1. Slide 2- 1

2. Chapter 2 Polynomial, Power, and Rational Functions

3. 2.1 Linear and Quadratic Functions and Modeling

4. Slide 2- 4 Quick Review

5. Slide 2- 5 What you’ll learn about Polynomial Functions Linear Functions and Their Graphs Average Rate of Change Linear Correlation and Modeling Quadratic Functions and Their Graphs Applications of Quadratic Functions … and why Many business and economic problems are modeled by linear functions. Quadratic and higher degree polynomial functions are used to model some manufacturing applications.

6. Slide 2- 6 Polynomial Function

7. Slide 2- 7 Polynomial Functions of No and Low Degree NameFormDegree Zero Functionf(x) = 0Undefined Constant Functionf(x) = a (a ≠ 0)0 Linear Functionf(x)=ax + b (a ≠ 0)1 Quadratic Functionf(x)=ax 2 + bx + c (a ≠ 0)2

8. Slide 2- 8 Example Finding an Equation of a Linear Function

9. Slide 2- 9 Average Rate of Change

10. Slide 2- 10 Constant Rate of Change Theorem A function defined on all real numbers is a linear function if and only if it has a constant nonzero average rate of change between any two points on its graph.

11. Slide 2- 11 Characterizing the Nature of a Linear Function Point of ViewCharacterization Verbalpolynomial of degree 1 Algebraic f(x) = mx + b (m ≠ 0) Graphicalslant line with slope m and y-intercept b Analyticalfunction with constant nonzero rate of change m: f is increasing if m > 0, decreasing if m < 0; initial value of the function = f(0) = b

12. Slide 2- 12 Properties of the Correlation Coefficient, r 1. -1 ≤ r ≤ 1 2. When r > 0, there is a positive linear correlation. 3. When r < 0, there is a negative linear correlation. 4. When |r| ≈ 1, there is a strong linear correlation. 5. When |r| ≈ 0, there is weak or no linear correlation.

13. Slide 2- 13 Linear Correlation

14. Slide 2- 14 Regression Analysis 1. Enter and plot the data (scatter plot). 2. Find the regression model that fits the problem situation. 3. Superimpose the graph of the regression model on the scatter plot, and observe the fit. 4. Use the regression model to make the predictions called for in the problem.

15. Slide 2- 15 Example Transforming the Squaring Function

16. Slide 2- 16 Example Transforming the Squaring Function

17. Slide 2- 17 The Graph of f(x)=ax 2

18. Slide 2- 18 Vertex Form of a Quadratic Equation Any quadratic function f(x) = ax 2 + bx + c, a ≠ 0, can be written in the vertex form f(x) = a(x – h) 2 + k The graph of f is a parabola with vertex (h,k) and axis x = h, where h = -b/(2a) and k = c – ah 2. If a > 0, the parabola opens upward, and if a < 0, it opens downward.

19. Slide 2- 19 Find the vertex and the line of symmetry of the graph y = (x – 1) 2 + 2 Domain Range (- ,  ) [2,  ) Vertex (1,2) x = 1

20. Slide 2- 20 Find the vertex and the line of symmetry of the graph y = -(x + 2) 2 - 3 Domain Range (- ,  ) (- ,-3] Vertex (-2,-3) x = -2

21. Slide 2- 21 Let f(x) = x 2 + 2x + 4. (a) Write f in standard form. (b) Determine the vertex of f. (c) Is the vertex a maximum or a minimum? Explain f(x) = x 2 + 2x + 4 f(x) = (x + 1) 2 + 3 Vertex (-1,3) opens up (-1,3) is a minimum + 1- 1

22. Slide 2- 22 Let f(x) = 2x 2 + 6x - 8. (a) Write f in standard form. (b) Determine the vertex of f. (c) Is the vertex a maximum or a minimum? Explain f(x) = 2(x + 3/2) 2 - 25/2 Vertex (-3/2,-25/2) opens up (-3/2,25/2) is a minimum + 9/4 - 9/2f(x) = 2(x 2 + 3x ) - 8

23. Slide 2- 23 If we perform completing the square process on f(x) = ax 2 + bx + c and write it in standard form, we get

24. Slide 2- 24 So the vertex is

25. Slide 2- 25 To get the coordinates of the vertex of any quadratic function, simply use the vertex formula. If a > 0, the parabola open up and the vertex is a minimum. If a < 0, the parabola opens down and the parabola is a maximum.

26. Slide 2- 26 Example Finding the Vertex and Axis of a Quadratic Function

27. Slide 2- 27 Characterizing the Nature of a Quadratic Function Point of View Characterization

28. Slide 2- 28 Vertical Free-Fall Motion

29. 2.2 Power Functions and Modeling

30. Slide 2- 30 Quick Review

31. Slide 2- 31 What you’ll learn about Power Functions and Variation Monomial Functions and Their Graphs Graphs of Power Functions Modeling with Power Functions … and why Power functions specify the proportional relationships of geometry, chemistry, and physics.

32. Slide 2- 32 Power Function Any function that can be written in the form f(x) = k·x a, where k and a are nonzero constants, is a power function. The constant a is the power, and the k is the constant of variation, or constant of proportion. We say f(x) varies as the a th power of x, or f(x) is proportional to the a th power of x.

33. Slide 2- 33 Example Analyzing Power Functions

34. Slide 2- 34 Monomial Function Any function that can be written as f(x) = k or f(x) = k·x n, where k is a constant and n is a positive integer, is a monomial function.

35. Slide 2- 35 Example Graphing Monomial Functions

36. Slide 2- 36 Graphs of Power Functions For any power function f(x) = k·x a, one of the following three things happens when x < 0. f is undefined for x < 0. f is an even function. f is an odd function.

37. Slide 2- 37 Graphs of Power Functions

38. 2.3 Polynomial Functions of Higher Degree with Modeling

39. Slide 2- 39 Quick Review

40. Slide 2- 40 What you’ll learn about Graphs of Polynomial Functions End Behavior of Polynomial Functions Zeros of Polynomial Functions Intermediate Value Theorem Modeling … and why These topics are important in modeling and can be used to provide approximations to more complicated functions, as you will see if you study calculus.

41. Slide 2- 41 The Vocabulary of Polynomials

42. Slide 2- 42 Example Graphing Transformations of Monomial Functions

43. Slide 2- 43 Cubic Functions

44. Slide 2- 44 Quartic Function

45. Slide 2- 45 Local Extrema and Zeros of Polynomial Functions A polynomial function of degree n has at most n – 1 local extrema and at most n zeros.

46. Slide 2- 46 Leading Term Test for Polynomial End Behavior

47. Slide 2- 47 Example Applying Polynomial Theory

48. Slide 2- 48 Example Finding the Zeros of a Polynomial Function

49. Slide 2- 49 Multiplicity of a Zero of a Polynomial Function

50. Slide 2- 50 Example Sketching the Graph of a Factored Polynomial

51. Slide 2- 51 Intermediate Value Theorem If a and b are real numbers with a < b and if f is continuous on the interval [a,b], then f takes on every value between f(a) and f(b). In other words, if y 0 is between f(a) and f(b), then y 0 =f(c) for some number c in [a,b].

52. 2.4 Real Zeros of Polynomial Functions

53. Slide 2- 53 Quick Review

54. Slide 2- 54 What you’ll learn about Long Division and the Division Algorithm Remainder and Factor Theorems Synthetic Division Rational Zeros Theorem Upper and Lower Bounds … and why These topics help identify and locate the real zeros of polynomial functions.

55. Slide 2- 55 Remainder Check: Quotient * Divisor + Remainder = Dividend Division

56. Slide 2- 56 Division Algorithm for Polynomials

57. Slide 2- 57 Dividing Polynomials Long division of polynomials is similar to long division of whole numbers. dividend = (quotient divisor) + remainder The result is written in the form: quotient + When you divide two polynomials you can check the answer using the following:

58. Slide 2- 58 + 2 Example: Divide x 2 + 3x – 2 by x + 1 and check the answer. x x 2 + x 2x2x– 2 2x + 2 – 4– 4 remainder Check: 1. 2. 3. 4. 5. 6. correct (x + 2) quotient (x + 1) divisor + (– 4) remainder = x 2 + 3x – 2 dividend Answer: x + 2 + – 4– 4

59. Slide 2- 59 Example: Divide 4x + 2x 3 – 1 by 2x – 2 and check the answer. Write the terms of the dividend in descending order. 1. x2x2 2. 2x 3 – 2x 2 3. 2x22x2 + 4x 4. + x 5. 2x 2 – 2x 6. 6x6x – 1 7. + 3 8. 6x – 6 9. 5 Check: (x 2 + x + 3)(2x – 2) + 5 = 4x + 2x 3 – 1 Answer: x 2 + x + 3 5 Since there is no x 2 term in the dividend, add 0x 2 as a placeholder.

60. Slide 2- 60 x x 2 – 2x – 3x + 6 – 3 – 3x + 6 0 Answer: x – 3 with no remainder. Check: (x – 2)(x – 3) = x 2 – 5x + 6 Example: Divide x 2 – 5x + 6 by x – 2.

61. Slide 2- 61 Example: Divide x 3 + 3x 2 – 2x + 2 by x + 3 and check the answer. x2x2 x 3 + 3x 2 0x20x2 – 2x – 2 – 2x – 6 8 Check: (x + 3)(x 2 – 2) + 8 = x 3 + 3x 2 – 2x + 2 Answer: x 2 – 2 + 8 + 2 Note: the first subtraction eliminated two terms from the dividend. Therefore, the quotient skips a term. + 0x

62. Slide 2- 62 16 Synthetic division is a shorter method of dividing polynomials. This method can be used only when the divisor is of the form x – a. It uses the coefficients of each term in the dividend. Example: Divide 3x 2 + 2x – 1 by x – 2 using synthetic division. 3 2 – 1 2 Since the divisor is x – 2, a = 2. 3 1. Bring down 3 2. (2 3) = 6 6 815 3. (2 + 6) = 8 4. (2 8) = 16 5. (–1 + 16) = 15 coefficients of quotient remainder value of a coefficients of the dividend 3x + 8Answer: 15

63. Slide 2- 63 Example: Divide x 3 – 3x + 4 by x + 3 using synthetic division. Notice that the degree of the first term of the quotient is one less than the degree of the first term of the dividend. remainder a coefficients of quotient – 3– 3 Since, x – a = x + 3, a = – 3. 1 0 – 3 4 1 – 3– 3 – 3– 3 9– 18 6– 14 coefficients of dividend = x 2 – 3x + 6 – 14 Insert zero coefficient as placeholder for the missing x 2 term.

64. Slide 2- 64 Remainder Theorem: The remainder of the division of a polynomial f (x) by x – a is f (a). Example: Using the remainder theorem, evaluate f(x) = x 4 – 4x – 1 when x = 3. 9 1 0 0 – 4 – 1 3 1 3 39 6927 2368 The remainder is 68 at x = 3, so f (3) = 68. You can check this using substitution:f(3) = (3) 4 – 4(3) – 1 = 68. value of x

65. Slide 2- 65 Example: Using synthetic division and the remainder theorem, evaluate f (x) = x 2 – x at x = – 2. 6 1 – 1 0 – 2– 2 1 – 2– 2 – 3– 36 Then f (– 2) = 6 and (– 2, 6) is a point on the graph of f(x) = x 2 – x. f(x) = x 2 – x x y 2 4 (– 2, 6) remainder

66. Slide 2- 66 Example Using Polynomial Long Division

67. Slide 2- 67 Example Using Polynomial Long Division

68. Slide 2- 68 Remainder Theorem

69. Slide 2- 69 Example Using the Remainder Theorem

70. Slide 2- 70 Factor Theorem

71. Slide 2- 71 Example Using Synthetic Division

72. Slide 2- 72 Rational Zeros Theorem

73. Slide 2- 73 Upper and Lower Bound Tests for Real Zeros

74. Slide 2- 74 Show 2x - 3 is a factor of 6x 2 + x – 15 (x = 3/2) 3/2 6 1 -15 6 9 10 15 0 Note: Since the remainder is 0, 2x - 3 is a factor of 6x 2 + x – 15

75. Slide 2- 75 Factor x 4 – 3x 3 – 5x 2 + 3x + 4 are possible factors P(1) = 0 1 1 -3 -5 3 4 1 1 -2 -7 -4 0 x 3 – 2x 2 – 7x - 4 so x 4 – 3x 3 – 5x 2 + 3x + 4 = (x 3 – 2x 2 – 7x – 4)(x – 1)

76. Slide 2- 76 Factor x 3 – 2x 2 – 7x - 4 are possible factors P(-1) = 0 -1 1 -2 -7 -4 1 -3 3 -4 4 0 x 2 – 3x – 4 so x 4 – 3x 3 – 5x 2 + 3x + 4 = (x 2 – 3x – 4)(x – 1)(x + 1)

77. Slide 2- 77 Factor x 3 – 2x 2 – 7x - 4 are possible factors P(-1) = 0 -1 1 -2 -7 -4 1 -3 3 -4 4 0 x 2 – 3x – 4 so x 4 – 3x 3 – 5x 2 + 3x + 4 = (x 2 – 3x – 4)(x – 1)(x + 1)

78. Slide 2- 78 x 4 – 3x 3 – 5x 2 + 3x + 4 = (x 2 – 3x – 4)(x – 1)(x + 1) or (x – 4)(x + 1)(x – 1)(x + 1)

79. Slide 2- 79 Factor x 4 – 8x 3 +17x 2 + 2x - 24 are possible factors P(4) = 0 4 1 -8 17 2 -24 1 4 -4 -16 1 4 6 24 0 x 3 – 4x 2 + x + 6 so x 4 – 8x 3 + 17x 2 + 2x - 24 = (x 3 – 4x 2 + x + 6)(x – 4)

80. Slide 2- 80 Factor x 3 – 4x 2 + x + 6 P(2) = 0 2 1 -4 1 6 1 2 -2 -4 -3 -6 0 x 2 – 2x – 3 so x 4 – 8x 3 +17x 2 + 2x - 24 = (x 2 – 2x – 3)(x – 4)(x - 2) are possible factors

81. Slide 2- 81 x 4 – 8x 3 +17x 2 + 2x - 24 = (x 2 – 2x – 3)(x – 4)(x - 2) or (x – 3)(x + 1)(x – 4)(x - 2)

82. Slide 2- 82 Show x 3 – 3x 2 + 5 = 0 has no rational roots  1,  5 only possible roots P(1) = 1 3 – 3(1) 2 + 5 = 3 P(-1) = (-1) 3 – 3(-1) 2 + 5 = 1 P(5) = (5) 3 – 3(5) 2 + 5 = 55 P(-5) = (-5) 3 – 3(-5) 2 + 5 = -195 Since none of the possible roots give zero in the remainder theorem, there are no rational roots.

83. Slide 2- 83 Example Finding the Real Zeros of a Polynomial Function

84. Slide 2- 84 Example Finding the Real Zeros of a Polynomial Function

85. Slide 2- 85 Example Finding the Real Zeros of a Polynomial Function

86. 2.5 Complex Zeros and the Fundamental Theorem of Algebra

87. Slide 2- 87 Quick Review

88. Slide 2- 88 What you’ll learn about Two Major Theorems Complex Conjugate Zeros Factoring with Real Number Coefficients … and why These topics provide the complete story about the zeros and factors of polynomials with real number coefficients.

89. Slide 2- 89 Fundamental Theorem of Algebra A polynomial function of degree n has n complex zeros (real and nonreal). Some of these zeros may be repeated.

90. Slide 2- 90 Linear Factorization Theorem

91. Slide 2- 91 Fundamental Polynomial Connections in the Complex Case The following statements about a polynomial function f are equivalent if k is a complex number: 1. x = k is a solution (or root) of the equation f(x) = 0 2. k is a zero of the function f. 3. x – k is a factor of f(x).

92. Slide 2- 92 Example Exploring Fundamental Polynomial Connections

93. Slide 2- 93 Complex Conjugate Zeros

94. Slide 2- 94 Example Finding a Polynomial from Given Zeros

95. Slide 2- 95 Factors of a Polynomial with Real Coefficients Every polynomial function with real coefficients can be written as a product of linear factors and irreducible quadratic factors, each with real coefficients.

96. Slide 2- 96 Find an equation of a polynomial with roots of 2i and 3. (x – 2i)(x + 2i)(x – 3) = 0 (x 2 – 2ix + 2ix – 4i 2 )(x – 3) = 0 (x 2 + 4)(x – 3) = 0 x 3 – 3x 2 + 4x – 12 = 0

97. Slide 2- 97 Find an equation of a polynomial with roots of 1- i and -2. (x – (1 - i))(x – (1 + i))(x + 2) = 0 (x 2 – x - ix – x + 1 + i +ix -i - i 2 )(x + 2) = 0 (x 2 – 2x + 2)(x + 2) = 0 x 3 + 2x 2 –2x 2 - 4x + 2x + 4= 0 (x – 1 + i)(x – 1 - i)(x + 2) = 0 x 3 - 2x + 4 = 0

98. Slide 2- 98 2.5 Complex Zeros and the Fundamental Theorem of Algebra Page 230

99. Slide 2- 99 2.5 Complex Zeros and the Fundamental Theorem of Algebra (cont’d) Page 230

100. Slide 2- 100 Example Factoring a Polynomial

101. Slide 2- 101 Example Factoring a Polynomial

102. 2.6 Graphs of Rational Functions

103. Slide 2- 103 Quick Review

104. Slide 2- 104 What you’ll learn about Rational Functions Transformations of the Reciprocal Function Limits and Asymptotes Analyzing Graphs of Rational Functions … and why Rational functions are used in calculus and in scientific applications such as inverse proportions.

105. Slide 2- 105 Rational Functions

106. Slide 2- 106 Example Finding the Domain of a Rational Function

107. Slide 2- 107 3 Rules for Asymptotes for Rational Functions

108. Slide 2- 108 Graph a Rational Function

109. Slide 2- 109 Graph a Rational Function

110. Slide 2- 110 Example Finding Asymptotes of Rational Functions

111. Slide 2- 111 Example Graphing a Rational Function -3 1 2 -----  ++++++ 0 ----  ++++++

112. Slide 2- 112 Sketch the graph of f(0) = -3 Solve x – 3 = 0; x = 3 Solve x + 1 = 0; x = -1 and sketch the asymptote. Horizontal asymptote, y = 1 -3 1 +++  --------- 0 ++++++++ Example Graphing a Rational Function

113. Slide 2- 113 Sketch the graph of f(0) = -2 Solve x + 2 = 0; x = -2 Solve x 2 - 1 = 0; x = -1 x = 1 and sketch the asymptotes. Horizontal asymptote, y = 0 -2 -1 1 ----  +++ 0 ----  ++++++ Example Graphing a Rational Function

114. Slide 2- 114 Graphs of Rational Functions Page 239

115. Slide 2- 115 Graphs of Rational Functions Page 239

116. Slide 2- 116 Graphs of Rational Functions Page 239

117. Slide 2- 117 Graphs of Rational Functions Page 239

118. 2.7 Solving Equations in One Variable

119. Slide 2- 119 Quick Review

120. Slide 2- 120 What you’ll learn about Solving Rational Equations Extraneous Solutions Applications … and why Applications involving rational functions as models often require that an equation involving fractions be solved.

121. Slide 2- 121 Extraneous Solutions When we multiply or divide an equation by an expression containing variables, the resulting equation may have solutions that are not solutions of the original equation. These are extraneous solutions. For this reason we must check each solution of the resulting equation in the original equation.

122. Slide 2- 122 Example Solving by Clearing Fractions The LCD is x Multiply both sides by x Subtract 3x from both sides (x – 2)(x – 1) = 0 x = 2 or x = 1 Check Both solutions are correct

123. Slide 2- 123 Example Eliminating Extraneous Solutions The LCD is (x – 1)(x - 3) (x – 1)(1) + 2x(x - 3) = 2 2x 2 – 5x – 3 = 0 (2x + 1)(x - 3) = 0 x = -½ or x = 3 Check: x = 3 is not defined x = -½ is the only solution x – 1 + 2x 2 - 6x = 2

124. Slide 2- 124 Example Eliminating Extraneous Solutions The LCD is (x)(x + 2) (x + 2)(x - 3) + 3x + 6 = 0 x 2 + 2x = 0 x(x + 2) = 0 x = 0 or x = -2 Check: x = 0 is not defined x = -2 is not defined No Solution x 2 – x – 6 + 3x + 6 = 0 Solve the equation

125. Slide 2- 125 Example Finding a Minimum Perimeter

126. Slide 2- 126 Example Acid Problem a.Pure acid is added to 78 oz of a 63% acid solution. Let x be the amount (in ounces) of pure acid added. Find an algebraic representation for C(x), the concentration of acid as a function of x. Determine how much pure acid should be added so that the mixture is at least 83% acid. 1.00x + (.63)(78) = C(x)(x + 78) 1.00x + 49.14 >.83x + 64.74.17x > 15.6x > 91.76

127. 2.8 Solving Inequalities in One Variable

128. Slide 2- 128 Quick Review

129. Slide 2- 129 What you’ll learn about Polynomial Inequalities Rational Inequalities Other Inequalities Applications … and why Designing containers as well as other types of applications often require that an inequality be solved.

130. Slide 2- 130 Polynomial Inequalities

131. Slide 2- 131 Example Finding where a Polynomial is Zero, Positive, or Negative -34 (-)(-) 2 (+)(-) 2 (+)(+) 2 negative positive

132. Slide 2- 132 Example Solving a Polynomial Inequality Graphically

133. Slide 2- 133 Example Solving a Polynomial Inequality Graphically

134. Slide 2- 134 Example Creating a Sign Chart for a Rational Function -31 (-) (-)(-) negative positive (-) (+)(-) (+) (+)(+) (+) (+)(-) negative 0und. 0

135. Slide 2- 135 Example Solving an Inequality Involving a Radical 2 (-)(+)(+)(+) undefined positivenegative 00

136. Slide 2- 136 Chapter Test

137. Slide 2- 137 Chapter Test

138. Slide 2- 138 Chapter Test

139. Slide 2- 139 Chapter Test Solutions

140. Slide 2- 140 Chapter Test Solutions

141. Slide 2- 141 Chapter Test Solutions