Chapter 5: Exercises with Solutions

Chapter 5Quadratic Functionsy20 f(x)=(x+2) 2 −16y20 f(x)=(x+2) 2 −16x10x10(−2,−16)27. First, complete the square:f(x) = −x 2 − 2x + 8= −(x 2 + 2x − 8)= −(x 2 + 2x + 1 − 1 − 8)= − (( x 2 + 2x + 1 ) − 1 − 8 )()= − (x + 1) 2 − 9= − (x + 1) 2 + 9Read off the vertex as (h, k) = (−1, 9). The axis of symmetry is a vertical line throughthe vertex with equation x = −1.To find the x-intercepts algebraically, set y = 0 and factor.By the zero product property, either0 = −x 2 − 2x + 80 = −(x 2 + 2x − 8)Solve these linear equations independently.So the x-intercepts are (−4, 0) and (2, 0).0 = −(x + 4)(x − 2)x + 4 = 0 or x − 2 = 0.x = −4 or x = 2Lastly, to find the y-intercept, set x = 0 in the equation and solve for y:y = −x 2 − 2x + 8y = −0 2 − 2(0) + 8y = 8Version: Fall 2007