Question

show all asymptotes and "holes"
f(x)=x^2-4/x^2-x-6

Answer 1

\[f(x)=x^2-4/(x^2-x-6)=(x+2)(x-2)/(x=2)(x-3)\]In many cases, we simplify this to\[f(x)=(x+2)/(x-3)\]using some ideas from real analysis about removable discontinuities. I suspect your teacher want you to say that the vertical asymptote is at x=3 (put there by the factor of (x-3) in the denominator), and it has a "hole" at x=-2(put there by the factor of (x-2)/(x-2) in the function).