91Ó°ÊÓ

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

which may also written as,

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

The domain of $Q\left(x\right)$is $\left\{x|x≠Â\pm 2\right\}$.

Now we will write $Q$in lowest terms. Here, we can see that there are no common factors between numerator and denominator. Therefore,$Q$is in its lowest terms.

Therefore, by solving we get, $x=Â\pm 1$. From here we find the only real finite zero of the numerator as $Â\pm 1$.Hence, the $x$-intercept of $Q$is $Â\pm 1$. The $y-$intercept can be found by determining the value of $Q$when $x=0$and here the value of $Q$is $\frac{1}{4}$for$x=0$. Here, the$y-$intercept is,$\frac{1}{4}$.

The obtained vertical asymptotes are at$x=Â\pm 2$.

In this graph there is no horizontal asymptote. So, let us calculate the oblique asymptotes.

This can be found by dividing the numerator by denominator and the part except the remainder is the asymptote.

In $Q$when we do this procedure, we find that $Q$can be again written as,

$y=\frac{x^4-1}{x^2-4}$

Therefore, the function has neither horizontal nor oblique as the asymptote.