Trigonometry Examples

$y = - 3 \tan (\frac{1}{2} x)$

Step 1

Combine $\frac{1}{2}$ and $x$ .

$y = - 3 \tan (\frac{x}{2})$

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 $y = - 3 \tan (\frac{x}{2})$ . 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 $y = - 3 \tan (\frac{x}{2})$ .

$\frac{x}{2} = - \frac{π}{2}$

Step 3

Since the expression on each side of the equation has the same denominator, the numerators must be equal.

$x = - π$

Step 4

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

$\frac{x}{2} = \frac{π}{2}$

Step 5

Since the expression on each side of the equation has the same denominator, the numerators must be equal.

$x = π$

Step 6

The basic period for $y = - 3 \tan (\frac{x}{2})$ will occur at $(- π, π)$ , where $- π$ and $π$ are vertical asymptotes.

$(- π, π)$

Step 7

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

Tap for more steps...

Step 7.1

$\frac{1}{2}$ is approximately $0.5$ which is positive so remove the absolute value

$\frac{π}{\frac{1}{2}}$

Step 7.2

Multiply the numerator by the reciprocal of the denominator.

$π \cdot 2$

Step 7.3

Move $2$ to the left of $π$ .

$2 π$

Step 8

The vertical asymptotes for $y = - 3 \tan (\frac{x}{2})$ occur at $- π$ , $π$ , and every $2 π n$ , where $n$ is an integer.

$x = π + 2 π n$

Step 9

Tangent only has vertical asymptotes.

No Horizontal Asymptotes

No Oblique Asymptotes

Vertical Asymptotes: $x = π + 2 π n$ where $n$ is an integer

Step 10