Find the Vertex -x^2-12x-43

$- x^{2} - 12 x - 43$

Step 1

Rewrite the equation in vertex form.

Complete the square for $- x^{2} - 12 x - 43$ .

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

$a = - 1$

$b = - 12$

$c = - 43$

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

$d = \frac{- 12}{2 \cdot - 1}$

Simplify the right side.

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

Factor $2$ out of $- 12$ .

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

Move the negative one from the denominator of $\frac{- 6}{- 1}$ .

$d = - 1 \cdot - 6$

Rewrite $- 1 \cdot - 6$ as $- - 6$ .

$d = - - 6$

Multiply $- 1$ by $- 6$ .

$d = 6$

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

$e = - 43 - \frac{{(- 12)}^{2}}{4 \cdot - 1}$

Simplify the right side.

Simplify each term.

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

$e = - 43 - \frac{144}{4 \cdot - 1}$

Multiply $4$ by $- 1$ .

$e = - 43 - \frac{144}{- 4}$

Divide $144$ by $- 4$ .

$e = - 43 - - 36$

Multiply $- 1$ by $- 36$ .

$e = - 43 + 36$

Add $- 43$ and $36$ .

$e = - 7$

Substitute the values of $a$ , $d$ , and $e$ into the vertex form $- {(x + 6)}^{2} - 7$ .

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

Set $y$ equal to the new right side.

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

Step 2

Use the vertex form, $y = a {(x - h)}^{2} + k$ , to determine the values of $a$ , $h$ , and $k$ .

$a = - 1$

$h = - 6$

$k = - 7$

Step 3

Find the vertex $(h, k)$ .

$(- 6, - 7)$

Step 4

Find the Vertex -x^2-12x-43

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