Find the Vertex -x^2+12x-20

$- x^{2} + 12 x - 20$

Step 1

Rewrite the equation in vertex form.

Complete the square for $- x^{2} + 12 x - 20$ .

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

$a = - 1$

$b = 12$

$c = - 20$

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 $\frac{6}{- 1}$ .

$d = - 1 \cdot 6$

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

$e = - 20 - \frac{12^{2}}{4 \cdot - 1}$

Simplify the right side.

Simplify each term.

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

$e = - 20 - \frac{144}{4 \cdot - 1}$

Multiply $4$ by $- 1$ .

$e = - 20 - \frac{144}{- 4}$

Divide $144$ by $- 4$ .

$e = - 20 - - 36$

Multiply $- 1$ by $- 36$ .

$e = - 20 + 36$

Add $- 20$ and $36$ .

$e = 16$

Substitute the values of $a$ , $d$ , and $e$ into the vertex form $- {(x - 6)}^{2} + 16$ .

$- {(x - 6)}^{2} + 16$

Set $y$ equal to the new right side.

$y = - {(x - 6)}^{2} + 16$

Step 2

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

$a = - 1$

$h = 6$

$k = 16$

Step 3

Find the vertex $(h, k)$ .

$(6, 16)$

Step 4

Find the Vertex -x^2+12x-20

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