The most important step in a problem like this is *factoring* as many terms as you can. In your case, the denominator is already fully factored, but the top can be further factored.

If you plug in the polynomial in the numerator into the quadratic formula, #(-b+-sqrt(b^2-4ac))/(2a)#, you get solutions to the equation to be weird un-whole numbers which makes sense given the #c# value. Now, replace the numerator with the factored version (which is more helpful if it's with whole numbers but it still works either way).

#f(x)=((x-2.737034...)(x+2.070368...))/(x-3)#

You can easily identify a hole if the same factor is in both the numerator and the denominator, which isn't the case in this equation so there shouldn't be any holes.

There is, however, an asymptote: at #x=3#, the graph diverges to positive and negative infinity, making #x=3# an undefined point in the function. You can tell it's an asymptote and not a hole because it doesn't have a twin factor in the numerator. Factors in the denominator are asymptotes, factors in the numerator are solutions, and factors that are shared by both the numerator and the denominator are holes.