Question #ea797

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Question #ea797

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Answer 1 · 2017-01-27T22:24:53

The vertex form of the equation of a vertically oriented parabola is:
$y = a {(x - h)}^{2} + k$ $y = a(x - h)^2 + k$ where a is the coefficient of the $x^{2}$ $x^2$ term and $(h, k)$ $(h,k)$ is the vertex

Explanation:

Given: $y = - 2 x^{2} - 16 x - 32$ $y = -2x^2 - 16x -32$

Please observe that $a = - 2$ $a = -2$

Add zero to the right side of the equation in form of $a h^{2} - a h^{2}$ $ah^2 - ah^2$ :

$y = - 2 x^{2} - 16 x + - 2 h^{2} - - 2 h^{2} - 32$ $y = -2x^2 - 16x + -2h^2 - -2h^2 -32$

Factor -2 out of the first three terms on the right:

$y = - 2 (x^{2} + 8 x + h^{2}) + 2 h^{2} - 32$ $y = -2(x^2 + 8x + h^2) + 2h^2 -32$

Using the pattern ${(x - h)}^{2} = x^{2} - 2 h x + h^{2}$ $(x - h)^2 = x^2 - 2hx + h^2$ , we set the middle term in the right side of the pattern equal to middle term in the equation:

$- 2 h x = 8 x$ $-2hx = 8x$

$h = - 4$ $h = -4$

Substitute the left side of the pattern into the equation:

$y = - 2 {(x - h)}^{2} + 2 h^{2} - 32$ $y = -2(x - h)^2 + 2h^2 -32$

Substitute - 4 for h:

$y = - 2 {(x - - 4)}^{2} + 2 {(- 4)}^{2} - 32$ $y = -2(x - -4)^2 + 2(-4)^2 -32$

Combine the constant terms:

$y = - 2 {(x - - 4)}^{2}$ $y = -2(x - -4)^2$

This is the vertex form.

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Question #ea797

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