The curve of a quadratic function f(x)= - (x+h)^2+k+2 touches the x – axis at (– 2, 0).
(i) Find the value of h and of k.
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Answer:
You may recall studying quadratic equations in Intermediate Algebra. In this section, we review those equations in the context of our next family of functions: the quadratic functions. Definition 2.5. A quadratic function is a function of the form f(x) = ax2 +bx+c, where a, b and c are real numbers with a= 0. The domain of a quadratic function is (−∞,∞). The most basic quadratic function is f(x) = x2, whose graph appears below. Its shape should look familiar from Intermediate Algebra –it is called a parabola. The point (0,0) is called the vertex of the parabola. In this case, the vertex is a relative minimum and is also the where the absolute minimum value of f can be found. y (−2,4) (−1,1) 1 2 3 4 −2 −1 (0, 0) f(x) = x2 (2, 4) (1, 1) 1 2 x Much like many of the absolute value functions in Section 2.2, knowing the graph of f(x) = x2 enables us to graph an entire family of quadratic functions using transformations. Example 2.3.1. Graph the following functions starting with the graph of f(x) = x2 and using transformations. Find the vertex, state the range and find the x-and y-intercepts, if any exist. 1. g(x) = (x+2)2 −3 Solution. 2. h(x) = −2(x−3)2 +1 1. Since g(x) = (x + 2)2 −3 = f(x+2)−3, Theorem 1.7 instructs us to first subtract 2 from each of the x-values of the points on y = f(x). This shifts the graph of y = f(x) to the left 2 units and moves (−2,4) to (−4,4), (−1,1) to (−3,1), (0,0) to (−2,0), (1,1) to (−1,1) and (2, 4) to (0,4). Next, we subtract 3 from each of the y-values of these new points. This moves the graph down 3 units and moves (−4,4) to (−4,1), (−3,1) to (−3,−2), (−2,0) to (−2,3), (−1,1) to (−1,−2) and (0,4) to (0,1). We connect the dots in parabolic fashion to get.From the graph, we see that the vertex has moved from (0,0) on the graph of y = f(x) to (−2,−3) on the graph of y = g(x). This sets [−3,∞) as the range of g. We see that the graph of y = g(x) crosses the x-axis twice, so we expect two x-intercepts. To find these, we set y = g(x) = 0 and solve. Doing so yields the equation (x + 2)2 − 3 = 0, or (x +2)2 = 3. Extracting square roots gives x+2 = ±√ are (−2 −√ 3, 0) ≈ (−3.73,0) and (−2 + √ 3, or x = −2±√ 3. Our x-intercepts 3, 0) ≈ (−0.27,0). The y-intercept of the graph, (0, 1) was one of the points we originally plotted, so we are done. 2. Following Theorem 1.7 once more, to graph h(x) = −2(x−3)2+1 = −2f(x−3)+1, we first start by adding 3 to each of the x-values of the points on the graph of y = f(x). This effects a horizontal shift right 3 units and moves (−2,4) to (1,4), (−1,1) to (2,1), (0,0) to (3,0), (1, 1) to (4,1) and (2,4) to (5,4). Next, we multiply each of our y-values first by −2 and then add 1 to that result. Geometrically, this is a vertical stretch by a factor of 2, followed by a reflection about the x-axis, followed by a vertical shift up 1 unit. This moves (1,4) to (1,−7), (2, 1) to (2,−1), (3,0) to (3,1), (4,1) to (4,−1) and (5,4) to (5,−7).The vertex is (3,1) which makes the range of h (−∞,1].