# How do you identify the period and asympotes for y=-tan(pi/2theta)?

Jan 4, 2018

#### Answer:

Period is 2 and asymptotes occur at $x = 1 + 2 n$, $n \in \mathbb{Z}$.

#### Explanation:

The natural period of $y = \tan \left(\theta\right)$ is $\pi$ and the function naturally has an asymptote halfway through its first period, so at $\frac{\pi}{2}$.

The period of $y = \tan \left(B \theta\right)$ is found by computing $\frac{\pi}{B}$. So the period of $y = - \tan \left(\frac{\pi}{2} \theta\right)$ is $\frac{\pi}{\frac{\pi}{2}} = 2$.

Since tangent has an asymptote halfway through its period, this function will have an asymptote at $x = \frac{2}{2} = 1$. The asymptote will repeat every period, so in general there are asymptotes at $x = 1 + 2 n$ where $n \in \mathbb{Z}$ (n is an integer).