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

1 Answer
turksvids
Jan 4, 2018

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).

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