1. ## Finding asymptotes

How do I find all of the asymptotes (vertical, horizontal, oblique) of the function (x^3 + 1) / (x^2 - 2x + 2) ? For the vertical asymptote I believe it's 1 + i but I'm not too sure about that. Can anyone help me out with that asymptote and whatever other asymptotes this formula might have? Thanks!

2. ## Re: Finding asymptotes

Originally Posted by Hamzawesome
How do I find all of the asymptotes (vertical, horizontal, oblique) of the function (x^3 + 1) / (x^2 - 2x + 2) ? For the vertical asymptote I believe it's 1 + i but I'm not too sure about that. Can anyone help me out with that asymptote and whatever other asymptotes this formula might have? Thanks!
Since$\displaystyle f(x)$ is defined for all$\displaystyle x\in R$ it follows that $\displaystyle f(x)$ has no vertical asymptote .

Since $\displaystyle \lim_{x \to \infty} f(x) = +\infty$ it follows that $\displaystyle f(x)$ has no horizontal asymptote .

However , $\displaystyle f(x)$ could has an oblique asymptote...

3. ## Re: Finding asymptotes

How did you come up with the answer of 1+i? Are you aware that it is a complex root of denominator?

You can find the oblique asymptote by using synthetic division and disregarding the remainder since the denominator is one degree less than the numerator.

4. ## Re: Finding asymptotes

Originally Posted by princeps
Since$\displaystyle f(x)$ is defined for all$\displaystyle x\in R$ it follows that $\displaystyle f(x)$ has no vertical asymptote .

Since $\displaystyle \lim_{x \to \infty} f(x) = +\infty$ it follows that $\displaystyle f(x)$ has no horizontal asymptote .

However , $\displaystyle f(x)$ could has an oblique asymptote...
For the oblique asymptote I got y=x+1, after using long division. But for the vertical and horizontal asymptotes I just got "none" as you inferred.

5. ## Re: Finding asymptotes

You're oblique asymptote is not correct, normally it has to be $\displaystyle y=x+2$, recheck your calculations about $\displaystyle \lim_{x\to \infty} f(x)-ax$