What is the vertex form of #y=x^2+8x-7#?

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Answer 1 · 2016-08-19T07:17:23

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

Given -

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

The vertex form of the equation is -

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

Where

#a# is the coefficient of #x^2#
#h# is the #x# coordinate of thevertex
#k# is the #y# coordinate of the vertex

Vertex-

#x=(-b)/(2a)=(-8)/2=-4#

At #x=-4#

#y=(-4)^2+8(-4)-7#
#y=16-32-7=-23#

Then-

#a=1#
#h=-4#
#k=-23#

Plug in the values in the formula

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

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