Find the Vertex -x^2-6x+4

$- x^{2} - 6 x + 4$

Step 1

Rewrite the equation in vertex form.

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

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

$a = - 1$

$b = - 6$

$c = 4$

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

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

Simplify the right side.

Simplify each term.

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

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

Multiply $4$ by $- 1$ .

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

Divide $36$ by $- 4$ .

$e = 4 - - 9$

Multiply $- 1$ by $- 9$ .

$e = 4 + 9$

Add $4$ and $9$ .

$e = 13$

Substitute the values of $a$ , $d$ , and $e$ into the vertex form $- {(x + 3)}^{2} + 13$ .

$- {(x + 3)}^{2} + 13$

Set $y$ equal to the new right side.

$y = - {(x + 3)}^{2} + 13$

Step 2

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

$a = - 1$

$h = - 3$

$k = 13$

Step 3

Find the vertex $(h, k)$ .

$(- 3, 13)$

Step 4

Find the Vertex -x^2-6x+4

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