SOLUTION: 2sec(t)tan(t) + sec^(2)(t) What (if any) are the zeros, holes, and vertical asymptotes?

Algebra ->  Trigonometry-basics -> SOLUTION: 2sec(t)tan(t) + sec^(2)(t) What (if any) are the zeros, holes, and vertical asymptotes?      Log On


   



Question 393007: 2sec(t)tan(t) + sec^(2)(t)
What (if any) are the zeros, holes, and vertical asymptotes?

Answer by robertb(5830) About Me  (Show Source):
You can put this solution on YOUR website!
The expression could be written as %282sint%29%2F%28cost%29%5E2+%2B+1%2F%28cost%29%5E2+=+%282sint+%2B+1%29%2F%28cost%29%5E2.
Find out first the values of t where the top is equal to the bottom, i.e., 2sint++%2B+1+=+%28cost%29%5E2:
2sint++%2B+1+=+1+-+%28sint%29%5E2
==> 2sint++=+-%28sint%29%5E2, or sint(2 + sint) = 0.
==> sint = 0 ==> t =0, +/-pi+, +/- 2pi+, +/- 3pi,...
At these t values the bottom, %28cost%29%5E2 is not equal to zero. The top, 2sint+%2B+1 is also not equal to 0 (it is equal to 1). Therefore there are no "holes" in the graph.
There are vertical asymptotes at t values where %28cos+t%29%5E2 = 0, namely
t = +/-pi%2F2, +/- 3pi%2F2+, +/- 5pi%2F2,...
The x-intercepts are located at t =7pi%2F6 +/-2n%2Api+ and t =11pi%2F6 +/- 2n%2Api+. (These are the t values where 2sint + 1 = 0).