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

1 Answer
Jan 4, 2018

Answer:

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).