# Asymptope

• May 18th 2009, 10:17 AM
anon_404
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 (Thinking)

• May 18th 2009, 10:20 AM
Rapha
Hi.

Quote:

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

Quote:

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

Yes
Quote:

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
• May 18th 2009, 11:29 AM
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?
• May 18th 2009, 11:41 AM
stapel
Quote:

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. (Wink)