Write in Standard Form (x-5)^2=-8(y+1)

${(x - 5)}^{2} = - 8 (y + 1)$

Step 1

Solve for $y$ .

Rewrite the equation as $- 8 (y + 1) = {(x - 5)}^{2}$ .

$- 8 (y + 1) = {(x - 5)}^{2}$

Divide each term in $- 8 (y + 1) = {(x - 5)}^{2}$ by $- 8$ and simplify.

Divide each term in $- 8 (y + 1) = {(x - 5)}^{2}$ by $- 8$ .

$\frac{- 8 (y + 1)}{- 8} = \frac{{(x - 5)}^{2}}{- 8}$

Simplify the left side.

Cancel the common factor of $- 8$ .

Cancel the common factor.

$\frac{- 8 (y + 1)}{- 8} = \frac{{(x - 5)}^{2}}{- 8}$

Divide $y + 1$ by $1$ .

$y + 1 = \frac{{(x - 5)}^{2}}{- 8}$

Simplify the right side.

Move the negative in front of the fraction.

$y + 1 = - \frac{{(x - 5)}^{2}}{8}$

Subtract $1$ from both sides of the equation.

$y = - \frac{{(x - 5)}^{2}}{8} - 1$

Step 2

To write a polynomial in standard form, simplify and then arrange the terms in descending order.

$y = a x^{2} + b x + c$

Step 3

Reorder terms.

$y = - (\frac{1}{8} \cdot {(x - 5)}^{2}) - 1$

Step 4

Remove parentheses.

$y = - \frac{1}{8} \cdot {(x - 5)}^{2} - 1$

Step 5