91Ó°ÊÓ

Consider the given graph,

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

The graph has one x-intercept, $0$. As this crosses, the x-axis, it is an odd multiplicity. So, one possibility for the numerator,

localid="1646070026847" $p\left(x\right)=x$.

Consider the given question,

The denominator of a rational function in lowest terms determines the vertical asymptotes of its graph.

The vertical asymptotes are $x=-1,1$.

As $R\left(x\right)$approaches $∞$to the left of localid="1646070034364" $x=-1$and $R\left(x\right)$approaches $-∞$to the right of localid="1646070037273" $x=-1$, then localid="1646070039760" $x+1$is a factor of odd multiplicity of localid="1646070042346" $q\left(x\right)$.

Similarly, localid="1646070047642" $x-1$is also a factor of odd multiplicity in localid="1646070045100" $q\left(x\right)$.

So, one possibility for the denominator is given below,

localid="1646070050871" $$\begin{gathered} q\left(x\right)=\left(x+1\right)\left(x-1\right) \\ q\left(x\right)=x^2-1 \end{gathered}$$

Consider the given question,

The horizontal asymptote of the given graph is the x-axis. 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 $-1$. Hence, a possible rational function is given below,

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