Chapter 5: Exercises with Solutions

Chapter 5Quadratic FunctionsSo the x-intercepts are (−4, 0) and (2, 0).Lastly, 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 = −8So the y-intercept is (0, −8). Finally, put this all together to make the graph.y10 f(x)=(x+1) 2 −9x=−1(−4,0) (2,0)x10(−2,−8)(−1,−9)(0,−8)To 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= [−9, ∞).y10 f(x)=(x+1) 2 −9y10 f(x)=(x+1) 2 −9x10x10(−1,−9)25. First, complete the square:f(x) = x 2 + 4x − 12= x 2 + 4x + 4 − 4 − 12= ( x 2 + 4x + 4 ) − 4 − 12= (x + 2) 2 − 16Version: Fall 2007