# Why is the graph of $f(x) = \frac{x^3}{x^2 - 1}$ approaching negative infinity below $x = -1$?

I'm studying calculus at elementary level .

While drawing a graph, I had little confusing thing.

The function is $f(x) = \dfrac{x^3}{x^2-1}$ ($x$ is real number)

and I did this in wolfram alpha ,

and I wonder why the graph near below $x= -1$ is going to minus infinity.

I think it's going to plus infinity because from the $f(x)$ if we say $x$ goes to nearly $-1$ then denominator term goes to minus infinity and since numerator term is $x^3$ then it's still minus. So we have minus in denominator term and numerator so it's plus.

This was what I was thinking and I don't get that it's wrong.

Could you explain it?

• Please read this tutorial on how to typeset mathematics on this site. – N. F. Taussig Apr 16 '16 at 1:46

I see you already got it, but for the benefit of others who may come here with the same question, and just to be thorough, the reason is that since $f(x) = \dfrac{x^3}{x^2-1}$, then we can do polynomial long division (divide $x^3$ by $x^2-1$) to see that $y=x$ is the oblique asymptote of $f(x)$. Since asymptotes describe "end behavior" then $f(x)$ is going to "behave like" the line $y=x$ as $|x|$ gets really large, i.e., as $x$ gets really far away from $0$ in either direction, i.e., as $x \to \pm\infty$.