How do you write in vertex form for #y=x^2 + 8x - 7# by completing the square?

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Answer 1 · 2015-06-13T22:20:34

#y = x^2+8x-7 = (x+4)^2-23#

#x^2+8x-7# is of the form #ax^2+bx+c#

with #a=1#, #b=8# and #c=-7#.

Notice that in general:

#a(x+b/(2a))^2 = ax^2+bx+b^2/(4a)#

In our case #b/(2a) = 8/2 = 4#, so we want #(x+4)^2#

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

So

#y = x^2+8x-7#

#= x^2+8x+16 - 16 - 7#

#= (x+4)^2 - 16 - 7#

#=(x+4)^2 - 23#