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

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

Step 1

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

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

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

Subtract $4$ from both sides of the equation.

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

Step 2

Complete the square.

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

$a = 1$

$b = 4$

$c = 0 = 8 x - 4$

Consider the vertex form of a parabola.

$a {(x + d)}^{2} + e = 8 x - 4$

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 ${(y + 2)}^{2} - 4$ .

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

Step 3

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

Add $4$ to both sides of the equation.

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

Combine the opposite terms in $8 x - 4 + 4$ .

Add $- 4$ and $4$ .

${(y + 2)}^{2} = 8 x + 0$

Add $8 x$ and $0$ .

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

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

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