What is the vertex form of #y=x^2 - 8x + 16 #?

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Answer 1 · 2016-02-18T02:29:26

It is #y=(x-4)^2#

The vertex form of a parabola's equation is generally expressed as :

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

Hence the given parabola can be written as follows

#y=(x-4)^2#

so it is #a=1# , #h=4#, #k=0#

So the vertex is #(h=4,k=0)#

graph{(x-4)^2 [-1.72, 12.33, -0.69, 6.333]}