Precalculus Examples

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

Step 1

Since $y$ is on the right side of the equation, switch the sides so it is on the left side of the equation.

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

Step 2

Add $6$ to both sides of the equation.

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

Step 3

Complete the square.

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Step 3.1

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

$a = - 1$

$b = 2$

$c = 0 = x + 6$

Step 3.2

Consider the vertex form of a parabola.

$a {(x + d)}^{2} + e = x + 6$

Step 3.3

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

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Step 3.3.1

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

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

Step 3.3.2

Simplify the right side.

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Step 3.3.2.1

Cancel the common factor of $2$ .

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Step 3.3.2.1.1

Cancel the common factor.

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

Step 3.3.2.1.2

Rewrite the expression.

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

Step 3.3.2.1.3

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

$d = - 1 \cdot 1$

Step 3.3.2.2

Multiply $- 1$ by $1$ .

$d = - 1$

Step 3.4

Find the value of $e$ using the formula $e = c - \frac{b^{2}}{4 a}$ .

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Step 3.4.1

Substitute the values of $c$ , $b$ and $a$ into the formula $e = c - \frac{b^{2}}{4 a}$ .

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

Step 3.4.2

Simplify the right side.

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Step 3.4.2.1

Simplify each term.

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Step 3.4.2.1.1

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

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

Step 3.4.2.1.2

Multiply $4$ by $- 1$ .

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

Step 3.4.2.1.3

Divide $4$ by $- 4$ .

$e = 0 - - 1$

Step 3.4.2.1.4

Multiply $- 1$ by $- 1$ .

$e = 0 + 1$

Step 3.4.2.2

Add $0$ and $1$ .

$e = 1$

Step 3.5

Substitute the values of $a$ , $d$ , and $e$ into the vertex form $- {(y - 1)}^{2} + 1$ .

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

Step 4

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

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Step 4.1

Subtract $1$ from both sides of the equation.

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

Step 4.2

Subtract $1$ from $6$ .

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

Step 5

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

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Step 5.1

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

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

Step 5.2

Simplify the left side.

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Step 5.2.1

Dividing two negative values results in a positive value.

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

Step 5.2.2

Divide ${(y - 1)}^{2}$ by $1$ .

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

Step 5.3

Simplify the right side.

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Step 5.3.1

Simplify each term.

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Step 5.3.1.1

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

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

Step 5.3.1.2

Rewrite $- 1 \cdot x$ as $- x$ .

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

Step 5.3.1.3

Divide $5$ by $- 1$ .

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

Step 6

Factor.

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Step 6.1

Factor $- 1$ out of $- x$ .

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

Step 6.2

Rewrite $- 5$ as $- 1 (5)$ .

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

Step 6.3

Factor $- 1$ out of $- (x) - 1 (5)$ .

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