Find the Standard Form of the Parabola x^2+14x-12y+13=0

$x^{2} + 14 x - 12 y + 13 = 0$

Step 1

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

Add $12 y$ to both sides of the equation.

$x^{2} + 14 x + 13 = 12 y$

Subtract $13$ from both sides of the equation.

$x^{2} + 14 x = 12 y - 13$

Step 2

Complete the square.

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

$a = 1$

$b = 14$

$c = 0 = 12 y - 13$

Consider the vertex form of a parabola.

$a {(x + d)}^{2} + e = 12 y - 13$

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

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

$d = \frac{14}{2 \cdot 1}$

Cancel the common factor of $14$ and $2$ .

Factor $2$ out of $14$ .

$d = \frac{2 \cdot 7}{2 \cdot 1}$

Cancel the common factors.

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

$d = \frac{2 \cdot 7}{2 (1)}$

Cancel the common factor.

$d = \frac{2 \cdot 7}{2 \cdot 1}$

Rewrite the expression.

$d = \frac{7}{1}$

Divide $7$ by $1$ .

$d = 7$

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

$e = 0 - \frac{14^{2}}{4 \cdot 1}$

Simplify the right side.

Simplify each term.

Raise $14$ to the power of $2$ .

$e = 0 - \frac{196}{4 \cdot 1}$

Multiply $4$ by $1$ .

$e = 0 - \frac{196}{4}$

Divide $196$ by $4$ .

$e = 0 - 1 \cdot 49$

Multiply $- 1$ by $49$ .

$e = 0 - 49$

Subtract $49$ from $0$ .

$e = - 49$

Substitute the values of $a$ , $d$ , and $e$ into the vertex form ${(x + 7)}^{2} - 49$ .

${(x + 7)}^{2} - 49 = 12 y - 13$

Step 3

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

Add $49$ to both sides of the equation.

${(x + 7)}^{2} = 12 y - 13 + 49$

Add $- 13$ and $49$ .

${(x + 7)}^{2} = 12 y + 36$

Step 4

Factor $12$ out of $12 y + 36$ .

Factor $12$ out of $12 y$ .

${(x + 7)}^{2} = 12 (y) + 36$

Factor $12$ out of $36$ .

${(x + 7)}^{2} = 12 y + 12 \cdot 3$

Factor $12$ out of $12 y + 12 \cdot 3$ .

${(x + 7)}^{2} = 12 (y + 3)$

Find the Standard Form of the Parabola x^2+14x-12y+13=0

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