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

1 Answer
Dec 17, 2016

Answer:

Horizontal: #larr y = 0 rarr#.

Vertical: #uarr x = 0 darr#.

Explanation:

As #x to +-oo, y to 0#. So, y = 0 is an asymptote.

As #x to 0, y to +_oo#. 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 = +-1#.

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