Find the Vertex Form y=-2x^2-12x+3

$y = - 2 x^{2} - 12 x + 3$

Step 1

Complete the square for $- 2 x^{2} - 12 x + 3$ .

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

$a = - 2$

$b = - 12$

$c = 3$

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 $d = \frac{b}{2 a}$ .

$d = \frac{- 12}{2 \cdot - 2}$

Simplify the right side.

Cancel the common factor of $- 12$ and $2$ .

Factor $2$ out of $- 12$ .

$d = \frac{2 \cdot - 6}{2 \cdot - 2}$

Cancel the common factors.

Factor $2$ out of $2 \cdot - 2$ .

$d = \frac{2 \cdot - 6}{2 (- 2)}$

Cancel the common factor.

$d = \frac{2 \cdot - 6}{2 \cdot - 2}$

Rewrite the expression.

$d = \frac{- 6}{- 2}$

Cancel the common factor of $- 6$ and $- 2$ .

Factor $- 2$ out of $- 6$ .

$d = \frac{- 2 (3)}{- 2}$

Cancel the common factors.

Factor $- 2$ out of $- 2$ .

$d = \frac{- 2 \cdot 3}{- 2 \cdot 1}$

Cancel the common factor.

$d = \frac{- 2 \cdot 3}{- 2 \cdot 1}$

Rewrite the expression.

$d = \frac{3}{1}$

Divide $3$ by $1$ .

$d = 3$

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 $e = c - \frac{b^{2}}{4 a}$ .

$e = 3 - \frac{{(- 12)}^{2}}{4 \cdot - 2}$

Simplify the right side.

Simplify each term.

Raise $- 12$ to the power of $2$ .

$e = 3 - \frac{144}{4 \cdot - 2}$

Multiply $4$ by $- 2$ .

$e = 3 - \frac{144}{- 8}$

Divide $144$ by $- 8$ .

$e = 3 - - 18$

Multiply $- 1$ by $- 18$ .

$e = 3 + 18$

Add $3$ and $18$ .

$e = 21$

Substitute the values of $a$ , $d$ , and $e$ into the vertex form $- 2 {(x + 3)}^{2} + 21$ .

$- 2 {(x + 3)}^{2} + 21$

Step 2

Set $y$ equal to the new right side.

$y = - 2 {(x + 3)}^{2} + 21$

Find the Vertex Form y=-2x^2-12x+3

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