Precalculus Examples

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

Step 1

Solve for $y$ .

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Step 1.1

Rewrite the equation as $- 4 y - 8 = x^{2} + 8 x$ .

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

Step 1.2

Add $8$ to both sides of the equation.

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

Step 1.3

Divide each term in $- 4 y = x^{2} + 8 x + 8$ by $- 4$ and simplify.

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Step 1.3.1

Divide each term in $- 4 y = x^{2} + 8 x + 8$ by $- 4$ .

$\frac{- 4 y}{- 4} = \frac{x^{2}}{- 4} + \frac{8 x}{- 4} + \frac{8}{- 4}$

Step 1.3.2

Simplify the left side.

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Step 1.3.2.1

Cancel the common factor of $- 4$ .

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Step 1.3.2.1.1

Cancel the common factor.

$\frac{- 4 y}{- 4} = \frac{x^{2}}{- 4} + \frac{8 x}{- 4} + \frac{8}{- 4}$

Step 1.3.2.1.2

Divide $y$ by $1$ .

$y = \frac{x^{2}}{- 4} + \frac{8 x}{- 4} + \frac{8}{- 4}$

Step 1.3.3

Simplify the right side.

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Step 1.3.3.1

Simplify each term.

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Step 1.3.3.1.1

Move the negative in front of the fraction.

$y = - \frac{x^{2}}{4} + \frac{8 x}{- 4} + \frac{8}{- 4}$

Step 1.3.3.1.2

Cancel the common factor of $8$ and $- 4$ .

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Step 1.3.3.1.2.1

Factor $4$ out of $8 x$ .

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

Step 1.3.3.1.2.2

Move the negative one from the denominator of $\frac{2 x}{- 1}$ .

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

Step 1.3.3.1.3

Rewrite $- 1 \cdot (2 x)$ as $- (2 x)$ .

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

Step 1.3.3.1.4

Multiply $2$ by $- 1$ .

$y = - \frac{x^{2}}{4} - 2 x + \frac{8}{- 4}$

Step 1.3.3.1.5

Divide $8$ by $- 4$ .

$y = - \frac{x^{2}}{4} - 2 x - 2$

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.

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

Step 4

Remove parentheses.

$y = - \frac{1}{4} x^{2} - 2 x - 2$

Step 5