A f x x2 2 e f x 2x2 4x 1 b f x x

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Unformatted text preview: = h, we get y = k , so (h, k ) is on the graph. If x = h, then x − h = 0 and so (x − h)2 is a positive number. If a > 0, then a(x − h)2 is positive, and so y = a(x − h)2 + k is always a number larger than k . That means that when a > 0, (h, k ) is the lowest point on the graph and thus the parabola must open upwards, making (h, k ) the vertex. A similar argument shows that if a < 0, (h, k ) is the highest point on the graph, so the parabola opens downwards, and (h, k ) is also the vertex in this case. Alternatively, we can apply the machinery in Section 1.8. The vertex of the parabola y = x2 is easily seen to be the origin, (0, 0). We leave it to the reader to convince oneself that if we apply any of the transformations in Section 1.8 (shifts, reflections, and/or scalings) to y = x2 , the vertex of the resulting parabola will always be the point the graph corresponding to (0, 0). To obtain the formula f (x) = a(x − h)2 + k , we start with g (x) = x2 and first define g1 (x) = ag (x) = ax2 . This is results in a vertical scaling and/or reflection.2 Since w...
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This note was uploaded on 05/03/2013 for the course MATH Algebra taught by Professor Wong during the Fall '13 term at Chicago Academy High School.

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