How do I find the vertex of $y = {(x + 7)}^{2}$ $y=(x+7)^2$ ?

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How do I find the vertex of $y = {(x + 7)}^{2}$ $y=(x+7)^2$ ?

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Answer 1 · 2014-10-26T01:19:55

The equation is in vertex form ( $y = a {(x - h)}^{2} + k$ $y = a(x-h)^2 + k$ ) where the vertex is (h, k). Since the original form has you subtracting h, h more than likely has an opposite sign then what is in the equation. In the example, 7 is being added so therefore h = - 7. Since nothing is being added to the equation, k = 0. Therefore the vertex of $y = {(x + 7)}^{2}$ $y = (x +7)^2$ is $(- 7, 0)$ $(-7, 0)$ .

Here are some other examples to help you:

$y = - 2 {(x - 4)}^{2} - 3$ $y = -2 (x - 4)^2 - 3$ has h = 4 and k= -3, so the vertex is $(4, - 3)$ $(4,-3)$ .

$y = 4 {(x + 1)}^{2} + 5$ $y = 4(x + 1)^2 + 5$ has h = -1 and k = 5, so the vertex is $(- 1, 5)$ $(-1,5)$ .

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How do I find the vertex of $y = {(x + 7)}^{2}$ $y=(x+7)^2$ ?

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