Calculus Examples

$y = \tan (x - 2)$

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 $y = \tan (x - 2)$ . 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 $y = \tan (x - 2)$ .

$x - 2 = - \frac{π}{2}$

Step 2

Add $2$ to both sides of the equation.

$x = - \frac{π}{2} + 2$

Step 3

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

$x - 2 = \frac{π}{2}$

Step 4

Add $2$ to both sides of the equation.

$x = \frac{π}{2} + 2$

Step 5

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

$(- \frac{π}{2} + 2, \frac{π}{2} + 2)$

Step 6

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

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Step 6.1

The absolute value is the distance between a number and zero. The distance between $0$ and $1$ is $1$ .

$\frac{π}{1}$

Step 6.2

Divide $π$ by $1$ .

$π$

Step 7

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

$x = - \frac{π}{2} + 2 + π n$

Step 8

Tangent only has vertical asymptotes.

No Horizontal Asymptotes

No Oblique Asymptotes

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

Step 9