1. ## Asymptope

Have i correctly found an asymptope for this function?

$\displaystyle \frac{x^3-3x^2+1}{x^3}$

As x approaches 0 from left and right i get
$\displaystyle \frac{1}{0}$ so infinity therefor there is a vertical asymptope at 0.

When graphing this, would this mean x never = 0?

Can anything else be determined about the graph of this function from what i have? if what i have is correct

2. Hi.

Originally Posted by anon_404
Have i correctly found an asymptope for this function?

$\displaystyle \frac{x^3-3x^2+1}{x^3}$

As x approaches 0 from left and right i get
$\displaystyle \frac{1}{0}$ so infinity therefor there is a vertical asymptope at 0.
Thats right

Originally Posted by anon_404
When graphing this, would this mean x never = 0?

Yes
Originally Posted by anon_404
Can anything else be determined about the graph of this function from what i have?

No, except: the graph f(x) is not symmetric. f is continious on $\displaystyle (-\infty, 0)$ and $\displaystyle (0, \infty)$

Yours, Rapha

3. As x approaches infinity, i have found the limit to be 1, have i calculated the correct limit? if so does this mean there is a horizontal aymptope at y = 1?

4. Originally Posted by anon_404
As x approaches infinity, i have found the limit to be 1, have i calculated the correct limit? if so does this mean there is a horizontal aymptope at y = 1?
Yes.