What is the vertex of #y= -1/16(2x-4)^2+8#?

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Answer 1 · 2015-12-30T08:00:24

#(2,8)#

This is almost in vertex form, except for that there is a #2# being multiplied by the #x#.

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

#y=-1/16(2x-4)(2x-4)+8#

#y=-1/4(x-2)^2+8#

(Since the #2x-4# term is squared, a #2# is factored from each term.)

This is now in vertex form.

The center is at #(h,k)rarr(2,8)#.

graph{-1/16(2x-4)^2+8 [-13.78, 14.7, -2.26, 11.98]}