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.

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