Algebra 2 Module 5 Review

5.1 - Cubic Functions For 1-2, describe, in words, the transformations applied to the graph of f(x) = x³ to produce the graph of g(x). Then graph g(x) with at least 3 specific points. 1) g(x) = (x - 3)³ + 2 Transformation(s): Reflection over x-axis Translation right 3 and up 2 [Image of a graph of g(x) = (x - 3)³ + 2] 2) g(x) = -3(x + 2)³ - 2 Transformation(s): Reflection over x-axis, Vertical Stretch by a factor of 3, Translation left 2, down 2 [Image of a graph of g(x) = -3(x + 2)³ - 2] For 3-4, given the general equation f(x) = a(x - h)³ + k, write the specific equation for the graph. 3) [Image of a graph] f(x) = a(x - 1)³ + 2 1 = a(1)³ + 2 a = -1 f(x) = -(x - 1)³ + 2 4) [Image of a graph] f(x) = a(x - 3)³ + 1 2 = a(2.5 - 3)³ + 1 1 = a(-0.5)³ a = -8 f(x) = -8(x - 3)³ + 1 For 5-6, write the specific equation of the form g(x) = a(x-h)³ + k after the given transformations of the graph of f(x) = x³. 5) A reflection across the x-axis, followed by a translation 11 units up and 7 units to the left. g(x) = -(x+7)³ + 11 6) A vertical stretch by a factor of 6, followed by a translation 9 units to the right and 3 units down. g(x) = 6(x-9)³ - 3 **5.2 - Polynomial Functions** For 1-3, identify whether each function graphed has an odd or even degree and a positive or negative leading coefficient.   1) Degree: Even Leading Coeff:   Neg 2) Degree: Even Leading Coeff: Pos 3) Degree: Odd Leading Coeff: Pos For 4-5, use a graphing calculator to determine the number of turning points and the number and type (global or local) of any maximum   and/or minimum values. 4) f(x) = x(x-4)² Turning Points: 2 Global Maximum(s): 0 Local Maximum(s): 1 Global Minimum(s): 0 Local Minimum(s): 1 5) f(x) = -x²(x-2)(x+1) Turning Points: 3 Global Maximum(s): 1 (Note: This is not a global maximum) Local Maximum(s): 1 Global Minimum(s): 0 Local Minimum(s): 2 For 6-7, graph each function without a calculator. State the degree, end behavior, x- and y-intercepts, and the intervals where the function is positive or negative.   6) f(x) = 3x(x-1)(x+3) Degree: 3rd End Behavior: As x -> ∞, f(x) -> ∞ As x -> -∞, f(x) -> -∞ x-intercept(s): x = 1, x = -3 y-intercept: y = -3 or (0, -3) Positive: (-∞, -3) U (0, 1) U (3, ∞) Negative: (-3, 0) U (1, 3) [Image of a graph of f(x) = 3x(x-1)(x+3)] 7) f(x) = (x+3)²(x-3)(x-1) Degree: 4th End Behavior: As x -> ∞, f(x) -> ∞ As x -> -∞, f(x) -> ∞ x-intercept(s): x = -3, 3, 1 y-intercept: y = -27 or (0, -27) Positive: (-∞, -3) U (1, ∞) Negative: (-3, 1) U (3, ∞) [Image of a graph of f(x) = (x+3)²(x-3)(x-1)] For 8-11, write the equation of each graph in intercept form, with integer x-intercepts. Assume the leading coefficient, a, is either +1 or -1.   8) f(x) = (x+3)(x+1)(x-1)(x-3) 9) f(x) = -(x+4)(x+1)(x-1)(x-1) [Image of two graphs] 10) f(x) = (x+5)(x+3)(x-1)(x-4)² [Image of a graph of f(x) = (x+5)(x+3)(x-1)(x-4)²] For 12-13, write a function, V(x), for the volume of the box that could be made by cutting an x in. by x in. square from each corner of the given rectangular piece of cardboard. Then, graph your function to answer the questions that follow. Approximate decimal answers to the nearest 10th. 12) 8 in. x 15 in. V(x) = (8-2x)(15-2x)(x) A. What is the maximum volume? V ≈ 90.7 in³ B. What value of x maximizes the volume? x ≈ 1.7 inches C. What are the dimensions of the box with the maximum volume? L: 11.6 in W: 4.6 in H: 1.7 in D. What are the domain and range of the function? D: [0, 4] R: [0, 90.7] 13) 19 in. x 15 in. V(x) = (19-2x)(15-2x)(x) A. What is the maximum volume? V ≈ 352.7 in³ B. What value of x maximizes the volume? x ≈ 2.8 inches C. What are the dimensions of the box with the maximum volume? L: 13.4 in W: 9.4 in H: 2.8 in D. What are the domain and range of the function? D: [0, 7.5] R: [0, 352.7]