Find the Standard Form of the Parabola y^2+6y+3x-3=0

$y^{2} + 6 y + 3 x - 3 = 0$

Step 1

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

Subtract $3 x$ from both sides of the equation.

$y^{2} + 6 y - 3 = - 3 x$

Add $3$ to both sides of the equation.

$y^{2} + 6 y = - 3 x + 3$

Step 2

Complete the square.

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

$a = 1$

$b = 6$

$c = 0 = - 3 x + 3$

Consider the vertex form of a parabola.

$a {(x + d)}^{2} + e = - 3 x + 3$

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

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

$d = \frac{6}{2 \cdot 1}$

Cancel the common factor of $6$ and $2$ .

Factor $2$ out of $6$ .

$d = \frac{2 \cdot 3}{2 \cdot 1}$

Cancel the common factors.

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

$d = \frac{2 \cdot 3}{2 (1)}$

Cancel the common factor.

$d = \frac{2 \cdot 3}{2 \cdot 1}$

Rewrite the expression.

$d = \frac{3}{1}$

Divide $3$ by $1$ .

$d = 3$

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

$e = 0 - \frac{6^{2}}{4 \cdot 1}$

Simplify the right side.

Simplify each term.

Raise $6$ to the power of $2$ .

$e = 0 - \frac{36}{4 \cdot 1}$

Multiply $4$ by $1$ .

$e = 0 - \frac{36}{4}$

Divide $36$ by $4$ .

$e = 0 - 1 \cdot 9$

Multiply $- 1$ by $9$ .

$e = 0 - 9$

Subtract $9$ from $0$ .

$e = - 9$

Substitute the values of $a$ , $d$ , and $e$ into the vertex form ${(y + 3)}^{2} - 9$ .

${(y + 3)}^{2} - 9 = - 3 x + 3$

Step 3

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

Add $9$ to both sides of the equation.

${(y + 3)}^{2} = - 3 x + 3 + 9$

Add $3$ and $9$ .

${(y + 3)}^{2} = - 3 x + 12$

Step 4

Factor.

Factor $3$ out of $- 3 x + 12$ .

Factor $3$ out of $- 3 x$ .

${(y + 3)}^{2} = 3 (- x) + 12$

Factor $3$ out of $12$ .

${(y + 3)}^{2} = 3 (- x) + 3 (4)$

Factor $3$ out of $3 (- x) + 3 (4)$ .

${(y + 3)}^{2} = 3 (- x + 4)$

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

${(y + 3)}^{2} = 3 (- (x) + 4)$

Rewrite $4$ as $- 1 (- 4)$ .

${(y + 3)}^{2} = 3 (- (x) - 1 (- 4))$

Factor $- 1$ out of $- (x) - 1 (- 4)$ .

${(y + 3)}^{2} = 3 (- (x - 4))$

Multiply $- 1$ by $3$ .

${(y + 3)}^{2} = - 3 (x - 4)$

Find the Standard Form of the Parabola y^2+6y+3x-3=0

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