Precalculus Examples

${(x - 1)}^{2} + 8 (y + 2) = 0$

Step 1

Isolate $y$ to the left side of the equation.

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Step 1.1

Simplify each term.

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Step 1.1.1

Apply the distributive property.

${(x - 1)}^{2} + 8 y + 8 \cdot 2 = 0$

Step 1.1.2

Multiply $8$ by $2$ .

${(x - 1)}^{2} + 8 y + 16 = 0$

Step 1.2

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

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Step 1.2.1

Subtract ${(x - 1)}^{2}$ from both sides of the equation.

$8 y + 16 = - {(x - 1)}^{2}$

Step 1.2.2

Subtract $16$ from both sides of the equation.

$8 y = - {(x - 1)}^{2} - 16$

Step 1.3

Divide each term in $8 y = - {(x - 1)}^{2} - 16$ by $8$ and simplify.

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Step 1.3.1

Divide each term in $8 y = - {(x - 1)}^{2} - 16$ by $8$ .

$\frac{8 y}{8} = \frac{- {(x - 1)}^{2}}{8} + \frac{- 16}{8}$

Step 1.3.2

Simplify the left side.

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Step 1.3.2.1

Cancel the common factor of $8$ .

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Step 1.3.2.1.1

Cancel the common factor.

$\frac{8 y}{8} = \frac{- {(x - 1)}^{2}}{8} + \frac{- 16}{8}$

Step 1.3.2.1.2

Divide $y$ by $1$ .

$y = \frac{- {(x - 1)}^{2}}{8} + \frac{- 16}{8}$

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.

$y = - \frac{{(x - 1)}^{2}}{8} + \frac{- 16}{8}$

Step 1.3.3.1.2

Divide $- 16$ by $8$ .

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

Step 1.4

Reorder terms.

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

Step 2

Use the vertex form, $y = a {(x - h)}^{2} + k$ , to determine the values of $a$ , $h$ , and $k$ .

$a = - \frac{1}{8}$

$h = 1$

$k = - 2$

Step 3

Find the vertex $(h, k)$ .

$(1, - 2)$

Step 4