Answer
$ y $ is continuous in the domain
$ R-\{1,3\}$
Work Step by Step
Given $$ y=\frac{x+1}{x^2-4x+3}$$
Find the zeros of the denominator
$ x^2-4x+3=0 \Rightarrow (x-3)(x-1)=0\\$
$ \Rightarrow (x-3)=0 \Rightarrow x=3$
$ \Rightarrow (x-1)=0\Rightarrow x=1$
So, $ y $ is continuous in the domain
$ R-\{1,3\}$
(All values except 1 and 3).
