i need the answer analitically, not graphed
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A hole is created when the function is undefined at one point, but is defined very near to the point. A hole is most often associated with the denominator of a fraction taking on the value of zero for one value of the variable. So (x-5) is an example of such a denominator. However, this would simply generate an asymptote at x=5, because values close to 5 generate very large positive or negative values, and to avoid this we can introduce the factor x-5 into the numerator as well. A zero at x=-1 implies x+1 is a factor in the numerator. An asymptote at x=8 may be because x-8 is a factor in the denominator.

These are the ingredients, so let's put something together: f(x)=(x+1)(x-5)/((x-8)(x-5)=(x^2-4x-5)/(x^2-13x+40). When  x=5, f(x) tends to (x+1)/(x-8), so values close to 5 are close to 6/-3=-2, but the function is actually undefined at exactly x=5, so there's a hole. When x=8 f(x) tends to infinity and values near to x=8 send the function to very high positive or negative values. The asymptote separates the large positive from the large negative: when x is a little bigger than 8, f(x) is large and positive; when x is a little smaller than 8 it's large and negative.

 

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