How do you find the vertex of $y = - 2 {(x + 1)}^{2} + 7$ $y = -2(x+1)^2 +7$ ?

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Answer 1 · 2016-01-03T17:19:46

Once you're in vertex form (y = a ${(x - p)}^{2}$ $(x - p)^2$ + q), it is very easy to find the placement of the vertex

In vertex form (y = a ${(x - p)}^{2}$ $(x - p)^2$ + q), the vertex is found at (p,q).

So, the vertex of your quadratic function is (-1, 7)

How do you find the vertex of y=−2(x+1)2+7y = -2(x+1)^2 +7?