SECTION 3.3 Quadratic Functions and Their Properties 163 Figure 20 ( ) = − + f x x x6 9 2 x y 3 6 (0, 9) Axis of symmetry x 5 3 6 (6, 9) (3, 0) (a) For f x x x a b 6 9, 1, 6, 2 ( ) = − + = = − and c 9. = Because a 1 0, = > the parabola is concave up. The x-coordinate of the vertex is h b a 2 6 2 1 3 = − = − − ⋅ = The y-coordinate of the vertex is k f 3 3 6 3 9 0 2 ( ) = = − ⋅ + = The vertex is 3, 0 ( ). The axis of symmetry is the line x 3. = The y-intercept is f 0 9. ( ) = Since the vertex 3, 0 ( ) lies on the x-axis, the graph touches the x-axis at the x-intercept. By using the axis of symmetry and the y-intercept at 0, 9 ( ), we can locate the additional point 6, 9 ( ) on the graph. See Figure 20. (b) The domain of f is the set of all real numbers. Based on the graph, the range of f is the interval 0, . [ )∞ (c) The function f is decreasing on the interval , 3 ( ] −∞ and increasing on the interval 3, . [ )∞ (d) The graph of f is above the x-axis everywhere except at the vertex 3, 0 ( ). So, f x 0 ( ) > on , 3 3, ( ) ( ) −∞ ∪ ∞ or for x x 3, 3, < > and f x( ) is never negative. Solution Now Work PROBLEM 51 Graphing a Quadratic Function Using Its Vertex, Axis, and Intercepts (a) Graph f x x x 2 1 2 ( ) = + + by determining whether the graph is concave up or concave down and by finding its vertex, axis of symmetry, y-intercept, and x-intercepts, if any. (b) Find the domain and the range of f . (c) Determine where f is increasing and where it is decreasing. (d) Determine where f x 0 ( ) > and where f x 0. ( ) < EXAMPLE 5 Figure 21 ( ) = + + f x x x 2 1 2 –1 1 1 2 x y Axis of symmetry x 5 – 1 – 4 1 – 2 (0, 1) , 1 ( ) – 1 – 4 7 – 8 , ( ) – NOTE In Example 5, since the vertex lies in Quadrant II and the parabola is concave up, we can conclude that the graph of the quadratic function has no x-intercepts. j Solution (a) For ( ) = + + f x x x 2 1, 2 we have = = a b 2, 1, and c 1. = Because = > a 2 0, the parabola is concave up. The x-coordinate of the vertex is = − = − h b a2 1 4 The y-coordinate of the vertex is ( ) ( ) = − = ⋅ + − + = k f 1 4 2 1 16 1 4 1 7 8 The vertex is ( ) − 1 4 , 7 8 . The axis of symmetry is the line = − x 1 4 . The y-intercept is f 0 1. ( ) = The x-intercept(s), if any, satisfy the equation x x 2 1 0. 2 + + = The discriminant b ac 4 1 4 2 1 7 0. 2 2 − = − ⋅ ⋅ = − < This equation has no real solutions, which means the graph has no x-intercepts. Use the point 0, 1 ( ) and the axis of symmetry = − x 1 4 to locate the additional point ( ) − 1 2 , 1 on the graph. See Figure 21. (b) The domain of f is the set of all real numbers. Based on the graph, the range of f is the interval 7 8 , .) ⎡ ∞ ⎣ ⎢ (c) The function f is decreasing on the interval (−∞ − ⎤ ⎦ ⎥ , 1 4 and is increasing on the interval ) ⎡ − ∞ ⎢ ⎣ 1 4 , . (d) The graph of f is always above the x-axis. So, f x 0 ( ) > on the interval , ( ) −∞ ∞ or for all real numbers x. Now Work PROBLEM 55
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