Find the horizontal and vertical asymptotes of the curve: $\frac{(9 +x^4)} { (x^2-x^4)}=-1$ I may just be tired but I set up the limit for $\frac{(9 +x^4)}{ (x^2-x^4)}= -1$ I then factored the bottom and got $x=2$ and $x=-2$, however, the program I am using says that those are not the vertical asymptotes. Can someone give me a math jargon less explanation what I am doing wrong? 
 A: As x gets suuuper duper huge, what is going to happen? Well, let's try it with 1,000,000. Then 1,000,000,000. Then 1 x 10^12. Etc. Essentially, what you will find is that it approaches -1. That is because when x gets enormous, all that is really going to matter is the x^4 term, because it will DWARF the other terms. 
We notice that there is a positive and negative x^4, so the horizontal asymptote when x is super big in the positive direction is going to be -1. The same goes for the negative direction. 
The vertical asymptotes are wherever the equation is not defined for a value of x. This is the same as saying where the equation is 0 in the denominator. So, you just need to find the values of x which make the denominator 0 and the function undefined. To do this, you find the solutions to the equation in the denominator, aka the solutions to x^2 - x^4. This is because the solutions to the denominator make the denominator 0, which makes the function as a whole undefined. Hint: there are three of them, all different. 
The program as a whole is interpreting y as the one horizontal asymptote and x as the three vertical asymptotes. so when you find the vertical asymptotes, put a different value of x in for each one. 
Idk how you got 2 or -2; when you put them into the function, you get 25/12, a totally normal point. If you get totally stuck, desmos it.
