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# Asymptotes of Secant, Cosecant, and Cotangent - Problem 2

###### 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 find the x intercepts and asymptotes of y equals negative one quarter co-secant of pi over 3x minus pi over 3. Now first of all let’s make a little substitution just to make this easier, I’ll call this theta and so my function is -1/4 co-secant theta and that’s the same us -1/4 over sine theta.

Now this function is never going to equal 0, the only way it can equal 0 is if the numerator equals 0 and it's going to ever do that so there are no x intercepts here. However the denominator can equal 0 and that’s going to give us vertical asymptotes. So the vertical asymptotes happen when sine theta equals 0 which is at the integer multiples of pi.

Now we made a substitution theta is pi over 3 x minus pi over 3. So let me re-substitute, pi over 3x minus pi over 3 equals n theta. To figure out what the equation of the asymptote is I’m going to have to solve this for x and so I add pi over 3 and then I multiply everything by 3 over pi.

So x equals nPi times 3 over pi is going to be, pi’s cancel, 3n and pi over 3 times 3 over pi is 1. So x equals 3n plus 1.

This is the pattern of the vertical asymptotes, so the vertical asymptotes are going to be x equals; when n is 0 we are going to get x equals 1. When n is 1 we’ll get x equals 4.when n is 2,7 we will get 10 and so on. But we will also get x equals -2, -5, -8 and so on. So all the vertical asymptotes are 3 units apart and they start at x equals 1.

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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.

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