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4. Contents Lesson 9-1Multiplying and Dividing Rational Expressions Lesson 9-2Adding and Subtracting Rational Expressions Lesson 9-3Graphing Rational Functions Lesson 9-4Direct, Joint, and Inverse Variation Lesson 9-5Classes of Functions Lesson 9-6Solving Rational Equations and Inequalities

5. Lesson 1 Contents Example 1Simplify a Rational Expression Example 2Use the Process of Elimination Example 3Simplify by Factoring Out –1 Example 4Multiply Rational Expressions Example 5Divide Rational Expressions Example 6Polynomials in the Numerator and Denominator Example 7Simplify a Complex Fraction

6. Example 1-1a Simplify Look for common factors. 1 1 Factor. Simplify. Answer:

7. Example 1-1b Under what conditions is this expression undefined? A rational expression is undefined if the denominator equals zero. To find out when this expression is undefined, completely factor the denominator. Answer: The values that would make the denominator equal to 0 are –7, 3, and –3. So the expression is undefined at y = –7, y = 3, and y = –3. These values are called excluded values.

8. Example 1-1c a.Simplify b.Under what conditions is this expression undefined? Answer: Answer: undefined for x = –5, x = 4, x = –4

9. Example 1-2a Multiple-Choice Test Item For what values of p isundefined? A 5 B –3, 5 C 3, –5 D 5, 1, –3 Read the Test Item You want to determine which values of p make the denominator equal to 0.

10. Example 1-2b Solve the Test Item Look at the possible answers. Notice that the p term and the constant term are both negative, so there will be one positive solution and one negative solution. Therefore, you can eliminate choices A and D. Factor the denominator. Factor the denominator. Solve each equation. Answer:B Zero Product Property or

11. Example 1-2c Multiple-Choice Test Item For what values of p isundefined? A –5, –3, –2 B –5 C 5 D –5, –3 Answer:D

12. Example 1-3a Simplify Factor the numerator and the denominator. Simplify. Answer:or –a or 1 1 a 1

13. Example 1-3b Simplify Answer: –x

14. Example 1-4a Simplify Simplify. Answer:Simplify. Factor. 1111111 1111111

15. Example 1-4b Simplify Factor. 1111111 1111111 1 Answer:Simplify.

16. Example 1-4c Simplify each expression. a. b. Answer:

17. Example 1-5a Simplify Answer:Simplify. Factor. 1111111 1111111 Multiply by the reciprocal of divisor.

18. Example 1-5b Simplify Answer:

19. Example 1-6a Simplify Multiply by the reciprocal of the divisor. 1 –11 1 11 Answer:Simplify.

20. Example 1-6b Simplify Multiply by the reciprocal of the divisor. Simplify.Answer: Factor. 1 11 1

21. Example 1-6c Answer: 1 Simplify each expression. a. b. Answer:

22. Example 1-7a Simplify Express as a division expression. Multiply by the reciprocal of divisor.

23. Example 1-7b Factor. 11–1 1 11 Simplify.Answer:

24. Example 1-7c Simplify Answer:

25. End of Lesson 1

26. Lesson 2 Contents Example 1LCM of Monomials Example 2LCM of Polynomials Example 3Monomial Denominators Example 4Polynomial Denominators Example 5Simplify Complex Fractions Example 6Use a Complex Fraction to Solve a Problem

27. Example 2-1a Find the LCM of 15a 2 bc 3, 16b 5 c 2, and 20a 3 c 6. Factor the first monomial. Factor the second monomial. Factor the third monomial.

28. Example 2-1b Use each factor the greatest number of times it appears as a factor and simplify. Answer:

29. Example 2-1c Find the LCM of 6x 2 zy 3, 9x 3 y 2 z 2, and 4x 2 z. Answer: 36x 3 y 3 z 2

30. Example 2-2a Find the LCM of x 3 – x 2 – 2x and x 2 – 4x + 4. Factor the first polynomial. Factor the second polynomial. Answer: Use each factor the greatest number of times it appears as a factor.

31. Example 2-2b Find the LCM of x 3 + 2x 2 – 3x and x 2 + 6x + 9. Answer:

32. Example 2-3a Simplify The LCD is 42a 2 b 2. Find equivalent fractions that have this denominator. Simplify each numerator and denominator. Add the numerators. Answer:

33. Example 2-3b Simplify Answer:

34. Example 2-4a Simplify Factor the denominators. The LCD is 6(x – 5). Subtract the numerators.

35. Example 2-4b Distributive Property Combine like terms. Simplify. 1 1 Answer:

36. Example 2-4c Simplify Answer:

37. Example 2-5a Simplify The LCD of the numerator is ab. The LCD of the denominator is b.

38. Example 2-5b Simplify the numerator and denominator. Write as a division expression. Multiply by the reciprocal of the divisor. 1 1

39. Example 2-5c Simplify.Answer:

40. Example 2-5d Simplify Answer:

41. Example 2-6a Coordinate Geometry Find the slope of the line that passes throughand Definition of slope

42. Example 2-6b The LCD of the numerator is 3k. The LCD of the denominator is 2k. Write as a division expression. Simplify. Answer: The slope is

43. Example 2-6c Coordinate Geometry Find the slope of the line that passes throughand Answer:

44. End of Lesson 2

45. Lesson 3 Contents Example 1Vertical Asymptotes and Point Discontinuity Example 2Graph with a Vertical Asymptote Example 3Graph with Point Discontinuity Example 4Use Graphs of Rational Functions

46. Example 3-1a Determine the equations of any vertical asymptotes and the values of x for any holes in the graph of First factor the numerator and denominator of the rational expression.

47. Example 3-1b Answer: The function is undefined for x = –2 and –3. Since x = –3 is a vertical asymptote and x = –2 is a hole in the graph. 1 1

48. Example 3-1c Determine the equations of any vertical asymptotes and the values of x for any holes in the graph of Answer: vertical asymptote: x = –5 ; hole: x = –3

49. Answer: Example 3-2a Graph The function is undefined for x = –1. Since is in its simplest form, x = –1 is a vertical asymptote. Draw the vertical asymptote.

50. Example 3-2b Make a table of values. xf (x)f (x) –41.33 –3 1.5 –2 2 0 0 1 0.5 20.67 30.75 Answer: Plot the points and draw the graph.

51. Example 3-2c As |x| increases, it appears that the y values of the function get closer and closer to 1. The line with the equation f (x) = 1 is a horizontal asymptote of the function. Answer:

52. Example 3-2d Graph Answer:

53. Example 3-3a Graph Notice thatorTherefore, the graph of is the graph of with a hole at

54. Example 3-3b Answer:

55. Example 3-3c Graph Answer:

56. Example 3-4a Transportation A train travels at one velocity V 1 for a given amount of time t 1 and then another velocity V 2 for a different amount of time t 2. The average velocity is given by Let t 1 be the independent variable and let V be the dependent variable. Draw the graph if V 1 = 50 miles per hour, V 2 = 30 miles per hour, and t 2 = 1 hour.

57. Example 3-4b Answer: The function is The vertical asymptote is Graph the vertical asymptote and the function. Notice that the horizontal asymptote is

58. Example 3-4c What is the V -intercept of the graph? Answer:The V -intercept is 30.

59. Example 3-4d What values of t 1 and V are meaningful in the context of the problem? Answer:In the problem context, time and velocity are positive values. Therefore, positive values of t 1 and V values between 30 and 60 are meaningful.

60. Example 3-4e Transportation A train travels at one velocity V 1 for a given amount of time t 1 and then another velocity V 2 for a different amount of time t 2. The average velocity is given by a.Let t 1 be the independent variable and let V be the dependent variable. Draw the graph if V 1 = 60 miles per hour, V 2 = 30 miles per hour, and t 2 = 1 hour.

61. Example 3-4f Answer:

62. b. What is the V -intercept of the graph? c.What values of t 1 and V are meaningful in the context of the problem? Example 3-4g Answer:The V -intercept is 30. Answer: t 1 is positive and V is between 30 and 60.

63. End of Lesson 3

64. Lesson 4 Contents Example 1Direct Variation Example 2Joint Variation Example 3Inverse Variation Example 4Use Inverse Variation

65. Example 4-1a If y varies directly as x and y = –15 when x = 5, find y when x = 3. Use a proportion that relates the values. Direct proportion y 1 = –15 ; x 1 = 5, and x 2 = 3 Cross multiply.

66. Example 4-1b Simplify. Divide each side by 5. Answer:When x = 3, the value of y is –9.

67. Example 4-1c If y varies directly as x and y = 12 when x = –3, find y when x = 7. Answer: –28

68. Example 4-2a Suppose y varies jointly as x and z. Find y when x = 10 and z = 5 if y = 12 when x = 3 and z = 8. Use a proportion that relates the values. Joint variation y 1 = 12, x 1 = 3, z 1 = 8, x 2 = 10, and z 2 = 5 Cross multiply.

69. Example 4-2b Simplify. Divide each side by 24. Answer:When x = 10 and z = 5, y = 25.

70. Example 4-2c Suppose y varies jointly as x and z. Find y when x = 3 and z = 2 if y = 11 when x = 5 and z = 22. Answer:

71. Example 4-3a If a varies inversely as b and a = –6 when b = 2, find a when b = –7. Use a proportion that relates the values. Inverse variation a 1 = –6 ; b 1 = 2, and b 2 = –7 Cross multiply.

72. Example 4-3b Simplify. Answer:When b = –7, a is Divide each side by –7.

73. Example 4-3c If a varies inversely as b and a = 3 when b = 8, find a when b = 6. Answer: 4

74. Example 4-4a Space The apparent length of an object is inversely proportional to one’s distance from the object. Earth is about 93 million miles from the Sun and Venus is about 67 million miles away. How much larger would the diameter of the Sun appear on Venus than on Earth? ExploreYou know that the apparent diameter of the Sun varies inversely with the distance from the Sun. You also know the distance from the Sun to Venus and from the Sun to Earth. You want to determine how much larger the diameter of the Sun appears on Venus than on Earth.

75. Example 4-4b PlanLet the apparent diameter of the Sun from Earth equal 1 unit and the apparent diameter of the Sun from Venus equal v. Then use a proportion that relates the values. Solve Substitution

76. Example 4-4c Cross multiply. Divide each side by 67 million miles. Simplify.

77. Example 4-4d Answer: From Venus, the diameter of the Sun will appear about 1.39 times as large as it appears from Earth. ExamineSince the distance between the Sun and Earth is between 1 and 2 times the distance between the Sun and Venus, the answer seems reasonable.

78. Example 4-4e Space Mars is about 142.5 million miles from the Sun and Earth is about 93 million miles away. How much smaller would the diameter of the Sun appear on Mars than on Earth? Answer:about 0.65 of the size

79. End of Lesson 4

80. Lesson 5 Contents Example 1Identify a Function Given the Graph Example 2Match Equation with Graph Example 3Identify a Function Given its Equation

81. Example 5-1a Identify the type of function represented by the graph. Answer:The graph is a V shape. So, it is an absolute value function.

82. Example 5-1b Identify the type of function represented by the graph. Answer:The graph is a parabola, so it is a quadratic function.

83. Example 5-1c Identify the type of function represented by each graph. a.b. Answer:greatest integer function Answer:inverse variation

84. Example 5-2a Shipping Charges A chart gives the shipping rates for an Internet company. They charge $3.50 to ship less than 1 pound, $3.95 for 1 pound and over up to 2 pounds, and $5.20 for 2 pounds and over up to 3 pounds. Which graph depicts these rates? a.b.c.

85. a.b.c. Example 5-2b The shipping rate is constant for x values from 0 to 1. Then it jumps at x = 1 and remains constant until x = 2. The graph jumps again at x = 2 and remains constant until x = 3. Answer:The graph of this function looks like steps, so this is c, the step or greatest integer function.

86. Example 5-2c a.b.c. A ball is thrown into the air. The path of the ball is represented by the equation Which graph best represents this situation? Answer: b; quadratic function

87. Example 5-3a Identify the type of function represented by Then graph the equation. Since the equation has no x-intercept, it is the constant function. Determine some points on the graph and graph it. Answer:

88. Example 5-3b Identify the type of function represented by Then graph the equation. Since the equation includes an expression with a square root, it is a square root function. Plot some points and use what you know about square root graphs to graph it. Answer:

89. Example 5-3c Identify the type of function represented by each equation. Then graph the equation. a.b. Answer: rational function Answer: direct variation

90. End of Lesson 5

91. Lesson 6 Contents Example 1Solve a Rational Equation Example 2Elimination of a Possible Solution Example 3Work Problem Example 4Rate Problem Example 5Solve a Rational Inequality

92. Example 6-1a SolveCheck your solution. The LCD for the three denominators is Original equation Multiply each side by 24(3 – x).

93. Example 6-1b 1 1 11 1 6 Simplify. Add.

94. Example 6-1c Check Original equation Simplify. The solution is correct.

95. Example 6-1d Answer: The solution is –45.

96. Example 6-1e Answer: Solve

97. Example 6-2a SolveCheck your solution. The LCD is Original equation Multiply by the LCD, (p 2 – 1). p – 1 1 1 1

98. Example 6-2b Distributive Property Simplify. Add (2p 2 – 2p + 1) to each side.

99. Example 6-2c Factor. or Zero Product Property Solve each equation. Divide each side by 3.

100. Example 6-2d Check Original equation Simplify.

101. Since p = –1 results in a zero in the denominator, eliminate –1. Answer: The solution is p = 2. Example 6-2e Simplify. Original equation

102. Example 6-2f Answer: Solve

103. Example 6-3a Mowing Lawns Tim and Ashley mow lawns together. Tim working alone could complete the job in 4.5 hours, and Ashley could complete it alone in 3.7 hours. How long does it take to complete the job when they work together? In 1 hour, Tim could completeof the job. In 1 hour, Ashley could completeof the job.

104. Example 6-3b In t hours, Tim could completeor of the job. In t hours, Ashley could completeorof the job. Part completed by Timplus part completed by Ashleyequalsentire job. 1

105. Example 6-3c Solve the equation. Original equation Multiply each side by 16.65. Distributive Property Simplify.

106. Example 6-3d Simplify. Divide each side by 8.2. Answer: It would take them about 2 hours working together.

107. Example 6-3e Cleaning Libby and Nate clean together. Nate working alone could complete the job in 3 hours, and Libby could complete it alone in 5 hours. How long does it take to complete the job when they work together? Answer: about 2 hours

108. Example 6-4a Swimming Janine swims for 5 hours in a stream that has a current of 1 mile per hour. She leaves her dock and swims upstream for 2 miles and then back to her dock. What is her swimming speed in still water? WordsThe formula that relates distance, time, and rate is VariablesLet r be her speed in still water. Then her speed with the current is r + 1 and her speed against the current is r – 1.

109. Example 6-4b Time going with the currentplus time going against the currentequals total time. 5 Equation Solve the equation. Original equation

110. Example 6-4c Multiply each side by r 2 – 1. Distributive Property r + 1r – 1 1 1 Simplify. Subtract 4r from each side.

111. Example 6-4d Use the Quadratic Formula to solve for r. Quadratic Formula x = r, a = 5, b = –4, and c = –5 Simplify.

112. Example 6-4e Simplify. Use a calculator. Answer:Since the speed must be positive, the answer is about 1.5 miles per hour.

113. Example 6-4f Swimming Lynne swims for 1 hour in a stream that has a current of 2 miles per hour. She leaves her dock and swims upstream for 3 miles and then back to her dock. What is her swimming speed in still water? Answer:about 6.6 mph

114. Example 6-5a Solve Step 1Values that make the denominator equal to 0 are excluded from the denominator. For this inequality the excluded value is 0. Step 2Solve the related equation. Related equation

115. Example 6-5b Multiply each side by 9s. Simplify. Add. Divide each side by 6.

116. Example 6-5c Step 3Draw vertical lines at the excluded value and at the solution to separate the number line into regions. Now test a sample value in each region to determine if the values in the region satisfy the inequality.

117. Example 6-5d Test is a solution.

118. Example 6-5e is not a solution. Test

119. Example 6-5f is a solution. Test

120. Example 6-5g Answer: The solution

121. Example 6-5h Solve Answer:

122. End of Lesson 6

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