Answer
$f(x)=\dfrac{x^2+x+1}{x^2-4x+3}$
Work Step by Step
Let us consider that the function $f(x)$ is defined om some interval $(a, \infty)$.Then , we have $\lim\limits_{n\to \infty} f(x)=L$
This implies that the value of the function $f(x)$ can be defined arbitrarily close to the limit $L$ by taking $x$ sufficiently large.
Then, the line $y=L$ is called a horizontal asymptote of the function $y=f(x)$
The equations of vertical and horizontal asymptotes are:
$x=1,3$ and $y=1$
The function can be written as:
$f(x)=\dfrac{p(x)}{(x-1)(x-3)}=\dfrac{p(x)}{x^2-4x+3}$
or, $f(x)=\dfrac{x^2+x+1}{x^2-4x+3}$