Find the Standard Form of the Parabola x^2-8x-y+12=0

$x^{2} - 8 x - y + 12 = 0$

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

Add $y$ to both sides of the equation.

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

Subtract $12$ from both sides of the equation.

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

Complete the square.

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

$a = 1$

$b = - 8$

$c = 0 = y - 12$

Consider the vertex form of a parabola.

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

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 ${(x - 4)}^{2} - 16$ .

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

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

Add $16$ to both sides of the equation.

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

Add $- 12$ and $16$ .

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

Find the Standard Form of the Parabola x^2-8x-y+12=0

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