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 (Thinking)