EXAMPLE 7 Graph f (x) x2 6x 3 using the vertex formula. Include the intercepts. Solution a 1, b 6, c 3. Since a 1, the graph opens upward. The x-coordinate, h, of the vertex is f x x2 x x y h 3. Then, the y-coordinate, k, of the vertex is k f (3). f (x) x2 6x 3 f (3) (3)2 6(3) 3 9 18 3 6 The vertex is (3, 6). The axis of symmetry is x 3. Find more points on the graph of f (x) x2 6x 3, then use the axis of symmetry to fi nd other points on the parabola. 3 5 1 2 5 2 2 5 5 h b 2a (6) 2(1) x f(x) 4 5 5 2 6 3 To fi nd the x-intercepts, let f (x) 0 and solve for x. 0 x2 6x 3 (6) 2(6)2 4(1)(3) x 2(1) 6 2 Solve using the quadratic formula. x 6 124 2 6 216 2 Simplify. x 3 16 The x-intercepts are (3 16, 0) and (3 16, 0). We can see from the graph that the y-intercept is (0, 3). The domain is (q, q); the range is 6, q). YOU TRY 6 Graph f (x) x2 8x 13 using the vertex formula. Include the intercepts. Using Technology In Section 7.4 we said that the solutions of the equation x2 x 6 0 are the x-intercepts of the graph of y x2 x 6. The x-intercepts are also called the zeros of the equation since they are the values of x that make y 0. Enter x2 x 6 in Y1 then fi nd the x-intercepts shown on the graph by pressing 2nd TRACE and then selecting 2:zero. Move the cursor to the left of an x-intercept using the right arrow key and press ENTER. Move the cursor to the right of the x-intercept using the right arrow key 662 CHAPTER 10 Quadratic Equations and Functions www.mhhe.com/messersmith
messersmith_power_intermediate_algebra_1e_ch4_7_10
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