Find the Vertex f(x)=x^2+6x-3

$f (x) = x^{2} + 6 x - 3$

Rewrite the equation in vertex form.

Complete the square for $x^{2} + 6 x - 3$ .

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

$a = 1$

$b = 6$

$c = - 3$

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

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

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

Factor $2$ out of $6$ .

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

Cancel the common factors.

Factor $2$ out of $2 \cdot 1$ .

$d = \frac{2 \cdot 3}{2 (1)}$

Cancel the common factor.

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

Rewrite the expression.

$d = \frac{3}{1}$

Divide $3$ by $1$ .

$d = 3$

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

$e = - 3 - \frac{6^{2}}{4 \cdot 1}$

Simplify the right side.

Simplify each term.

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

$e = - 3 - \frac{36}{4 \cdot 1}$

Multiply $4$ by $1$ .

$e = - 3 - \frac{36}{4}$

Divide $36$ by $4$ .

$e = - 3 - 1 \cdot 9$

Multiply $- 1$ by $9$ .

$e = - 3 - 9$

Subtract $9$ from $- 3$ .

$e = - 12$

Substitute the values of $a$ , $d$ , and $e$ into the vertex form ${(x + 3)}^{2} - 12$ .

${(x + 3)}^{2} - 12$

Set $y$ equal to the new right side.

$y = {(x + 3)}^{2} - 12$

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

$a = 1$

$h = - 3$

$k = - 12$

Find the vertex $(h, k)$ .

$(- 3, - 12)$

Find the Vertex f(x)=x^2+6x-3

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