Trigonometry Examples

$y = \tan (9 x)$

Step 1

For any $y = \tan (x)$ , vertical asymptotes occur at $x = \frac{π}{2} + n π$ , where $n$ is an integer. Use the basic period for $y = \tan (x)$ , $(- \frac{π}{2}, \frac{π}{2})$ , to find the vertical asymptotes for $y = \tan (9 x)$ . Set the inside of the tangent function, $b x + c$ , for $y = a \tan (b x + c) + d$ equal to $- \frac{π}{2}$ to find where the vertical asymptote occurs for $y = \tan (9 x)$ .

$9 x = - \frac{π}{2}$

Step 2

Divide each term in $9 x = - \frac{π}{2}$ by $9$ and simplify.

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Step 2.1

Divide each term in $9 x = - \frac{π}{2}$ by $9$ .

$\frac{9 x}{9} = \frac{- \frac{π}{2}}{9}$

Step 2.2

Simplify the left side.

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Step 2.2.1

Cancel the common factor of $9$ .

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Step 2.2.1.1

Cancel the common factor.

$\frac{9 x}{9} = \frac{- \frac{π}{2}}{9}$

Step 2.2.1.2

Divide $x$ by $1$ .

$x = \frac{- \frac{π}{2}}{9}$

Step 2.3

Simplify the right side.

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Step 2.3.1

Multiply the numerator by the reciprocal of the denominator.

$x = - \frac{π}{2} \cdot \frac{1}{9}$

Step 2.3.2

Multiply $- \frac{π}{2} \cdot \frac{1}{9}$ .

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Step 2.3.2.1

Multiply $\frac{1}{9}$ by $\frac{π}{2}$ .

$x = - \frac{π}{9 \cdot 2}$

Step 2.3.2.2

Multiply $9$ by $2$ .

$x = - \frac{π}{18}$

Step 3

Set the inside of the tangent function $9 x$ equal to $\frac{π}{2}$ .

$9 x = \frac{π}{2}$

Step 4

Divide each term in $9 x = \frac{π}{2}$ by $9$ and simplify.

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Step 4.1

Divide each term in $9 x = \frac{π}{2}$ by $9$ .

$\frac{9 x}{9} = \frac{\frac{π}{2}}{9}$

Step 4.2

Simplify the left side.

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Step 4.2.1

Cancel the common factor of $9$ .

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Step 4.2.1.1

Cancel the common factor.

$\frac{9 x}{9} = \frac{\frac{π}{2}}{9}$

Step 4.2.1.2

Divide $x$ by $1$ .

$x = \frac{\frac{π}{2}}{9}$

Step 4.3

Simplify the right side.

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Step 4.3.1

Multiply the numerator by the reciprocal of the denominator.

$x = \frac{π}{2} \cdot \frac{1}{9}$

Step 4.3.2

Multiply $\frac{π}{2} \cdot \frac{1}{9}$ .

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Step 4.3.2.1

Multiply $\frac{π}{2}$ by $\frac{1}{9}$ .

$x = \frac{π}{2 \cdot 9}$

Step 4.3.2.2

Multiply $2$ by $9$ .

$x = \frac{π}{18}$

Step 5

The basic period for $y = \tan (9 x)$ will occur at $(- \frac{π}{18}, \frac{π}{18})$ , where $- \frac{π}{18}$ and $\frac{π}{18}$ are vertical asymptotes.

$(- \frac{π}{18}, \frac{π}{18})$

Step 6

The absolute value is the distance between a number and zero. The distance between $0$ and $9$ is $9$ .

$\frac{π}{9}$

Step 7

The vertical asymptotes for $y = \tan (9 x)$ occur at $- \frac{π}{18}$ , $\frac{π}{18}$ , and every $\frac{π n}{9}$ , where $n$ is an integer.

$x = \frac{π}{18} + \frac{π n}{9}$

Step 8

Tangent only has vertical asymptotes.

No Horizontal Asymptotes

No Oblique Asymptotes

Vertical Asymptotes: $x = \frac{π}{18} + \frac{π n}{9}$ where $n$ is an integer

Step 9