Chapter 5: Exercises with Solutions

Section 5.4The Quadratic Formulaf(x) = −2x 2 − 16x − 32f(0) = −2(0) 2 − 16(0) − 32f(0) = −32So the y-intercept is (0, −32).Now calculate an additional point and mirror it over, and then complete the graph.x=−450yx y = −2(x + 4) 2−2−8(−6,−8)(−4,0)(−2,−8)x10(−8,−32)(0,−32)f(x)=−2(x+4) 2To 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].50y50yx10(−4,0)x10f(x)=−2(x+4) 2f(x)=−2(x+4) 237. The graph of a quadratic function has exactly two x-intercepts when the discriminantis positive. For this function f(x) = kx 2 − 3x + 5, a = k, b = −3 and c = 5, sothe discriminant is D = b 2 − 4ac = (−3) 2 − 4(k)(5) = 9 − 20k. In order for there to betwo x-intercepts, this must be positive. So, set it greater than zero and solve for thevalue of k.Version: Fall 2007