Precalculus Examples

$y = \tan (x + \frac{π}{2})$

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

$x + \frac{π}{2} = - \frac{π}{2}$

Step 2

Move all terms not containing $x$ to the right side of the equation.

Tap for more steps...

Step 2.1

Subtract $\frac{π}{2}$ from both sides of the equation.

$x = - \frac{π}{2} - \frac{π}{2}$

Step 2.2

Combine the numerators over the common denominator.

$x = \frac{- π - π}{2}$

Step 2.3

Subtract $π$ from $- π$ .

$x = \frac{- 2 π}{2}$

Step 2.4

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

Tap for more steps...

Step 2.4.1

Factor $2$ out of $- 2 π$ .

$x = \frac{2 (- π)}{2}$

Step 2.4.2

Cancel the common factors.

Tap for more steps...

Step 2.4.2.1

Factor $2$ out of $2$ .

$x = \frac{2 (- π)}{2 (1)}$

Step 2.4.2.2

Cancel the common factor.

$x = \frac{2 (- π)}{2 \cdot 1}$

Step 2.4.2.3

Rewrite the expression.

$x = \frac{- π}{1}$

Step 2.4.2.4

Divide $- π$ by $1$ .

$x = - π$

Step 3

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

$x + \frac{π}{2} = \frac{π}{2}$

Step 4

Move all terms not containing $x$ to the right side of the equation.

Tap for more steps...

Step 4.1

Subtract $\frac{π}{2}$ from both sides of the equation.

$x = \frac{π}{2} - \frac{π}{2}$

Step 4.2

Combine the numerators over the common denominator.

$x = \frac{π - π}{2}$

Step 4.3

Subtract $π$ from $π$ .

$x = \frac{0}{2}$

Step 4.4

Divide $0$ by $2$ .

$x = 0$

Step 5

The basic period for $y = \tan (x + \frac{π}{2})$ will occur at $(- π, 0)$ , where $- π$ and $0$ are vertical asymptotes.

$(- π, 0)$

Step 6

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

Tap for more steps...

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 $y = \tan (x + \frac{π}{2})$ occur at $- π$ , $0$ , and every $x = - π + π n$ , where $n$ is an integer.

$x = - π + π n$

Step 8

Tangent only has vertical asymptotes.

No Horizontal Asymptotes

No Oblique Asymptotes

Vertical Asymptotes: $x = - π + π n$ where $n$ is an integer

Step 9