6.10 Functions and Their Graphs 393 Note that the domain of the function in Example 7, the values that can be used for x, is the set of all real numbers, .R The range of the function, the resulting values obtained for y, is the set of all real numbers greater than or equal to −1. When graphing parabolas, if you think you need additional points to graph the equation, you can always substitute values for x and determine the corresponding values of y and plot those points. For example, if you substituted 6 for x, the corresponding value of y is 8. Thus, you could plot the point (6, 8) as we did in Fig. 6.46. 3. y-coordinate of vertex: Substitute = x 3, and solve for y. = − + = − + = − + = − y x x6 8 (3) 6(3)8 9188 1 2 2 Thus, the vertex is at − (3, 1). 4. y-intercept: Substitute = x 0 and solve for y. = − + = − + = y x x y 6 8 (0) 6(0) 8 8 2 2 Thus, the y-intercept is at (0, 8). 5. x-intercepts: Substitute = y 0 and solve for x. = − + − + = x x x x 0 6 8, or 6 8 0 2 2 We can solve this equation by factoring. − + = − − = − = − = = = x x x x x x x x 6 8 0 ( 4)( 2) 0 4 0 or 2 0 4 2 2 Thus, the x-intercepts are (4, 0) and (2, 0). a) Plot the vertex (3, 1), − the axis of symmetry = x 3, the y-intercept (0, 8), and the x-intercepts (4, 0) and (2, 0). Then graph the equation (Fig. 6.46). 7 Now try Exercise 57 y 4 5 3 2 8 7 6 1 2221 4 5 (3, 21) 6 7 1 2 x 21 y 5 x2 2 6x 1 8 3 9 Figure 6.46 Example 8 Domain and Range of a Quadratic Function a) Graph the quadratic function = − − + f x x x ( ) 2 8. 2 b) Determine the domain and range of the function. Solution a) We rewrite the function as = − − + y x x2 8, 2 and use the steps outlined in the general procedure to graph the function. 1. Since = − a 1, which is less than 0, the parabola opens downward. 2. Axis of symmetry: = − = − − − = − = − x b a2 ( 2) 2( 1) 2 2 1. Thus, the equation of the axis of symmetry is = − x 1.
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