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# Asymptotes of Secant, Cosecant, and Cotangent - Problem 1

###### Norm Prokup

###### Norm Prokup

**Cornell University**

PhD. in Mathematics

Norm was 4th at the 2004 USA Weightlifting Nationals! He still trains and competes occasionally, despite his busy schedule.

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.

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###### Norm Prokup

PhD. in Mathematics, University of Rhode Island

B.S. in Mechanical Engineering, Cornell University

He uses really creative examples for explaining tough concepts and illustrates them perfectly on the whiteboard. It's impossible to get lost during his lessons.

Thiswas EXCELLENT! I am a math teacher and have been looking for an easy/logical way to explain the lateral area of a cone to my students and this was incredibly helpful, thank you very much!”

I just learned more In 3 minutes of polygons here than I do in 3 weeks in my math class”

Hahaha, his examples are the same problems of my math HW!”

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