Copyright 2006 Pearson Education Inc Publishing as Pearson
Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Slide 8 - 1
8 8. 1 8. 2 8. 3 8. 4 8. 5 8. 6 8. 7 8. 8 8. 9 Quadratic Functions and Equations Quadratic Equations The Quadratic Formula Applications Involving Quadratic Equations Studying Solutions of Quadratic Equations Reducible to Quadratic Functions and Their Graphs More About Graphing Quadratic Functions Problem Solving and Quadratic Functions Polynomial and Rational Inequalities Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley
8. 7 More About Graphing Quadratic Functions n Completing the Square n Finding Intercepts Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley
Completing the Square 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. Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Slide 8 - 4
Example Graph: Solution y f (x) = x 2 – 2 x – 1 6 (x 2 = – 2 x) – 1 = (x 2 – 2 x + 1 – 1) – 1 = (x 2 – 2 x + 1) – 1 = (x – 1)2 – 2 The vertex is at (1, – 2). 5 4 3 2 1 -5 -4 -3 -2 -1 1 -1 -2 2 3 4 5 x -3 -4 -5 Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Slide 8 - 5
Example Graph: Solution y f (x) = – 2 x 2 + 6 x – 3 6 = – 2(x 2 5 4 3 – 3 x) – 3 = – 2(x 2 – 3 x + 9/4 – 9/4) – 3 2 1 = – 2(x 2 – 3 x + 9/4) – 3 + 18/4 = – 2(x – 3/2)2 + 3/2 The vertex is at (3/2, 3/2). -5 -4 -3 -2 -1 1 -1 -2 2 3 4 5 x -3 -4 -5 Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Slide 8 - 6
The Vertex of a Parabola The vertex of a parabola given by f (x) = ax 2 + bx + c is The x-coordinate of the vertex is –b/(2 a). The axis of symmetry is x = -b/(2 a). The second coordinate of the vertex is most commonly found by computing Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Slide 8 - 7
Finding Intercepts For any function f, the y-intercept occurs at f (0). Thus for f (x) = ax 2 + bx + c, the yintercept is simply (0, c). To find xintercepts, we look for points where y = 0 or f (x) = 0. Thus, for f (x) = ax 2 + bx + c, the xintercepts occur at those x-values for which ax 2 + bx + c = 0 Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Slide 8 - 8
y - intercept y 6 5 4 3 f (x) = ax 2 + bx + c 2 x - intercepts 1 -5 -4 -3 -2 -1 1 -1 -2 2 3 4 5 x -3 -4 -5 Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Slide 8 - 9
Example Find the x- and y-intercepts of the graph of f (x) = x 2 + 3 x – 1. Solution The y-intercept is simply (0, f (0)), or (0, – 1). To find the x-intercepts, we solve the equation: x 2 + 3 x – 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. 3, 0) and (0. 3, 0). Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Slide 8 - 10
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