Find the Standard Form of the Parabola y^2-4x+4y-16=0

$y^{2} - 4 x + 4 y - 16 = 0$

Step 1

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

Add $4 x$ to both sides of the equation.

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

Add $16$ to both sides of the equation.

$y^{2} + 4 y = 4 x + 16$

Step 2

Complete the square.

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

$a = 1$

$b = 4$

$c = 0 = 4 x + 16$

Consider the vertex form of a parabola.

$a {(x + d)}^{2} + e = 4 x + 16$

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

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

Step 3

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

Add $4$ to both sides of the equation.

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

Add $16$ and $4$ .

${(y + 2)}^{2} = 4 x + 20$

Step 4

Factor $4$ out of $4 x + 20$ .

Factor $4$ out of $4 x$ .

${(y + 2)}^{2} = 4 (x) + 20$

Factor $4$ out of $20$ .

${(y + 2)}^{2} = 4 x + 4 \cdot 5$

Factor $4$ out of $4 x + 4 \cdot 5$ .

${(y + 2)}^{2} = 4 (x + 5)$

Find the Standard Form of the Parabola y^2-4x+4y-16=0

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