# Asymptope

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• May 18th 2009, 10:17 AM
anon_404
Asymptope
Have i correctly found an asymptope for this function?

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

As x approaches 0 from left and right i get
$\frac{1}{0}$ 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?

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

As x approaches 0 from left and right i get
$\frac{1}{0}$ 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 $(-\infty, 0)$ and $(0, \infty)$

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)