Week 7 Recap Session

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Q 1. Find the coordinates of the vertex for the parabola defined by the given quadratic function. f(x) = (x – 4)2 – 42. Graph of the quadratic function. f(x) = -2x2 – 5x + 43. For the quadratic function, f(x) = -2x2 + 8x + 7 a. Determine, without graphing, whether the function has a minimum value or a maximum value; b. Find the minimum or maximum value and determine where it occurs; and Location: x ? -b/2a = -8/2(-2) = -8/-4 : x = 2 c. Identify the function’s domain and its range.4. Use the Leading Coefficient Test to determine the end behavior of the graph of the given polynomial function. f(x) = 5x4 + 7x2 - x + 9.

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1.f(x) = (x – 4)2 – 4 f(x) = x2 – 8x + 16 – 4 f(x) = x2 – 8x + 12 -> a = 1, b = -8, x = 12 Vertex (-b/2a, f(-b/2a)) x = -b/2a = -(-8)/2(1) = 8/2 = 4 f(4) = (4)2 – 8(4) + 12 = 16 – 32 + 12 = -4 Vertex: (4, -4) 2.a = -2: it opens downward b = -5 c = 4 Vertex (-5/4, 57/8) X = -b/2a = -(-5)/2(-2) = - 5/4 f(-5/4) = -2(-5/4)2 – 5(-5/4) + 4 = -2(25/16) + 25/4 + 4 = -25/8 + 25/4 + 4 LCD = 8 = (-25+2(25)+32)/8 = 57/8

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