Find the Vertex Form y=-2x^2+x-7

$y = - 2 x^{2} + x - 7$

Step 1

Complete the square for $- 2 x^{2} + x - 7$ .

Use the form $a x^{2} + b x + c$ , to find the values of $a$ , $b$ , and $c$ .

$a = - 2$

$b = 1$

$c = - 7$

Consider the vertex form of a parabola.

$a {(x + d)}^{2} + e$

Find the value of $d$ using the formula $d = \frac{b}{2 a}$ .

Substitute the values of $a$ and $b$ into the formula $d = \frac{b}{2 a}$ .

$d = \frac{1}{2 \cdot - 2}$

Simplify the right side.

Multiply $2$ by $- 2$ .

$d = \frac{1}{- 4}$

Move the negative in front of the fraction.

$d = - \frac{1}{4}$

Find the value of $e$ using the formula $e = c - \frac{b^{2}}{4 a}$ .

Substitute the values of $c$ , $b$ and $a$ into the formula $e = c - \frac{b^{2}}{4 a}$ .

$e = - 7 - \frac{1^{2}}{4 \cdot - 2}$

Simplify the right side.

Simplify each term.

One to any power is one.

$e = - 7 - \frac{1}{4 \cdot - 2}$

Multiply $4$ by $- 2$ .

$e = - 7 - \frac{1}{- 8}$

Move the negative in front of the fraction.

$e = - 7 - - \frac{1}{8}$

Multiply $- - \frac{1}{8}$ .

Multiply $- 1$ by $- 1$ .

$e = - 7 + 1 (\frac{1}{8})$

Multiply $\frac{1}{8}$ by $1$ .

$e = - 7 + \frac{1}{8}$

To write $- 7$ as a fraction with a common denominator, multiply by $\frac{8}{8}$ .

$e = - 7 \cdot \frac{8}{8} + \frac{1}{8}$

Combine $- 7$ and $\frac{8}{8}$ .

$e = \frac{- 7 \cdot 8}{8} + \frac{1}{8}$

Combine the numerators over the common denominator.

$e = \frac{- 7 \cdot 8 + 1}{8}$

Simplify the numerator.

Multiply $- 7$ by $8$ .

$e = \frac{- 56 + 1}{8}$

Add $- 56$ and $1$ .

$e = \frac{- 55}{8}$

Move the negative in front of the fraction.

$e = - \frac{55}{8}$

Substitute the values of $a$ , $d$ , and $e$ into the vertex form $- 2 {(x - \frac{1}{4})}^{2} - \frac{55}{8}$ .

$- 2 {(x - \frac{1}{4})}^{2} - \frac{55}{8}$

Step 2

Set $y$ equal to the new right side.

$y = - 2 {(x - \frac{1}{4})}^{2} - \frac{55}{8}$

Find the Vertex Form y=-2x^2+x-7

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