<obligatory moan about LaTeX not being working yet>

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 >=0,

or:

m^2 >= -4n.

RonL