Find the Vertex Form x=y^2+6y+1

$x = y^{2} + 6 y + 1$

Step 1

Complete the square for $y^{2} + 6 y + 1$ .

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

$a = 1$

$b = 6$

$c = 1$

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

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

Simplify the right side.

Simplify each term.

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

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

Multiply $4$ by $1$ .

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

Divide $36$ by $4$ .

$e = 1 - 1 \cdot 9$

Multiply $- 1$ by $9$ .

$e = 1 - 9$

Subtract $9$ from $1$ .

$e = - 8$

Substitute the values of $a$ , $d$ , and $e$ into the vertex form ${(y + 3)}^{2} - 8$ .

${(y + 3)}^{2} - 8$

Step 2

Set $x$ equal to the new right side.

$x = {(y + 3)}^{2} - 8$