Answer
$$\eqalign{
& a.{\text{ No horizontal asymptotes}} \cr
& b.{\text{ No vertical asymptotes}} \cr} $$
Work Step by Step
$$\eqalign{
& f\left( x \right) = \frac{{{x^2} - 4x + 3}}{{x - 1}} \cr
& \cr
& {\text{Factor the numerator}} \cr
& f\left( x \right) = \frac{{\left( {x - 3} \right)\left( {x - 1} \right)}}{{x - 1}} \cr
& {\text{Cancel the common factors}} \cr
& f\left( x \right) = x - 3, x\ne3 \cr
& \cr
& a.{\text{ Calculate }}\mathop {\lim }\limits_{x \to \infty } f\left( x \right){\text{ and }}\mathop {\lim }\limits_{x \to - \infty } f\left( x \right) \cr
& \mathop {\lim }\limits_{x \to \infty } f\left( x \right)\,\,\, \Rightarrow \,\,\,\,\mathop {\lim }\limits_{x \to \infty } \left( {x - 3} \right) \cr
& \mathop {\lim }\limits_{x \to \infty } \left( {x - 3} \right) = \infty \cr
& \cr
& \mathop {\lim }\limits_{x \to \infty } f\left( x \right)\,\,\, \Rightarrow \,\,\,\,\mathop {\lim }\limits_{x \to - \infty } \left( {x - 3} \right) \cr
& \mathop {\lim }\limits_{x \to \infty } \left( {x - 3} \right) = - \infty \cr
& \cr
& {\text{The limits are not finite}},{\text{ so the graph of the function has no }} \cr
& {\text{horizontal asymptotes}}{\text{.}} \cr
& \cr
& b.{\text{ Find the vertical asymptotes}} \cr
& {\text{The function }}f\left( x \right) = x - 3, x\ne 3{\text{ is a straight line, so the graph of the }} \cr
& {\text{function has no vertical asymptotes}}{\text{.}} \cr
& \cr
& a.{\text{ No horizontal asymptotes}} \cr
& b.{\text{ No vertical asymptotes}} \cr} $$
