For f to have a vertical asymtote requires that x^2-mx-n have at least
one real root. This is equivalent to the discriminanat being >=0.
The discriminant is the term under the square root sign in the quadratic
formula, so in this case the disciminant is: m^2+4n, and so f has vetical
asymtots if (and only if):
m^2 >= -4n.