# How do you find the vertical, horizontal or slant asymptotes for m(x) = (1-x^2)/(x^3)?

Dec 17, 2016

Horizontal: $\leftarrow y = 0 \rightarrow$.

Vertical: $\uparrow x = 0 \downarrow$.

#### Explanation:

As $x \to \pm \infty , y \to 0$. So, y = 0 is an asymptote.

As $x \to 0 , y \to {+}_{\infty}$. So, x = 0 is the other asymptote.

There is no quotient in the division #y=(1-x^2)/x^3. So, there is no

slant asymptote.

Interestingly, the horizontal asymptote

y = 0 cuts the graph at $x = \pm 1$.

graph{yx^3-1+x^2=0 [-10, 10, -5, 5]}