Find the Vertex Form f(x)=-x^2-4x-6

$f (x) = - x^{2} - 4 x - 6$

Step 1

Write $f (x) = - x^{2} - 4 x - 6$ as an equation.

$y = - x^{2} - 4 x - 6$

Step 2

Complete the square for $- x^{2} - 4 x - 6$ .

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

$a = - 1$

$b = - 4$

$c = - 6$

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

$d = \frac{- 4}{2 \cdot - 1}$

Simplify the right side.

Cancel the common factor of $- 4$ and $2$ .

Factor $2$ out of $- 4$ .

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

Move the negative one from the denominator of $\frac{- 2}{- 1}$ .

$d = - 1 \cdot - 2$

Rewrite $- 1 \cdot - 2$ as $- - 2$ .

$d = - - 2$

Multiply $- 1$ by $- 2$ .

$d = 2$

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

$e = - 6 - \frac{{(- 4)}^{2}}{4 \cdot - 1}$

Simplify the right side.

Simplify each term.

Cancel the common factor of ${(- 4)}^{2}$ and $4$ .

Rewrite $- 4$ as $- 1 (4)$ .

$e = - 6 - \frac{{(- 1 (4))}^{2}}{4 \cdot - 1}$

Apply the product rule to $- 1 (4)$ .

$e = - 6 - \frac{{(- 1)}^{2} \cdot 4^{2}}{4 \cdot - 1}$

Raise $- 1$ to the power of $2$ .

$e = - 6 - \frac{1 \cdot 4^{2}}{4 \cdot - 1}$

Multiply $4^{2}$ by $1$ .

$e = - 6 - \frac{4^{2}}{4 \cdot - 1}$

Factor $4$ out of $4^{2}$ .

$e = - 6 - \frac{4 \cdot 4}{4 \cdot - 1}$

Move the negative one from the denominator of $\frac{4}{- 1}$ .

$e = - 6 - (- 1 \cdot 4)$

Multiply $- (- 1 \cdot 4)$ .

Multiply $- 1$ by $4$ .

$e = - 6 - - 4$

Multiply $- 1$ by $- 4$ .

$e = - 6 + 4$

Add $- 6$ and $4$ .

$e = - 2$

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

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

Step 3

Set $y$ equal to the new right side.

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

Find the Vertex Form f(x)=-x^2-4x-6

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