Find the Standard Form of the Parabola y^2-12x+8y=-40

$y^{2} - 12 x + 8 y = - 40$

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

$y^{2} + 8 y = - 40 + 12 x$

Complete the square.

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

$a = 1$

$b = 8$

$c = 0 = - 40 + 12 x$

Consider the vertex form of a parabola.

$a {(x + d)}^{2} + e = - 40 + 12 x$

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

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

$d = \frac{8}{2 \cdot 1}$

Cancel the common factor of $8$ and $2$ .

Factor $2$ out of $8$ .

$d = \frac{2 \cdot 4}{2 \cdot 1}$

Cancel the common factors.

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

$d = \frac{2 \cdot 4}{2 (1)}$

Cancel the common factor.

$d = \frac{2 \cdot 4}{2 \cdot 1}$

Rewrite the expression.

$d = \frac{4}{1}$

Divide $4$ by $1$ .

$d = 4$

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

$e = 0 - \frac{8^{2}}{4 \cdot 1}$

Simplify the right side.

Simplify each term.

Raise $8$ to the power of $2$ .

$e = 0 - \frac{64}{4 \cdot 1}$

Multiply $4$ by $1$ .

$e = 0 - \frac{64}{4}$

Divide $64$ by $4$ .

$e = 0 - 1 \cdot 16$

Multiply $- 1$ by $16$ .

$e = 0 - 16$

Subtract $16$ from $0$ .

$e = - 16$

Substitute the values of $a$ , $d$ , and $e$ into the vertex form ${(y + 4)}^{2} - 16$ .

${(y + 4)}^{2} - 16 = - 40 + 12 x$

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

Add $16$ to both sides of the equation.

${(y + 4)}^{2} = - 40 + 12 x + 16$

Add $- 40$ and $16$ .

${(y + 4)}^{2} = 12 x - 24$

Factor $12$ out of $12 x - 24$ .

Factor $12$ out of $12 x$ .

${(y + 4)}^{2} = 12 (x) - 24$

Factor $12$ out of $- 24$ .

${(y + 4)}^{2} = 12 x + 12 \cdot - 2$

Factor $12$ out of $12 x + 12 \cdot - 2$ .

${(y + 4)}^{2} = 12 (x - 2)$

Find the Standard Form of the Parabola y^2-12x+8y=-40

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