Chapter 5: Exercises with Solutions

Chapter 5Quadratic FunctionsRead off the vertex as (h, k) = (3, −2). The axis of symmetry is a vertical line throughthe vertex with equation x = 3.To find the y-intercept, set x = 0 in the equation and solve for y:So the y-intercept is (0, −11).f(x) = −x 2 + 6x − 11f(0) = −(0) 2 + 6(0) − 11f(0) = −11Now calculate an additional point and mirror it over, and then complete the graph.20yx=3x y = −(x − 3) 2 − 24 −3(3,−2)(2,−3)(4,−3)x10(0,−11)(6,−11)f(x)=−(x−3) 2 −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= (−∞, −2].20y20yx10(3,−2)x10f(x)=−(x−3) 2 −2f(x)=−(x−3) 2 −233. The graph of a quadratic function has exactly one x-intercept when the discriminantis zero. For this function f(x) = x 2 − 4x + 4k, a = 1, b = −4 and c = 4k, so thediscriminant is D = b 2 − 4ac = (−4) 2 − 4(1)(4k) = 16 − 16k. In order for there to beVersion: Fall 2007