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Welcome to Interactive Chalkboard Algebra 2 Interactive Chalkboard Copyright © by The McGraw-Hill Companies, Inc. Send all inquiries to: GLENCOE DIVISION.

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Presentation on theme: "Welcome to Interactive Chalkboard Algebra 2 Interactive Chalkboard Copyright © by The McGraw-Hill Companies, Inc. Send all inquiries to: GLENCOE DIVISION."— Presentation transcript:

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2 Welcome to Interactive Chalkboard Algebra 2 Interactive Chalkboard Copyright © by The McGraw-Hill Companies, Inc. Send all inquiries to: GLENCOE DIVISION Glencoe/McGraw-Hill 8787 Orion Place Columbus, Ohio 43240

3 Splash Screen

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

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

6 Example 1-1a 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.

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

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

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

10 Example 1-1e 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.

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

12 Example 1-2a 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.

13 Example 1-2b Original function Replace r with 5. Simplify. Original function Replace r with 6. Simplify.

14 Example 1-2c 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.

15 Example 1-2d 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.

16 Example 1-2e 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.

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

18 Example 1-3a FindOriginal function Replace x with y 3. Answer: Property of powers

19 Example 1-3b 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.

20 Example 1-3c 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

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

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

23 Example 1-4a 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...

24 Example 1-4b 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...

25 Example 1-4c 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...

26 Example 1-4d 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...

27 Example 1-4e 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...

28 Example 1-4f 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...

29 End of Lesson 1

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

31 Answer: Example 2-1a Graphby making a table of values. xf(x)f(x) –45 –3–4 –2–3 – –19

32 Answer: Example 2-1b Graphby 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.

33 Example 2-1c xf (x)f (x) –3–8 –21 – Graphby making a table of values. Answer:

34 Example 2-2a 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. xf (x)f (x) –29 –1 01 1–3 2–7 319 change in signs

35 Example 2-2b 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.

36 Example 2-2c 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.

37 Example 2-3a GraphEstimate the x -coordinates at which the relative maximum and relative minimum occur. Make a table of values and graph the function. xf (x) –2–19 – –3 3– zero at x = –1 zero between x = 1 and x = 2 zero between x = 3 and x = 4 indicates a relative maximum indicates a relative minimum

38 Example 2-3b 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. xf (x) –2–19 – –3 3–

39 Example 2-3c GraphEstimate 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.

40 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. Example 2-4a Make a table of values for weeks 0 through 7. Plot the points and connect with a smooth curve.

41 Answer: Example 2-4b nw(n)w(n)

42 Example 2-4c 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.

43 Example 2-4d What trends in the patients 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.

44 Example 2-4e 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:

45 Example 2-4f 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.

46 End of Lesson 2

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

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

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

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

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

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

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

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

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

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

57 Example 3-2e 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.

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

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

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

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

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

63 Example 3-3d Solve Answer: –8, –27

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

65 Example 3-4b 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. Zero Product Property or Solve each equation.

66 Example 3-4c Solve Answer: 25

67 End of Lesson 3

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

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

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

71 Example 4-1c Iffind f (3). Answer: 34

72 Example 4-2a Show thatis 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

73 Example 4-2b 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,

74 Example 4-2c CheckYou can see that the graph of the related functioncrosses the x -axis at 3, –6, and –1. Thus,

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

76 Example 4-3a 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)

77 Example 4-3b 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.

78 Example 4-3c 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.

79 End of Lesson 4

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

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

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

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

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

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

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

87 Example 5-1g Solve each equation. State the number and type of roots. a. b. c. 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

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

89 Example 5-2a 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 –

90 Example 5-2b 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. 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. x1

91 Example 5-2c Answer: Number of Positive Real Zeros Number of Negative Real Zeros Number of Imaginary Zeros Total

92 Example 5-2d 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.

93 Example 5-3a 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 no yes

94 Example 5-3b 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. x1–124 –3 1–414–38 –2 1–38–12 –1 1–240 Each row in the table shows the coefficients of the depressed polynomial and the remainder.

95 Example 5-3c 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.

96 Example 5-3d Simplify.

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

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

99 Example 5-4a Short-Response Test Item Write a polynomial function of least degree with integer coefficients whose zeros include 4 and 4 – i. Read the Test ItemIf 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.

100 Example 5-4b Solve the Test ItemWrite 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.

101 Example 5-4c Square x – 4 and replace i 2 with –1. Simplify. Multiply using the Distributive Property. Combine like terms.

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

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

104 End of Lesson 5

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

106 Example 6-1a 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,

107 Example 6-1b List all of the possible rational zeros of Since the coefficient of x 4 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.

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

109 Example 6-2a 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.

110 Example 6-2b Write the equation for volume. Formula for volume Multiply. Subtract 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.

111 Example 6-2c 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. p12–8– –5– – So, the zero is 10. The other dimensions are 10 – 2 or 8 feet and or 14 feet.

112 Example 6-2d CheckVerify that the dimensions are correct. Answer:

113 Example 6-2e 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:

114 Example 6-3a 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.

115 Example 6-3b 0–15– –6– – –1911 p q Since f (2) = 0, you know that x = 2 is a zero. The depressed polynomial is

116 Example 6-3c 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 –15–1331 p q There is another zero at x = 3. The depressed polynomial is

117 Example 6-3d 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.

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

119 End of Lesson 6

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

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

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

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

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

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

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

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

128 Example 7-3a 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:

129 Example 7-3b To findevaluate 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.

130 Example 7-3c 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:

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

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

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

134 Example 7-4d Evaluateand x = –2. Function from part a Replace x with –2. Simplify.

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

136 Example 7-4f a.Findand and b.Evaluateand x = 1. Answer: and Answer: and

137 Example 7-5a 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. ExploreLet x = her income per paycheck, r (x) = her income after the deduction for retirement, t (x) = her income after tax.

138 Example 7-5b PlanWrite equations for r (x) and t (x). $100 is deducted for retirement. The tax rate is 4%. SolveIf Tracie has her retirement deducted before taxes, then her net income is represented by Replace x with 1500 in

139 Example 7-5c Replace x with 1400 in If Tracie has her retirement deducted after taxes, then her net income is represented by Replace x with 1500 in

140 Example 7-5d Replace x with 1440 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. ExamineThe answer makes sense. Since the taxes are being applied to a smaller amount, less taxes will be deducted from her paycheck.

141 Example 7-5e 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.

142 End of Lesson 7

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

144 Example 8-1a 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)}.

145 Example 8-1b 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.

146 Example 8-1c 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.

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

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

149 Example 8-2c Answer:The inverse of is

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

151 Example 8-2e Answer:

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

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

154 Example 8-3b Answer:The functions are inverses since both andequal x.

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

156 End of Lesson 8

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

158 Example 9-1a GraphState 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

159 Example 9-1b Make a table of values to graph the function yx

160 Example 9-1c 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.

161 Example 9-1d GraphState the domain, range, and x- and y- intercepts. Answer: domain: x 1 range: y 0 x -intercept: 1 y -intercept: none

162 Example 9-2a 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.

163 Example 9-2b The function is Make a table of values and graph the function. av Answer: The domain is a 0 and the range is v 0.

164 Example 9-2c 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.

165 Example 9-2d 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

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

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

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

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

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

171 Example 9-3e a.Graph Answer:

172 Example 9-3f b.Graph Answer:

173 End of Lesson 9

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