Find the Standard Form of the Parabola x^2+6x+8y+1=0

$x^{2} + 6 x + 8 y + 1 = 0$

Step 1

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

Subtract $8 y$ from both sides of the equation.

$x^{2} + 6 x + 1 = - 8 y$

Subtract $1$ from both sides of the equation.

$x^{2} + 6 x = - 8 y - 1$

Step 2

Complete the square.

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

$a = 1$

$b = 6$

$c = 0 = - 8 y - 1$

Consider the vertex form of a parabola.

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

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

${(x + 3)}^{2} - 9 = - 8 y - 1$

Step 3

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

Add $9$ to both sides of the equation.

${(x + 3)}^{2} = - 8 y - 1 + 9$

Add $- 1$ and $9$ .

${(x + 3)}^{2} = - 8 y + 8$

Step 4

Factor.

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

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

${(x + 3)}^{2} = 8 (- y) + 8$

Factor $8$ out of $8$ .

${(x + 3)}^{2} = 8 (- y) + 8 (1)$

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

${(x + 3)}^{2} = 8 (- y + 1)$

Factor $- 1$ out of $- y$ .

${(x + 3)}^{2} = 8 (- (y) + 1)$

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

${(x + 3)}^{2} = 8 (- (y) - 1 (- 1))$

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

${(x + 3)}^{2} = 8 (- (y - 1))$

Multiply $- 1$ by $8$ .

${(x + 3)}^{2} = - 8 (y - 1)$

Find the Standard Form of the Parabola x^2+6x+8y+1=0

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