Thread: Vertical Asymptotes

R(x) = x^3 - 2x^2 -7/ x^6 - 2x^5 + 6x^2 -4. How many vertical asymptotes are possible?

Originally Posted by tricey36

R(x) = x^3 - 2x^2 -7/ x^6 - 2x^5 + 6x^2 -4. How many vertical asymptotes are possible?

Generally you need to factor the numerator and denominator because if there is a cancellation then there is no asymptote for that particular factor. However they are asking for how many are possible, so let's talk about that.

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.

-Dan