Chapter 5: Exercises with Solutions
Chapter 5: Exercises with Solutions
Chapter 5: Exercises with Solutions
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Section 5.3Zeros of the QuadraticRead off the vertex as (h, k) = (−2, −16).through the vertex <strong>with</strong> equation x = −2.To find the x-intercepts algebraically, set y = 0 and factor.By the zero product property, either0 = x 2 + 4x − 12Solve these linear equations independently.So the x-intercepts are (−6, 0) and (2, 0).0 = (x + 6)(x − 2)x + 6 = 0 or x − 2 = 0.x = −6 or x = 2The axis of symmetry is a vertical lineLastly, to find the y-intercept, set x = 0 in the equation and solve for y:So the y-intercept is (0, −12).y = x 2 + 4x − 12y = 0 2 + 4(0) − 12y = −12Finally, put this all together to make the graph.y20 f(x)=(x+2) 2 −16x=−2(−6,0) (2,0)x10(−4,−12)(0,−12)(−2,−16)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= [−16, ∞).Version: Fall 2007