Find a rational function that might have the given graph.

Consider the given graph,

The numerator of a rational function $P\left(x\right)=\frac{p\left(x\right)}{q\left(x\right)}$ in lowest terms determines the x-intercepts of its graph.

There are two zeros of the graph at x-axis at points $\left(1,0\right),\left(3,0\right)$ and another one at y-axis at $\left(0,1\right)$.

Now, we can say the x-intercept at $1$ crosses the x-axis, so it is an odd multiplicity and $3$ also crosses the x-axis, so it is an odd multiplicity.

Hence, one possibility for the numerator is localid="1646145497110" role="math" $p\left(x\right)=\left(x^2+a\right)\left(x-1\right)\left(x-3\right)$.

Consider the given question,

The horizontal asymptote of the given graph is $y=1$. Thus, the degree of the numerator must be equal to the degree of the denominator and that the quotient of the leading coefficients must be $\frac{1}{1}$. Then,

$R\left(x\right)=\frac{1\times \left(x^2+a\right)\times \left(x-3\right)\left(x-1\right)}{1{\left(x-2\right)}^2{\left(x+1\right)}^2}......\left(i\right)$

Substitute $\left(0,1\right)$in equation (i),

role="math" localid="1646145489003" $$\begin{gathered} 1=\frac{\left(0^2+a\right)\left(0-3\right)\left(0-1\right)}{{\left(0-2\right)}^2{\left(0+1\right)}^2} \\ 1=\frac{3a}{4} \\ a=\frac{4}{3} \end{gathered}$$

Substitute value of a in equation (i),

$R\left(x\right)=\frac{\left(x^2+\frac{4}{3}\right)\left(x-3\right)\left(x-1\right)}{{\left(x-2\right)}^2{\left(x+1\right)}^2}$