Find the Vertex Form f(x)=-x^2-6x+7

$f (x) = - x^{2} - 6 x + 7$

Step 1

Write $f (x) = - x^{2} - 6 x + 7$ as an equation.

$y = - x^{2} - 6 x + 7$

Step 2

Complete the square for $- x^{2} - 6 x + 7$ .

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

$a = - 1$

$b = - 6$

$c = 7$

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

$d = \frac{- 6}{2 \cdot - 1}$

Simplify the right side.

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

Factor $2$ out of $- 6$ .

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

Move the negative one from the denominator of $\frac{- 3}{- 1}$ .

$d = - 1 \cdot - 3$

Rewrite $- 1 \cdot - 3$ as $- - 3$ .

$d = - - 3$

Multiply $- 1$ by $- 3$ .

$d = 3$

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

$e = 7 - \frac{{(- 6)}^{2}}{4 \cdot - 1}$

Simplify the right side.

Simplify each term.

Raise $- 6$ to the power of $2$ .

$e = 7 - \frac{36}{4 \cdot - 1}$

Multiply $4$ by $- 1$ .

$e = 7 - \frac{36}{- 4}$

Divide $36$ by $- 4$ .

$e = 7 - - 9$

Multiply $- 1$ by $- 9$ .

$e = 7 + 9$

Add $7$ and $9$ .

$e = 16$

Substitute the values of $a$ , $d$ , and $e$ into the vertex form $- {(x + 3)}^{2} + 16$ .

$- {(x + 3)}^{2} + 16$

Step 3

Set $y$ equal to the new right side.

$y = - {(x + 3)}^{2} + 16$