I want to talk about the asymptotes and x intercepts of this function y equals 5 secant 1/6 x. Now in order to analyze both the asymptotes in the intercepts I want to make a little substitution I’m going to call this theta.
And so this function becomes 5 secant theta and of course 5 secant theta is the same as 5 over cosine theta. Now if you look at this function this is never going to equal 0 the only way it’s going to equal is if the numerator equals 0 and the numerator is 5.
So this is never going to equal 0 and that means no x intercepts. Now what about vertical asymptotes, we will have vertical asymptotes when cosine theta equals 0 and we already know that that happens at theta equals pi over 2 plus and pi.
Now theta was the substitution let’s put 1/6 x back in there. 1/6 x equals pi over 2 plus n pi. So to find what x has to be I multiply everything by 6 and I get x equals 3 pi plus 6 n pi. So that means that the vertical asymptotes are x equals 3 pi, 9 pi, 15 pi and so on. -3 pi, -9 pi, -15 pi and so on.