Question

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

Answer 1

Period is 2 and asymptotes occur at `x=1+2n`, `n in ZZ`.

Explanation:

The natural period of `y=tan(theta)` is `pi` and the function naturally has an asymptote halfway through its first period, so at `pi/2`.

The period of `y=tan(B theta)` is found by computing `pi/B`. So the period of `y=-tan(pi/2 theta)` is `pi/(pi/2) = 2`.

Since tangent has an asymptote halfway through its period, this function will have an asymptote at `x=2/2=1`. The asymptote will repeat every period, so in general there are asymptotes at `x=1+2n` where `n in ZZ` (n is an integer).