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

##### 1 Answer
Nov 3, 2014

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

by factoring out $- 2$,

$\implies y + 1 = - 2 \left({x}^{2} + \frac{1}{2} x\right)$

by adding and subtracting $\frac{1}{16}$ in the parentheses on the right,
(Note: ${\left(\frac{1}{2} \div i \mathrm{de} 2\right)}^{2} = \frac{1}{16}$)

$\implies y + 1 = - 2 \left({x}^{2} + \frac{1}{2} + \frac{1}{16} - \frac{1}{16}\right)$

by distributing $- 2$ to $- \frac{1}{16}$,

$y + 1 = - 2 \left({x}^{2} + \frac{1}{2} x + \frac{1}{16}\right) + \frac{1}{8}$

since ${x}^{2} + \frac{1}{2} x + \frac{1}{16} = {\left(x + \frac{1}{4}\right)}^{2}$,

$y + 1 = - 2 {\left(x + \frac{1}{4}\right)}^{2} + \frac{1}{8}$

by subtracting $1$,

$y = - 2 {\left(x + \frac{1}{4}\right)}^{2} - \frac{7}{8}$,

which is in vertex form.

I hope that this was helpful.