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3. Lesson 7-1 Polynomial Functions
Lesson 7-2 Graphing Polynomial Functions
Lesson 7-3 Solving Equations Using Quadratic Techniques
Lesson 7-4 The Remainder and Factor Theorems
Lesson 7-5 Roots and Zeros
Lesson 7-6 Rational Zero Theorem
Lesson 7-7 Operations on Functions
Lesson 7-8 Inverse Functions and Relations
Lesson 7-9 Square Root Functions and Inequalities
Contents

4. Example 1 Find Degrees and Leading Coefficients
Example 2 Evaluate a Polynomial Function
Example 3 Functional Values of Variables
Example 4 Graphs of Polynomial Functions
Lesson 1 Contents

5. State the degree and leading coefficient of. in one variable
State the degree and leading coefficient of in one variable. If it is not a polynomial in one variable, explain why.
Answer: This is a polynomial in one variable. The degree is 3 and the leading coefficient is 7.
Example 1-1a

6. State the degree and leading coefficient of. in one variable
State the degree and leading coefficient of in one variable. If it is not a polynomial in one variable, explain why.
Answer: This is not a polynomial in one variable. It contains two variables, a and b.
Example 1-1b

7. State the degree and leading coefficient of. in one variable
State the degree and leading coefficient of in one variable. If it is not a polynomial in one variable, explain why.
Answer: This is not a polynomial in one variable. The term 2c–1 is not of the form ancn, where n is a nonnegative integer.
Example 1-1c

8. Rewrite the expression so the powers of y are in decreasing order.
State the degree and leading coefficient of in one variable. If it is not a polynomial in one variable, explain why.
Rewrite the expression so the powers of y are in decreasing order.
Answer: This is a polynomial in one variable with degree of 4 and leading coefficient 1.
Example 1-1d

9. Answer: degree 3, leading coefficient 3
State the degree and leading coefficient of each polynomial in one variable. If it is not a polynomial in one variable, explain why. a. b.
Answer: degree 3, leading coefficient 3
Answer: This is not a polynomial in one variable. It contains two variables, x and y.
Example 1-1e

10. Answer: degree 3, leading coefficient 1
Answer: This is not a polynomial in one variable. The term 3a–1 is not of the form ancn, where n is nonnegative.
Answer: degree 3, leading coefficient 1
Example 1-1f

11. Find the values of f (4), f (5), and f (6).
Nature Refer to Example 2 on page 347 of your textbook. A sketch of the arrangement of hexagons shows a fourth ring of 18 hexagons, a fifth ring of 24 hexagons, and a sixth ring of 30 hexagons.
Show that the polynomial function gives the total number of hexagons when
Find the values of f (4), f (5), and f (6).
Original function
Replace r with 4.
Simplify.
Example 1-2a

12. Original function Replace r with 5. Simplify. Original function
Example 1-2b

13. Answer: These match the function values for respectively.
From the information given in Example 2 of your textbook, you know that the total number of hexagons for three rings is 19. So, the total number of hexagons for four rings is or 37, five rings is or 61, and six rings is or 91.
Answer: These match the function values for respectively.
Example 1-2c

14. Find the total number of hexagons in a honeycomb with 20 rings.
Nature Refer to Example 2 on page 347 of your textbook. A sketch of the arrangement of hexagons shows a fourth ring of 18 hexagons, a fifth ring of 24 hexagons, and a sixth ring of 30 hexagons.
Find the total number of hexagons in a honeycomb with 20 rings.
Original function
Replace r with 20.
Answer:
Simplify.
Example 1-2d

15. Nature A sketch of the arrangement of hexagons shows a seventh ring of 36 hexagons, an eighth ring of 42 hexagons, and a ninth ring of 48 hexagons.
a. Show that the polynomial function gives the total number of hexagons when Recall that the total number of hexagons in six rings is 91.
Answer: f (7) = 127; f (8) = 169; f (9) = 217; the total number of hexagons for seven rings is or 127, eight rings is or 169, and nine rings is or 217. These match the functional values for r = 7, 8, and 9, respectively.
Example 1-2e

16. b. Find the total number of hexagons in a honeycomb with 30 rings.
Answer: 2611
Example 1-2f

17. Find Original function Replace x with y 3. Answer: Property of powers
Example 1-3a

18. To evaluate b(2x – 1), replace m in b(m) with 2x – 1.
Find
To evaluate b(2x – 1), replace m in b(m) with 2x – 1.
Original function
Replace m with 2x – 1.
Evaluate 2(2x – 1)2.
Simplify.
Example 1-3b

19. Distributive Property
To evaluate 3b(x), replace m with x in b(m), then multiply the expression by 3.
Original function
Replace m with x.
Distributive Property
Example 1-3c

20. Now evaluate b(2x – 1) – 3b(x).
Replace b(2x – 1) and 3b(x) with evaluated expressions.
Simplify.
Answer:
Example 1-3d

21. a. Find b. Find
Answer:
Answer:
Example 1-3e

22.  describe the end behavior,
For the graph,
 describe the end behavior,
 determine whether it represents an odd-degree or an even-degree function, and
 state the number of real zeros.
Answer: 
 It is an even-degree polynomial function.
 The graph does not intersect the x-axis, so the function has no real zeros. .
Example 1-4a

23.  describe the end behavior,
For the graph,
 describe the end behavior,
 determine whether it represents an odd-degree or an even-degree function, and
 state the number of real zeros.
Answer: 
 It is an odd-degree polynomial function.
 The graph intersects the x-axis at one point, so the function has one real zero. .
Example 1-4b

24.  describe the end behavior,
For the graph,
 describe the end behavior,
 determine whether it represents an odd-degree or an even-degree function, and
 state the number of real zeros.
Answer: 
 It is an even-degree polynomial function.
 The graph intersects the x-axis at two points, so the function has two real zeros. .
Example 1-4c

25.  describe the end behavior,
For each graph, a.
 describe the end behavior,
 determine whether it represents an odd-degree or an even-degree function, and
 state the number of real zeros.
Answer: 
 It is an even-degree polynomial function.
 The graph intersects the x-axis at two points, so the function has two real zeros. .
Example 1-4d

26.  describe the end behavior,
For each graph, b.
 describe the end behavior,
 determine whether it represents an odd-degree or an even-degree function, and
 state the number of real zeros.
Answer: 
 It is an odd-degree polynomial function.
 The graph intersects the x-axis at three points, so the function has three real zeros. .
Example 1-4e

27.  describe the end behavior,
For each graph, c.
 describe the end behavior,
 determine whether it represents an odd-degree or an even-degree function, and
 state the number of real zeros.
Answer: 
 It is an even-degree polynomial function.
 The graph intersects the x-axis at one point, so the function has one real zero. .
Example 1-4f

28. End of Lesson 1

29. Example 1 Graph a Polynomial Function
Example 2 Locate Zeros of a Function
Example 3 Maximum and Minimum Points
Example 4 Graph a Polynomial Model
Lesson 2 Contents

30. making a table of values. –4 5 –3 –2 –1 2
Graph by
making a table of values. x
f(x) –4 5 –3 –2 –1 2 1
–19
Answer:
Example 2-1a

31. making a table of values.
Graph by
making a table of values.
This is an odd degree polynomial with a negative leading coefficient, so f (x)  + as x  – and f (x)  – as x  +. Notice that the graph intersects the x-axis at 3 points indicating that there are 3 real zeros.
Answer:
Example 2-1b

32. making a table of values. –3 –8 –2 1 –1 2
Graph by
making a table of values. x
f (x) –3 –8 –2 1 –1 2 4 17
Answer:
Example 2-1c

33. Determine consecutive values of x between which each real zero of the function is located. Then draw the graph.
Make a table of values. Since f (x) is a 4th degree polynomial function, it will have between 0 and 4 zeros, inclusive. x
f (x) –2 9 –1 1 –3 2 –7 3 19
change in signs
change in signs
change in signs
change in signs
Example 2-2a

34. Look at the value of f (x) to locate the zeros
Look at the value of f (x) to locate the zeros. Then use the points to sketch the graph of the function.
Answer:
There are zeros between x = –2 and –1, x = –1 and 0, x = 0 and 1, and x = 2 and 3.
Example 2-2b

35. There are zeros between x = –1 and 0, x = 0 and 1, and x = 3 and 4.
Determine consecutive values of x between which each real zero of the function is located. Then draw the graph.
Answer:
There are zeros between x = –1 and 0, x = 0 and 1, and x = 3 and 4.
Example 2-2c

36. Make a table of values and graph the function.
Graph Estimate the x-coordinates at which the relative maximum and relative minimum occur.
Make a table of values and graph the function. x
f (x) –2
–19 –1 5 1 2 –3 3 –4 4 30
zero at x = –1
indicates a relative maximum
zero between x = 1 and x = 2
indicates a relative minimum
zero between x = 3 and x = 4
Example 2-3a

37. Answer: The value of f (x) at x = 0 is greater than the surrounding points, so it is a relative maximum. The value of f (x) at x = 3 is less than the surrounding points, so it is a relative minimum. x
f (x) –2
–19 –1 5 1 2 –3 3 –4 4 30
Example 2-3b

38. Graph Estimate the x-coordinates at which the relative maximum and relative minimum occur.
Answer: The value of f (x) at x = 0 is less than the surrounding points, so it is a relative minimum. The value of f (x) at x = –2 is greater than the surrounding points, so it is a relative maximum.
Example 2-3c

39. Health The weight w, in pounds, of a patient during a 7-week illness is modeled by the cubic equation where n is the number of weeks since the patient became ill.
Graph the equation.
Make a table of values for weeks 0 through 7. Plot the points and connect with a smooth curve.
Example 2-4a

40. n
w(n)
110 1
109.5 2
108.4 3
107.3 4
106.8 5
107.5 6 7
114.9
Answer:
Example 2-4b

41. Describe the turning points of the graph and its end behavior.
Answer: There is a relative minimum at week 4. For the end behavior, w (n) increases as n increases.
Example 2-4c

42. What trends in the patient’s weight does the graph suggest?
Answer: The patient lost weight for each of 4 weeks after becoming ill. After 4 weeks, the patient started to gain weight and continues to gain weight.
Example 2-4d

43. Weather The rainfall r, in inches per month, in a Midwestern town during a 7-month period is modeled by the cubic equation where m is the number of months after March 1.
a. Graph the equation.
Answer:
Example 2-4e

44. b. Describe the turning. points of the graph. and its end behavior. c
b. Describe the turning points of the graph and its end behavior c. What trends in the amount of rainfall received by the town does the graph suggest?
Answer: There is a relative maximum at Month 2, or May. For the end behavior, r (m) decreases as m increases.
Answer: The rainfall increased for two months following March. After two months, the amount of rainfall decreased for the next five months and continues to decrease.
Example 2-4f

45. End of Lesson 2

46. Example 1 Write an Expression in Quadratic Form
Example 2 Solve Polynomial Equations
Example 3 Solve Equations with Rational Exponents
Example 4 Solve Radical Equations
Lesson 3 Contents

47. Write in quadratic form, if possible.
Answer:
Example 3-1a

48. Write in quadratic form, if possible.
Answer:
Example 3-1b

49. Write in quadratic form, if possible.
Answer: This cannot be written in quadratic form since
Example 3-1c

50. Write in quadratic form, if possible.
Answer:
Example 3-1d

51. Write each expression in quadratic form, if possible. a. b. c.
Answer:
Answer:
Answer: This cannot be written in quadratic form since
Answer:
Example 3-1e

52. Write the expression on the left in quadratic form.
Solve .
Original equation
Write the expression on the left in quadratic form.
Factor the trinomial.
Factor each difference of squares.
Example 3-2a

53. Use the Zero Product Property.or
Answer: The solutions are –5, –2, 2, and 5.
Example 3-2b

54. Check. The graph of. shows
Check The graph of shows that the graph intersects the x-axis at –5, –2, 2, and 5.
Example 3-2c

55. This is the sum of two cubes.
Solve .
Original equation
This is the sum of two cubes.
Sum of two cubes formula with a = x and b = 6 or
Zero Product Property
Example 3-2d

56. Replace a with 1, b with –6, and c with 36.
The solution of the first equation is –6. The second equation can be solved by using the Quadratic Formula.
Quadratic Formula
Replace a with 1, b with –6, and c with 36.
Simplify.
Example 3-2e

57. Answer: The solutions of the original equation are
Simplify.
Answer: The solutions of the original equation are
Example 3-2f

58. Solve each equation. a. b.
Answer: –3, –1, 1, 3
Answer:
Example 3-2g

59. Write the expression on the left in quadratic form.
Solve
Original equation
Write the expression on the left in quadratic form.
Factor the trinomial.
Zero Product Property or
Example 3-3a

60. Isolate x on one side of the equation. or
Raise each side to the fourth power.
Simplify.
Example 3-3b

61. Check Substitute each value into the original equation.
Answer: The solution is 81.
Example 3-3c

62. Solve
Answer: –8, –27
Example 3-3d

63. Rewrite so that one side is zero.
Solve
Original equation
Rewrite so that one side is zero.
Write the expression on the left side in quadratic form.
Factor.
Example 3-4a

64. Zero Product Property or Solve each equation.
Answer: Since the square root of x cannot be negative, the equation has no solution. Thus, the only solution of the original equation is 9.
Example 3-4b

65. Solve
Answer: 25
Example 3-4c

66. End of Lesson 3

67. Example 1 Synthetic Substitution Example 2 Use the Factor Theorem
Example 3 Find All Factors of a Polynomial
Lesson 4 Contents

68. 3 10 41 164 654 If find f (4). Method 1 Synthetic Substitution
By the Remainder Theorem, f (4) should be the remainder when you divide the polynomial by x – 4.
Notice that there is no x term. A zero is placed in this position as a placeholder.
Answer: The remainder is 654. Thus, by using synthetic substitution, f (4) = 654.
Example 4-1a

69. Method 2 Direct Substitution
Replace x with 4.
Original function
Replace x with 4.
Simplify.
or 654
Answer: By using direct substitution, f (4) = 654.
Example 4-1b

70. If find f (3).
Answer: 34
Example 4-1c

71. Show that is a factor of Then find the remaining factors of the polynomial.
The binomial x – 3 is a factor of the polynomial if 3 is a zero of the related polynomial function. Use the factor theorem and synthetic division.
Example 4-2a

72. Since the remainder is 0, (x – 3) is a factor of the polynomial
Since the remainder is 0, (x – 3) is a factor of the polynomial. The polynomial can be factored as The polynomial is the depressed polynomial. Check to see if this polynomial can be factored.
Factor the trinomial.
Answer: So,
Example 4-2b

73. Check. You can see that the graph of the related function
Check You can see that the graph of the related function crosses the x-axis at 3, –6, and –1. Thus,
Example 4-2c

74. Show that is a factor of Then find the remaining factors of the polynomial.
Answer:
So, Since
Example 4-2d

75. Geometry The volume of a rectangular prism is given by Find the missing measures.
The volume of a rectangular prism is You know that one measure is x – 2, so x – 2 is a factor of V(x).
Example 4-3a

76. The quotient is . Use this to factor V(x).
Volume function
Factor.
Factor the trinomial
Answer: The missing measures of the prism are x + 4 and x + 5.
Example 4-3b

77. Answer: The missing measures of the prism are x – 2 and x + 5.
Geometry The volume of a rectangular prism is given by Find the missing measures.
Answer: The missing measures of the prism are x – 2 and x + 5.
Example 4-3c

78. End of Lesson 4

79. Example 1 Determine Number and Type of Roots
Example 2 Find Numbers of Positive and Negative Zeros
Example 3 Use Synthetic Substitution to Find Zeros
Example 4 Use Zeros to Write a Polynomial Function
Lesson 5 Contents

80. Solve State the number and type of roots.
Original equation
Add 10 to each side.
Answer: This equation has exactly one real root, 10.
Example 5-1a

81. Solve State the number and type of roots.
Original equation
Factor.
Zero Product Property or
Solve each equation.
Answer: This equation has two real roots, –8 and 6.
Example 5-1b

82. Solve State the number and type of roots.
Original equation
Factor out the GCF.
Use the Zero Product Property. or
Subtract 6 from each side.
Example 5-1c

83. Square Root Property
Answer: This equation has one real root at 0, and two imaginary roots at
Example 5-1d

84. Solve State the number and type of roots.
Original equation
Factor differences of squares.
Factor differences of squares. or
Zero Product Property
Example 5-1e

85. Solve each equation.
Answer: This equation has two real roots, –2 and 2, and two imaginary roots, 2i and –2i.
Example 5-1f

86. Solve each equation. State the number and type of roots.
Answer: This equation has exactly one root at –3.
Answer: This equation has exactly two roots, –3 and 4.
Answer: This equation has one real root at 0 and two imaginary roots at
Example 5-1g

87. d.
Answer: This equation has two real roots, –3 and 3, and two imaginary roots, 3i and –3i.
Example 5-1h

88. State the possible number of positive real zeros, negative real zeros, and imaginary zeros of
Since p(x) has degree 6, it has 6 zeros. However, some of them may be imaginary. Use Descartes Rule of Signs to determine the number and type of real zeros. Count the number of changes in sign for the coefficients of p(x).
yes
– to +
yes
+ to – no
– to – no
– to –
Example 5-2a

89. Since there are two sign changes, there are 2 or 0 positive real zeros
Since there are two sign changes, there are 2 or 0 positive real zeros. Find p(–x) and count the number of sign changes for its coefficients. x 1 no
– to – no
– to –
yes
– to +
yes
+ to –
Since there are two sign changes, there are 2 or 0 negative real zeros. Make a chart of possible combinations.
Example 5-2b

90. Answer: 2 6 4 Number of Positive Real Zeros
Number of Negative Real Zeros
Number of Imaginary Zeros
Total 2 6 4
Example 5-2c

91. State the possible number of positive real zeros, negative real zeros, and imaginary zeros of
Answer: The function has either 2 or 0 positive real zeros, 2 or 0 negative real zeros, and 4, 2, or 0 imaginary zeros.
Example 5-2d

92. Find all of the zeros of
Since f (x) has degree of 3, the function has three zeros. To determine the possible number and type of real zeros, examine the number of sign changes in f (x) and f (–x).
yes
yes no no no
yes
Example 5-3a

93. The function has 2 or 0 positive real zeros and exactly 1 negative real zero. Thus, this function has either 2 positive real zeros and 1 negative real zero or 2 imaginary zeros and 1 negative real zero.
To find the zeros, list some possibilities and eliminate those that are not zeros. Use a shortened form of synthetic substitution to find f (a) for several values of a. x 1 –1 2 4 –3 –4 14
–38 –2 8
–12
Each row in the table shows the coefficients of the depressed polynomial and the remainder.
Example 5-3b

94. Replace a with 1, b with –2, and c with 4.
From the table, we can see that one zero occurs at x = –1. Since the depressed polynomial, , is quadratic, use the Quadratic Formula to find the roots of the related quadratic equation
Quadratic Formula
Replace a with 1, b with –2, and c with 4.
Example 5-3c

95. Simplify.
Simplify.
Example 5-3d

96. Answer: Thus, this function has one real zero at –1 and two imaginary zeros at and The graph of the function verifies that there is only one real zero.
Example 5-3e

97. Find all of the zeros of
Answer:
Example 5-3f

98. Short-Response Test Item
Write a polynomial function of least degree with integer coefficients whose zeros include 4 and 4 – i.
Read the Test Item • If 4 – i is a zero, then 4 + i is also a zero, according to the Complex Conjugate Theorem. So, x – 4, x – (4 – i), and x – (4 + i) are factors of the polynomial function.
Example 5-4a

99. • Multiply the factors to find the polynomial function.
Solve the Test Item • Write the polynomial function as a product of its factors.
• Multiply the factors to find the polynomial function.
Write an equation.
Regroup terms.
Rewrite as the difference of two squares.
Example 5-4b

100. Square x – 4 and replace i2 with –1.
Simplify.
Multiply using the Distributive Property.
Combine like terms.
Example 5-4c

101. Answer: is a polynomial function of least degree with integral coefficients whose zeros are 4, 4 – i, and 4 + i.
Example 5-4d

102. Short-Response Test Item
Write a polynomial function of least degree with integer coefficients whose zeros include 2 and 1 + i.
Answer:
Example 5-4e

103. End of Lesson 5

104. Example 1 Identify Possible Zeros
Example 2 Use the Rational Zero Theorem
Example 3 Find All Zeros
Lesson 6 Contents

105. List all of the possible rational zeros of
If is a rational zero, then p is a factor of 4 and q is a factor of 3. The possible factors of p are 1, 2, and 4. The possible factors of q are 1 and 3.
Answer: So,
Example 6-1a

106. List all of the possible rational zeros of
Since the coefficient of x4 is 1, the possible zeros must be a factor of the constant term –15.
Answer: So, the possible rational zeros are 1, 3, 5, and 15.
Example 6-1b

107. List all of the possible rational zeros of each function. a.
Answer:
Answer:
Example 6-1c

108. Let x = the height, x – 2 = the width, and x + 4 = the length.
Geometry The volume of a rectangular solid is 1120 cubic feet. The width is 2 feet less than the height and the length is 4 feet more than the height. Find the dimensions of the solid.
Let x = the height, x – 2 = the width, and x + 4 = the length.
Example 6-2a

109. Write the equation for volume.
Formula for volume
Multiply.
Subtract 1120.
The leading coefficient is 1, so the possible integer zeros are factors of Since length can only be positive, we only need to check positive zeros.
Example 6-2b

110. The possible factors are 1, 2, 4, 5, 8, 10, 14, 16, 20, 28, 32, 35, 40, 56, 70, 80, 112, 140, 160, 224, 280, 560, and By Descartes’ Rule of Signs, we know that there is exactly one positive real root. Make a table and test possible real zeros. p 1 2 –8
–1120 3 –5
–1125 6
–1112 10 12
112
So, the zero is 10. The other dimensions are 10 – 2 or 8 feet and or 14 feet.
Example 6-2c

111. Check Verify that the dimensions are correct.
Answer:
Example 6-2d

112. Geometry The volume of a rectangular solid is 100 cubic feet
Geometry The volume of a rectangular solid is 100 cubic feet. The width is 3 feet less than the height and the length is 5 feet more than the height. Find the dimensions of the solid.
Answer:
Example 6-2e

113. The possible rational zeros are 1, 2, 3, 5, 6, 10, 15, and 30.
Find all of the zeros of
From the corollary to the Fundamental Theorem of Algebra, we know there are exactly 4 complex roots.
According to Descartes’ Rule of Signs, there are 2 or 0 positive real roots and 2 or 0 negative real roots.
The possible rational zeros are 1, 2, 3, 5, 6, 10, 15, and 30.
Make a table and test some possible rational zeros.
Example 6-3a

114. –15
–13 3 1 2 24 –6
–17 30 11
–19 p q
Since f (2) = 0, you know that x = 2 is a zero. The depressed polynomial is
Example 6-3b

115. There is another zero at x = 3. The depressed polynomial is
Since x = 2 is a positive real zero, and there can only be 2 or 0 positive real zeros, there must be one more positive real zero. Test the next possible rational zeros on the depressed polynomial. 5 6 1 3
–15
–13 p q
There is another zero at x = 3. The depressed polynomial is
Example 6-3c

116. Write the depressed polynomial.
Factor
Write the depressed polynomial.
Factor. or
Zero Product Property
There are two more real roots at x = –5 and x = –1.
Answer: The zeros of this function are –5, –1, 2, and 3.
Example 6-3d

117. Find all of the zeros of
Answer: –5, –3, 1, and 3
Example 6-3e

118. End of Lesson 6

119. Example 1 Add and Subtract Functions
Example 2 Multiply and Divide Functions
Example 3 Evaluate Composition of Relations
Example 4 Simplify Composition of Functions
Example 5 Use Composition of Functions
Lesson 7 Contents

120. Given , find
Addition of functions
and
Simplify.
Answer:
Example 7-1a

121. Subtraction of functions
Given , find
Subtraction of functions
and
Simplify.
Answer:
Example 7-1b

122. Given find each function.b.
Answer:
Answer:
Example 7-1c

123. Distributive Property
Given find
Product of functions
and
Distributive Property
Distributive Property
Simplify.
Answer:
Example 7-2a

124. Given find
Division of functions
Answer:
and
Example 7-2b

125. Since 4 makes the denominator 0, it is excluded from the domain of
Example 7-2c

126. Given find each function.b.
Answer:
Answer:
Example 7-2d

127. If f (x) = {(2, 6), (9, 4), (7, 7), (0, –1)} and g (x) = {(7, 0), (–1, 7), (4, 9), (8, 2)}, find and
To find , evaluate g (x) first. Then use the range of g as the domain of f and evaluate f (x).
Answer:
Example 7-3a

128. Answer: Since 6 is not in the domain of g, is undefined for x = 2.
To find evaluate f (x) first. Then use the range of f as the domain of g and evaluate g (x).
is undefined.
Answer: Since 6 is not in the domain of g, is undefined for x = 2.
Example 7-3b

129. If f (x) = {(1, 2), (0, –3), (6, 5), (2, 1)} and g (x) = {(2, 0), (–3, 6), (1, 0), (6, 7)}, find and
Answer:
Example 7-3c

130. Composition of functions
Find and and
Composition of functions
Replace g (x) with 2x – 1.
Substitute 2x – 1 for x in f (x).
Example 7-4a

131. Composition of functions
Evaluate (2x – 1)2.
Simplify.
Composition of functions
Replace f (x) with
Example 7-4b

132. Substitute for x in g (x).
Simplify.
Answer: So, and
Example 7-4c

133. Evaluate and x = –2. Function from part a Replace x with –2. Simplify.
Example 7-4d

134. Function from part a Replace x with –2. Simplify. Answer: So, and
Example 7-4e

135. a. Find and and b. Evaluate and x = 1. Answer: and Answer: and
Example 7-4f

136. Taxes Tracie Long has $100 deducted from every paycheck for retirement
Taxes Tracie Long has $100 deducted from every paycheck for retirement. She can have this deduction taken before state taxes are applied, which reduces her taxable income. Her state income tax is 4%. If Tracie earns $1500 every pay period, find the difference in her net income if she has the retirement deduction taken before or after state taxes.
Explore Let x = her income per paycheck, r (x) = her income after the deduction for retirement, t (x) = her income after tax.
Example 7-5a

137. $100 is deducted for retirement.
Plan Write equations for r (x) and t (x).
$100 is deducted for retirement.
The tax rate is 4%.
Solve If Tracie has her retirement deducted before taxes, then her net income is represented by
Replace x with in
Example 7-5b

138. Replace x with in
If Tracie has her retirement deducted after taxes, then her net income is represented by
Replace x with 1500 in
Example 7-5c

139. Replace x with in
Answer: and The difference is 1344 – 1340 or 4. So, her net income is $4 more if the retirement deduction is taken before taxes.
Examine The answer makes sense. Since the taxes are being applied to a smaller amount, less taxes will be deducted from her paycheck.
Example 7-5d

140. Taxes Brandi Smith has $200 deducted from every paycheck for retirement. She can have this deduction taken before state taxes are applied, which reduces her taxable income. Her state income tax is 10%. If Brandi earns $2200 every pay period, find the difference in her net income if she has the retirement deduction taken before or after state taxes.
Answer: Her net income is $20 more if she has the retirement deduction taken before her state taxes.
Example 7-5e

141. End of Lesson 7

142. Example 1 Find an Inverse Relation Example 2 Find an Inverse Function
Example 3 Verify Two Functions are Inverses
Lesson 8 Contents

143. Geometry The ordered pairs of the relation {(1, 3), (6, 3), (6, 0), (1, 0)} are the coordinates of the vertices of a rectangle. Find the inverse of this relation and determine whether the resulting ordered pairs are also the coordinates of the vertices of a rectangle.
To find the inverse of this relation, reverse the coordinates of the ordered pairs. The inverse of the relation is {(3, 1), (3, 6), (0, 6), (0, 1)}.
Example 8-1a

144. Answer: Plotting the points shows that the ordered pairs also describe the vertices of a rectangle. Notice that the graph of the relation and the inverse are reflections over the graph of y = x.
Example 8-1b

145. Geometry The ordered pairs of the relation {(–3, 4), (–1, 5), (2, 3), (1, 1), (–2, 1)} are the coordinates of the vertices of a pentagon. Find the inverse of this relation and determine whether the resulting ordered pairs are also the coordinates of the vertices of a pentagon.
Answer: {(4, –3), (5, –1), (3, 2), (1, 1), (1, –2)} These ordered pairs also describe the vertices of a pentagon.
Example 8-1c

146. Step 1 Replace f (x) with y in the original equation.
Find the inverse of
Step 1 Replace f (x) with y in the original equation.
Step 2 Interchange x and y.
Example 8-2a

147. Step 4 Replace y with f –1(x).
Step 3 Solve for y.
Inverse
Multiply each side by –2.
Add 2 to each side.
Step 4 Replace y with f –1(x).
Example 8-2b

148. Answer: The inverse of is
Example 8-2c

149. Graph the function and its inverse.
Graph both functions on the coordinate plane. The graph of is the reflection for over the line
Example 8-2d

150. Answer:
Example 8-2e

151. b. Graph the function and its inverse. Answer:
a. Find the inverse of
b. Graph the function and its inverse.
Answer:
Answer:
Example 8-2f

152. Determine whether and are inverse functions.
Check to see if the compositions of f (x) and g (x) are identity functions.
Example 8-3a

153. Answer: The functions are inverses since both and equal x.
Example 8-3b

154. Determine whether and are inverse functions.
Answer: The functions are inverses since both compositions equal x.
Example 8-3c

155. End of Lesson 8

156. Example 1 Graph a Square Root Function
Example 2 Solve a Square Root Problem
Example 3 Graph a Square Root Inequality
Lesson 9 Contents

157. Graph State the domain, range, and x- and y-intercepts.
Since the radicand cannot be negative, identify the domain.
Write the expression inside the radicand as  0.
Solve for x.
The x-intercept is
Example 9-1a

158. Make a table of values to graph the function.
2.83 6
2.55 5
2.23 4
1.87 3
1.14 2
0.71 1 y x
Example 9-1b

159. Answer:
From the graph, you can see that the domain is and the range is y  0. The x-intercept is There is no y-intercept.
Example 9-1c

160. x-intercept: 1 y-intercept: none
Graph State the domain, range, and x- and y-intercepts.
Answer:
domain: x  1
range: y  0
x-intercept: 1
y-intercept: none
Example 9-1d

161. Graph the function. State the domain and range.
Physics When an object is spinning in a circular path of radius 2 meters with velocity v, in meters per second, the centripetal acceleration a, in meters per second squared, is directed toward the center of the circle. The velocity v and acceleration a of the object are related by the function
Graph the function. State the domain and range.
Example 9-2a

162. The function is Make a table of values and graph the function.
Answer:
The domain is a  0 and the range is v  0. a v 1
1.41 2 3
2.45 4
2.83 5
3.16
Example 9-2b

163. What would be the centripetal acceleration of an object spinning along the circular path with a velocity of 4 meters per second?
Original equation
Replace v with 4.
Square each side.
Divide each side by 2.
Answer: The centripetal acceleration would be 8 meters per second squared.
Example 9-2c

164. a. Graph the function. State the domain and range.
Geometry The volume V and surface area A of a soap bubble are related by the function
a. Graph the function. State the domain and range.
Answer:
domain: A  0
range: V  0
Example 9-2d

165. b. What would the surface area be if the volume were 94 cubic units?
Answer: 100 units2
Example 9-2e

166. Graph
Graph the related equation Since the boundary is not included, the graph should be dashed.
Example 9-3a

167. The domain includes values for So the graph is to the right of
Select a point and test its ordered pair.
Test (0, 0).
false
Shade the region that does not include (0, 0).
Example 9-3b

168. Graph the related equation
The domain includes values for Select a point and test its ordered pair.
Example 9-3c

169. Shade the region that includes (4, 1).
Test (4, 1).
true
Shade the region that includes (4, 1).
Example 9-3d

170. a. Graph
Answer:
Example 9-3e

171. b. Graph
Answer:
Example 9-3f

172. End of Lesson 9

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