Precalculus Examples

${(y + 5)}^{2} = - 6 x + 12$

Step 1

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

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Step 1.1

Rewrite the equation as $- 6 x + 12 = {(y + 5)}^{2}$ .

$- 6 x + 12 = {(y + 5)}^{2}$

Step 1.2

Subtract $12$ from both sides of the equation.

$- 6 x = {(y + 5)}^{2} - 12$

Step 1.3

Divide each term in $- 6 x = {(y + 5)}^{2} - 12$ by $- 6$ and simplify.

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Step 1.3.1

Divide each term in $- 6 x = {(y + 5)}^{2} - 12$ by $- 6$ .

$\frac{- 6 x}{- 6} = \frac{{(y + 5)}^{2}}{- 6} + \frac{- 12}{- 6}$

Step 1.3.2

Simplify the left side.

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Step 1.3.2.1

Cancel the common factor of $- 6$ .

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Step 1.3.2.1.1

Cancel the common factor.

$\frac{- 6 x}{- 6} = \frac{{(y + 5)}^{2}}{- 6} + \frac{- 12}{- 6}$

Step 1.3.2.1.2

Divide $x$ by $1$ .

$x = \frac{{(y + 5)}^{2}}{- 6} + \frac{- 12}{- 6}$

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.

$x = - \frac{{(y + 5)}^{2}}{6} + \frac{- 12}{- 6}$

Step 1.3.3.1.2

Divide $- 12$ by $- 6$ .

$x = - \frac{{(y + 5)}^{2}}{6} + 2$

Step 1.4

Reorder terms.

$x = - \frac{1}{6} \cdot {(y + 5)}^{2} + 2$

Step 2

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

$a = - \frac{1}{6}$

$h = 2$

$k = - 5$

Step 3

Since the value of $a$ is negative, the parabola opens left.

Opens Left

Step 4

Find the vertex $(h, k)$ .

$(2, - 5)$

Step 5

Find $p$ , the distance from the vertex to the focus.

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Step 5.1

Find the distance from the vertex to a focus of the parabola by using the following formula.

$\frac{1}{4 a}$

Step 5.2

Substitute the value of $a$ into the formula.

$\frac{1}{4 \cdot (- \frac{1}{6})}$

Step 5.3

Simplify.

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Step 5.3.1

Cancel the common factor of $1$ and $- 1$ .

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Step 5.3.1.1

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

$\frac{- 1 (- 1)}{4 \cdot (- \frac{1}{6})}$

Step 5.3.1.2

Move the negative in front of the fraction.

$- \frac{1}{4 (\frac{1}{6})}$

Step 5.3.2

Combine $4$ and $\frac{1}{6}$ .

$- \frac{1}{\frac{4}{6}}$

Step 5.3.3

Cancel the common factor of $4$ and $6$ .

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Step 5.3.3.1

Factor $2$ out of $4$ .

$- \frac{1}{\frac{2 (2)}{6}}$

Step 5.3.3.2

Cancel the common factors.

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Step 5.3.3.2.1

Factor $2$ out of $6$ .

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

Step 5.3.3.2.2

Cancel the common factor.

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

Step 5.3.3.2.3

Rewrite the expression.

$- \frac{1}{\frac{2}{3}}$

Step 5.3.4

Multiply the numerator by the reciprocal of the denominator.

$- (1 (\frac{3}{2}))$

Step 5.3.5

Multiply $\frac{3}{2}$ by $1$ .

$- \frac{3}{2}$

Step 6

Find the focus.

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Step 6.1

The focus of a parabola can be found by adding $p$ to the x-coordinate $h$ if the parabola opens left or right.

$(h + p, k)$

Step 6.2

Substitute the known values of $h$ , $p$ , and $k$ into the formula and simplify.

$(\frac{1}{2}, - 5)$

Step 7

Find the axis of symmetry by finding the line that passes through the vertex and the focus.

$y = - 5$

Step 8

Find the directrix.

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Step 8.1

The directrix of a parabola is the vertical line found by subtracting $p$ from the x-coordinate $h$ of the vertex if the parabola opens left or right.

$x = h - p$

Step 8.2

Substitute the known values of $p$ and $h$ into the formula and simplify.

$x = \frac{7}{2}$

Step 9

Use the properties of the parabola to analyze and graph the parabola.

Direction: Opens Left

Vertex: $(2, - 5)$

Focus: $(\frac{1}{2}, - 5)$

Axis of Symmetry: $y = - 5$

Directrix: $x = \frac{7}{2}$

Step 10