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.