2. By completing the square, we can rewrite any polynomial ax 2 + bx + c in the form a(x – h) 2 + k. Once that has been done, the procedures discussed in Section 8.6 will enable us to graph any quadratic function.

3. Solution Example Graph: f (x) = x 2 – 2x – 1 f (x) + 1 = x 2 – 2x f (x) + 1 = (x 2 – 2x + 1) – 1 f (x) + 1 = (x 2 – 2x + ) f (x) + 1 = (x – 1) 2 – 1 The vertex is at (1, –2). f (x) = (x – 1) 2 – 2

4. Solution ExampleGraph:

7. Slide 8- 7 Copyright © 2008 Pearson Education, Inc. Publishing as Pearson Addison-Wesley

8. Finding Intercepts For any function f, the y-intercept occurs at f (0). Thus for f (x) = ax 2 + bx + c, the y-intercept is simply (0, c). To find x-intercepts, we look for points where y = 0 or f (x) = 0. Thus, for f (x) = ax 2 + bx + c, the x-intercepts occur at those x-values for which ax 2 + bx + c = 0. ax 2 + bx + c = 0

9. Example Solution Find the x- and y-intercepts of the graph of f (x) = x 2 + 3x – 1. The y-intercept is simply (0, f (0)), or (0, –1). To find the x-intercepts, we solve the equation: x 2 + 3x – 1 = 0. Since we are unable to solve by factoring, we use the quadratic formula to get If graphing, we would approximate to get (–3.303, 0) and (0.303, 0).