Find the Standard Form of the Parabola x=y^2+12y+38

$x = y^{2} + 12 y + 38$

Step 1

Since $y$ is on the right side of the equation, switch the sides so it is on the left side of the equation.

$y^{2} + 12 y + 38 = x$

Step 2

Subtract $38$ from both sides of the equation.

$y^{2} + 12 y = x - 38$

Step 3

Complete the square.

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

$a = 1$

$b = 12$

$c = 0 = x - 38$

Consider the vertex form of a parabola.

$a {(x + d)}^{2} + e = x - 38$

Find the value of $d$ using the formula $d = \frac{b}{2 a}$ .

Substitute the values of $a$ and $b$ into the formula $d = \frac{b}{2 a}$ .

$d = \frac{12}{2 \cdot 1}$

Cancel the common factor of $12$ and $2$ .

Factor $2$ out of $12$ .

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

Cancel the common factors.

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

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

Cancel the common factor.

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

Rewrite the expression.

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

Divide $6$ by $1$ .

$d = 6$

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

$e = 0 - \frac{12^{2}}{4 \cdot 1}$

Simplify the right side.

Simplify each term.

Raise $12$ to the power of $2$ .

$e = 0 - \frac{144}{4 \cdot 1}$

Multiply $4$ by $1$ .

$e = 0 - \frac{144}{4}$

Divide $144$ by $4$ .

$e = 0 - 1 \cdot 36$

Multiply $- 1$ by $36$ .

$e = 0 - 36$

Subtract $36$ from $0$ .

$e = - 36$

Substitute the values of $a$ , $d$ , and $e$ into the vertex form ${(y + 6)}^{2} - 36$ .

${(y + 6)}^{2} - 36 = x - 38$

Step 4

Move all terms not containing $y$ to the right side of the equation.

Add $36$ to both sides of the equation.

${(y + 6)}^{2} = x - 38 + 36$

Add $- 38$ and $36$ .

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

Find the Standard Form of the Parabola x=y^2+12y+38

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