What is the vertex form of #y=x^2/2+4x+8 #?

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Answer 1 · 2016-05-24T07:14:06

The vertex form is

#(x--4)^2=2(y-0)" "#with vertex at #(h, k)=(-4, 0)#

The given equation is

#y=1/2x^2+4x+8#

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

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

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

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

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

#2(y-0)=(x+4)^2#

#(x+4)^2=2(y-0)#

The vertex form is

#(x--4)^2=2(y-0)" "#with vertex at #(h, k)=(-4, 0)#

God bless...I hope the explanation is useful.