What is the vertex form of #y=2x^2+8x-5#?

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Answer 1 · 2016-08-22T07:10:22

#y=2(x+2)^2-13#

Given -

#y=2x^2+8x-5#

Find the vertex

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

At #x=-2#

#y=2(-2)^2+8(-2)-5=8-16-5=-13#

Quadratic equation in vertex form is -

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

Where -

#a=2#
#h=-2#
#k=-13#

Plug in the values

#y=2(x+2)^2-13#