Find the Vertex Form x=y^2-6y+9

$x = y^{2} - 6 y + 9$

Step 1

Complete the square for $y^{2} - 6 y + 9$ .

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

$a = 1$

$b = - 6$

$c = 9$

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

$e = 9 - \frac{{(- 6)}^{2}}{4 \cdot 1}$

Simplify the right side.

Simplify each term.

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

$e = 9 - \frac{36}{4 \cdot 1}$

Multiply $4$ by $1$ .

$e = 9 - \frac{36}{4}$

Divide $36$ by $4$ .

$e = 9 - 1 \cdot 9$

Multiply $- 1$ by $9$ .

$e = 9 - 9$

Subtract $9$ from $9$ .

$e = 0$

Substitute the values of $a$ , $d$ , and $e$ into the vertex form ${(y - 3)}^{2} + 0$ .

${(y - 3)}^{2} + 0$

Step 2

Set $x$ equal to the new right side.

$x = {(y - 3)}^{2} + 0$