Precalculus Examples

$y = \frac{1}{2} \tan (2 x)$

Step 1

Combine $\frac{1}{2}$ and $\tan (2 x)$ .

$y = \frac{\tan (2 x)}{2}$

Step 2

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 $y = \frac{\tan (2 x)}{2}$ . 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 $y = \frac{\tan (2 x)}{2}$ .

$2 x = - \frac{π}{2}$

Step 3

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

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Step 3.1

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

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

Step 3.2

Simplify the left side.

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Step 3.2.1

Cancel the common factor of $2$ .

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Step 3.2.1.1

Cancel the common factor.

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

Step 3.2.1.2

Divide $x$ by $1$ .

$x = \frac{- \frac{π}{2}}{2}$

Step 3.3

Simplify the right side.

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Step 3.3.1

Multiply the numerator by the reciprocal of the denominator.

$x = - \frac{π}{2} \cdot \frac{1}{2}$

Step 3.3.2

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

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Step 3.3.2.1

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

$x = - \frac{π}{2 \cdot 2}$

Step 3.3.2.2

Multiply $2$ by $2$ .

$x = - \frac{π}{4}$

Step 4

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

$2 x = \frac{π}{2}$

Step 5

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

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Step 5.1

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

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

Step 5.2

Simplify the left side.

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Step 5.2.1

Cancel the common factor of $2$ .

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Step 5.2.1.1

Cancel the common factor.

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

Step 5.2.1.2

Divide $x$ by $1$ .

$x = \frac{\frac{π}{2}}{2}$

Step 5.3

Simplify the right side.

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Step 5.3.1

Multiply the numerator by the reciprocal of the denominator.

$x = \frac{π}{2} \cdot \frac{1}{2}$

Step 5.3.2

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

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Step 5.3.2.1

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

$x = \frac{π}{2 \cdot 2}$

Step 5.3.2.2

Multiply $2$ by $2$ .

$x = \frac{π}{4}$

Step 6

The basic period for $y = \frac{\tan (2 x)}{2}$ will occur at $(- \frac{π}{4}, \frac{π}{4})$ , where $- \frac{π}{4}$ and $\frac{π}{4}$ are vertical asymptotes.

$(- \frac{π}{4}, \frac{π}{4})$

Step 7

The absolute value is the distance between a number and zero. The distance between $0$ and $2$ is $2$ .

$\frac{π}{2}$

Step 8

The vertical asymptotes for $y = \frac{\tan (2 x)}{2}$ occur at $- \frac{π}{4}$ , $\frac{π}{4}$ , and every $\frac{π n}{2}$ , where $n$ is an integer.

$x = \frac{π}{4} + \frac{π n}{2}$

Step 9

Tangent only has vertical asymptotes.

No Horizontal Asymptotes

No Oblique Asymptotes

Vertical Asymptotes: $x = \frac{π}{4} + \frac{π n}{2}$ where $n$ is an integer

Step 10