Find the Standard Form of the Parabola x^2+4x-8y+12=0

$x^{2} + 4 x - 8 y + 12 = 0$

Step 1

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

Add $8 y$ to both sides of the equation.

$x^{2} + 4 x + 12 = 8 y$

Subtract $12$ from both sides of the equation.

$x^{2} + 4 x = 8 y - 12$

Step 2

Complete the square.

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

$a = 1$

$b = 4$

$c = 0 = 8 y - 12$

Consider the vertex form of a parabola.

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

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

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

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

Cancel the common factor of $4$ and $2$ .

Factor $2$ out of $4$ .

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

Cancel the common factors.

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

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

Cancel the common factor.

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

Rewrite the expression.

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

Divide $2$ by $1$ .

$d = 2$

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

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

Simplify the right side.

Simplify each term.

Cancel the common factor of $4^{2}$ and $4$ .

Factor $4$ out of $4^{2}$ .

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

Cancel the common factors.

Factor $4$ out of $4 \cdot 1$ .

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

Cancel the common factor.

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

Rewrite the expression.

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

Divide $4$ by $1$ .

$e = 0 - 1 \cdot 4$

Multiply $- 1$ by $4$ .

$e = 0 - 4$

Subtract $4$ from $0$ .

$e = - 4$

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

${(x + 2)}^{2} - 4 = 8 y - 12$

Step 3

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

Add $4$ to both sides of the equation.

${(x + 2)}^{2} = 8 y - 12 + 4$

Add $- 12$ and $4$ .

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

Step 4

Factor $8$ out of $8 y - 8$ .

Factor $8$ out of $8 y$ .

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

Factor $8$ out of $- 8$ .

${(x + 2)}^{2} = 8 (y) + 8 (- 1)$

Factor $8$ out of $8 (y) + 8 (- 1)$ .

${(x + 2)}^{2} = 8 (y - 1)$

Find the Standard Form of the Parabola x^2+4x-8y+12=0

Related Questions

Support LCMGCF.com by disabling your adblocker.