Calculus Examples

$f (x) = 4 \tan (π x)$

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 $y = 4 \tan (π x)$ . 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 $y = 4 \tan (π x)$ .

$π x = - \frac{π}{2}$

Step 2

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

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Step 2.1

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

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

Step 2.2

Simplify the left side.

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Step 2.2.1

Cancel the common factor of $π$ .

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Step 2.2.1.1

Cancel the common factor.

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

Step 2.2.1.2

Divide $x$ by $1$ .

$x = \frac{- \frac{π}{2}}{π}$

Step 2.3

Simplify the right side.

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Step 2.3.1

Multiply the numerator by the reciprocal of the denominator.

$x = - \frac{π}{2} \cdot \frac{1}{π}$

Step 2.3.2

Cancel the common factor of $π$ .

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Step 2.3.2.1

Move the leading negative in $- \frac{π}{2}$ into the numerator.

$x = \frac{- π}{2} \cdot \frac{1}{π}$

Step 2.3.2.2

Factor $π$ out of $- π$ .

$x = \frac{π \cdot - 1}{2} \cdot \frac{1}{π}$

Step 2.3.2.3

Cancel the common factor.

$x = \frac{π \cdot - 1}{2} \cdot \frac{1}{π}$

Step 2.3.2.4

Rewrite the expression.

$x = \frac{- 1}{2}$

Step 2.3.3

Move the negative in front of the fraction.

$x = - \frac{1}{2}$

Step 3

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

$π x = \frac{π}{2}$

Step 4

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

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Step 4.1

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

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

Step 4.2

Simplify the left side.

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Step 4.2.1

Cancel the common factor of $π$ .

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Step 4.2.1.1

Cancel the common factor.

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

Step 4.2.1.2

Divide $x$ by $1$ .

$x = \frac{\frac{π}{2}}{π}$

Step 4.3

Simplify the right side.

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Step 4.3.1

Multiply the numerator by the reciprocal of the denominator.

$x = \frac{π}{2} \cdot \frac{1}{π}$

Step 4.3.2

Cancel the common factor of $π$ .

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Step 4.3.2.1

Cancel the common factor.

$x = \frac{π}{2} \cdot \frac{1}{π}$

Step 4.3.2.2

Rewrite the expression.

$x = \frac{1}{2}$

Step 5

The basic period for $y = 4 \tan (π x)$ will occur at $(- \frac{1}{2}, \frac{1}{2})$ , where $- \frac{1}{2}$ and $\frac{1}{2}$ are vertical asymptotes.

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

Step 6

Find the period $\frac{π}{| b |}$ to find where the vertical asymptotes exist.

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Step 6.1

$π$ is approximately $3.14159265$ which is positive so remove the absolute value

$\frac{π}{π}$

Step 6.2

Cancel the common factor of $π$ .

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Step 6.2.1

Cancel the common factor.

$\frac{π}{π}$

Step 6.2.2

Rewrite the expression.

$1$

Step 7

The vertical asymptotes for $y = 4 \tan (π x)$ occur at $- \frac{1}{2}$ , $\frac{1}{2}$ , and every $n$ , where $n$ is an integer.

$x = \frac{1}{2} + n$

Step 8

Tangent only has vertical asymptotes.

No Horizontal Asymptotes

No Oblique Asymptotes

Vertical Asymptotes: $x = \frac{1}{2} + n$ where $n$ is an integer

Step 9