Asymptotes of Secant, Cosecant, and Cotangent - Problem 3
I want to talk about the x intercepts and asymptotes of this function; y equals cotangent pi over 2 root x. And this is not your usual transformation, so it might be a little tricky. I don’t know if I’d want to graph this but let’s find the x intercepts and the asymptotes using the usual method.
Let’s set this equal to theta. We know that cotangent theta equals cosine theta over sine theta. So we know that cotangent will have x intercepts when cosine equals 0. Now cosine equals 0 when theta equals pi over 2 plus n pi.
Now let’s re substitute to find out what x has to be. Pi over 2 root x equals pi over 2 plus nPi. Let me first multiply everything by 2 over pi. So I’ll get 2 over pi times pi over 2 root x pi over 2 times 2 over pi 1 and pi times 2 over pi, the pi’s will cancel I’ll get 2n. So root x equals 1 plus 2 n and then x equals 1 plus 2n².
What do the x intercepts going to look like? Well remember that x has to be positive for this function, it’s not actually defined for negative values of x it's actually not defined at 0 either but when x equals we’ll see n equals 0 we get x equals 1² which is x equals 1. So 1,0 will be one x intercept when n equals 1 we get 1 plus 2, 3² which is 9, 9,0 when n equals 2 we get 4 plus 1, 5, squared which is 25 and so on. And these are all the odd perfect squares 1 ,9 25 the next one is 49 and so on.
What about the vertical asymptotes? Well we know that we’ll get the vertical asymptotes when sine is 0. So sine theta equals 0 when theta equals nPi. Now theta is pi over 2 root x. So first of all I’m going to multiply both sides by 2 over pi, 2 over pi times pi over 2 is 1, so I get root x equals, pi’s cancel here and I get 2n. And so x, I square both sides x equals 4n². So this is the equation of all, my vertical asymptotes.
Once again remember that I can’t have negative values of x although I can say that x equals 0 is a vertical asymptote because just saying that it’s a vertical asymptote doesn’t mean the function is defined there. When n equals 0 I do get x equals 0. So that’s my first vertical asymptote, x equals 0.
The second one happens when n equals 1 get 4 times 1² 4, x equals 4. And when n equal 2, I get 4 times 4, 16 and you can see that this is going to be all the even perfect squares 0, 4, 16 the next one will be 36, 64 and so on.
So these are the vertical asymptotes and these are the x intercepts. It’s a strange function it might be fun to graph this on your calculator to check it out but even though it's not a function we might want to graph by hand you can still see that the methods that we use for finding the x intercepts in the vertical asymptotes work.