Find the Standard Form of the Parabola x^2+10x+16y-7=0

$x^{2} + 10 x + 16 y - 7 = 0$

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

Subtract $16 y$ from both sides of the equation.

$x^{2} + 10 x - 7 = - 16 y$

Add $7$ to both sides of the equation.

$x^{2} + 10 x = - 16 y + 7$

Complete the square.

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

$a = 1$

$b = 10$

$c = 0 = - 16 y + 7$

Consider the vertex form of a parabola.

$a {(x + d)}^{2} + e = - 16 y + 7$

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

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

$d = \frac{10}{2 \cdot 1}$

Cancel the common factor of $10$ and $2$ .

Factor $2$ out of $10$ .

$d = \frac{2 \cdot 5}{2 \cdot 1}$

Cancel the common factors.

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

$d = \frac{2 \cdot 5}{2 (1)}$

Cancel the common factor.

$d = \frac{2 \cdot 5}{2 \cdot 1}$

Rewrite the expression.

$d = \frac{5}{1}$

Divide $5$ by $1$ .

$d = 5$

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

$e = 0 - \frac{10^{2}}{4 \cdot 1}$

Simplify the right side.

Simplify each term.

Raise $10$ to the power of $2$ .

$e = 0 - \frac{100}{4 \cdot 1}$

Multiply $4$ by $1$ .

$e = 0 - \frac{100}{4}$

Divide $100$ by $4$ .

$e = 0 - 1 \cdot 25$

Multiply $- 1$ by $25$ .

$e = 0 - 25$

Subtract $25$ from $0$ .

$e = - 25$

Substitute the values of $a$ , $d$ , and $e$ into the vertex form ${(x + 5)}^{2} - 25$ .

${(x + 5)}^{2} - 25 = - 16 y + 7$

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

Add $25$ to both sides of the equation.

${(x + 5)}^{2} = - 16 y + 7 + 25$

Add $7$ and $25$ .

${(x + 5)}^{2} = - 16 y + 32$

Factor.

Factor $16$ out of $- 16 y + 32$ .

Factor $16$ out of $- 16 y$ .

${(x + 5)}^{2} = 16 (- y) + 32$

Factor $16$ out of $32$ .

${(x + 5)}^{2} = 16 (- y) + 16 (2)$

Factor $16$ out of $16 (- y) + 16 (2)$ .

${(x + 5)}^{2} = 16 (- y + 2)$

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

${(x + 5)}^{2} = 16 (- (y) + 2)$

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

${(x + 5)}^{2} = 16 (- (y) - 1 (- 2))$

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

${(x + 5)}^{2} = 16 (- (y - 2))$

Multiply $- 1$ by $16$ .

${(x + 5)}^{2} = - 16 (y - 2)$

Find the Standard Form of the Parabola x^2+10x+16y-7=0

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