Answer
See graph.
Work Step by Step
Given $f(x)=\frac{x^2-1}{x^2-4x+3}=\frac{(x+1)(x-1)}{(x-3)(x-1)}=\frac{x+1}{x-3}, (x\ne1)$, we can identify: V.A. $x=3$, H.A. $y=1$, O.A. $none$, hole $(1,-1)$. See graph.
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