f(x) = (x+1)/(x2-4x-5)
f(x) = (x+1)/[(x-5)(x+1)]
f(x) is undefined at x = 5 and x = -1 (since they would cause a division by 0). However, we can cancel the (x+1) in the numerator with the (x+1) in the denominator:
f(x) = 1/(x-5)
Now we have a vertical asymptote at x = 5. However f(x) is still undefined at x = -1, only now we have a "hole" (undefined point) there rather than a vertical asymptote. The coordinates of the hole are (-1,f(-1)). Plug x = -1 into the simplified f(x) (= 1/(x-5)) to find the value of f(-1).