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

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

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Answer 1 · 2014-11-03T05:33:46

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

by factoring out $- 2$ $-2$ ,

$\Rightarrow y + 1 = - 2 (x^{2} + \frac{1}{2} x)$ $=> y+1=-2(x^2+1/2x)$

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

$\Rightarrow y + 1 = - 2 (x^{2} + \frac{1}{2} + \frac{1}{16} - \frac{1}{16})$ $=> y+1=-2(x^2+1/2+1/16-1/16)$

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

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

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

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

by subtracting $1$ $1$ ,

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

which is in vertex form.

I hope that this was helpful.

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

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