# What are the asymptote(s) and hole(s), if any, of # f(x) =1/x^2-1/(1-x)+x/(3-x)#?

Vertical asymptotes at

Asymptotes and holes are present due to the fact that the denominator of any fraction cannot be 0, since division by zero is impossible.

Since there are no cancelling factors, the non permissible values are all vertical asymptotes.

Which is all the vertical asymptotes.

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The function f(x) = 1/x^2 - 1/(1-x) + x/(3-x) has two vertical asymptotes at x = 0 and x = 1. There are no horizontal asymptotes. The function has a hole at x = 3.

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