1. asymptote and holes

Write the equations of the asymptotes and give the coordinates of all holes for f(x).

f(x) = [-3(x-4)(x+2)^2(x+1)(x-5)] / [(x+3)^2(x-1)(x-5)]

2. Originally Posted by Mr_Green
Write the equations of the asymptotes and give the coordinates of all holes for f(x).

f(x) = [-3(x-4)(x+2)^2(x+1)(x-5)] / [(x+3)^2(x-1)(x-5)]
$f(x) = \frac{-3(x-4)(x+2)^2(x+1)(x-5)}{(x+3)^2(x-1)(x-5)} = \frac{-3(x-4)(x+2)^2(x+1)}{(x+3)^2(x-1)}$ $x \neq 5$

Horizontal Asymptotes:
As the degree of the numerator is 1 greater than the degree of the denominator there are no horizontal asymptotes.

Vertical Asymptotes:
These occur where the denominator is 0. However have a care, since your original expression had a factor of x - 5 in both the numerator and denominator, which cancels. Hence there is no vertical asymptote at x = 5.

Otherwise the denominator is 0 at x = -3 and x = 1. These are your vertical asymptotes.

For "holes" I am assuming you mean a place where the function is not defined. As I mentioned I had cancelled a x - 5 in the numerator and denominator. This would have caused $f(5) = \frac{0}{0}$ which is undefined. So x = 5 is not in the domain of f(x). Thus x = 5 is a "hole" (if I have the definition correct.)

As it's always a good idea to graph the function to check your results I have attached a graph at the bottom of this post. (It doesn't show the "hole.")

-Dan