What is the vertex of $y=-1/2(x-4)^2-7$ ?

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Answer 1 · 2014-11-05T16:54:35

Vertex Form of Quadratic Function

$y=a(x-h)^2+k$ ,

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

Let us look at the posted problem.

Since

$a=-1/2$ , $h=4$ , and $k=-7$ ,

the vertex is $(4,-7)$ .

I hope that this was helpful.

What is the vertex of y=-1/2(x-4)^2-7y=-1/2(x-4)^2-7?