odd) or ()+ 3 (even), ()2 (odd)

   hxxxfxx gxxfxx gxx

b.    ()() (even), 2 () (odd) 2 xx hxxxfx xx gxx

107.  1 2 x fx x    fx is defined if 1 0 2 x x  and 20. x 

2022 xxx 

Thus, the values 2 and 2 are not in the domain of f 1

0 2 x x  if 10 x  and 20, x  or if 10 x  and 20. x 

Case 1: 10 x  and 20. x  10111 xxx  20222 xxx 

Thus, 10 x  and 2011. xx 

Case 2: 10 x  and 20. x     101,11, xx  

   203,22, xx  

Thus, 10 x  and 20 x      ,22,. 

The domain of f is      ,21,12,.  

 1121.gxxxx 

Writing g as a piecewise function, we have

if 20 0if 01 21if 12 xxx gxx xx

2.8 Critical Thinking/Discussion/Writing

109. a. The domain of ()fx is (,0)[1,).  

b. The domain of ()gx is [0, 2].

c. The domain of ()() fxgx  is [1, 2].

d. The domain of () () fx gx is [1, 2).

110. a. The domain of f is (,0).  The domain of ff  is  because 1 1 ff x   and the denominator is the square root of a negative number.

b. The domain of f is (,1)  . The domain of ff  is (,0)  because 1 1 1 1 ff x   and the denominator must be greater than 0. If x = 0, then the denominator = 0.

111. a. The sum of two even functions is an even function. ()() and ()() fxfxgxgx ()()()()()() fgxfxgxfxgx  ()(). fgx

112.

b. The sum of two odd functions is an odd function.

()() and ()() fxfxgxgx 

()()()()()() fgxfxgxfxgx  ()(). fgx

c. The sum of an even function and an odd function is neither even nor odd.

() even()() and () odd fxfxfxgx

()()()() gxgxfxgx    ()(),fxgx  which is neither even nor odd.

d. The product of two even functions is an even function.

()() and ()() fxfxgxgx

()()()()()() fgxfxgxfxgx  ()(). fgx

e. The product of two odd functions is an even function.

()() and ()() fxfxgxgx 

()()()()()() fgxfxgxfxgx  ()(). fgx

f. The product of an even function and an odd function is an odd function.

() even()() and () odd fxfxfxgx

()() ()()()() gxgx fxgxfxgxfgx

a. ()() and ()() fxfxgxgx 

fgxfgx  is odd.

b. ()() and ()() fxfxgxgx

fgxfgxfgx

fgx  is even.

c. () odd()() and fxfxfx 

 () even()() gxgxgxfgx

fgxfgxfgx   is even.

d. () even()() and () odd fxfxfxgx

()() gxgxfgxfgx

is even.

2.8 Getting Ready for the Next Section

113. a. Yes, R defines a function. b.

2,3,1,1,3,1,1,2 S

No, S does not define a function since the first value 1 maps to two different second values, 1 and 2.

114. The slope of 25 1, 52 PP   while the slope of y = x is 1. Since the slopes are the negative reciprocals, the lines are perpendicular. The midpoint of PP  is 255277 2222,,,

which lies on the line y = x. Thus, y = x is the perpendicular bisector of PP 

115. 3 2332 2 x xyxyy 

116. 221,011 xyyxyxy  117. 2222 2 4,04 4 xyxxy xy  

118. 111 233232 1 23 xxx yyy y x 

2.9 Inverse Functions

2.9 Practice Problems

1.  2 1 fxx is not one-to-one because the horizontal line y = 1 intersects the graph at two different points.

2. a.  1 123 f  b.   94 f 

Because

,fgxgfxx  the two functions are inverses.

4. The graph of 1 f is the reflection of the graph of f about the line y = x

5.   23fxx is a one-to-one function, so the function has an inverse. Interchange the variables and solve for y:

6. Interchange the variables and solve for y:

7.  3 fxx x

The function is not defined if the denominator is zero, so the domain is

,33,

The range of the function is the same as the domain of the inverse, thus the range is

8. G is one-to-one because the domain is restricted, so an inverse exists.

2 21 1,0. Interchange the variables and solve for : 1,01. Gxyxx y xyyyGxx

9. From the text, we have 11 33 5 dp 111650 333597 5 d 

The bell was 3597 feet below the surface when the gauge failed.

2.9 Concepts and Vocabulary

1. If no horizontal line intersects the graph of a function f in more than one point, the f is a one-to-one function.

2. A function f is one-to-one if different x-values correspond to different y-values.

3. If   3 fxx  , then  1 1 3 fxx 

4. The graphs of a function f and its inverse 1 f are symmetric about the line y = x

5. True

6. True. For example, the inverse of   fxx  is  1 fxx  .

7. False.  1 fx means the inverse of f.

8. True

2.9 Building Skills

9. One-to-one 10. Not one-to-one 11. Not one-to-one 12. One-to-one

13. Not one-to-one 14. Not one-to-one

15. One-to-one 16. Not one-to-one

17. 1 (2)7(7)2 ff 18. 1 (4)7(7)4ff 19. 1 (1)2(2)1 ff

20. 1 (3)5(5)3ff

21. 1 ()() fabfba  22. 1 ()() fcdfdc 

23.      11337337337ffff  24.      11252525ffff  

25. a. (3)2(3)33 f 

b. Using the result from part (a), 1 (3)3. f 

c. 

11191919ffff

26. a. 3 (2)28 f 

b. Using the result from part (a), 1 (8)2 f 

c. 

11151515ffff

11272727ffff

27. a. 3 (1)112 f 

b. Using the result from part (a), 1 (2)1 f 

28. a.   3 3 3 (1)21111

2 362(3) 32 3 33 221(3) 1 333 3626 5 3 23 5 3 xxx fgxxxx xxx xxx xx xx x xx x

32322(1) 2 111 32323(1) 3 111 3222 5 1 3233 5 1 xxx gfxxxx xxx xxx xx xx x xx x

43. a. One-to-one

b. 1 ()153. Interchange the variables and solve for :153 151 ()5. 33 fxyx yxy x yfxx  

d. Domain of f:   ,  ; x-intercept of f: 5;

y-intercept of f: 15 domain of 1 : f   ,  ; x-intercept of 1 : f 15; y-intercept of 1 : f 5

44. a. One-to-one

b. 1 ()25. Interchange the variables and solve for :25 515 (). 222 gxyx yxy x ygxx   

d. Domain of g:   , 

x-intercept of g: 5 2 y-intercept of g: 5 domain of 1 : g   ,  ; x-intercept of 1 : g 5; y-intercept of 1 : g 5 2

45. a. Not one-to-one

46. a. Not one-to-one

47. a. One-to-one

b. 12 ()3 Interchange the variables and solve for :3 3()(3). fxyx yxy xyyfxx

  

d. Domain of f:   0,  ; x-intercept of f: none; y-intercept of f: 3 domain of 1 : f   3,  ; x-intercept of 1 :3 f ; y-intercept of 1 : f none

48. a. One-to-one

 

b. 12 ()4. Interchange the variables and solve for :4 4()(4) fxyx yxy xyyfxx



d. Domain of f:   0,  ; x-intercept of f: 16; y-intercept of f: 4 domain of 1 : f   ,4 ; x-intercept of 1 : f 4; y-intercept of 1 : f 16

49. a. One-to-one

d. Domain of g:   ,  ; x-intercept of g: 1; y-intercept of g: 1 domain of 1 : g   ,  ; x-intercept of 1 : g 1; y-intercept of 1 : g 1

50. a. One-to-one

b. 3 3 313 ()1. Interchange the variables and solve for :1 1()1. hxyx yxy xyygxx

  

d. Domain of h:   ,  ; x-intercept of h: 1; y-intercept of h: 1 domain of 1 : h   ,  ; x-intercept of 1 : h 1; y-intercept of 1 : h 1

51. a. One-to-one



b. 1 1 (). Interchange the variables 1 1 and solve for :(1)1 1 111 1()1. fxy x yxxy y yyfxx xxx

  

b. 3 3 313 ()1. Interchange the variables and solve for :1 1()1 gxyx yxy xyygxx

  

d. Domain of f:     ,11,  

x-intercept of f: none; y-intercept of f: 1 domain of 1 : f     ,00,  

x-intercept of 1 : f 1

y-intercept of 1 : f none

52. a. One-to-one

b. 1 ()1. Interchange the variables 11 and solve for :1 gxy x y yxx yy

  1 11(1)1 11 (). 11 xyyxyyyx ygx xx

 

d. Domain of g:     ,00, 

x-intercept of g: 1; y-intercept of g: none domain of 1 : g     ,11, 

x-intercept of 1 : g none

y-intercept of 1 : g 1

53. a. One-to-one



b.   2 2 12 ()21. Interchange the variables and solve for :21 2121 ()2143 fxyx yxy xyxy yfxxxx







d. Domain of f:   1,  ; x-intercept of f: none; y-intercept of f: 3

Domain of 1 : f   2,  ;

x-intercept of 1 : f 3

y-intercept of 1 : f none

54. a. One-to-one



b.   2 2 12

()12. Interchange the variables and solve for : 1212 12 ()1221 fxyx y xyxy xy yfxxxx

  

d. Domain of f:   2,  ; x-intercept of f: 1; y-intercept of f: 12

Domain of 1 : f   1, 

x-intercept of 1 : f 12

y-intercept of 1 : f 1

In exercises 55 and 56, use the fact that the range of f is the same as the domain of 1 f

55. Domain: (,2)(2,)  

Range: (,1)(1,)  

56. Domain: (,1)(1,)  

Range: (,3)(3,)  

57. 1 (). Interchange the variables 2 1 and solve for :21 2 fxyx x y yxxyxy y     21(1)21xyyxyxx  1 21 (). 1 yfxx x  

Domain of f: (,2)(2,)  

Range of f: (,1)(1,).  

58. 2 (). Interchange the variables 1 2 and solve for :2 1 2(1)2 x gxy x y yxxyxy y xyyxyxx

1 22 (). 11 xx ygx xx

Domain of g: (,1)(1,) 

Range of g: (,1)(1,).  

59. 1 12 (). Interchange the variables 1 12 and solve for : 1 1221 1 (2)1(). 2 fxyx x y yx y xxyyxyyx yxxyfxx x



Domain of f: (,1)(1,)  

Range of f: (,2)(2,).  

60. 1 1 (). Interchange the variables 3 1 and solve for :31 3 31(1)31 31 (). 1 hxyx x y yxxyxy y xyyxyxx yhxx x 



Domain of h: (,3)(3,)  

Range of h: (,1)(1,).  

61. f is one-to-one since the domain is restricted, so an inverse exists.



 2 2 ,0. Interchange the variables and solve for : ,0. fxyxx y xyyxx



62. g is one-to-one since the domain is restricted, so an inverse exists.



 2 2 ,0. Interchange the variables and solve for : ,0. gxyxx y xyyxx



63. f is one-to-one since the domain is restricted, so an inverse exists.

  ,0. Interchange the variables and solve for :,0. fxyxxx yyxx

 

64. g is one-to-one since the domain is restricted, so an inverse exists.

  ,0. Interchange the variables and solve for :,0. gxyxxx yyxx

 

65. f is one-to-one since the domain is restricted, so an inverse exists.



 2 2 1,0. Interchange the variables and solve for : 11,1. fxyxx y xyyxx



66. g is one-to-one since the domain is restricted, so an inverse exists.



 2 2 5,0. Interchange the variables and solve for : 55,5. gxyxx y xyyxx



67. f is one-to-one since the domain is restricted, so an inverse exists.



 2 2 2,0. Interchange the variables and solve for : 22,2. fxyxx y xyyxx



68. g is one-to-one since the domain is restricted, so an inverse exists.



 2 2 1,0. Interchange the variables and solve for : 11,1. gxyxx y xyyxx



2.9 Applying the Concepts

69. a. 1 ()273 ()273(). KCC CKKKC  

This represents the Celsius temperature corresponding to a given Kelvin temperature.

b. (300)30027327C C 

c. (22)22273295K K 

70. a. The two points are (212, 373) and (32, 273). The rate of change is 3732731005 212321809  273(32)52297 99 bb  52297 (). 99 KFF

b. 5229722975 9999 92297 922975() 55 KFKF KFFKK  

This represents the Fahrenheit temperature corresponding to a given Kelvin temperature.

c. 52297 (98.6)(98.6)310K 99 K 

71. a. (())(273)92297 55 99(273)2297 555 91609 32 555 FKCC C CC   

b. 52297 (())273 99 522972457 99 5160 99 CKFF F F   

72. 95160 (())323232 599 59160160160 (())3295999 FCxxx x CFxxx x           Therefore, F and C are inverses of each other.

73. a. ()0.75, Exx  where x represents the number of dollars ()1.25, Dxx  where x represents the number of euros.

b. (())0.75(1.25)0.9375. EDxxxx  Therefore, the two functions are not inverses.

c. She loses money either way.

74. a. 40.0540.05 wxwx  2080.xw This represents the food sales in terms of his hourly wage.

b. 20(12)80$160 x 

75. a. 740.05$60 xx  . This means that if food sales ≤ $60, he will receive the minimum hourly wage. If food sales > $60, his wages will be based on food sales. 40.05if 60 7if 60 xx w x   

b. The function does not have an inverse because it is constant on (0, 60), and it is not one-to-one.

c. If the domain is restricted to [60,),  the function has an inverse.

76. a. 2 1.11. 1.11 T Tll     This shows the length as the function of the period.

b. 2 2 3.2 ft 1.11 l

c. 1.11709.3 sec T 

77. a. 21 1 8() 864 V VxxVxVx  This represents the height of the water in terms of the velocity.

b. (i)   2 1 3014.0625 ft 64 x  (ii)   2 1 206.25 ft 64 x 

78. a. 2 642 yxx  has no inverse because it is not one-to-one across its domain, [0, 32]. (It fails the horizontal line test.)

However, if the domain is restricted to [0, 16], the function is one-to-one, and it has an inverse.

22 2 6422640 64648 4 644096864210242 44 3210242 2 yxxxxy y x yy x y

1024200512. yy 

(Because y is a number of feet, it cannot be negative.) This is the range of the original function. The domain of the original function is [0, 16], which is the range of the inverse.

The range of 3210242 2 y x   is [16, 32], so this is not the inverse. The range of 3210242 , 2 y x  0512, y  is [0, 16], so this is the inverse.

Note that the bottom half of the graph is the inverse.

b. (i) 6440968(32) 0.51 ft 4 x 

(ii) 6440968(256) 4.69 ft 4 x 

(iii) 6440968(512) 16 ft 4 x 

79. a. The function represents the amount she still owes after x months.

b. 36,000600. Interchange the variables and solve for : 36,000600 yx yxy  

60036,00060 600 x yxy  1 1 ()60. 600 fxx 

This represents the number of months that have passed from the first payment until the balance due is $x

c. 1 60(22,000)23.3324 months 600 y 

There are 24 months remaining.

80. a. To find the inverse, solve 2 8321200xpp for p: 2 2 83212000 32(32)4(8)(1200) 2(8) 32102438,40032 16 32323737632422336 16 16 1 222336 4 ppx x p x xx x

Because the domain of the original function is (0, 2], its range is [1168, 1200).

So the domain of the inverse is [1168, 1200), and its range is (0, 2]. The range of 1 222336 4 px is (2, 4], so it is not the inverse. The range of 1 222336,11681200, 4 pxx  is (0, 2], so it is the inverse. This gives the price of computer chips in terms of the demand x

Note that the bottom half of the graph is the inverse.

b. 1 22(1180.5)2336$0.75 4 p 

2.9 Beyond the Basics



81. ((3))(1)3,((5))(3)5, and ((2))(4)2(()) for each . ((1))(3)1,((3))(5)3, and ((4))(2)4(()) for each . fgffgf fgffgxxx gfggfg gfggfxxx





 So, f and g are inverses.

82. ((2))(1)2,((0))(2)0, ((3))(3)3, and ((2))(1)2(()) for each . ((1))(2)1,((2))(0)2, ((3))(3)3, and ((4))(1)4 (()) for each . So and are inve fgffgf fgf fgffgxx x gfggfg gfggfg gfxxx fg 

  rses.

83. a.

b. f is not one-to-one

c. Domain: [–2, 2]; range: [0, 2]

84. a. Domain: (,2)[3,)  . Note that the domain is not (,2)(2,)  because  2 x  for 23. x 

b. The function is not one-to-one. The function is constant on each interval [n, n + 1), n an integer.

85. a. f satisfies the horizontal line test.

b. 1 1. Interchange the variables 1 1 and solve for :1 1 1 111 1 y x yx y xyxyx y       1 (1) () 11 xyyxyxx yfxxx xx 

c. Domain of f: (,1)(1,);   range of f: (,1)(1,).  

86. a. g satisfies the horizontal line test.

b. 2 2 2222 12 1. Interchange the variables and solve for :1 11 ()1 yx yxy xyyx ygxx

c. Domain of f = range of f: [0, 1]

87. a. (i) 1 ()21. Interchange the variables and solve for : 21 11 () 22 fxyx yxy yfxx

(ii) 1 ()34. Interchange the variables and solve for : 34 14 () 33 gxyx yxy ygxx   

(iii)

(iv) 

234167fgxxx 

 321461gfxxx   (v)

1 67. Interchange the variables and solve for : 17 67 66 fgxyx y xyfgxx

1 61. Interchange the variables and solve for : 11 61 66 gfxyx y

11 1141 2332 12111 63266 fgxx xx

11 1114 3223 11417 66366 gfxx

b. (i) 

1 11 17 66fgxx gfx

   (ii) 

1 11 11 66gfxx fgx

88. a. (i) 1 ()23. Interchange the variables and solve for : 23 13 () 22 fxyx yxy yfxx

(ii) 3 3 1 3 ()1. Interchange the variables and solve for : 1 ()1 gxyx yxy ygxx

(iii)    33 21321fgxxx  (iv)  3 32 (23)1 8365426 gfxx xxx

(i)

1 3 11 13 1 22gfxx fgx

2.9 Critical Thinking/Discussion/Writing

89. No. For example, 3 () fxxx  is odd, but it does not have an inverse, because (0)(1),ff  so it is not one-to-one.

90. Yes. The function  (0,1) f  is even, and it has an inverse:  1 (1,0). f 

91. Yes, because increasing and decreasing functions are one-to-one.

92. a. {(1,1),(0,0),(1,1)} R  b. {(1,1),(0,0),(1,2)} R 

2.9 Getting Ready for the Next Section

93.  2 1234xxxx 

94.  2 5623xxxx 

95.  2 2842xxxx 

96.  2 71025xxxx 

97.  2 7120 340 3040 34 xx xx xx xx

Solution: {3, 4}

98.

99.

2 60 230 2030 23 xx xx xx xx

Solution: {–2, 3}

 2 3720 3120 31020 12 3 xx xx xx x x    

Solution:  1 3 2,

100. 2 410xx

Use the quadratic formula. a = 1, b = –4, c = 1

  2 2 4 2 44411 21 412423 23 22 bbac x a  

 

 

Solution:  23,23 

101.

 2 23 yx

Start with the graph of  2 , fxx  then shift it two units left and three units down.

102.  2 13 yx

Start with the graph of  2 , fxx  then shift it one unit right and three units up.

103.  2 12 yx

Start with the graph of  2 , fxx  then shift it one unit left. Reflect the graph across the xaxis and then shift it two units up

104.  2 31 yx

Start with the graph of  2 , fxx  then shift it three units right. Reflect the graph across the x-axis and then shift it one unit down.

Chapter 2 Review Exercises

Building Skills

1. False. The midpoint is 33111 22,(0,6).

2. False. The equation is a circle with center (2,3) and radius 5.

3. True

4. False. A graph that is symmetric with respect to the origin is the graph of an odd function. A graph that is symmetric with respect to the y-axis is the graph of an even function.

5. False.

The slope is 4/3 and the y-intercept is 3.

6. False. The slope of a line that is perpendicular to a line with slope 2 is –1/2.

7. True

8. False. There is no graph because the radius cannot be negative.

9. a. 22 (,)(13)(35)25dPQ 

b. 3(1)53,(1,4) 22 M 

c. 351 132 m 

10. a. 22 (,)(3(3))(15)62dPQ 

b. 335(1),(0,2) 22 M 

c. 15 1 3(3) m 

11. a. 22 (,)(94)(8(3))52dPQ 

b. 493(8)1311 ,, 2222 M  

c. 8(3) 1 94 m 

12. a. 22 (,)(72)(83)202dPQ 

b. 2(7)3(8)55 ,, 2222 M 

c. 8311 729 m 

13. a. 22 (,)(52)(2(7))34DPQ 

b. 257(2)79 ,, 2222 M 

c. 2(7)5 523 m 

14. a. 22 (,)(10(5))(34)274dPQ 

b. 5104(3)51 ,, 2222 M

c. 347 10(5)15 m 

15.

22 22 22 ,203568 ,300534 ,320334

Using the Pythagorean theorem, we have

ACCB

Alternatively, we can show that AC and CB are perpendicular using their slopes. 0550(3)3 ; 3033(2)5

ACCB mm mmACCBABC

1, so is a right triangle.

22 22 22 (,)(,) (5)9(4)49 (5)9(4)49 1034865 3131

 



The point is ,0. 1818 dACdBC xx xx xxxx x 

20. 22 2 22 2 (3,2),(2,1),(0,) (,)(0(3))((2)) (2)9 (,)(0(2))((1)) (1)4 ABCy dACy y dBCy y     

22 22 22 (,)(,) (2)9(1)4 (2)9(1)4 41325

4 The point is 0,4. dACdBC yy yy yyyy y 

21. Not symmetric with respect to the x-axis; symmetric with respect to the y-axis; not symmetric with respect to the origin.

dAB dCD dAC dBD





16. 22 22 22 22 (,)(41)(82)35 (,)(107)(5(1))35 (,)(71)(12)35 (,)(104)(58)35





The four sides are equal, so the quadrilateral is a rhombus.

22. Not symmetric with respect to the x-axis; not symmetric with respect to the y-axis; symmetric with respect to the origin.

23. Symmetric with respect to the x-axis; not symmetric with respect to the y-axis; not symmetric with respect to the origin.

AB dAO dBO



17. 22 22 (6,3),(4,5) (,)(60)(30)45 (,)(40)(50)41



 (4, 5) is closer to the origin.

24. Symmetric with respect to the x-axis; symmetric with respect to the y-axis; symmetric with respect to the origin.

25. x-intercept: 4; y-intercept: 2; not symmetric with respect to the x-axis; not symmetric with respect to the y-axis; not symmetric with respect to the origin.

ABC dAC dBC 



18. 22 22 (6,4),(5,10),(2,3) (,)(2(6))(34)65 (,)(25)(310)58

 (5, 10) is closer to (2, 3).



ABCx dACx x dBCx x



19. 22 2 22 2 (5,3),(4,7),(,0) (,)((5))(03) (5)9 (,)(4)(07) (4)49







26. x-intercept: 4; y-intercept: –3; not symmetric with respect to the x-axis; not symmetric with respect to the y-axis; not symmetric with respect to the origin.

27. x-intercept: 0; y-intercept: 0; not symmetric with respect to the x-axis; symmetric with respect to the y-axis; not symmetric with respect to the origin.

28. x-intercept: 0; y-intercept: 0; symmetric with respect to the x-axis; not symmetric with respect to the y-axis; not symmetric with respect to the origin.

29. x-intercept: 0; y-intercept: 0; not symmetric with respect to the x-axis; not symmetric with respect to the y-axis; symmetric with respect to the origin.

30. x-intercept: 0; y-intercept: 0; not symmetric with respect to the x-axis; not symmetric with respect to the y-axis; symmetric with respect to the origin.

31. No x-intercept; y-intercept: 2; not symmetric with respect to the x-axis; symmetric with respect to the y-axis; not symmetric with respect to the origin.

32. x-intercepts: –1, 1; y-intercept: 1; not symmetric with respect to the x-axis; symmetric with respect to the y-axis; not symmetric with respect to the origin.

33. x-intercepts: –4, 4; y-intercepts: –4, 4; symmetric with respect to the x-axis; symmetric with respect to the y-axis; symmetric with respect to the origin.

34. x-intercepts: 2,2; y-intercept: –4 not symmetric with respect to the x-axis symmetric with respect to the y-axis not symmetric with respect to the origin.

35. 22 (2)(3)25 xy

36. The center of the circle is the midpoint of the diameter.  5(5)24 0,3. 22 M     

The length of the radius is the distance from the center to one of the endpoints of the diameter = 22 (50)(23)26.  The equation of the circle is 22(3)26.xy

37. The radius is 2, so the equation of the circle is 22 (2)(5)4. xy

38. 2 25102. 5 xyxy  Line with slope 2/5 and y-intercept –2.

39. 5 152105. 252 xy xyxy  Line with slope 5/2 and y-intercept –5.

40. Circle with center (–1, 3) and radius 4.

41. 22 22 2440 2144414 xyxy xxyy   22 (1)(2)9. xy Circle with center (1, –2) and radius 3.

42. 2222 2222 3366022 2121(1)3. xyxxxy xxyxy

  Circle with center (1, 0) and radius 3.

43. 22(1)24yxyx 

44. 5055 0222;5myx 

45. 73 2;32(1)5 11 25 mbb yx  

46. 1 x 

47. a. 323;323yxmyxm 

The slopes are equal, so the lines are parallel.

b. 35735; xym 532053 xym 

The slopes are neither equal nor negative reciprocals, so the lines are neither parallel nor perpendicular.

c. 0; axbycmab  0 bxaydmba 

The slopes are negative reciprocals, so the lines are perpendicular.

d. 11 2(3); 33yxm  53(3)3yxm 

The slopes are neither equal nor negative reciprocals, so the lines are neither parallel nor perpendicular.

48. a. The equation with x-intercept 4 passes through the points (0, 2) and (4, 0), so its slope is 021 402  Thus, the slope of the line we are seeking is also 1 2 . The line passes through (0, 1), so its equation is  11

101. 22 yxyx 

b. The slope of the line we are seeking is 2 and the line passes through the origin, so its equation is y 0 = 2(x 0), or y = 2x

49. Domain: {–1, 0, 1, 2}; range: {–1, 0, 1, 2}.

This is a function.

50. Domain: {0, 3}; range: {–2, –1, 1, 2}.

This is not a function.

51. Domain: (,); range: (,).

This is a function.

52. Domain: [2,); range: [0,).

This is a function.

53. Domain: [0.2,0.2]; range: [0.2,0.2].

This is not a function.

54. Domain: (,0]; range: [0,).

This is a function.

55. Domain: {1}; range: (,). 

This is not a function.

56. Domain: (,); range: {2}. 

This is a function.

57. Domain: (,); range: [0,).

This is a function.

58. Domain: [1,); range: (,).

This is not a function.

59. (2)3(2)15 f 

60. 2 (2)(2)22 g 

61. ()43141 fxxx 

62. 2 ()2222 gxxx

63. 

  2 1(1)(1) 3(1)1123 fgfg  

64.

 2 1(1)(1) 3(1)1(1)21 fgfg  

65.       2 2(2)(2) 3(2)1(2)210 fgfg  

66.



 2 0(0)(0) 023(0)12 gfgf  

2 3332122 fg  

2 23(2)1223 gf   69.    22 32135fgxxx  70.

2 2 312961gfxxxx   71.       331194ffxxx   72.    2 2422242ggxxxx 

73. ()3()1331 fahahah

75.

76.

77. Domain: (,); range: {3}.

78. Domain: (,); range: [2,). Decreasing on (,0);  increasing on (0,).

79.

2 Domain: ,;range: 0, 3

80. Domain: [6,6]; range: [0,6]. Increasing on (6,0) ; decreasing on (0,6).

81. Domain: (,); range: [1,). Decreasing on (,0);  increasing on (0,). 

82. Domain: (,); range: [0,). Decreasing on (,0);  increasing on (0,). 

83. The graph of g is the graph of f shifted one unit left.

84. The graph of g is the graph of f shifted one unit right, stretched vertically by a factor of 2, then shifted three units up.

85. The graph of g is the graph of f shifted two units right, and then reflected in the x-axis.

86. The graph of g is the graph of f shifted one unit left, then two units down.

87. 2424 ()()()() fxxxxxfx 

() is even. fx Not symmetric with respect to the x-axis; symmetric with respect to the y-axis; not symmetric with respect to the origin.

88. 33 ()()()() fxxxxxfx 

() is odd. fx Not symmetric with respect to the x-axis; not symmetric with respect to the y-axis; symmetric with respect to the origin.

89. ()33() fxxxfx 

() is even. fx Not symmetric with respect to the x-axis; symmetric with respect to the y-axis; not symmetric with respect to the origin.

90. ()35() or ()() fxxfxfxfx  is neither even nor odd. Not symmetric with respect to the x-axis; not symmetric with respect to the y-axis; not symmetric with respect to the origin.

91. ()() or ()() fxxfxfxfx  is neither even nor odd. Not symmetric with respect to the x-axis; not symmetric with respect to the y-axis; not symmetric with respect to the origin.

92. 2 ()()() is odd. fxfxfx x 

Not symmetric with respect to the x-axis; not symmetric with respect to the y-axis; symmetric with respect to the origin.

93.  2 2 ()4() where () and ()4. fxxfxghx gxxhxx   

94.    50 2 502 ()2() where () and ()2. gxxxgxfhx fxxhxxx   

95.  3 ()() where 25 3 () and (). 25 x hxhxfgx x fxxgxx x     

96.  3 3 ()(21)5() where ()5 and ()21. HxxHxfgx fxxgxx   

97. f(x) is one-to-one. ()2.fxyx 1

Interchange the variables and solve for : 22(). y xyyxfx 

98. f(x) is one-to-one. ()23. fxyx 1

Interchange the variables and solve for : 13 23(). 22 y xyyxfx 

99. f(x) is one-to-one. 3 ()2. fxyx 31 3

Interchange the variables and solve for : 22(). y xyyxfx 

100. f(x) is one-to-one. 3 ()81. fxyx 3 3 1 3

Interchange the variables and solve for : 1 81 8 1 1(). 2 y x

101. 1 1 (),2. 2

Interchange the variables and solve for . 1 21 2 21(1)21 2121 () 11

Domain of : (,2)(2,) Range of : (,1)(1,) x fxyx x y y xxyxy y xyyxyxx xxyyfx xx f f

102. 1 23 (),1. 1

Interchange the variables and solve for . 23 23 1 23(2)3 3 () 2

Domain of : (,1)(1,) Range of : (,2)(2,) x fxyx x y y xxyxy y xyyxyxx yfxx x f f

103. a. (3,3),(2,0),(0,1),(3,4).

Find the equation of each segment: 0(3) 3.03(2)6. 2(3)

The equation of is 36. AB ABCD mbb AByx

 101 ;1. 0(2)2 1

The equation of is 1. 2 BC mb BCyx   41 1;1. 30

The equation of is 1. CD mb CDyx  

So, 36if 32 1 ()1if 20 2 1if 03 xx fxxx xx

b. Domain: [–3,3]; range: [–3,4]

c. x-intercept: –2; y-intercept: 1

e. f.

k. f is one-to-one because it satisfies the horizontal line test.

Applying the Concepts

104. a. rate of change (slope) = 25.9519.2 0.45. 2510  19.20.45(10)14.7. bb 

The equation is 0.4514.7.Pd

b. The slope represents the amount of increase in pressure (in pounds per square inch) as the diver descends one foot deeper. The y-intercept represents the pressure at the surface of the sea.

c. 2 0.45(160)14.786.7 lb/in. P 

d. 104.70.4514.7200 feet d 

105. a.  173,00054,000 rate of change slope 223,00087,000 0.875   54,0000.875(87,000) b  22,125. b 

The equation is 0.87522,125.Cw

b. The slope represents the cost to dispose of one pound of waste. The x-intercept represents the amount of waste that can be disposed with no cost. The y-intercept represents the fixed cost.

c. 0.875(609,000)22,125$510,750 C 

d. 1,000,0000.87522,125 1,168,142.86 pounds w w  

106. a. At 60 mph = 1 mile per minute, so if the speedometer is correct, the number of minutes elapsed is equal to the number of miles driven.

b. The odometer is based on the speedometer, so if the speedometer is incorrect, so is the odometer.

107. a. 2 (2)10055(2)3(2)$198. f  She started with $100, so she won $98.

b. She was winning at a rate of $49/hour.

c. 2 0100553(20)(35) tttt  20,53.tt Since t represents the amount of time, we reject 53. t  Chloe will lose all her money after playing for 20 hours.

d. $100/20 = $5/hour.

108. If 100500 x  , then the sales price per case is $4 – 0.2(4) = $3.20. The first 100 cases cost $400. 4if 0100 ()3.280if 100500 3180if 500 xx fxxx

109.a.

110. a. Revenue = number of units × price per unit: 2 32 (50005010)(100.5) 5125300050,000 xpttt ttt

b. 100.5220.pttp  2 2 ()(220) 500050(220)10(220) 407008000, which is the

pp pp

number of toys made at price p. The revenue is

 2 407008000ppp

32 407008000. ppp 

111. a.

7.4474363.88yx

b. [70, 90, 2] by [150, 300, 25]

c.   7.447476363.88202 y  A player whose height is 76 inches weighs about 202 pounds.

Chapter 2 Practice Test A

1. The endpoints of the diameter are ( 2, 3) and ( 4, 5), so the center of the circle is  2(4) 35 22,3,4. C      

The length of the diameter is    2 2 4253822. 

Therefore, the length of the radius is 2. The equation of the circle is  22 342.xy

2. To test if the graph is symmetric with respect to the y-axis, replace x with –x: 22 3()2()1321, xxyxxy  which is not the same as the original equation, so the graph is not symmetric with respect to the y-axis. To test if the graph is symmetric with respect to the x-axis, replace y with –y:

22 32()1321, xxyxxy  which is the same as the original equation, so the graph is symmetric with respect to the x-axis. To test if the graph is symmetric with respect to the origin, replace x with –x and y with –y: 22 3()2()()1321, xxyxxy  which is not the same as the original equation, so the graph is not symmetric with respect to the origin.

3. 2 0(3)(1)0 or 3 or 1 xxxxxx  2 0(03)(01)0.yy  The x-intercepts are 0, 3, and –1; the y-intercept is 0.

4. Intercepts: 2 2 2213 2213 yyy xxx  

5. 71(2)9bb The equation is y = x + 9.

6. 7 8274 2 xyyx  the slope of the line is 4. 14(2)9. bb  So the equation is 49.yx

7.       2 2(2)(2) 2(2)123(2)2 (3)(12)36 fgfg 

8. 2 ((2))(2(2)3)(1)12(1)1 gfgg

9.  222 4322 432 (2)2(2) 4424 424 ffxxxxx xxxxx xxxx

10. a. 3 (1)(1)23 f 

b. 3 (0)022 f 

c. 2 (1)12(1)1 f 

11. 101; xxx must also be greater than or equal to 0, so the domain is [0, 1).

12.     2(4)72(1)7 (4)(1) 2 413 ff  

13. 44 22 33 ()2()2() () fxxxfx xx 

f(x) is even.

14. Increasing on (,0) and (2,); decreasing on (0,2).

15. Shift the graph of yx  three units to the right, then stretch the graph vertically by a factor of 2, and then shift the resulting graph four units up.

16. 22 2525(25)0(25) 025522.5 seconds tt tt 

17.  1 2772ff

18. 2 (). Interchange the variables 1 2 and solve for : 1 fxyx x y yx y 

20. a. (230)0.25(230)30$87.50 C  b. 57.500.2530110 miles mm 

Chapter 2 Practice Test B

1. To test if the graph is symmetric with respect to the y-axis, replace x with –x: 2222,xyxy  which is the same as the original equation, so the graph is symmetric with respect to the y-axis. To test if the graph is symmetric with respect to the xaxis, replace y with –y: 2222,xyxy  which is the same as the original equation, so the graph is symmetric with respect to the x-axis. To test if the graph is symmetric with respect to the origin, replace x with –x, and y with –y: 2222,xyxy  which is the same as the original equation, so the graph is symmetric with respect to the origin. The answer is D.

2. 22 093;099. xxyy 

The x-intercepts are ±3; the y-intercept is –9. The answer is B.

3. D 4. D 5. C

6. Suppose the coordinates of the second point are (a, b). Then 12 23 b a  Substitute each of the points given into this equation to see which makes it true. The answer is C.

7. Find the slope of the original line: 5 6352. 3 xyyx  The slope is 2. The equation of the line with slope 2, passing through (–1, 2) is 22(1).yx

The answer is D.

8.    22 32513. fgxxx  

The answer is B.

9.      2 22 43 222 88 ffxxxxx xxx   

The answer is A.

10. 1(1)2 (1). 1(1) aa ga aa  

The answer is C.

19. ()1001000Axx

11. 101; xx  x must also be greater than or equal to 0, so the domain is [0, 1].

The answer is A.

12.  223463100 5205,2 xxxx xxx

The answer is D.

13. A 14. A 15. B

16. D 17. C

18. (). Interchange the variables 32 and solve for : 32 fxyx x y yx y

The answer is C.

19. 5190;5(70)190160.wxw

The answer is B.

20. 500.225125. mm

The answer is A.

Cumulative Review Exercises (Chapters P–2)

1. a. 23 32662 23493 xyxyy yxyxx

b. 11 11 111 1 11 xyxyxy xyyxxy xyxy

2. a. 2 215(25)(3) xxxx 

b. 322 2 248(2)4(2) (4)(2) xxxxxx xx 

3. a. 75108192536383

b. 22 12(1)(2)(2)(1) 12(1)(2) (2)(2) (1)(2) 2 (1)(2) xxxxxx xxxx xxxx xx x xx

4. a. 112323 23 43 232323  

b. 115252 52 54 525252   

5. a. 3753124 xxx 

b. 13 There is no solution. 11xx 

6. a. 2 30(3)0 0or 3 xxxx xx  

b. 2 3100(5)(2)0 5 or 2 xxxx xx  

7. a. 2 114(2)(3) 230 2(2) 123123 44 xxx i xx    

b. 22 3 41290(23)0 2 xxxx 

8. a.     680420 416 or 24 xxxx xxxx

b. 2 11 10210. 1 Let . xx xx ux x

2 2 2 2 2

9. a. 2 22 2 312131(21) 314414720 7(7)4(4)(2) 717 . If 2(4)8 xxxx xxxxx x

while 717 210.281 8

22 2 2 2 2 1221 1221 1442121 43421 43421 1624916(21) 162493216

Check to make sure that neither solution is extraneous. The solution set is {0,89}.

10. a. 25118(,8) xx

b. 3453(,3) xx 

11. a. 323502804. xxx 

The solution set is (0, 4).

b. 512742623. xxx  The solution set is [–3, –2].

12. a. 2177217 62834 xx xx   The solution set is [–3, 4].

b. 2352354 or 2351. xxx xx  

The solution set is (,1][4,). 

13. 22 (,)(25)(2(2))5dAC  22 (,)(26)(25)5dBC  Since the lengths of the two sides are equal, the triangle is isosceles.

15. First, find the equation of the circle with center (2, –1) and radius determined by (2, –1) and (–3, –1): 22 2(3))(1(1))5 r 

The equation is 222 (2)(1)5. xy Now check to see if the other three points satisfy the equation: 22222 (22)(41)555,  222222 (52)(31)5345  (true because 3, 4, 5 is a Pythagorean triple), and 222222 (62)(21)5435. 

Since all the points satisfy the equation, they lie on the circle.

16. 22 22 22 6490 649. Now complete both squares: 6944994 xyxy xxyy xxyy  

 22 (3)(2)4. xy

The center is (3, –2) and the radius is 2. 17. 35yx

18. The x-intercept is 4, so (4, 0) satisfies the equation. To write the equation in slopeintercept form, find the y-intercept: 02(4)8bb

The equation is 28.yx

19. The slope of the perpendicular line is the negative reciprocal of the slope of the original line. The slope of the original line is 2, so the slope of the perpendicular is –1/2. Now find the y-intercept of the perpendicular: 1 1(2)0. 2 bb  The equation of the perpendicular is 1 . 2 yx 

14.

20. The slope of the parallel line is the same as the slope of the original line, 2. Now find the yintercept of the parallel line: 12(2) b 

5. b  The equation of the parallel line is 25.yx

21. The slope of the perpendicular line is the negative reciprocal of the slope of the original line. The slope of the original line is 7(1) 4 53  , so the slope of the perpendicular is –1/4. The perpendicular bisector passes through the midpoint of the original segment. The midpoint is 3517 22,(4,3).



Use this point and the slope to find the y-intercept: 1 3(4)4 4 bb  . The equation of the perpendicular bisector is 1 4. 4 yx

22. The slope is undefined because the line is vertical. Because it passes through (5, 7), the equation of the line is x = 5.

23. Use the slope formula to solve for x: 511 22(5)62106 5 2 xx x x

24. The line through (x, 3) and (3, 7) has slope –2 because it is perpendicular to a line with slope 1/2. Use the slope formula to solve for x: 37 22(3)432 3 5 xx x x

29. Let x = the number of books initially purchased, and 1650 x  the cost of each book. Then x – 16 = the number of books sold, and 1650 16 x  the selling price of each book. The profit = the selling price – the cost, so 2 16501650 10 16 16501650(16)10(16) 1650165026,40010160 xx xxxx xxxx



  2 2 1016026,4000 1626400(60)(44)0 xx xxxx



 60,44.xx Reject –44 because there cannot be a negative number of books. So she bought 60 books.

30. Let x = the monthly note on the 1.5 year lease, and 1.5(12)x = 18x = the total expense for the 1.5 year lease. Then x – 250 = the monthly note on the 2 year lease, and 2(12)(x – 250) = 24x – 6000 the total expenses for the 2 year lease. Then 1824600021,000 xx 4227,000642.86. xx So the monthly note for the 1.5 year lease is $642.86, and the monthly note for the 2 year lease is $642.86 – 250 = $392.86.

31. a. The domain of f is the set of all values of x which make 10 x  (because the square root of a negative number is not a real value.) So 1 or [1,) x  in interval notation is the domain.

b. 0132;013yyx  31918. xxx  The x-intercept is 8, and the y-intercept is –2.

c. (1)1133 f 

d. ()013013

198. In interval notation, this is(8,).

32. a. 2 2 (2)(2)2;(0)00; (2)24 ff f 

b. f decreases on (,0)  and increases on (0,). 

33. a.  11 222 22 2 fgxx xx xx  

Because 0 is not in the domain of g, it must be excluded from the domain of   . fg 

Because 2 is not in the domain of f, any values of x for which g(x) = 2 must also be excluded from the domain of  2 :21,fgx x   so 1 is excluded also. The domain of   fg  is (,0)(0,1)(1,).  

b.  2 2(2)24. 1 2 gfxxx x 

Because 2 is not in the domain of f, it must be excluded from the domain of   . gf 

Because 0 is not in the domain of g, any values of x for which f(x) = 0 must also be excluded from the domain of   gf 

However, there is no value for x which makes f(x) = 0. So the domain of   gf  is (,2)(2,).  

Chapter 2 Graphs and Functions

2.1 The Coordinate Plane

2.1 Practice Problems

3. ( ) ( ) ( ) ( ) 1122 ,5,2;,4,1 xyxy=-=-

( )( ) () ( ) () () 22 2121 22 2 2 4512 1121.4 dxxyy =-+=---+=+-=»

4. ( ) ( ) ( ) ( ) ( ) () 1122 33 ,6,2;,2,0 ,1,5 xyxy xy ===

( )( ) ( )() ()() 22 12121 22 22 2602 8268 dxxyy =-+=--+=-+-= ( )( ) ()() ()() 22 23131 22 22 1652 5334 dxxyy =-+=-+=-+= ( )( ) () ( ) () ()() 22 33232 22 22 1250 3534 dxxyy =-+=--+=+=

Yes, the triangle is an isosceles triangle.

We are asked to find the distance between the points A(60, 0) and B(0, 60). ( )( ) ()() () 22 222 (,)600060 6060260 60284.85 ft dAB =-+=+-= =»

6. ()21 56113 ,, 2222 M æ-+-ö+æö ==-ç÷ ç÷ èø èø

2.1 Concepts and Vocabulary

1. A point with a negative first coordinate and a positive second coordinate lies in the second quadrant.

2. Any point on the x-axis has second coordinate 0.

3. The distance between the points ( )11 , Pxy and ( )22 , Qxy is given by the formula () ( )( ) 22 2121 ,.dPQxxyy =-+-

4. The coordinates of the midpoint M(x, y) of the line segment joining ( )11 , Pxy and ( )22 , Qxy are given by () 1212 ,,. 22 xxyy xy ++ æö = ç÷ èø

5. True

6. False The point (7, 4) is 4 units to the right and 6 units below the point (3, 2).

7. False. Every point in quadrant II has a negative x-coordinate.

8. True.

2.1 Building Skills

9. (2, 2): Q1; (3, 1): Q4; ( 1, 0): x-axis ( 2, 5): Q3; (0, 0): origin; ( 7, 4): Q2 (0, 3): y-axis; ( 4, 2): Q2

10.a. Answers will vary. Sample answer: (2,0),-(1,0),(0,0),(1,0),(2,0)The y-coordinate is 0.

The set of all points of the form (x, 1) is a horizontal line that intersects the y-axis at 1.

11.a. If the x-coordinate of a point is 0, the point lies on the y-axis.

The set of all points of the form (–1, y) is a vertical line that intersects the x-axis at –1.

12.a. A vertical line that intersects the x-axis at –3.

b. A horizontal line that intersects the y-axis at 4.

13.a. 0 y > b. 0 y <

c. 0 x < d. 0 x >

14.a. Quadrant III b. Quadrant I

c. Quadrant IV d. Quadrant II

In Exercises 15 24, use the distance formula, ( )( ) 22 2121 dxxyy =-+- and the midpoint formula, 1212(,),22 xxyy xy ++ æö = ç÷ èø.

15.a. 222 (22)(51)44 d =-+-==

b. () 2215 ,2,322 M ++ æö == ç÷ èø

16.a. 222 (23)(55)(5)5 d =--+-=-=

b. () 3(2)55,0.5,522 M +-+ æö == ç÷ èø

17.a. 22 22 (2(1))(3(5)) 3213 d =--+--=+=

b. ( ) 125(3),0.5,422 M -+-+æö ==ç÷ èø

18.a. 22 22 (7(4))(91) (3)(10)109 d =---+-=-+-=

b. ( ) 4(7)1(9),5.5,422 M -+-+æö ==-ç÷ èø

19.a. 22 22 (3(1))(6.51.5) 4(8)8045 d =--+-=+-==

b. ( ) 131.5(6.5),1,2.522 M -++æö ==ç÷ èø

20.a. 22 22 (10.5)(10.5) (0.5)(1.5)2.5510 22 d =-+-=+-===

b. ( ) 0.510.5(1),0.75,0.2522 M ++æö ==ç÷ èø

21.a. ( )222 22(54)11 d =-+-==

b. () 2245 ,2,4.522 M æö ++ == ç÷ èø

22.a. 22 2 (()())() (2)2 dvwvwtt ww =+--+==

b. () ()() ,,22 vwvwtt Mvt -+++ æö == ç÷ èø

23.a. ( )( ) ( ) 22 2222 2222 2 ()() 22 24222 2()2 dkttk ktkttktk ttkkttkk tktk =-+=-++-+ =-+=-+ =-=-

b. , 22 tkkt M ++ æö = ç÷ èø

24.a. ( )( ) ( ) 22 2222 22 22 2 ()() 22 242 22 2()2 dnmmn nmnmmmnn mmnn mmnn mnmn =--+-=+++++ =++ =++ =+=+

b. ()() , 22 , 22 mnnm M mnnm +-+æö = ç÷ èø æö = ç÷ èø

25. P = (–1, –2), Q = (0, 0), R = (1, 2) ( )( ) ( )( ) 22 22 22 22 (,)0(1)0(2)5 (,)(10)(20)5 (,)1(1)2(2) 242025

dPQ

dQR

dPR =--+--= =-+-= =--+-=+==

Because d(P, Q) + d(Q, R) = d(P, R), the points are collinear.

26. P = (–3, –4), Q = (0, 0), R = (3, 4) ( )( ) ( )( ) 22 22 22 22 (,)0(3)0(4)255 (,)(30)(40)255 (,)3(3)4(4) 6810010

dPQ

dQR dPR =--+--== =-+-== =--+-=+==

Because d(P, Q) + d(Q, R) = d(P, R), the points are collinear.

27. P = (4, –2), Q = (1, 3), R = (–2, 8) ()( ) 22 22 (,)143(2)34 (,)(21)(83)34 dPQ dQR =-+--= =--+-= ( )( ) 22 22 (,)248(2) (6)10136234

dPR =--+-=-+==

Because d(P, Q) + d(Q, R) = d(P, R), the points are collinear.

28. It is not possible to arrange the points in such a way so that d(P, Q) + d(Q, R) = d(P, R), so the points are not collinear.

29. P = (–1, 4), Q = (3, 0), R = (11, –8) ( )() ( )( ) 22 22 22 22 (,)3(1)0442 (,)(113)((8)0)82 (,)11(1)84 (12)(12)288122 dPQ dQR dPR =--+-= =-+--= =--+-=+-==

Because d(P, Q) + d(Q, R) = d(P, R), the points are collinear.

30. It is not possible to arrange the points in such a way so that d(P, Q) + d(Q, R) = d(P, R), so the points are not collinear.

31. It is not possible to arrange the points in such a way so that d(P, Q) + d(Q, R) = d(P, R), so the points are not collinear.

32. P = (1, 7), Q = (–3, 7.5), R = (–7, 8) ( )( ) ( )() 22 22 22 22 (,)317.5716.25 (,)(7(3))(87.5) 16.25 (,)7187 (8)165216.25

dPQ

dQR

dPR =--+-= =---+= =--+=-+==

Because d(P, Q) + d(Q, R) = d(P, R), the points are collinear.

33. First, find the midpoint M of PQ () 4008 ,2,422 M -++ æö ==ç÷ èø

Now find the midpoint R of PM () () 4204 ,3,222 R æ-+-ö + ==ç÷ èø

Finally, find the midpoint S of MQ () 2048 ,1,622 S -++ æö ==ç÷ èø

Thus, the three points are ( 3, 2), ( 2, 4), and ( 1, 6).

34. First, find the midpoint M of PQ () ( ) 412 816 ,4,422 M æ+-ö -+ ==ç÷ èø

Now find the midpoint R of PM. () () 44 84 ,2,022 R æ+-ö -+ ==ç÷ èø

Finally, find the midpoint S of MQ. () ( ) 416412 ,10,822 S æ-+-ö + ==ç÷ èø

Thus, the three points are ( 2, 0), (4, 4), and (10, 8)

dPQ

35. ( )() ( )() ( )() 22 22 22 (,)1(5)4517 (,)4(1)1432 (,)4(5)1517

dQR

dPR =---+-=

The triangle is isosceles.

36. ()() ( )() ( )() 22 22 22 (,)63625 (,)165652 (,)13525 dPQ

dQR

dPR

The triangle is an isosceles triangle.

dPQ

37. ( )() ( )() ( )() 22 22 22 (,)0(4)7817 (,)305713 (,)3(4)5810

dQR

dPR =--+-= =--+-=

The triangle is scalene.

dPQ

dQR

41. ( )( ) ( ) () ( ) ( ) 22 22 (,)111(1)22 (,)3(1)31 32313231 822 dPQ dQR =--+--= =---+-=-++++ == () ( ) ( ) ( ) 22 (,)313(1) 32313231 822 dPR =--+--=+++-+ ==

The triangle is equilateral.

38. ( )( ) ( )( ) ( )() 22 22 22 (,)161672 (,)5(1)3(1)42 (,)5636130

dPR =--+--= =---+--= =--+-=

The triangle is scalene.

39. ()( ) ()( ) ()( ) 22 22 22 (,)909(1)145 (,)591(9)229 (,)501(1)29 dPQ dQR

dPR =-+---= =-+--= =-+--=

The triangle is scalene.

40. ( )() ()( ) ( )( ) 22 22 22 (,)4(4)5465 (,)042565 (,)0(4)24213 dPQ dQR dPR =--+-= =-+--= =--+--=

The triangle is isosceles.

42. ( )( ) ()( ) 22 2 2 (,)1.5(0.5)1(1) 5 3 (,)31(1.5)1 2 dPQ dQR =---+-= æö =---+-ç÷ èø () () ( ) ( ) ()( ) () () ( ) ( ) 2 2 2 2 313312.25 3 31 4 32313332.25 1.753 5 31(0.5) (,)3 2(1) 31310.25 3 31 4 3231310.25 1.753 5 dPR æö -+-+ èø = æö +-+ç÷ èø -++-+ = += = æö +--ç÷ èø æö -+-+ èø = æö +++ç÷ èø =-++-+ ++ = The triangle is equilateral.

43. First find the lengths of the sides: ( )( ) ( )( ) ( )( ) ( )( ) 22 22 22 22 (,)173(12)17 (,)14(1)11317 (,)221441117 (,)2274(12)17 dPQ dQR dRS dSP =--+--= =--+-= =-+--= =-+---= (continuedon next page)

(continued)

All the sides are equal, so the quadrilateral is either a square or a rhombus. Now find the length of the diagonals: ( )( ) ( )( ) 22 22 (,)14711(12)172 (,)22(1)43172 dPR dQS =-+--= =--+--=

The diagonals are equal, so the quadrilateral is a square.

44. First find the lengths of the sides: ()( ) ()( ) 22 22 (,)9811(10)2 (,)8912(11)2 dPQ dQR =-+---= =-+---= ()( ) ()( ) 22 22 (,)7811(12)2 (,)8710(11)2 dRS dSP =-+---= =-+---=

All the sides are equal, so the quadrilateral is either a square or a rhombus. Now find the length of the diagonals. ()( ) ()( ) 22 22 (,)8812(10)2 (,)7911(11)2 dPR dQS =-+---= =-+---=

The diagonals are equal, so the quadrilateral is a square.

45. ()( ) 22 2 22 2 522(1) 449 541325413 04120(6)(2) 2 or6 x xx xxxx xxxx xx =-+--

46. ( )( ) 22 2 2 2 2 132(10)(3) 14469 6153 1696153 06160(8)(2) 8 or2 y yy yy yy yyyy yy =--+-=+++ =++Þ =++ =+-Þ=+-Þ =-=

47. P = (–5, 2), Q = (2, 3), R = (x, 0) (R is on the x-axis, so the y-coordinate is 0). ( )() 22 22 (,)(5)02 (,)(2)(03) dPRx dQRx =--+=-+-

( )() 2222 2222 22 (5)02(2)(03) (5)(02)(2)(03) 10254449 1029413 1416 8 7 xx xx xxxx xx x x --+-=-+++-=-++++=-++ +=-+ ==-

The coordinates of R are 8 ,07 æöç÷ èø

48. P = (7, –4), Q = (8, 3), R = (0, y) (R is on the y-axis, so the x-coordinate is 0). ()( ) 22 22 (,)07(4) (,)(08)(3) dPRy dQRy =-+-=-+()( ) 22 22 07(4) (08)(3) y y -+-=-+( )22 22 49(4)64(3) 498166469 865673 148 4 7 yy yyyy yy y y +--=++++=+-+ +=-+ = =

The coordinates of R are 4 0, 7 æö ç÷ èø.

2.1 Applying the Concepts

49.

50. ( ) 20102016308324 , 22 2013,316 M ++ æö = èø = The population is 2013 was about 316 million.

55. 2014 is the midpoint of the initial range, so ( ) 20122016326425 , 22 2014,375.5 M ++ æö = ç÷ èø = Americans spent about $376 billion on prescription drugs in 2014.

56. 2014 is the midpoint of the initial range, so ( ) 2012201624973696 , 22 2014,3096.5 M ++ æö = ç÷ èø = There were about 3097 million Internet users in 2014.

57. Percentage of Android sales in June 2013: 51.5%

58. Percentage of iPhone sales in December 2012: 49.7%

59. Android sales were at a maximum in June 2014.

60. iPhone sales were at a maximum in December 2012.

61. Denote the diagonal connecting the endpoints of the edges a and b by d. Then a, b, and d form a right triangle. By the Pythagorean theorem, 222abd += . The edge c and the diagonals d and h also form a right triangle, so 222 cdh += . Substituting 2 d from the first equation, we obtain 2222 abch ++=

b. 22 22 (,)(800200)(1200400) 1000 (,)(2000800)(3001200) 1500 dDM dMP =-+= =-+= The distance traveled by the pilot = 1000 + 1500 = 2500 miles.

c. 22 (,)(2000200)(300400) 3,250,00032510000 10032510051350013 1802.78 miles dDP =-+==× ==×= »

62.a.

63. First, find the initial length of the rope using the Pythagorean theorem: 22 241026 c =+= . After t seconds, the length of the rope is 26 – 3t. Now find the distance from the boat to the dock, x, using the Pythagorean theorem again and solving for x: 222 22 22 2 (263)10

So we have 5 2.50 2 x x + =Þ= and 4 5.5 2 y + =Þ 7 y =

The coordinates of D are (0, 7).

6761569100

5761569

5761569 tx ttx ttx ttx -=+ -+=+ -+= -+=

2.1 Beyond the Basics

64. The midpoint of the diagonal connecting (0, 0) and (a + b, c) is ,.22 abc + æö ç÷ èø The midpoint of the diagonal connecting (a, 0) and (b, c) is also ,.22 abc + æö ç÷ èø Because the midpoints of the two diagonals are the same, the diagonals bisect each other.

65.a. If AB is one of the diagonals, then DC is the other diagonal, and both diagonals have the same midpoint. The midpoint of AB is ( ) 2534 ,3.5,3.522 ++ æö =ç÷ èø . The midpoint of

DC = 38 (3.5,3.5),. 22 xy++ æö = ç÷ èø

So we have 3 3.54 2 x x + =Þ= and 8 3.5 2 y + =Þ 1 y =-

The coordinates of D are (4, –1).

b. If AC is one of the diagonals, then DB is the other diagonal, and both diagonals have the same midpoint. The midpoint of AC is ( ) 2338 ,2.5,5.522 ++ æö =ç÷ èø . The midpoint of

DB = 54 (2.5,5.5),. 22 xy++ æö = ç÷ èø

c. If BC is one of the diagonals, then DA is the other diagonal, and both diagonals have the same midpoint. The midpoint of BC is () 5348 ,4,622 ++ æö =ç÷ èø . The midpoint of DA is 23 (4,6),. 22 xy++ æö = ç÷ èø So we have 2 46 2 x x + =Þ= and 3 69 2 y y + =Þ=

The coordinates of D are (6, 9).

66. The midpoint of the diagonal connecting (0, 0) and (x, y) is , 22 xy æö ç÷ èø. The midpoint of the diagonal connecting (a, 0) and (b, c) is ,.22 abc + æö ç÷ èø Because the diagonals bisect each other, the midpoints coincide. So 22 xab + =Þ xab =+ , and 22 yc yc=Þ= Therefore, the quadrilateral is a parallelogram.

67.a. The midpoint of the diagonal connecting (1, 2) and (5, 8) is () 1528 ,3,5.22 ++ æö =ç÷ èø

The midpoint of the diagonal connecting (–2, 6) and (8, 4) is 2864 ,(3,5).22 -++ æö =ç÷ èø Because the midpoints are the same, the figure is a parallelogram.

b. The midpoint of the diagonal connecting (3, 2) and (x, y) is 32 ,.22 xy++ æö ç÷ èø The midpoint of the diagonal connecting (6, 3) and (6, 5) is (6, 4). So 3 69 2 x x + =Þ= and 2 46 2 y y + =Þ=

68. Let P(0, 0), Q(a, 0), R(a + b, c), and S(b, c) be the vertices of the parallelogram.

22 (0)(00). PQRSaa ==-+-=

( )22 22 ()(0) . QRPSabac bc ==+-+=+

The sum of the squares of the lengths of the sides = 222 2(). abc ++ 22 (,)(). dPRabc =++

()2 2 (,)(0). dQSabc =-+-

The sum of the squares of the lengths of the diagonals is ( ) ( )2222()() abcabc +++-+= 222222 22 aabbcaabbc ++++-++= 222222 2222(). abcabc ++=++

69. Let P(0, 0), Q(a, 0), and R(0, b) be the vertices of the right triangle. The midpoint M of the hypotenuse is ,.22 ab æö ç÷ èø 22 2222 (,)022 222 ab dQMa abab æöæö =-+-ç÷ç÷ èøèø

æöæö+ =+-= ç÷ç÷ èøèø 22 22 22 22 22 22 (,)022 222 (,)0022 222 ab dRMb abab ab dPM abab

æöæö=-+-ç÷ç÷ èøèø

æöæö+ =-+= ç÷ç÷ èøèø

æöæö=-+-ç÷ç÷ èøèø

æöæö+ =-+-= ç÷ç÷ èøèø

70. Let P(0, 0), Q(a, 0), and R(0, a) be the vertices of the triangle.

Using the Pythagorean theorem, we have 2222222 12 22 caacaca acc =+Þ=Þ=Þ

71. Since ABC is an equilateral triangle and O is the midpoint of AB, then the coordinates of A are ( a, 0).

AB = AC = AB = 2a. Using triangle BOC and the Pythagorean theorem, we have ()2 22222 22222 2 433 BCOBOCaaOC aaOCaOCOCa =+Þ=+Þ =+Þ=Þ=

Thus, the coordinates of C are ( ) 0,3a and the coordinates of D are ( ) 0,3. a -

72.

(continuedon next page)

continued)

Thus, we have 1 2 MPPQ =

Similarly, we can show that 1 2 MQPQ = Thus, M is the midpoint of PQ, and () 1212 ,,22 xxyy xy ++ æö = ç÷ èø.

2.1 Critical Thinking/Discussion/Writing

73.a. y-axis

b. x-axis

74.a. The union of the x- and y-axes

b. The plane without the x- and y-axes

75.a. Quadrants I and III

b. Quadrants II and IV

76.a. The origin

b. The plane without the origin

77.a. Right half-plane

b. Upper half-plane

78. Let (x, y) be the point. The point lies in if

Quadrant I x > 0 and y > 0

Quadrant II x < 0 and y > 0

Quadrant III x < 0 and y < 0

Quadrant IV x > 0 and y < 0

2.1 Getting Ready for the Next Section

79.a. 22 22 111121 224442 xy æöæö +=+=+== ç÷ç÷ èøèø

b. 22 22 2222 1 2244 xy æöæö +=+=+= ç÷ç÷ èøèø

80.a. ()()() () () 2 222 2 2 121112 234913 xy-++=é--ù++ ëû =-+=+=

b. ()()()() 2222 22 124122 3491625 xy-++=-++ =+=+=

81.a. 3 223 110 2323 y x xy+=+=+=-= b. 3 443 110 4343 y x xy +=+=+=-+= -

82.a. () 1212 1212 112 xy xy +=+=+ =-+-=b. 3232 112 3232 xy xy +=+=+=+= 83. 2 222 2 6 663 2 69 xxxx xxæö -+=-+ ç÷ èø =-+ 84. () 2 222 2 8 884 2 816 xxxx xxæö -+=-+ç÷ èø =-+ 85. 2 22239 333 24yyyyyy æö +=++=++ ç÷ èø 86. 2 22525 55 24yyyy æö ++=++ ç÷ èø 87. 2 2 22 24 aa xaxxaxæö -+=-+ ç÷ èø 88. 2 2 22 24 yy xxyxxy æö ++=++ ç÷ èø

2.2 Graphs of Equations

2.2 Practice Problems

1. 2 1 yx=-+

x y = x2 +1 (x, y)

2 y = ( 2)2 + 1 ( 2, 3)

1 y = ( 1)2 + 1 ( 1, 0)

0 y = (0)2 + 1 (0, 1)

1 y = (1)2 + 1 (1, 0)

2 y = (2)2 + 1 (2, 3)

2. To find the x-intercept, let y = 0, and solve the equation for x: 2 0232 xx =+-Þ ( )() 1 0212or2 2 xxxx =-+Þ==- . To find the y-intercept, let x = 0, and solve the equation for y: ()() 2 203022.yy =+-Þ=-

The x-intercepts are 1 2 and 2; the y-intercept is 2

3. To test for symmetry about the y-axis, replace x with –x to determine if (–x, y) satisfies the equation. ()2 22211xyxy --=Þ-= , which is the same as the original equation. So the graph is symmetric about the y-axis.

4. x-axis: ()3 223 , xyxy =-Þ=- which is not the same as the original equation, so the equation is not symmetric with respect to the x-axis.

y-axis: ()2 323 , xyxy -=Þ= which is the same as the original equation, so the equation is symmetric with respect to the y-axis.

origin: ()() 23 23 , xyxy -=-Þ=- which is not the same as the original equation, so the equation is not symmetric with respect to the origin.

5. 4277324ytt=-++

a. First, find the intercepts. If t = 0, then y = 324, so the y-intercept is (0, 324). If y = 0, then we have ()() ()()() 42 42 22 2 077324 773240 8140 99409,9,2 tt

So, the t-intercepts are ( 9, 0) and (9, 0) Next, check for symmetry.

t-axis: 4277324ytt -=-++ is not the same as the original equation, so the equation is not symmetric with respect to the t-axis.

y-axis: () () 4277324ytt =--+-+Þ

4277324,ytt=-++ which is the same as the original equation. So the graph is symmetric with respect to the y-axis.

origin: () () 4277324ytt -=--+-+Þ

4277324,ytt -=-++ which is not the same as the original equation. So the graph is not symmetric with respect to the origin. Now, make a table of values. Since the graph is symmetric with respect to the y-axis, if (t, y) is on the graph, then so is ( t, y). However, the graph pertaining to the physical aspects of the problem consists only of those values for t ≥ 0

t y = t4 +77t2 +324 (t, y)

0 324 (0, 324)

1 400 (1, 400)

2 616 (2, 616)

3 936 (3, 936)

4 1300 (4, 1300)

5 1624 (5, 1624)

6 1800 (6, 1800)

7 1696 (7, 1696)

8 1156 (8, 1156)

9 0 (9, 0) (continuedon next page)

(continued)

2.2 Concepts and Vocabulary

1. The graph of an equation in two variables, such as x and y, is the set of all ordered pairs (a, b) that satisfy the equation.

2. If (–2, 4) is a point on a graph that is symmetric with respect to the y-axis, then the point (2, 4) is also on the graph.

3. If (0, 5) is a point of a graph, then 5 is a y- intercept of the graph.

c. The population becomes extinct after 9 years.

6. The standard form of the equation of a circle is 222()() xhykr -+-= (h, k) = (3, 6) and r = 10

The equation of the circle is 22 (3)(6)100. xy-++=

7. ()() ()() 22 2136,2,1,6 xyhkr -++=Þ=-=

This is the equation of a circle with center (2, 1) and radius 6

8. 22 46120xyxy++--=Þ 224612xxyy++-=

Now complete the square: 2244691249xxyy+++-+=++Þ ()() 22 2325xy++-=

This is a circle with center ( 2, 3) and radius 5.

4. An equation in standard form of a circle with center (1, 0) and radius 2 is ()2 2 14xy-+=

5. False. The equation of a circle has both an 2 x -term and a 2 y -term. The given equation does not have a 2 y -term.

6. False. The graph below is an example of a graph that is symmetric about the x-axis, but does not have an x-intercept.

7. False. The center of the circle with equation ()() 22 349xy+++= is ( ) 3,4.

8. True

2.2 Building Skills

In exercises 9 14, to determine if a point lies on the graph of the equation, substitute the point’s coordinates into the equation to see if the resulting statement is true.

9. on the graph: (–3, –4), (1, 0), (4, 3); not on the graph: (2, 3)

10. on the graph: (–1, 1), (1, 4), 5 ,03 æöç÷ èø; not on the graph: (0, 2)

11. on the graph: (3, 2), (0, 1), (8, 3); not on the graph: (8, –3)

12. on the graph: (1, 1), 1 2, 2 æö ç÷ èø; not on the graph: (0, 0), 1 3,3 æöç÷ èø

13. on the graph: (1, 0), ( ) ( )2,3,2,3 - ; not on the graph: (0, –1)

14. Each point is on the graph.

36.

For exercises 37–46, read the answers directly from the given graphs.

37. x-intercepts: –1, 1 y-intercepts: none symmetries: y-axis

38 x-intercepts: none y-intercepts: –1, 1 symmetries: x-axis

39. x-intercepts: ,0, ppy-intercepts: 0 symmetries: origin

40. x-intercepts: , 22 ppy-intercepts: 2 symmetries: y-axis

41. x-intercepts: –3, 3 y-intercepts: –2, 2 symmetries: x-axis, y-axis, origin

42. x-intercepts: –2, 2 y-intercepts: –3, 3 symmetries: x-axis, y-axis, origin

43. x-intercepts: –2, 0, 2 y-intercepts: 0 symmetries: origin

44. x-intercepts: –2, 0, 2 y-intercepts: 0 symmetries: origin

45. x-intercepts: –2, 0, 2 y-intercepts: 0, 3 symmetries: y-axis

46. x-intercepts: 0, 3 y-intercepts: –2, 0, 2 symmetries: x-axis

51. To find the x-intercept, let y = 0, and solve the equation for x: 34(0)124. xx +=Þ= To find the y-intercept, let x = 0, and solve the equation for y: 3(0)4123. yy +=Þ= The x-intercept is 4; the y-intercept is 3.

52. To find the x-intercept, let y = 0, and solve the equation for x: 5 23(0)5. 2 xx+=Þ= To find the y-intercept, let x = 0, and solve the equation for y: 5 2(0)35. 3 yy +=Þ= The x-intercept is 52; the y-intercept is 53

53. To find the x-intercept, let y = 0, and solve the equation for x: 0 15. 53 x x +=Þ= To find the y-intercept, let x = 0, and solve the equation for y: 0 13. 53 y y +=Þ= The x-intercept is 5; the y-intercept is 3.

54. To find the x-intercept, let y = 0, and solve the equation for x: 0 12. 23 x x -=Þ= To find the y-intercept, let x = 0, and solve the equation for y: 0 13. 23 y y -=Þ=- The x-intercept is 2; the y-intercept is 3.

55. To find the x-intercept, let y = 0, and solve the equation for x: 2 02. 1 x x x + =Þ=-To find the y-intercept, let x = 0, and solve the equation for y: 02 2. 01 y + ==-The xintercept is –2; the y-intercept is –2.

56. To find the x-intercept, let y = 0, and solve the equation for x: 02 2. 01 xx=Þ=+ To find the y-intercept, let x = 0, and solve the equation for y: 2 02. 1 y y y=Þ= + The xintercept is –2; the y-intercept is 2.

57. To find the x-intercept, let y = 0, and solve the equation for x: 2 0684 or xxx =-+Þ=

2. x = To find the y-intercept, let x = 0, and solve the equation for y: 2 06(0)8 y =-+Þ

8. y = The x-intercepts are 2 and 4; the y-intercept is 8.

58. To find the x-intercept, let y = 0, and solve the equation for x: 2 05(0)66.xx =-+Þ= To find the y-intercept, let x = 0, and solve the equation for y: 2 0562 or yyy =-+Þ=

3. y = The x-intercept is 6; the y-intercepts are 2 and 3.

59. To find the x-intercept, let y = 0, and solve the equation for x: 22042.xx+=Þ=± To find the y-intercept, let x = 0, and solve the equation for y: 22 042. yy +=Þ=± The x-intercepts are –2 and 2; the y-intercepts are –2 and 2.

60. To find the x-intercept, let y = 0, and solve the equation for x: ()2 2 10913 2 or4 xx xx -+=Þ-=±Þ =-=

To find the y-intercept, let x = 0, and solve the equation for y: ()2 222 019198 822 yyy y -+=Þ+=Þ=Þ =±=±

The x-intercepts are –2 and 4; the y-intercepts are 22.±

61. To find the x-intercept, let y = 0, and solve the equation for x: 2 093. xx =-Þ=± To find the y-intercept, let x = 0, and solve the equation for y: 2 903.yy=-Þ= The x-intercepts are –3 and 3; the y-intercept is 3.

62. To find the x-intercept, let y = 0, and solve the equation for x: 2 011. xx =-Þ=± To find the y-intercept, let x = 0, and solve the equation for y: 2 01 y =-Þ no solution. The x-intercepts are –1 and 1; there is no y-intercept.

63. To find the x-intercept, let y = 0, and solve the equation for x: (0)1x =Þ no solution. To find the y-intercept, let x = 0, and solve the equation for y: (0)1 y =Þ no solution. There is no x-intercept; there is no y-intercept.

64. To find the x-intercept, let y = 0, and solve the equation for x: 22 011 xx =+Þ=-Þ there is no real solution. To find the y-intercept, let x = 0, and solve the equation for y: 2 011.yy=+Þ= There is no x-intercept; the y-intercept is 1.

65. 2 1 yx -=+ is not the same as the original equation, so the equation is not symmetric with respect to the x-axis.

2 ()1yx=-+Þ 2 1 yx=+ , so the equation is symmetric with respect to the y-axis.

2 ()1yx -=-+Þ 2 1 yx -=+ , is not the same as the original equation, so the equation is not symmetric with respect to the origin.

66. 22 ()11xyxy =-+Þ=+ , so the equation is symmetric with respect to the x-axis.

2 1 xy -=+ is not the same as the original equation, so the equation is not symmetric with respect to the y-axis.

2 ()1xy -=-+Þ 2 1 xy -=+ is not the same as the original equation, so the equation is not symmetric with respect to the origin.

67. 3 yxx-=+ is not the same as the original equation, so the equation is not symmetric with respect to the x-axis.

3() yxx=--Þ 3 yxx=--Þ

3 ()yxx =-+ is not the same as the original equation, so the equation is not symmetric with respect to the y-axis.

33() yxxyxx -=--Þ-=--Þ

33 (), yxxyxx -=-+Þ=+ so the equation is symmetric with respect to the origin.

68. 3 2 yxx-=- is not the same as the original equation, so the equation is not symmetric with respect to the x-axis.

3 2()() yxx=---Þ 3 2 yxx=-+Þ

3 2()yxx =-- is not the same as the original equation, so the equation is not symmetric with respect to the y-axis.

332()()2 yxxyxx -=---Þ-=-+Þ

332()2 yxxyxx -=--Þ=- , so the equation is symmetric with respect to the origin.

69. 4252 yxx-=+ is not the same as the original equation, so the equation is not symmetric with respect to the x-axis.

425()2() yxx =-+-Þ 4252 yxx =+ , so the equation is symmetric with respect to the y-axis.

442 5()2()52 yxxyxx -=-+-Þ-=+ is not the same as the original equation, so the equation is not symmetric with respect to the origin.

70. 64232 yxxx -=-++ is not the same as the original equation, so the equation is not symmetric with respect to the x-axis.

642 3()2()() yxxx =--+-+-Þ 64232 yxxx =-++ , so the equation is symmetric with respect to the y-axis.

642 3()2()() yxxx -=--+-+-Þ

64232 yxxx -=-++ is not the same as the original equation, so the equation is not symmetric with respect to the origin.

71. 5332 yxx-=-+ is not the same as the original equation, so the equation is not symmetric with respect to the x-axis.

533()2() yxx =--+-Þ 5332 yxx =- is not the same as the original equation, so the equation is not symmetric with respect to the y-axis.

5353 3()2()32 yxxyxx -=--+-Þ-=-Þ

5353 (32)32 yxxyxx -=--+Þ=-+ , so the equation is symmetric with respect to the origin.

72. 2 2 yxx-=- is not the same as the original equation, so the equation is not symmetric with respect to the x-axis.

22() yxx=---Þ 2 2 yxx =- , so the equation is symmetric with respect to the y-axis.

22() yxx -=---Þ 2 2 yxx-=- is not the same as the original equation, so the equation is not symmetric with respect to the origin.

73. 2222 ()2()121 xyxyxyxy -+-=Þ-= is not the same as the original equation, so the equation is not symmetric with respect to the x-axis.

2222 ()2()121 xyxyxyxy -+-=Þ-= is not the same as the original equation, so the equation is not symmetric with respect to the y-axis.

22 ()()2()()1 xyxy --+--=Þ 22 21xyxy+= , so the equation is symmetric with respect to the origin.

74. 2222()1616xyxy+-=Þ+= , so the equation is symmetric with respect to the x-axis.

2222 ()1616 xyxy -+=Þ+= , so the equation is symmetric with respect to the y-axis.

2222 ()()1616 xyxy -+-=Þ+= , so the equation is symmetric with respect to the origin.

For exercises 75–78, use the standard form of the equation of a circle, 222()() xhykr -+-= .

75. Center (2, 3); radius = 6

76. Center (–1, 3); radius = 4

77. Center (–2, –3); radius = 11

78. Center 13 , 22 æöç÷ èø; radius = 3 2

79. 22(1)4xy+-=

80. 22 (1)1 xy-+= 81. 22 (1)(2)2 xy++-=

82. 22 (2)(3)7 xy+++=

83. Find the radius by using the distance formula: 22 (13)(5(4))97 d =--+--=

The equation of the circle is 22 (3)(4)97. xy-++=

84. Find the radius by using the distance formula: 22 (12)(15)255 d =--+-== . The equation of the circle is 22 (1)(1)25. xy++-=

85. The circle touches the x-axis, so the radius is 2. The equation of the circle is 22 (1)(2)4. xy-+-=

86. The circle touches the y-axis, so the radius is 1. The equation of the circle is 22 (1)(2)1 xy-+-=

87. Find the diameter by using the distance formula:

22 (37)(64)104226 d =--+-==

So the radius is 26 . Use the midpoint formula to find the center: () 7(3)46,2,522 M +-+ æö == ç÷ èø . The equation of the circle is 22 (2)(5)26 xy-+-=

88. Find the center by finding the midpoint of the diameter: () 2835,5,122 C +-+ æö == èø

Find the length of the radius by finding the length of the diameter and dividing that by 2. 22 (28)(35)10010 d =-+--==

Thus, the length of the radius is 5, and the equation of the circle is 22 (5)(1)25. xy-+-=

89.a. 22 2240xyxy+---=Þ

22224xxyy-+-=

Now complete the square: 222121411xxyy-++-+=++Þ

22 (1)(1)6 xy-+-= . This is a circle with center (1, 1) and radius 6

b. To find the x-intercepts, let y = 0 and solve for x:

()

() 222 2 (1)(01)6116 151515 xx xxx -+-=Þ-+=Þ -=Þ-=±Þ=±

Thus, the x-intercepts are ( )15,0 + and ( ) 15,0. -

To find the y-intercepts, let x = 0 and solve for y:

()

() 222 2 (01)(1)6116 151515 yy yyy -+-=Þ+-=Þ -=Þ-=±Þ=±

Thus, the y-intercepts are ( )0,15 + and ( ) 0,15. -

90.a. 22 42150xyxy+---=Þ

224215xxyy-+-=

Now complete the square: 2244211541xxyy-++-+=++Þ

22 (2)(1)20. xy-+-= This is a circle with center (2, 1) and radius 25

b. To find the x-intercepts, let y = 0 and solve for x: 22 (2)(01)20 x -+-=Þ () () 22 2120219 219219 xx xx -+=Þ-=Þ -=±Þ=±

Thus, the x-intercepts are ( )219,0 + and ( ) 219,0. -

To find the y-intercepts, let x = 0 and solve for y: 22 (02)(1)20 y -+-=Þ ()2 2 4(1)20116 143,5 yy yy +-=Þ-=Þ -=±Þ=-

Thus, the y-intercepts are (0, 3) and (0, 5)

91.a. 22 2240 xyy++=Þ

22 2(2)0 xyy++=Þ 22 20.xyy++=

Now complete the square: 22222101(1)1.xyyxy +++=+Þ++=

This is a circle with center (0, –1) and radius 1.

b. To find the x-intercepts, let y = 0 and solve for x: 222(01)100xxx ++=Þ=Þ=

Thus, the x-intercept is (0, 0).

To find the y-intercepts, let x = 0 and solve for y: 22 0(1)1110,2 yyy ++=Þ+=±Þ=-

Thus, the y-intercepts are (0, 0) and (0, 2)

92.a. 22 3360 xyx++=Þ

22 3(2)0 xyx++=Þ 2220.xxy++= Now complete the square: 2222 2101(1)1.xxyxy +++=+Þ++=

This is a circle with center (–1, 0) and radius 1.

b. To find the x-intercepts, let y = 0 and solve for x: 22 (1)01110,2 xxx ++=Þ+=±Þ=-

Thus, the x-intercepts are (0, 0) and ( 2, 0).

To find the y-intercepts, let x = 0 and solve for y: 222 (01)100 yyy ++=Þ=Þ=

Thus, the y-intercept is (0, 0).

93.a. 222200.xyxxxy+-=Þ-+=

Now complete the square: 2211 0 44xxy-++=+Þ 2 2 11 . 24xy æö-+=ç÷ èø This is a circle with center 1 ,02 æö ç÷ èø and radius 1 2

b. To find the x-intercepts, let y = 0 and solve for x: 2 2 1111 0 2422 xx æö-+=Þ-=±Þç÷ èø

0,1. x = Thus, the x-intercepts are (0, 0) and (1, 0). To find the y-intercepts, let x = 0 and solve for y: 2 22 1111 0 2444 yy æö-+=Þ+=Þç÷ èø

2 00.yy=Þ=

Thus, the y-intercept is (0, 0).

94.a. 2222101.xyxy++=Þ+=- The radius cannot be negative, so there is no graph.

b. There are no intercepts.

2.2 Applying the Concepts

95. The distance from P(x, y) to the x-axis is x while the distance from P to the y-axis is y

So the equation of the graph is . xy =

96. The distance from P(x, y) to (1, 2) is 22 (1)(2) xy-+- while the distance from P to (3, –4) is 22 (3)(4) xy-++

So the equation of the graph is 2222 (1)(2)(3)(4) xyxy -+-=-++Þ 2222 (1)(2)(3)(4) xyxy -+-=-++Þ (continuedon next page)

(continued)

222144xxyy-++-+= 2269816xxyy-++++Þ

2456825 xyxy --+=-++Þ 15 42012. 33xyyx -=Þ=-

98. If you jog at 6 mph for 10 minutes, then you have traveled ( )1 6 61 = mile. So the graph starts at (0, 0) and increases to (10, 1). Resting for 10 minutes takes the graph to (20, 1). It will take 20 minutes to walk one mile at 3 mph back to the starting point. Thus, the graph passes through (40, 0).

99.a. July 2018 is represented by t = 0, so March 2018 is represented by t = –4. The monthly profit for March is determined by

2 0.5(4)3(4)8$12 P =----+= million.

b. July 2018 is represented by t = 0, so October 2018 is represented by t = 3. So the monthly profit for October is determined by

2 0.5(3)3(3)8$5.5 P =--+=- million. This is a loss.

Because t = 0 represents July 2018, t = –6 represents January 2018, and t = 5 represents December 2018.

d. To find the t-intercept, set P = 0 and solve for t: 2 00.538 tt =--+Þ 2 3(3)4(0.5)(8)325 2(0.5)1

2 or8 t ±---± == =-

The t-intercepts represent the months with no profit and no loss. In this case, t = 8 makes no sense in terms of the problem, so we disregard this solution. t = 2 represents Sept 2018.

e. To find the P-intercept, set t = 0 and solve for P: 2 0.5(0)3(0)88.PP =--+Þ= The P-intercept represents the profit in July 2018.

100.a.

b. To find the P-intercept, set t = 0 and solve for P: 2 0.002(0)0.51(0)17.5 P =-++Þ 17.5. P = The P-intercept represents the number of female college students (in millions) in 2005.

101.a. t Height = 2 16128320 tt -++ 0 320 feet 1 432 feet 2 512 feet

3 560 feet 4 576 feet 5 560 feet 6 512 feet

c. 010 t ££

d. To find the t-intercept, set y = 0 and solve for t:

2 016128320 tt =-++Þ 2 016(820) tt =---Þ

0(10)(2) tt =-+Þ10 or2.tt==-

The graph does not apply if t < 0, so the t-intercept is 10. This represents the time when the object hits the ground. To find the y-intercept, set t = 0 and solve for y:

2 16(0)128(0)320320.yy =-++Þ=

This represents the height of the building. 102.a.

b. 060 t ££

c. The total time of the experiment is 60 minutes or 1 hour.

2.2 Beyond the Basics

103. 22 22 22 22 42200 4220 44212041 (2)(1)25 xyxy xxyy xxyy xy +-+-=Þ -++=Þ -++++=++Þ -++=

So this is the graph of a circle with center (2, –1) and radius 5. The area of this circle is 25. p 22 42310xyxy+-+-=Þ 224231xxyy-++=Þ 22 22 44213141 (2)(1)36 xxyy xy -++++=++Þ -++=

Section 2.2 Graphs of Equations 183

So, this is the graph of a circle with center (2, –1) and radius 6. The area of this circle is 36. p Both circles have the same center, so the area of the region bounded by the two circles equals 362511. ppp -=

104. Using the hint, we know that the center of the circle will have coordinates (0, k).

Use the Pythagorean theorem to find k. 222224516259 3 kkk k +=Þ+=Þ=Þ =±

The equations of the circles are ()2 22 35. xy+±=

2.2 Critical Thinking/Discussion/Writing

105. The graph of 2 2 yx = is the union of the graphs of 2 = yx and 2. =yx

106. Let (x, y) be a point on the graph. The graph is symmetric with regard to the x-axis, so the point (x, –y) is also on the graph. Because the graph is symmetric with regard to the y-axis, the point (–x, y) is also on the graph. Therefore the point (–x, –y) is on the graph, and the graph is symmetric with respect to the origin. The graph of 3 yx = is an example of a graph that is symmetric with respect to the origin but is not symmetric with respect to the x- and y-axes.

107.a. First find the radius of the circle: 22 (,)(60)(81)85dAB =-+-=Þ

85 2 r =

The center of the circle is 6018 , 22 ++ æö =ç÷ èø 9 3,. 2 æö ç÷ èø

So the equation of the circle is 2 2 985 (3). 24 xy æö -+-= ç÷ èø

To find the x-intercepts, set y = 0, and solve for x: 2 2 22 985 (3)024 8185 (3)691 44 x xxx æö -+-=Þ ç÷ èø -+=Þ-+=Þ

2 680xx-+=

The x-intercepts are the roots of this equation.

b. First find the radius of the circle: 22 22 (,)(0)(1) (1) dABab ab =-+=+-Þ 22 (1) . 2 ab r +=

The center of the circle is 011 ,, 2222 abab +++ æöæö = ç÷ç÷ èøèø

So the equation of the circle is 22 221(1) 224 abab xy ++-æöæö-+-=ç÷ç÷ èøèø

To find the x-intercepts, set y = 0 and solve for x: 22 22 2 222 2 22222 2 2 01(1) 224 (1)(1) 444 442121 4440 0 abab x abab xax xaxabbabb xaxb xaxb ++-æöæö -+-=ç÷ç÷ èøèø ++-++= -++++=+-+ -+= -+=

The x-intercepts are the roots of this equation.

c. a = 3 and b = 1. Approximate the roots of the equation by drawing a circle whose diameter has endpoints A(0, 1) and B(3, 1).

The center of the circle is 3 ,12 æö ç÷ èø and the radius is 3 2 . The roots are approximately (0.4, 0) and (2.6, 0).

108.a. The coordinates of the center of each circle are (r, r) and (3r, r).

b. To find the area of the shaded region, first find the area of the rectangle shown in the figure below, and then subtract the sum of the areas of the two sectors, A and B.

2.2 Getting Ready for the Next Section

2.3 Lines

2.3 Practice Problems 1. () 358 6713 m ==A slope of 8 13 - means that the value of y decreases 8 units for every 13 units increase in x. 2. ( ) 2 2,3,3Pm--=() ()

3. () 4610 5 312 m ===

Use either point to determine the equation of the line Using ( 3, 4), we have () () ()453453 4515511 yxyx yxyx --=é--ùÞ+=+Þ ëû +=+Þ=+

Using ( 1, 6), we have () ()651651 655511 yxyx yxyx -=é--ùÞ-=+Þ ëû -=+Þ=+

4. ( ) ( ) ( ) ()() 11 320 point-slope form 320 3223 yymxxyx yx yxyx -=-Þ--=--=-Þ +=Þ=-

5. The slope is 2 3 - and the y-intercept is 4. The line goes through (0, 4), so locate a second point by moving two units down and three units right. Thus, the line goes through (3, 2).

6. x = 3. The slope is undefined, and there is no y-intercept. The x-intercept is 3 y = 7. The slope is 0, and the y-intercept is 7.

7. First, solve for y to write the equation in slope-intercept form: 34244324 xyyx +=Þ=-+Þ 3 6 4 yx=-+ . The slope is 3 4 - , and the y-intercept is 6. Find the x-intercept by setting y = 0 and solving the equation for x: 33 0668 44 =-+Þ=Þ= xxx . Thus, the graph passes through the points (0, 6) and (8, 0).

8. Use the equation 2.665Hx=+ . ( ) () 1 2 2.64365176.8 2.64465179.4 H H =+= =+=

The person is between 176.8 cm and 179.4 cm tall, or 1.768 m and 1.794 m.

9.a. Parallel lines have the same slope, so the slope of the line is 3744 2533 m ===

Using the point-slope form, we have () () 4 5231542 3 3154843230 yxyx yxxy -=é--ùÞ-=+Þ ëû

b. The slopes of perpendicular lines are negative reciprocals. Write the equation 4510 xy++= in slope-intercept form to find its slope: 4510 xy++=Þ 41 541 55yxyx =--Þ=-- . The slope of a line perpendicular to this lines is 5 4 .

Using the point-slope form, we have () ()()()5 434453 4 41651554310 yxyx yxxy --=-Þ+=-Þ +=-Þ--=

10. Because 2016 is 10 years after 2006, set x = 10. Then y = 0.44(10) + 6.70 = 11.1 There were 11.1 million registered motorcycles in the U.S. in 2016.

2.3 Concepts and Vocabulary

1. The slope of a horizontal line is 0 ; the slope of a vertical line is undefined.

2. The slope of the line passing through the points ( )11 , Pxy = and ( )22 , Qxy = is given by the formula 21 21 . yy m xx= -

3. Every line parallel to the line 32yx=- has slope, m, equal to 3 .

4. Every line perpendicular to the line 32yx=- has slope, m, equal to 1 3 . -

5. False. The slope of the line 1 4 5 yx=-+ is equal to 1 4 . -

6. False. The y-intercept of the line 23yx=- is equal to –3.

7. True

8. True

2.3 Building Skills

9. 734 ;413 m==the graph is rising.

10. 044 2022; m ===-the graph is falling.

11. 2(2)00; 268 m === the graph is horizontal.

12. 7(4)11slope is 3(3)0undefined; m ==Þ the graph is vertical.

13. 3.525.5 30.52.52.2; m ===-the graph is falling.

14. 3(2)11 231 m === ; the graph is rising.

15. () 514 4 1221 m=== +; the graph is rising.

16. () () 330333 1313232 m=== +-; the graph is rising.

17. 3 l 18. 2 l 19. 4 l 20. 1 l

21. 1 l passes through the points (2, 3) and ( 5, 4). 1 437 1. 527 m === l

22. 2 l is a horizontal line, so it has slope 0.

23. 3 l passes through the points (2, 3) and (0, 1). 3 13 2. 02 m ==l

24. 4 l passes through the points ( 3, 3) and (0, 1) () 4 134 033 m ==- l

25. ( ) 0,5;3 m = 35yx=+

26. ( ) 0,9;2 m =29yx=-+

27. 1 4 2 yx=+

28. 1 4 2 yx=-+

29. 33 1(2)13 22 3 4 2 yxyx yx -=--Þ-=-+Þ =-+

30. 222 (1) 555yxyx =+Þ=+

31. 40(5)404yxyy +=-Þ+=Þ=-

32. Because the slope is undefined, the graph is vertical. The equation is 5. x =

33. 01 1 10 m==-. The y-intercept is (0, 1), so the equation is 1. yx=-+

34. 31 2 10 m==. The y-intercept is (0, 1), so the equation is 21.yx=+

35. 33 0 3(1) m== Because the slope = 0, the line is horizontal. Its equation is 3. y =

36. 716 2(5)7 m== . Now write the equation in point-slope form, and then solve for y to write the equation in slope-intercept form.

6 1(5) 7 -=+Þyx 630 1 77 -=+Þyx

637 77 =+yx

37. 1(1)2 1(2)3 m == . Now write the equation in point-slope form, and then solve for y to write the equation in slope-intercept form.

2 1(2) 3 +=+Þyx 24 1 33 +=+Þyx 21 33 =+yx

38. 9(3)6 6(1)7 m ==- . Now write the equation in point-slope form, and then solve for y to write the equation in slope-intercept form.

6 3(1) 7 +=-+Þyx 66 3 77 +=--Þyx 627 77 =--yx

39. 17 27 44 0112 22 m===- . Now write the equation in point-slope form, and then solve for y to write the equation in slope-intercept form. 77 22 22 -=-Þ=-+ yxyx

40. 3(7)10 440 m ==Þthe slope is undefined. So the graph is a vertical line. The equation is 4. x =

41. 5 x = 42. 1.5 y =

43. 0 y = 44. 0 x =

45. 14 y = 46. 25yx=+

47. 2 4 3 yx=-- 48. 63yx=--

49. 404 ;0(3)3 m== 4 4 3 yx=+

50. 202 ;0(5)5 m ==2 2 5 yx=--

51. 7 y = 52. 4 x =

53. 5 y =- 54. 3 x =-

55. 32yx=-

The slope is 3 and the y-intercept is (0, –2). 2 03232 3 xxx =-Þ=Þ=

The x-intercept is ( ) 2 3 ,0.

56. 23yx=-+

The slope is –2 and the y-intercept is (0, 3). 3 02323 2 xxx =-+Þ=Þ=

The x-intercept is ( ) 3 2 ,0.

57. 1 240242. 2 xyyxyx +-=Þ=-+Þ=-+

The slope is 12 - , and the y-intercept is (0, 2).

To find the x-intercept, set y = 0 and solve for x: 2(0)404xx +-=Þ=

58. 1 39933 3 xyxyyx =-Þ+=Þ=+

The slope is 13, and the y-intercept is (0, 3).

To find the x-intercept, set y = 0 and solve for x: 3(0)99xx=-Þ=- .

59. 3 32603623 2 xyxyxy -+=Þ+=Þ+= .

The slope is 32, and the y-intercept is (0, 3).

To find the x-intercept, set y = 0 and solve for x: 32(0)60362 xxx -+=Þ=-Þ=-

60. 24152154 xyxy =-+Þ-=-Þ 115 24xy-+= . The slope is 12 - , and the y-intercept is 154 . To find the x-intercept, set y = 0 and solve for x: 24(0)15 x =-+Þ 152 x =

61. 505xx-=Þ= . The slope is undefined, and there is no y-intercept. The x-intercept is 5.

62. 5 250 2 yy+=Þ=- . The slope is 0, and the y-intercept is 52 - . This is a horizontal line, so there is no x-intercept.

63. 0 x = . The slope is undefined, and the y-intercepts are the y-axis. This is a vertical line whose x-intercept is 0.

64. 0 y = . The slope is 0, and the x-intercepts are the x-axis. This is a horizontal line whose y-intercept is 0.

For exercises 65–68, the two-interecepts form of the equation of a line is 1. xy ab +=

65. 1 43 xy+= 66. 1 32 xy -+=

67. 236 2361; 66632 xyxy xy+=Þ+=Þ+=

x-intercept = 3; y-intercept = 2

68. 341203412 3412 121212431; xyxy xyxy -+=Þ-=-Þ-=Þ-+=

x-intercept = –4; y-intercept = 3

69. 23 231211 121264 xyxy xy-=Þ-=Þ-=

The x-intercept is 6 and the y-intercept is –4.

70. 34 341211 121243 xyxy xy-=Þ-=Þ-=

The x-intercept is 4 and the y-intercept is –3.

71. 52 521011 101025 xyxy xy -+=Þ-+=Þ-+=

The x-intercept is –2 and the y-intercept is 5.

72. 45 452011 202054 xyxy xy -+=Þ-+=Þ-+=

The x-intercept is –5 and the y-intercept is 4.

73. 945 1. 725 m===The equation of the line through (2, 4) and (7, 9) is 41(2)yx-=-Þ y = x + 2. Check to see if (–1, 1) satisfies the equation by substituting x = –1 and y = 1: 11211. =-+Þ= So (–1, 1) lies on the line.

74. 325 1. 275 m === The equation of the line through (7, 2) and (2, –3) is 21(7)yx-=-Þ 5. yx=- Check to see if (5, 1) satisfies the equation by substituting x = 5 and y = 1: 15510. =-Þ¹ So (5, 1) does not lie on the line.

75. The given line passes through the points (0, 3) and (4, 0), so its slope is 3 4 - Any line parallel to this line will have the same slope. The line that passes through the origin and is parallel to the given line has equation

3 4 .yx =-

76. From exercise 75, the slope of the given line is 3 4 - Any line perpendicular to this line will have slope 4 3 The line that passes through the origin and is perpendicular to the given line has equation 4 3 .yx =

77. The red line passes through the points (–2, 0) and (0, 3), so its slope is 3 2 . The blue line passes through (4, 2) and has the same slope, so its equation is () 3 2 3 2 2424312 2384 yxyx yxyx -=-Þ-=-Þ =-Þ=-

78. The red line passes through the points (–2, 0) and (0, 3), so its slope is 3 2 The green line passes through (4, 2) and has slope 2 3 , - so its equation is () 2 3 214 33 243628 3214 yxyx yxyx -=--Þ-=-+Þ =-+Þ=-+

79. The slope of 31yx=- is 3. The slope of 32yx=+ is also 3. The lines are parallel.

80. The slope of 22yx=+ is 2. The slope of 22yx=-+ is –2. The lines are neither parallel nor perpendicular.

81. The slope of 24yx=- is 2. The slope of 1 2 4 yx=-+ is 1 2 . - The lines are perpendicular.

82. The slope of 31yx=+ is 3. The slope of 1 3 1 yx=- is 1 3. The lines are neither parallel nor perpendicular.

83. The slope of 387 xy+= is 38 - , while the slope of 570 xy-= is 57. The lines are neither parallel nor perpendicular.

84. The slope of 1023 xy+= is –5. The slope of 51 xy+=- is also –5, so the lines are parallel.

85. The slope of 48xy=+ is 14 . The slope of 41yx=-+ is 4 - , so the lines are perpendicular.

86. The slope of 31yx=+ is 3. The slope of 620 yx+= is 13 - . The lines are perpendicular.

87. Both lines are vertical lines. The lines are parallel.

88. The slope of 237 xy+= is 23 - , while 2 y = is a horizontal line. The lines are neither parallel nor perpendicular.

89. The equation of the line through (2, –3) with slope 3 is ( ) 332336yxyx +=-Þ+=-Þ

39.yx=-

90. The equation of the line through (–1, 3) with slope –2 is () ( ) ()321321 yxyx -=---Þ-=-+Þ 32221.yxyx -=--Þ=-+

91. A line perpendicular to a line with slope 1 2has slope 2. The equation of the line through (–1, 2) with slope 2 is () ( ) 221yx-=--Þ () 221222 24. yxyx yx -=+Þ-=+Þ =+

92. A line perpendicular to a line with slope 1 3 has slope –3. The equation of the line through (2, –1) with slope –3 is ( ) ( )132yx--=--Þ () 132136 35. yxyx yx +=--Þ+=-+Þ =-+

93. The slope of the line joining (1, –2) and (–3, 2) is ( )22 1. 31 =- The equation of the line through (–2, –5) with slope –1 is () () ( ) ()5252 527. yxyx yxyx --=---Þ+=-+Þ +=--Þ=--

94. The slope of the line joining (–2, 1) and (3, 5) is () 514 325= The equation of the line through (1, 2) with slope 4 5 is ()()()4 215241 5 51044546 46 55 yxyx yxyx yx -=-Þ-=-Þ -=-Þ=+Þ =+

95. The slope of the line joining (–3, 2) and (–4, –1) is () 12 3. 43 =

A line perpendicular to this line has slope 1 3.The equation of the line through (1, –2) with slope 1 3 - is () ()()()1 21321 3 36135 15 33 yxyx

96. The slope of the line joining (2, 1) and (4, –1) is 11 1. 42 =-

A line perpendicular to this line has slope 1 The equation of the line through (–1, 2) with slope 1 is ( ) 21213. yxyxyx -=--Þ-=+Þ=+

97. The slope of the line 65yx=+ is 6. The lines are parallel, so the slope of the new line is also 6. The equation of the line with slope 6 and y-intercept 4 is 64.yx=+

98. The slope of the line 1 2 5 yx=-+ is 1 2 - The lines are parallel, so the slope of the new line is also 1 2 - The equation of the line with slope 1 2 - and y-intercept 2 is 1 2 2. yx=-+

99. The slope of the line 65yx=+ is 6. The lines are perpendicular, so the slope of the new line is 1 6 . - The equation of the line with slope 1 6 - and y-intercept 4 is 1 6 4. yx=-+

100. The slope of the line 1 2 5 yx=-+ is 1 2 - The lines are perpendicular, so the slope of the new line is 2. The equation of the line with slope 2 and y-intercept –4 is 24.yx=-

101. The slope of x + y = 1 is 1 The lines are parallel, so they have the same slope. The equation of the line through (1, 1) with slope 1 is 1(1)yx-=--Þ 11yx-=-+Þ 2 yx=-+ .

102. The slope of 237 xy -+= is 23. The lines are parallel, so they have the same slope. The equation of the line through (1, 0) with slope 23 is 222 0(1)333yxyx -=-Þ=-

103. The slope of 3918 xy-= is 13. The lines are perpendicular, so the slope of the new line is 3 - . The equation of the line through (–2, 4) with slope 3 - is 43((2))yx-=---Þ 43632.yxyx -=--Þ=--

104. The slope of 214 xy -+= is 2. The lines are perpendicular, so the slope of the new line is 12. - The equation of the line through (0, 2) with slope 12 - is 1 2 2. yx=-+

2.3 Applying the Concepts

105.a. The y-intercept represents the initial expenses.

b. The x-intercept represents the point at which the teacher breaks even, i.e., the expenses equal the income.

c. The teacher’s profit if there are 16 students in the class is $640.

d. The slope of the line is ( ) 6407501390695 160168 ==The equation of the line is 695 750. 8 Pn=-

106.a. The y-intercept represents the initial prepaid amount.

b. The x-intercept represents the total number of minutes the cellphone can be used.

c. The slope of the line is 015151 750755=-=-

The equation of the line is 1 15. 5 Pt=-+

d. The cost per minute is 1 $20 5 =¢.

107. slope = rise41 run4010 Þ=

108. 4 miles = 21,120 feet. |slope| = rise run Þ 200025 21,120264 =

109. 8 in. in two weeks Þ the plant grows 4 in. per week. John wants to trim the hedge when it grows 6 in., so he should trim it every 6 4 1.5 weeks10 =»days.

110. 2 min.min.231 12.4 min. 5 in.31 in.5 x x × =Þ==

The water will overflow in about 12 min.

111.a. x = the number of weeks; y = the amount of money in the account after x weeks; 7130yx=+

b. The slope is the amount of money deposited each week; the y-intercept is the initial deposit.

112.a. x = the number of sessions of golf; y = the yearly payment to the club; 351000yx=+

b. The slope is the cost per golf session; the y-intercept is the yearly membership fee.

113.a. x = the number of months owed to pay off the refrigerator; y = the amount owed; 15600yx=-+

b. The slope is the amount that the balance due changes per month; the y-intercept is the initial amount owed.

114.a. x = the number of rupees; y = the number of dollars equal to x rupees. 1 0.019802 50.5 yxx ==

b. The slope is the number of dollars per rupee. The y-intercept is the number of dollars for 0 rupees.

115.a. x = the number of years after 2010; y = the life expectancy of a female born in the year 2010 + x; 0.1780.8yx=+

b. The slope is the rate of increase in life expectancy; the y-intercept is the current life expectancy.

116.a. 1400(2)14,000$11,200 v =-+=

b. 1400(6)14,000$5600 v =-+=

To find when the tractor will have no value, set v = 0 and solve the equation for t: 0140014,00010 years tt =-+Þ=

117. There are 30 days in June. For the first 13 days, you used data at a rate of 435 33.5 13 » MB/day. At the same rate, you will use ( ) 33.517569.5 = MB for the rest of the month.

435 + 569.5 = 1004.5

So, you don’t need to buy extra data. You will have about 20 MB left.

118. For the first three hours, you traveled at 195 65 3mph. =

5201956532565 5 drttt t =Þ-=Þ=Þ = You will arrive at your destination five hours after 12 pm or 5 pm.

119. 540,000yx=+

120.a. 0.2530Cx=+

c. 0.25(60)30$45 y =+=

d. 47.750.253071miles xx =+Þ=

121.a. The two points are (100, 212) and (0, 32). So the slope is 212321809 . 10001005==The equation is 99 32(0)32 55 FCFC -=-Þ=+

b. One degree Celsius change in the temperature equals 95 degrees change in degrees Fahrenheit.

c. C 9 32 5 FC=+

40ºC 104F

25º C 77F

–5ºC 23F

–10ºC 14F

d. 9 100F3237.78C 5 CC °=+Þ=°

9 90F3232.22C 5 CC °=+Þ=°

9 75F3223.89C 5 CC °=+Þ=°

9 10F3223.33C 5 CC -°=+Þ=-°

9 20F3228.89C 5 CC -°=+Þ=-°

e. 9

97.6F3236.44C 5 CC °=+Þ=° ;

9 99.6F3237.56C 5 CC °=+Þ=°

f. Let x = F = C. Then 9 32 5 xx=+Þ

4 3240 5 xx -=Þ=- . At –40°, °F = °C.

122.a. The two points are (4, 210.20) and (10, 348.80). So the slope is 348.80210.20138.6 23.1 1046==. The equation is 348.823.1(10)yx -=-Þ 23.1117.8yx=+

b. The slope represents the cost of producing one modem. The y-intercept represents the fixed cost.

c. 23.1(12)117.8$395yy =+Þ=

123.a. The year 2005 is represented by t = 0, and the year 2011 is represented by t = 6. The points are (0, 2425) and (6, 4026). So the slope is 40262425 266.8 6» The equation is 2425266.8(0)yt -=-Þ 266.82425yt=+

c. The year 2008 is represented by t = 3. So 266.8(3)24253225.4yy =+Þ= . Note that there cannot be a fraction of a person, so. there were 3225 women prisoners in 2008.

d. The year 2017 is represented by t = 12. So 266.8(12)24255626.6yy =+Þ= . There will be 5627 women prisoners in 2017.

124.a. The two points are (5, 5.73) and (8, 6.27).

The slope is 6.275.730.54 0.18. 853== -

The equation is 5.730.18(5)Vx-=-Þ 0.184.83.Vx=+

b. The slope represents the monthly change in the number of viewers. The V-intercept represents the number of viewers when the show first started.

c. 0.18(11)4.836.81VV =+Þ= million

125. The independent variable t represents the number of years after 2000, with t = 0 representing 2000. The two points are (0, 11.7) and (5, 12.7). So the slope is 12.711.7 0.2 5= . The equation is

11.70.2(0)pt-=-Þ 0.211.7.pt=+ The year 2010 is represented by t = 10. 0.2(10)11.713.7%.pp =+Þ=

126. The year 2004 is represented by t = 0, so the year 2009 is represented by t = 5. The two points are (0, 82.7) and (5, 84.2). So the slope is 84.282.71.5 0.3 55== . The equation is 0.382.7.yt=+

127.a.

212.4yx=-+

[0, 6, 1] by [0, 12, 1]

c. The price in the table is given as the number of nickels. 35¢ = 7 nickels, so let x = 7. ( ) 2712.41.6 y =-+=-

Thus, no newspapers will be sold if the price per copy is 35¢. Note that this is also clear from the graph, which appears to cross the x-axis at approximately x = 6.

128.a.

0.0910.3yx»+

[0, 700, 100] by [0, 80, 10]

c. The advertising expenses in the table are given as thousands of dollars, so let x = 700. ( ) 0.0970010.373.3 y »+=

Sales are given in thousands, so approximately 73.3100073,300 ´= computers will be sold.

2.3 Beyond the Basics

129. 3 39312 1(2) c cc=Þ=-Þ=

130. The y-intercept is –4, so its coordinates are (0, –4). Substitute x = 0, y = –4 into the equation and solve for c ()() 32030420 1 42042 2 xcyc ccc --=Þ---=Þ -=Þ=Þ=

131.a. Let A = (0, 1), B = (1, 3), C = (–1, –1). 31134 2;210112ABBCmm===== 11 2 10 ACm ==

The slopes of the three segments are the same, so the points are collinear.

b. 22 22 (,)(10)(31)5 (,)(11)(13)25 =-+-= =--+--= dAB dBC

22 (,)(10)(11)5 =--+--= dAC

Because d(B, C) = d(A, B) + d(A, C), the three points are collinear.

132.a. Let A = (1, 0.5), B = (2, 0), C = (0.5, 0.75). 00.50.750 0.5;0.5210.52ABBC mm==-==-

0.750.5 0.5 0.51 ACm==-

The slopes of the three segments are the same, so the points are collinear.

b. 2 2 155 (,)(12)0242 dAB æö =-+-== ç÷ èø 2213 (,)20 24 4535 164 dBC æöæö=-+-ç÷ç÷ èøèø == 22131 (,)1242 55 164 dAC æöæö=-+-ç÷ç÷ èøèø ==

Because d(B, C) = d(A, B) + d(A, C), the three points are collinear.

133.a. 413842 ; 1125(1)3ABBC mm==-== .

The product of the slopes = –1, so ABBC ^

b. 22 22 22 (,)(11)(41)13 (,)(5(1))(84)52 (,)(51)(81)65 dAB dBC dAC =--+-= =--+-= =-+-=

( )( )( ) 222 (,)(,)(,), dABdBCdAC += so the triangle is a right triangle.

134. 2(1)3121 ; 1(4)5312 2132(1)1 ; 2352(4)2

ABBC CDAD mm mm ========-

So, and ABCDBCAD PP , and ABCD is a parallelogram.

For exercises 135 and 136, refer to the figures accompanying the exercises in your text.

135. AD and BC are parallel because they lie on parallel lines 1l and 2. l AB and CD are parallel because they are parallel to the x-axis. Therefore, ABCD is a parallelogram.

ABCD @ and ADCB @ because opposite sides of a parallelogram are congruent.

ABDCDB @ VV by SSS. Then

1 rise run BD m CD == and 2 rise . run BD m AB == Since

AB = CD, 12. BDBD mm CDAB ===

136. OKABLOV:V because OLAKd == and .BLOKc == Then, 1 rise run d m c == and 2 rise run cc m dd ===-12 1. dc mm cd æö ×=-=ç÷ èø

137. Let the quadrilateral ABCD be such that @and. ABCDABCD P Locate the points as shown in the figure.

Because ABCD P , the y-coordinates of C and D are equal. Because ABCD @ , the x-coordinates of the points are as shown in the figure. The slope of AD is dc The slope of BC is 0 . dd bcbc= +So .ADBC P

22 (,). dADdc =+ ( )2 222 (,)(). dBCdbcbdc =++-=+

So ADBC @

138. Let 112233 (,),(,),(,) and AxyBxyCxy 44 (,)Dxy be the vertices of the quadrilateral.

Then the midpoint 1M of AB is 1212 , 22 xxyy ++ æö ç÷ èø; the midpoint 2 M of BC is 2323 , 22 xxyy ++ æö ç÷ èø; the midpoint 3M of

CD is 3434 , 22 xxyy ++ æö ç÷ èø; and the midpoint 4 M of AD is 1414 ,.22 xxyy ++ æö ç÷ èø

The slope of 12MM is 23 12 13 23 1213 22 22 yy yy yy xx xx xx ++= ++-

The slope of 23MM is 2334 24 2334 24 22 . 22 yyyy yy xxxx xx ++= ++-

The slope of 34MM is 34 14 3113 34 14 3113 22 22 yy yy yyyy xx xx xxxx ++== ++ -

The slope of 14MM is 1214 24 1214 24 22 22 yyyy yy xxxx xx ++= ++-So 1234 MMMM P and 2314 MMMM P , and 1234MMMM is a parallelogram.

139. Let (x, y) be the coordinates of point B. Then 22 22 (,)12.5(2)(2) (2)(2)156.25 and 42 324(2)3(2)

42 . Substitute this into the first 33 AB dABxy

equation and solve for

Solve this equation using the quadratic formula: 2 1001004(25)(1306.25) 2(25) 10010,000130,625 50 100140,625100375

or5.5

Now find y by substituting the x-values into the slope formula: 42 12 39.52 y y=Þ=or 42 8. 35.52 y y=Þ=- So the coordinates of B are (9.5, 12) or (–5.5, –8).

1212 ()()()() yyyyxxxx --=---Þ

1212 ()()()()0. xxxxyyyy --+--=

141. 12012 505 OPm==-

Since the tangent line ℓ is perpendicular to OP , the slope of ℓ is the negative reciprocal of 12 5 - or 5 12 . Using the point-slope form, we have () () 55 125125 1212 yxyx -=é--ùÞ-=+Þ ëû 5255169 12 12121212yxyx -=+Þ=+

142. 11 11 0 0 OP yy m xx== -

Since the tangent line ℓ is perpendicular to OP , the slope of ℓ is the negative reciprocal of 1 1 y x or 1 1 x yUsing the point-slope form, we have ( ) 1 11 1 2222 11111111 x yyxx y yyyxxxxxyyxy -=--Þ -=-+Þ+=+

Since the equation of the circle is 222 xya += , we substitute 2 a for 22 11xy + to obtain 2 11 xxyya += .

143.

The family of lines has slope 2. The lines have different y-intercepts.

144.

The family of lines has y-intercept 2. The lines have different slopes.

145.

The lines pass through (1, 0). The lines have different slopes.

146.

The lines pass through ( 1, 2) The lines have different slopes.

2.3 Critical Thinking/Discussion/Writing

147.a.

This is a family of lines parallel to the line 2.yx =- They all have slope –2.

Section 2.3 Lines 199

This is a family of lines that passes through the point (0, –4). Their y-intercept is –4.

148. ( ) 11 1122 22 12211221 21 12 ymxb mxbmxb ymxb mxmxbbxmmbb bb x mm =+üÞ+=+Þ=+ý þ -=-Þ-=-Þ= -

a. If 12 0 mm>> and 12, bb > then 2112 1212 bbbb x mmmm ==-

b. If 12 0 mm>> and 12, bb < then 21 12 . bb x mm= -

c. If 12 0 mm<< and 12, bb > then 2112 1221 bbbb x mmmm ==

d. If 12 0 mm<< and 12, bb < then 2121 1221 bbbb x mmmm ==-

2.3 Getting Ready for the Next Section

149. ()() 2 40220 202 or 202 xxx xx xx -=Þ+-=Þ +=Þ=-=Þ= Solution: {–2, 2}

150. ()() 2 10110 101or 101 xxx xx xx -=Þ+-=Þ +=Þ=-=Þ= Solution: {–1, 1}

151. ()() 2 20 120 xx xx --= +-= 101 or202xxxx +=Þ=--=Þ= Solution: {–1, 2}

152. ()() 2 230 310 xx xx +-= +-= 303 101 xx xx +=Þ=-=Þ=

Solution: {–3, 1}

153. () ( ) ( )( )( )313133131 3 ahaaha h ++-+=++-+ =

154. () ( ) ( ) ( ) ( ) () ( )() 22 222 222 2 2121 22121 242121 42 aha aahha aahha ahh ++-+ =+++-+ =+++-+ =+

155. () ( ) ( ) 2222 2 2 2 22 2 aahha aha hh hah ahh hh ah -++-+++ = == =--

156. () () () ( ) () 1111 1 aah h hahahaah haah aah æö æö-+-== ç÷ç÷ èø+++ èø =+

157. ( )( ) 130xx--<

Solve the associated equation: ( )( ) 1301or3.xxxx --=Þ==

So, the intervals are ( ) ( ) ( ) ,1,1,3,and 1,.-¥¥

Interval Test point Value of ( )( )13xx Result

( ),1 -¥ 0 3 +

( )1,3 2 2 - –

( ) 3,¥ 4 3 +

The solution set is ( ) 1,3.

158. 2 230xx--³ ()() 2 230310xxxx--³Þ-+³

Now solve the associated equation: ( )( ) 3103 or1.xxxx -+=Þ==-

So, the intervals are

( ] [ ] [ ) ,1,1,3,and 3,.-¥--¥

Interval Test point Value of 2 23xx Result

( ],1-¥- –2 5 +

[ ]1,3 - 0 –3 –

[ ) 3,¥ 5 12 +

The solution set is ( ] [ ) ,13,.-¥-¥ U

2.4 Functions

2.4 Practice Problems

1.a. The domain of R is {2, 2, 3} and its range is {1, 2}. The relation R is a function because no two ordered pairs in R have the same first component.

b. The domain of S is {2, 3} and its range is {5, 2} The relation S is not a function because the ordered paired (3, 2) and (3, 5) have the same first component.

2. Solve each equation for y a. 22222121 xyxy-=Þ-=Þ

2 21; xy±-= not a function

b. ()1 25525; 2 xyxyxy -=Þ-=Þ-= a function

3. () 2 25 gxxx =-+

a. ()()() 2 020500 g =-+=

b. ()()() 2 121517 g -=--+-=-

c. ()()() ( ) 2 22 22 25 2255 24525 gxhxhxh xxhhxh xhxxhh +=-+++ =-++++ =--+-+

4. ( )( ) ( )() ()() lengthheight3122 22244 sq. units TLMSA ====

5.a. () 1 1 fx x =is not defined when 101 xx -=Þ= or when101. xx-<Þ< . Thus, the domain of f is ( ),1 -¥

b. () 2 23gxxx=+- is not defined when 2 230.xx+-<

Use the test point method to see that 2 230xx+-< on the interval (–3, 1). Thus, the domain of g is ( ] [ ) ,31,.-¥-¥ U

6. () 2 , fxx = domain X = [ 3, 3]

a. () 2 1010103.16fxxx=Þ=Þ=±»±

Since 103 > and 103,-<- neither solution is in the interval X = [ 3, 3] Therefore, 10 is not in the range of f

b. () 2 442fxxx=Þ=Þ=±

Since 3 < 2 < 2 < 3, 4 is in the range of f.

c. The range of f is the interval [0, 9] because for each number y in this interval, the number xy = is in the interval [ 3, 3]

The graph is not a function because a vertical line can be drawn through three points, as shown.

Domain: ( ) 3,;-¥ range: ( ] { }2,23-U

9. () 2 45yfxxx ==+-

a. Check whether the ordered pair (2, 7) satisfies the equation: () ? 2 72425 77 =+= 

The point (2, 7) is on the graph.

b. Let y = 8, then solve for x: ()() 22 845043 0313 or1 xxxx xxxx -=+-Þ=++Þ =++Þ=-=-

The points ( 3, 8) and ( 1, 8) lie on the graph.

c. Let x = 0, then solve for y: () 2 04055 y =+-=-

The y-intercept is 5.

d. Let y = 0, then solve for x: ()() 2 045051 5 or1 xxxx xx =+-Þ=+-Þ =-=

The x-intercepts are 5 and 1.

10. The range of C(t) is [6, 12). () () ( ) 11 22 1110611.989611.995CC=+=+»

11. From Example 11, we have 22500 APx =+ and 1200 PDx =- feet. If c = the cost on land, the total cost C is given by ( ) ( ) ( ) 22 22 1.3 1.35001200 1.35001200 CcPDcAP cxcx cxx =+ =++éù=++-êú ëû

12.a. ( ) 1200100,000Cxx=+

b. ( ) 2500 Rxx =

c. ( ) ( ) ( ) ( )25001200100,000 1300100,000 PxRxCx xx x ==-+ =-

d. The break-even point occurs when ( ) ( ) = CxRx . 1200100,0002500 100,000130077 xx xx += =Þ» Metro needs 77 shows to break even.

2.4 Basic Concepts and Skills

1. In the functional notation ( )yfx = , x is the independent variable.

2. If (2)7 f -= , then 2 is in the domain of the function f, and 7 is in the range of f.

3. If the point (9, 14) is on the graph of a function f, then (9)14. f =-

4. If (3, 7) and (3, 0) ar both points on a graph, then the graph cannot be the graph of a function.

5. False.

6. False. For example, if () 1 , fx x = then a = 1 and b = 1 are both in the domain of f However, a + b = 0 is not in the domain of f

7. True x = 7 and the square root function is defined for all positive numbers.

8. False. The domain of f is all real x for x > 2. Values of x ≤ –2 make the fraction undefined

2.4 Building Skills

9. Domain: {a, b, c}; range: {d, e}; function

10. Domain: {a, b, c}; range: {d, e, f}; function

11. Domain: {a, b, c}; range: {1, 2}; function

12. Domain: {1, 2, 3}; range: {a, b, c, d}; not a function

13. Domain: {0, 3, 8}; range:{ –3, –2, –1, 1, 2}; not a function

14. Domain: {–3, –1, 0, 1, 2, 3}; range: {–8, –3, 0, 1}; function

15. 22xyyx +=Þ=-+ ; a function

16. 11xyyx=-Þ=+ ; a function

17. 1 y x = ; a function

18. 1 1 xyy x =-Þ=- ; a function

19. 222 yxyxyx =Þ=±Þ=± ; not a function

20. or xyyxyx =Þ==- ; not a function

21. 1 25 y x =; a function

22. 2 1 1 y x =; a function

23. 2323 yxyx-=Þ=- ; a function

24. 3 35153 5 xyyx -=Þ=- ; a function

25. 2 88 xyyx +=Þ=±- ; not a function

26. 2 or xyyxyx =Þ==- ; not a function

27. 3 232 55 xyyx +=Þ=- ; a function

28. 3388 xyyx +=Þ=- ; a function

In exercises 29 32, 2 ()31, fxxx=-+ 2 (),gx x = and ()2. hxx =-

29. 2 (0)03(0)11 2 (0)(0) is undefined 0 f gg =-+= =Þ (0)202h =-= 2 22 ()31 ()()3()131 faaa fxxxxx =-+ -=---+=++

30. 2 2 (1)13(1)11;(1)2; 1 fg=-+=-== 2 (1)211;(); hga a =-== 2 2 22 ()gx x x ==

31. 2 (1)(1)3(1)15; f -=---+= 2 (1)(1) is undefined; 1 gg-=Þ-(1)2(1)3;()2; hhcc -=--==()2()2 hxxx -=--=+

32. 2 2 (4)43(4)15;(4)1; 4 fg=-+===

(4)242(4) is undefined; hh =-=-Þ 2 (2); 2 gk k += + 2 22 ()()3()1 2331 fakakak aakkak +=+-++ =++--+

33.a. 2 2(0) (0)0 40 f == -

b. 2 (1)2(1)223 4133 f === -

c. 2 2(2)4 (2)(2) is 420undefinedff ==Þ -

d. 2 2(2)4 (2)(2) 4(2)0isff -==Þundefined

e. 22()2()2 4()4 xx fx xx -==

34.a. 2 (0)2(0)04(0) is undefined gg =+-Þ

b. 2 (1)2(1)14(1) is undefined gg =+-Þ

c. 2 (2)2(2)244g =+-=

d. 2 (3)2(3)(3)465g -=-+--=-+

e. 2 2 ()2()()4 24 gxxx xx -=-+-=-+-

35. The width of each rectangle is 1. The height of the left rectangle is () 2 1123. f =+= The height of the right rectangle is () 2 2226. f =+=

()() ( ) ()() ( ) ()()() 1112 13169 sq. units Aff =+ =+=

36. The width of each rectangle is 1. The height of the left rectangle is () 2 0022. f =+= The height of the right rectangle is () 2 1123. f =+=

()() ( ) ()() ( ) ()()() 1011 12135 sq. units Aff =+ =+=

37. (,) -¥¥

38. (,) -¥¥

39. The denominator is not defined for x = 9. The domain is (,9)(9,) -¥¥ U

40. The denominator is not defined for x = –9. The domain is (,9)(9,) -¥--¥ U

41. The denominator is not defined for x = –1 or x = 1. The domain is (,1)(1,1)(1,). -¥--È¥ U

42. The denominator is not defined for x = –2 or x = 2. The domain is (,2)(2,2)(2,). -¥--¥ UU

43. The numerator is not defined for x < 3, and the denominator is not defined for x = –2. The domain is [3,) ¥

44. The denominator is not defined for x ≥ 4. The domain is (,4) -¥

45. The denominator equals 0 if x = –1 or x = –2. The domain is (,2)(2,1)(1,) -¥----¥ UU

46. The denominator equals 0 if x = –2 or x = –3. The domain is (,3)(3,2)(2,) -¥----¥ UU

47. The denominator is not defined for x = 0. The domain is (,0)(0,) -¥¥ U

48. The denominator is defined for all values of x. The domain is (,) -¥¥ .

49. a function

50. not a function

59. ()7hx = , so solve the equation 2 71 xx =-+ 2 60(3)(2)02xxxxx --=Þ-+=Þ=or 3. x =

51. a function

60. ()7Hx = , so solve the equation 2 78 xx =++ 2 10114(1)(1) 2(1) xxx -±++=Þ=Þ 13 2 x -±=Þ there is no real solution.

61.a. 2 12(11)711, =-++Þ=- which is false. Therefore, (1, 1) does not lie on the graph of f.

52. not a function

b. 22 2 12(1)72(1)6 (1)31313 xx xxx =-++Þ+=Þ +=Þ+=±Þ=-±

The points ( )13,1 and ( )13,1-+ lie on the graph of f.

c. ()2 20175yy =-++Þ=

The y-intercept is (0, 5).

53. not a function

d. () () () 22 2 0217721 7714 11 222 xx xx =-++Þ-=-+Þ =+Þ±=±=+Þ

14 1 2 x =-±

54. a function

55. (4)2;(1)1;(3)5;(5)7 ffff -=--===

56. (2)5;(1)4;(3)0;(4)5 gggg -==-==

57. (2)5;(1)4;(0)3;(1)4 hhhh -=--===

58. (1)4;(0)0;(1)4 fff-===-

The x-intercepts are 14 1,0 2 æö ç÷ èø and 14 1,0. 2 æö -+ ç÷ èø

62.a. () () 2 10321221012, =----Þ= which is false Therefore, ( 2, 10) does not lie on the graph of f

b. ()12fx = , so solve the equation 2 31212 xx --= 22 22 3121244 440(2)0 202 xxxx xxx xx --=Þ+=-Þ ++=Þ+=Þ +=Þ=-

c. () () 2 301200yy =--Þ=

The y-intercept is (0, 0).

d. () 2 0312034 0 or4 xxxx xx =--Þ=-+Þ ==-

The x-intercepts are (0, 0) and ( 4, 0)

63. Domain: [ 3, 2]; range: [ 3, 3]

64. Domain: [ 1, 3]; range: [ 2, 4]

65. Domain: [ ) 4, -¥ ; range: [ 2, 3]

66. Domain: ( ],4 -¥ ; range: [ 1, 3]

67. Domain: [ ) 3, -¥ ; range: [ ] { }1,43 U

68. Domain: ( ) [ ),11,4-¥-U

Range: ( 2, 4]

69. Domain: ( ] [ ] [ ) ,42,24, -¥-¥ UU

Range: [ ] { }2,23-U

70. Domain: ( ) [ ) ,21, -¥--¥ U

Range: ( ) , -¥¥

71. [ ) 9, -¥ 72. [–1, 7]

73. –3, 4, 7, 9 74. 6

75. ( ) ( ) ( ) 74,15,52fff-===

76. ( ) ( ) ( ) 44,17,33fff-=-==

77. { } [ ) 3.75,2.25,312, --¥ U

78. Æ

79. [ ) 9, -¥ 80. { } [ ]42,6 U

81. ( ) ( ) ( ) 41,13,34ggg-=-==

82. ()()55268gg--=--=

83. [ )9,5 84. [ ) 5, ¥ 2.4 Applying the Concepts

85. A function because there is only one high temperature per day.

86. A function because there is only one cost of a first-class stamp on January 1 each year.

Section 2.4 Functions 205

88. Not a function because people with a different name may have the same birthday.

89. 2 ();(4)16;AxxA== A(4) represents the area of a tile with side 4.

90. 33();(3)27 in.; VxxV== V(3) represents the volume of a cube with edge 3.

91. It is a function. 2 ()6;(3)54SxxS==

92. ();(59)1.5 39.37 x fxf=» meters

93.a. The domain is [0, 8].

b. 2 (2)128(2)16(2)192h =-=

2 (4)128(4)16(4)256h =-=

2 (6)128(6)16(6)192h =-=

c. 2 012816016(8) tttt =-Þ=-Þ 0 or8tt== . It will take 8 seconds for the stone to hit the ground. d.

94. After 4 hours, there are (0.75)(16) = 12 ml of the drug.

After 8 hours, there are (0.75)(12 + 16) = 21 ml. After 12 hours, there are (0.75)(21 + 16) = 27.75 ml.

After 16 hours, there are (0.75)(27.75 + 16) = 32.81 ml.

After 20 hours, there are (0.75)(32.81 + 16) = 36.61 ml.

95. ( ) 2 2828 2828 xyyx Pxxxx +=Þ==-=-

96. ( ) ( ) 2 6023030 3030 Pxyxyyx Axxxx ==+Þ=+Þ==-=-

97. Note that the length of the base = the width of the base = x 2 2 2 22 22 64 64 222 6464 222 128128256 22 Vlwhxhh x Slwlhwh xxx xx xx xxx ===Þ= =++ æöæö =++ç÷ç÷ èøèø =++=+

98.

a. 222222 22 xyryrx yrx +=Þ=-Þ =-

The length of the rectangle is 2x and its height is 22 yrx =() 22 22 22222 42 Plwxrx xrx =+=+=+-

b. 22 2 Alwxrx ==-

99. The piece with length x is formed into a circle, so 2. 2 x Cxrr p p ==Þ= Thus, the area of the circle is 2 2 2 . 24 xx Arpp pp æö === ç÷ èø

The piece with length 20 x is formed into a square, so ( ) 1 20420. 4 Pxssx =-=Þ=-

Thus, the area of the square is ( ) ( ) 2 2 2 11 2020. 416 sxx éù =-=êú ëû

The sum of the areas is ( ) 2 2 1 20 416 x Ax p =+-

100. The volume of the tank is 2 64, Vrh p == so 2 64 h r p = The top is open, so the surface area is given by 22 2 2 64 22 128 . rrhrr r r r pppp p p æö +=+ç÷ èø =+

101. The volume of the pool is 2 2 288 288.Vxhh x ==Þ=

The total area to be tiled is 2 2881152 44xhx x x æö == ç÷ èø

The cost of the tile is 11526912 6. xx æö =ç÷ èø

The area of the bottom of the pool is 2 , x so the cost of the cement is 2 2. x Therefore, the total cost is 2 6912 2.Cx x =+

102.

Using the Pythagorean theorem, we have ( )( ) ( )( ) 222 12 22 150030210030 150030210030 dtt dtt =-+-Þ éù =-+ëû

Using the distance formula we have

106.a. 2 ()(127525)127525 Rxxxxx =-=domain [1, 30]

The distance from A to P is 222525xx+=+ mi. At 4 mi/hr, it will take Julio 2 25 4 x + hr to row that distance. The distance from P to C is (8 x) mi, so it will take Julio 8 5 xhr to walk that distance. The total time it will take him to travel is 2 258 . 45 xx T +=+

b. 2 2 2 (1)1275(1)25(1)1250 (5)1275(5)25(5)5750 (10)1275(10)25(10)10,250

2 2 2 2 (15)1275(15)25(15)13,500 (20)1275(20)25(20)15,500 (25)1275(25)25(25)16,250 (30)1275(30)25(30)15,750 R R R R =-= =-= =-= =-=

105.a. (5)127525(5)1150.p =-= If 5000 TVs can be sold, the price per TV is $1150. (15)127525(15)900.p =-= If 15,000 TVs can be sold, the price per TV is $900. (30)127525(30)525.p =-= If 30,000 TVs can be sold, the price per TV is $525. b. c. 6501275256252525 xxx =-Þ-=-Þ= 25,000 TVs can be sold at $650 per TV.

This is the amount of revenue (in thousands of dollars) for the given number of TVs sold (in thousands).

d. 2 2 2 4700127525 511880 51514(1)(188) 2(1) 4 or47 xx xx x xx =-Þ -+=Þ ±=Þ == 47 is not in the domain, so 4000 TVs must be sold in order to generate revenue of 4.7 million dollars.

107.a. ()5.575,000Cxx=+ b. ()0.6(15)9 Rxxx ==

c. ()()()9(5.575,000) 3.575,000 PxRxCxxx x =-=-+ =-

d. The break-even point is when the profit is zero: 3.575,000021,429 xx -=Þ=

e. (46,000)3.5(46,000)75,000 $86,000 P == The company’s profit is $86,000 when 46,000 copies are sold.

108.a. ()0.5500,000;()5 CxxRxx =+= . The break-even point is when the profit is zero (when the revenue equals the cost): 50.5500,0004.5500,000 xxx =+Þ=Þ 111,111.11 x = . Because a fraction of a CD cannot be sold, 111,111 CD’s must be sold.

115. ()() fxgx ¹ because they have different domains. ()gx is not defined for x = –1, while ()fx is defined for all real numbers.

116. ()() fxgx ¹ because they have different domains. ()gx is not defined for x = 3, while ()fx is not defined for x = 3 or x = –2.

117. ()() fxgx = because ( ) ( )3103fg == and ( ) ( ) 5265.fg ==

118. ()() fxgx ¹ because ( ) 216 f = while ( ) 213. g =

b. ()()() 750,0005(0.5500,000) 1,250,0004.5277,778

PxRxCx xx xx ==-+ =Þ=

The company must sell 277,778 CDs in order to make a profit of $750,000.

2.4 Beyond the Basics

109. 2 4224 4 xxyxxyx y =Þ-=Þ=+Þ4242 ();yfxxx xx ++ =Þ=

Domain: 9 (,0)(0,).(4). 2 f -¥¥= U

110. 3223xyyyxy-=Þ-=-Þ 33 (2)3()22yxyfx xx -=-Þ=-Þ= 3

Domain: (,2)(2,).(4)2 f -¥¥= U

111. () 2 2 2 12 1 x xyxy x++=Þ=Þ + 2 22 (); Domain: 1(,);(4)17x fxf x=-¥¥=+

112. 2222 yxxyyxyx -=-Þ+=Þ () 2 222() 22 yxxyfxxx xx +=Þ=Þ= ++

Domain: 1 [0,);(4)9 f ¥=

113. ()() fxgx ¹ because they have different domains.

114. ()() fxgx ¹ because they have different domains.

119. 2 (2)15(2)231563 faaa==+-Þ=-Þ 3. a =

120. 222 (6)2866680 gbbbb==++Þ++=Þ (2)(4)02 or4.++=Þ=-=bbbb

121. 3(6)2 (6)001829 2(6) a haa b + ==Þ=+Þ=-(3)3(3)2(9) is undefinedhas a 2(3)zero h b +Þin the denominator. So 606. bb -=Þ=

122. ( ) 22 ()2323 fxxfxx=-Þ=()2 22 ()(23)4129 fxxxx =-=-+

123.

22 222 2 11111 ()1 gxxgx x xxx x æö =-Þ=-=ç÷ èø

22 22 111 ()0 gxgxx x xx æöæöæö +=-+-= ç÷ç÷ç÷ èøèøèø

124. 1 111()1 111 1 1 (1)(1) 121 (1)(1)2 1 x xx x fxf x xx x xx x xx xx xæö+=Þ=ç÷èø++ + + --+ + ==-=-++ +

125. 3 ()45

35(35)3(41) 3 ()414135(1220)(5(41))45 4141 (35)(123)17 (1220)(205)17 x fx x xxx xx ft xx x x x xxx x xx + =Þ++++ == ++--æö -ç÷ èø-

2.4 Critical Thinking/Discussion/Writing

126. Answers may vary. Sample answers are given

a. 2 yx=- b. 1 2 y x = -

c. 2 yx =- d. 1 2 y x = -

127.a. 2 0 axbxc++=

b. yc =

c. The equation will have no x-intercepts if 2 40bac-< .

d. It is not possible for the equation to have no y-intercepts because ().yfx =

128.a. () fxx = b. ()0fx =

c. () fxx =

d. 2 () fxx =- (Note: the point is the origin.)

e. ()1fx =

f. A vertical line is not a function.

129.a. {(a, 1), (b, 1)}

{(a, 1), (b, 2)}

{(a, 1), (b, 3)}

{(a, 3), (b, 1)}

{(a, 3), (b, 2)}

{(a, 3), (b, 3)}

{(a, 2), (b, 1)} {(a, 2), (b, 2)}

{(a, 2), (b, 3)}

There are nine functions from X to Y

b. {(1, a)}, {(2, a)}, {(3, a)}

{(1, a)}, {(2, a)}, {(3, b)}

{(1, a)}, {(2, b}, {(3, a)}

{(1, b)}, {(2, a)}, {(3, a)}

{(1, b)}, {(2, b}, {(3, a)}

{(1, b)}, {(2, a)}, {(3, b)}

{(1, b)}, {(2, b)}, {(3, b)}

There are eight functions from Y to X.

2.4 Getting Ready for the Next Section

131. 24122168 xxx -<Þ<Þ<

The solution set is ( ) ,8. -¥

132. ( ) 59715977 221or1 xxxx xxx +£+Þ+£+Þ £Þ£³

The solution set is [ ) 1,. ¥

133. 2 0 x >

Solve the associated equation: 2 00.xx=Þ=

So, the intervals are ( ) ( ) ,0 and 0,.-¥¥

Interval Test point Value of 2 x Result

( ),0 -¥ –1 1 + ( ) 0,¥ 1 1 +

The solution set is ( ) ( ) ,00, -¥¥ U

134. ( )( ) 350 xx -+³

Solve the associated equation: ( )( ) 3503 or5.xxxx -+=Þ==-

So, the intervals are ( ] [ ] [ ) ,5,5,3,and 3,.-¥--¥

Interval Test point Value of ( )( )35 xx -+ Result

( ],5-¥- –10 –65 –

[ ]5,3 - 0 15 +

[ ) 3,¥ 5 –20 –

The solution set is [ ] 5,3. -

For exercises 135–140, () 2 32,fxx =() 3, gxx=+ and () 2 2 1 hx x = +

135. () () () () 2 2 03203 0033 2 02 01 f g h =-= =+= == +

136. () () () () 2 2 13211 11342 2 11 11 f g h =-= =+== == +

137. () () () () () 2 2 2322385 22311 22 2 215 f g h -=--=-=-=-+== -== -+

138. () () ( ) ()() () () ( ) 22 22 22 2 13213221 3242241 1134 22 1 11211 2 22 fxxxx xxxx gxxx hx xx x xx +=-+=-++ =---=--+ +=++=+ +== +++++ = ++

139. () () () () () 22 22 323 3 22 11 fxxx gxx hx x x -=--=-=-+ -== -++

140. () ( ) () () 22 2 3223 3 2 1 fxxx gxx hx x -=--=-=-+ -=+

2.5 Properties of Functions

2.5 Practice Problems

1. The function is decreasing on ( ) ( ) 0,3,12,13, and ( ) 15,24; increasing on ( )3,12 and ( )13,15

2. Relative maxima of 3640 at x = 12 and 4070 at x = 15; relative minima of 40 at x = 3 and 3490 at x = 13. 3.

Relative minimum of 0 at x = 0 Relative maximum of 1 at x = 1

4. ( ) 2 11 vrr =-

[0, 13, 1] by [0, 250, 25]

Mrs. Osborn’s windpipe should be contracted to a radius of 7.33 mm for maximizing the airflow velocity.

5. () 2fxx =-

Replace x with x: ()() () 2 2 fxxxfx -=--=-=

Thus, the function is even.

6. () 3fxx =-

Replace x with x: ()() () 3 3 fxxxfx -=--==-

Thus, the function is odd.

7.a. 42 42 ()3()5() 35() () is even. gxxx xxfx gx -=--=-=Þ

b. ( ) 5353 53 ()4()2()42 42() () is odd. fxxxxx xxfx fx -=-+-=-=-+=-Þ

c. ( ) ( ) () () 2121 is neither even nor odd. hxxx hx hxh -=-+=-+ ¹ ¹-Þ

8. () () () ()() () 2 22 1;2,4 2123;41415 15312 6 422 fxxab ff fbfa ba =-== =-=-=-=-===-

The average rate of change is 6

9. ( ) ()() ()() ()() ()() () 1;2,,2 2121 1 112 22 12 1 2 fttabxx faf fbfxx fbfax x baxx x x =-==¹ ==-===-== ==-The average rate of change is 1

10. () () () () () () () 2 2 2 1008002000 01000800020002000 3100380032000500 fxxx f f =-+ =-+= =-+= ( ) ( ) 3050020001500 500 3033 ff===-

The average rate of change is –500, so the number of bacteria per cubic centimeter decreases by 500 each day after adding the chlorine.

11. () ()()() 2 2 22 3 3 23 fxxx fxhxhxh xxhhxh =-++=-+++=---++( ) ( ) ( )( ) 222 2 233 2 21 fxhfx h xxhhxhxx h xhhh xh h

2.5 Concepts and Vocabulary

1. A function f is decreasing if 12xx < implies that ( ) ( )12fxfx >

2. ( )fa is a relative maximum of f if there is an interval ( )12 , xx containing a such that ( ) ( )fafx ³ for every x in the interval ( )12,.xx

3. A function f is even if f( x) = f(x) for all x in the domain of f.

4. The average rate of change of f as x changes from x = a to x = b is ()() ,. fbfa ab ba¹ -

5. True

6. False. A relative maximum or minimum could occur at an endpoint of the domain of the function.

7. True

8. False. The graph of an odd function is symmetric with respect to the origin.

2.5 Building Skills

9. Increasing on ( ) , -¥¥

10. Decreasing on ( ) , -¥¥

11. Increasing on ( ) ,2, -¥ decreasing on ( ) 2, ¥

12. Decreasing on ( ) ,3, -¥ increasing on ( ) 3, ¥

13. Increasing on ( ) ,2,-¥- constant on ( ) 2,2, - increasing on ( ) 2, ¥

14. Decreasing on ( ) ,1,-¥- constant on ( ) 1,4, - decreasing on ( ) 4, ¥

15. Increasing on ( ),3-¥- and ( ) 1 2 -,2 , decreasing on ( )1 2 3, and ( ) 2, ¥

16. Increasing on ( ) ( ) 3,1,0,1, and ( ) 2,. ¥ Decreasing on ( ) ( ) ,3,1,0,-¥-- and ( ) 1,2.

17. No relative extrema

18. No relative extrema

19. (2, 10) is a relative maximum point and a turning point.

20. (3, 2) is a relative minimum point and a turning point.

21. Any point on (x, 2) is a relative maximum and a relative minimum point on the interval (–2, 2) Relative maximum at ( 2, 2); relative minimum at (2, 2). None of these points are turning points.

22. Any point on (x, 3) is a relative maximum and a relative minimum point on the interval (–1, 4). Relative maximum at (4, 3); relative minimum at ( 1, 3). None of these points are turning points.

23. (–3, 4) and (2, 5) are relative maxima points and turning points. ( ) 1 2 ,2 is a relative minimum and a turning point.