How do you find the vertex of # f(x)= -2x^2 + 16x +4#?

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Answer 1 · 2016-11-15T10:47:39

The vertex form is #=-2(x-4)^2+36#

Let's complete the squares

#f(x)=-2x^2+16x+4#

#=-2(x^2-8x)+4#

#=-2(x^2-8x+16)+4+32#

#=-2(x-4)^2+36#

The vertex form is #=-2(x-4)^2+36#

The vertex is #(4,36)#

The axis of symmetry is #x=4#

graph{-2(x-4)^2+36 [-69.2, 62.56, -1, 64.87]}