Find the Standard Form of the Parabola y^2+2y+4x-7=0

$y^{2} + 2 y + 4 x - 7 = 0$

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

Subtract $4 x$ from both sides of the equation.

$y^{2} + 2 y - 7 = - 4 x$

Add $7$ to both sides of the equation.

$y^{2} + 2 y = - 4 x + 7$

Complete the square.

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

$a = 1$

$b = 2$

$c = 0 = - 4 x + 7$

Consider the vertex form of a parabola.

$a {(x + d)}^{2} + e = - 4 x + 7$

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

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

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

Cancel the common factor of $2$ .

Cancel the common factor.

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

Rewrite the expression.

$d = 1$

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

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

Simplify the right side.

Simplify each term.

Raise $2$ to the power of $2$ .

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

Multiply $4$ by $1$ .

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

Cancel the common factor of $4$ .

Cancel the common factor.

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

Rewrite the expression.

$e = 0 - 1 \cdot 1$

Multiply $- 1$ by $1$ .

$e = 0 - 1$

Subtract $1$ from $0$ .

$e = - 1$

Substitute the values of $a$ , $d$ , and $e$ into the vertex form ${(y + 1)}^{2} - 1$ .

${(y + 1)}^{2} - 1 = - 4 x + 7$

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

Add $1$ to both sides of the equation.

${(y + 1)}^{2} = - 4 x + 7 + 1$

Add $7$ and $1$ .

${(y + 1)}^{2} = - 4 x + 8$

Factor.

Factor $4$ out of $- 4 x + 8$ .

Factor $4$ out of $- 4 x$ .

${(y + 1)}^{2} = 4 (- x) + 8$

Factor $4$ out of $8$ .

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

Factor $4$ out of $4 (- x) + 4 (2)$ .

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

Factor $- 1$ out of $- x$ .

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

Rewrite $2$ as $- 1 (- 2)$ .

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

Factor $- 1$ out of $- (x) - 1 (- 2)$ .

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

Multiply $- 1$ by $4$ .

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

Find the Standard Form of the Parabola y^2+2y+4x-7=0

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