Precalculus Examples

$y^{2} - 6 y = - x + 4$

Step 1

Rewrite the equation in vertex form.

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Step 1.1

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

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Step 1.1.1

Rewrite the equation as $- x + 4 = y^{2} - 6 y$ .

$- x + 4 = y^{2} - 6 y$

Step 1.1.2

Subtract $4$ from both sides of the equation.

$- x = y^{2} - 6 y - 4$

Step 1.1.3

Divide each term in $- x = y^{2} - 6 y - 4$ by $- 1$ and simplify.

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Step 1.1.3.1

Divide each term in $- x = y^{2} - 6 y - 4$ by $- 1$ .

$\frac{- x}{- 1} = \frac{y^{2}}{- 1} + \frac{- 6 y}{- 1} + \frac{- 4}{- 1}$

Step 1.1.3.2

Simplify the left side.

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Step 1.1.3.2.1

Dividing two negative values results in a positive value.

$\frac{x}{1} = \frac{y^{2}}{- 1} + \frac{- 6 y}{- 1} + \frac{- 4}{- 1}$

Step 1.1.3.2.2

Divide $x$ by $1$ .

$x = \frac{y^{2}}{- 1} + \frac{- 6 y}{- 1} + \frac{- 4}{- 1}$

Step 1.1.3.3

Simplify the right side.

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Step 1.1.3.3.1

Simplify each term.

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Step 1.1.3.3.1.1

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

$x = - 1 \cdot y^{2} + \frac{- 6 y}{- 1} + \frac{- 4}{- 1}$

Step 1.1.3.3.1.2

Rewrite $- 1 \cdot y^{2}$ as $- y^{2}$ .

$x = - y^{2} + \frac{- 6 y}{- 1} + \frac{- 4}{- 1}$

Step 1.1.3.3.1.3

Move the negative one from the denominator of $\frac{- 6 y}{- 1}$ .

$x = - y^{2} - 1 \cdot (- 6 y) + \frac{- 4}{- 1}$

Step 1.1.3.3.1.4

Rewrite $- 1 \cdot (- 6 y)$ as $- (- 6 y)$ .

$x = - y^{2} - (- 6 y) + \frac{- 4}{- 1}$

Step 1.1.3.3.1.5

Multiply $- 6$ by $- 1$ .

$x = - y^{2} + 6 y + \frac{- 4}{- 1}$

Step 1.1.3.3.1.6

Divide $- 4$ by $- 1$ .

$x = - y^{2} + 6 y + 4$

Step 1.2

Complete the square for $- y^{2} + 6 y + 4$ .

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Step 1.2.1

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

$a = - 1$

$b = 6$

$c = 4$

Step 1.2.2

Consider the vertex form of a parabola.

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

Step 1.2.3

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

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Step 1.2.3.1

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

$d = \frac{6}{2 \cdot - 1}$

Step 1.2.3.2

Simplify the right side.

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Step 1.2.3.2.1

Cancel the common factor of $6$ and $2$ .

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Step 1.2.3.2.1.1

Factor $2$ out of $6$ .

$d = \frac{2 \cdot 3}{2 \cdot - 1}$

Step 1.2.3.2.1.2

Move the negative one from the denominator of $\frac{3}{- 1}$ .

$d = - 1 \cdot 3$

Step 1.2.3.2.2

Multiply $- 1$ by $3$ .

$d = - 3$

Step 1.2.4

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

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Step 1.2.4.1

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

$e = 4 - \frac{6^{2}}{4 \cdot - 1}$

Step 1.2.4.2

Simplify the right side.

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Step 1.2.4.2.1

Simplify each term.

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Step 1.2.4.2.1.1

Raise $6$ to the power of $2$ .

$e = 4 - \frac{36}{4 \cdot - 1}$

Step 1.2.4.2.1.2

Multiply $4$ by $- 1$ .

$e = 4 - \frac{36}{- 4}$

Step 1.2.4.2.1.3

Divide $36$ by $- 4$ .

$e = 4 - - 9$

Step 1.2.4.2.1.4

Multiply $- 1$ by $- 9$ .

$e = 4 + 9$

Step 1.2.4.2.2

Add $4$ and $9$ .

$e = 13$

Step 1.2.5

Substitute the values of $a$ , $d$ , and $e$ into the vertex form $- {(y - 3)}^{2} + 13$ .

$- {(y - 3)}^{2} + 13$

Step 1.3

Set $x$ equal to the new right side.

$x = - {(y - 3)}^{2} + 13$

Step 2

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

$a = - 1$

$h = 13$

$k = 3$

Step 3

Find the vertex $(h, k)$ .

$(13, 3)$

Step 4