Find the Standard Form of the Parabola y^2-16y-12x+40=0

$y^{2} - 16 y - 12 x + 40 = 0$

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

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

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

Subtract $40$ from both sides of the equation.

$y^{2} - 16 y = 12 x - 40$

Complete the square.

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

$a = 1$

$b = - 16$

$c = 0 = 12 x - 40$

Consider the vertex form of a parabola.

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

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

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

$d = \frac{- 16}{2 \cdot 1}$

Cancel the common factor of $- 16$ and $2$ .

Factor $2$ out of $- 16$ .

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

Cancel the common factors.

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

$d = \frac{2 \cdot - 8}{2 (1)}$

Cancel the common factor.

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

Rewrite the expression.

$d = \frac{- 8}{1}$

Divide $- 8$ by $1$ .

$d = - 8$

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

$e = 0 - \frac{{(- 16)}^{2}}{4 \cdot 1}$

Simplify the right side.

Simplify each term.

Raise $- 16$ to the power of $2$ .

$e = 0 - \frac{256}{4 \cdot 1}$

Multiply $4$ by $1$ .

$e = 0 - \frac{256}{4}$

Divide $256$ by $4$ .

$e = 0 - 1 \cdot 64$

Multiply $- 1$ by $64$ .

$e = 0 - 64$

Subtract $64$ from $0$ .

$e = - 64$

Substitute the values of $a$ , $d$ , and $e$ into the vertex form ${(y - 8)}^{2} - 64$ .

${(y - 8)}^{2} - 64 = 12 x - 40$

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

Add $64$ to both sides of the equation.

${(y - 8)}^{2} = 12 x - 40 + 64$

Add $- 40$ and $64$ .

${(y - 8)}^{2} = 12 x + 24$

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

Factor $12$ out of $12 x$ .

${(y - 8)}^{2} = 12 (x) + 24$

Factor $12$ out of $24$ .

${(y - 8)}^{2} = 12 x + 12 \cdot 2$

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

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

Find the Standard Form of the Parabola y^2-16y-12x+40=0

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