Chapter 5: Exercises with Solutions

Chapter 5Quadratic FunctionsTo find the domain of f, mentally project every point of the graph onto the x-axis, asshown on the left below. This covers the entire x-axis, so the domain= (−∞, ∞). Tofind the range, mentally project every point of the graph onto the y-axis, as shown onthe right below. The shaded interval on the y-axis is range= [0, ∞).20yf(x)=(x−2) 220yf(x)=(x−2) 2x10(2,0)x1035. The graph of a quadratic function has exactly one x-intercept when the discriminantis zero. For this function f(x) = kx 2 − 16x − 32, a = k, b = −16 and c = −32,so the discriminant is D = b 2 − 4ac = (−16) 2 − 4(k)(−32) = 256 + 128k. In order forthere to be only one x-intercept, this must be zero. So, set it equal to zero and solvefor the value of k.Subtracting 256 from each side,Now divide both sides by 128:0 = D0 = 256 + 128k−256 = 128k−2 = kSo, if k = −2, then f(x) will have one x-intercept. So, replacing k with −2, we getthat f(x) = −2x 2 − 16x − 32.Complete the square to find the vertex.f(x) = −2x 2 − 16x − 32= −2(x 2 + 8x + 16)= −2(x + 4) 2Read off the vertex as (h, k) = (−4, 0). The axis of symmetry is a vertical line throughthe vertex with equation x = −4.To find the y-intercept, set x = 0 in the equation and solve for y:Version: Fall 2007