91Ó°ÊÓ

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

which may also written as,

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

The domain of $P$is, $\left\{x|x≠Â\pm 1\right\}$.

As $0$is in the domain, substitute $x=0$in the given equation.

$$\begin{gathered} P\left(0\right)=-\frac{1}{1} \\ =-1 \end{gathered}$$

So, the obtained intercepts are,$\left(0,-1\right)$.

$$\begin{gathered} \left(x+1\right)\left(x-1\right)=0 \\ x=Â\pm 1 \end{gathered}$$

The two lines $x=1,x=-1$are the two vertical asymptotes of the graph.

In the given function, the degree of the numerator is greater than the degree of the denominator. So, $P$is improper.

Let us use the long division method to find the horizontal asymptote.

Therefore, $P\left(x\right)=x^2+2+\frac{3}{\left(x+1\right)\left(x-1\right)}$

The obtained quotient is $x^2+2$. So, the graph does not have any horizontal or oblique asymptote.

The graph of $P$will approach the graph of$y=x^2+2$as$x→∞$.

$$\begin{gathered} x^2+2=x^2+2+\frac{3}{\left(x+1\right)\left(x-1\right)} \\ 0=\frac{3}{\left(x+1\right)\left(x-1\right)} \\ 0≠3 \end{gathered}$$

There exists no solution. So, the graph of$P$do not intersect$y=x^2+2$.