R(x) = x^3 - 2x^2 -7/ x^6 - 2x^5 + 6x^2 -4. How many vertical asymptotes are possible?
We set the denominator to zero and solve for x. This is a 6th degree polynomial so we know it will have 0, 2, 4, or 6 real zeros. The best way to do this would be to graph it. (You need to be careful: the rational root theorem only gives you rational zeros, not necessarily all of them.) From the looks of the graph I'd say there are only two.