How do you write $y+1=-2x^2-x$ in the vertex form?

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Answer 1 · 2014-11-03T05:33:46

$y+1=-2x^2-x$

by factoring out $-2$ ,

$=> y+1=-2(x^2+1/2x)$

by adding and subtracting $1/16$ in the parentheses on the right,
(Note: $(1/2 divide 2)^2=1/16$ )

$=> y+1=-2(x^2+1/2+1/16-1/16)$

by distributing $-2$ to $-1/16$ ,

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

since $x^2+1/2x+1/16=(x+1/4)^2$ ,

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

by subtracting $1$ ,

$y=-2(x+1/4)^2-7/8$ ,

which is in vertex form.

I hope that this was helpful.

How do you write y+1=-2x^2-xy+1=-2x^2-x in the vertex form?