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

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

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

Subtract $y$ from both sides of the equation.

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

Subtract $40$ from both sides of the equation.

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

Complete the square.

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

$a = 1$

$b = - 12$

$c = 0 = - y - 40$

Consider the vertex form of a parabola.

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

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

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

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

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 ${(x - 6)}^{2} - 36$ .

${(x - 6)}^{2} - 36 = - y - 40$

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

Add $36$ to both sides of the equation.

${(x - 6)}^{2} = - y - 40 + 36$

Add $- 40$ and $36$ .

${(x - 6)}^{2} = - y - 4$

Factor.

Factor $- 1$ out of $- y$ .

${(x - 6)}^{2} = - (y) - 4$

Rewrite $- 4$ as $- 1 (4)$ .

${(x - 6)}^{2} = - (y) - 1 (4)$

Factor $- 1$ out of $- (y) - 1 (4)$ .

${(x - 6)}^{2} = - (y + 4)$

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

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