How do you find the vertex of $y = 5 {(x + 2)}^{2} + 7$ $y = 5(x+2)^2 + 7$ ?

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Answer 1 · 2017-05-01T16:49:54

The vertex form of a parabola of this type is:

$y = a {(x - h)}^{2} + k [1]$ $y = a(x-h)^2 + k" [1]"$

where $(h, k)$ $(h,k)$ is the vertex.

Given:

$y = 5 {(x + 2)}^{2} + 7 [2]$ $y = 5(x+2)^2 + 7" [2]"$

We can convert equation [2] into the same form as equation [1] by writing $+ 2 as - - 2$ $+2" as" --2$ :

$y = 5 {(x - - 2)}^{2} + 7 [3]$ $y = 5(x--2)^2 + 7" [3]"$

We observe that the vertex is $(- 2, 7)$ $(-2,7)$

How do you find the vertex of y=5(x+2)2+7y = 5(x+2)^2 + 7?