Find the Vertex f(x)=-x^2+6x-10

$f (x) = - x^{2} + 6 x - 10$

Rewrite the equation in vertex form.

Complete the square for $- x^{2} + 6 x - 10$ .

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

$a = - 1$

$b = 6$

$c = - 10$

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 $\frac{3}{- 1}$ .

$d = - 1 \cdot 3$

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

$e = - 10 - \frac{6^{2}}{4 \cdot - 1}$

Simplify the right side.

Simplify each term.

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

$e = - 10 - \frac{36}{4 \cdot - 1}$

Multiply $4$ by $- 1$ .

$e = - 10 - \frac{36}{- 4}$

Divide $36$ by $- 4$ .

$e = - 10 - - 9$

Multiply $- 1$ by $- 9$ .

$e = - 10 + 9$

Add $- 10$ and $9$ .

$e = - 1$

Substitute the values of $a$ , $d$ , and $e$ into the vertex form $- {(x - 3)}^{2} - 1$ .

$- {(x - 3)}^{2} - 1$

Set $y$ equal to the new right side.

$y = - {(x - 3)}^{2} - 1$

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

$a = - 1$

$h = 3$

$k = - 1$

Find the vertex $(h, k)$ .

$(3, - 1)$