Trigonometry Examples

$y^{2} + 2 y = - x - 5$

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 $- x - 5 = y^{2} + 2 y$ .

$- x - 5 = y^{2} + 2 y$

Step 1.1.2

Add $5$ to both sides of the equation.

$- x = y^{2} + 2 y + 5$

Step 1.1.3

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

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Step 1.1.3.1

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

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

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.

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

Step 1.1.3.2.2

Divide $x$ by $1$ .

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

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 $\frac{y^{2}}{- 1}$ .

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

Step 1.1.3.3.1.2

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

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

Step 1.1.3.3.1.3

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

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

Step 1.1.3.3.1.4

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

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

Step 1.1.3.3.1.5

Multiply $2$ by $- 1$ .

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

Step 1.1.3.3.1.6

Divide $5$ by $- 1$ .

$x = - y^{2} - 2 y - 5$

Step 1.2

Complete the square for $- y^{2} - 2 y - 5$ .

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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 $c$ .

$a = - 1$

$b = - 2$

$c = - 5$

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 $d = \frac{b}{2 a}$ .

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

Step 1.2.3.2

Simplify the right side.

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Step 1.2.3.2.1

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

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Step 1.2.3.2.1.1

Factor $2$ out of $- 2$ .

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

Step 1.2.3.2.1.2

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

$d = - 1 \cdot - 1$

Step 1.2.3.2.2

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

$d = - - 1$

Step 1.2.3.2.3

Multiply $- 1$ by $- 1$ .

$d = 1$

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 $e = c - \frac{b^{2}}{4 a}$ .

$e = - 5 - \frac{{(- 2)}^{2}}{4 \cdot - 1}$

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 $- 2$ to the power of $2$ .

$e = - 5 - \frac{4}{4 \cdot - 1}$

Step 1.2.4.2.1.2

Multiply $4$ by $- 1$ .

$e = - 5 - \frac{4}{- 4}$

Step 1.2.4.2.1.3

Divide $4$ by $- 4$ .

$e = - 5 - - 1$

Step 1.2.4.2.1.4

Multiply $- 1$ by $- 1$ .

$e = - 5 + 1$

Step 1.2.4.2.2

Add $- 5$ and $1$ .

$e = - 4$

Step 1.2.5

Substitute the values of $a$ , $d$ , and $e$ into the vertex form $- {(y + 1)}^{2} - 4$ .

$- {(y + 1)}^{2} - 4$

Step 1.3

Set $x$ equal to the new right side.

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

Step 2

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

$a = - 1$

$h = - 4$

$k = - 1$

Step 3

Find the vertex $(h, k)$ .

$(- 4, - 1)$

Step 4