Results 1 to 5 of 5

Math Help - Finding asymptotes

  1. #1
    Newbie
    Joined
    Dec 2011
    Posts
    13

    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!
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Senior Member
    Joined
    Nov 2011
    From
    Crna Gora
    Posts
    420
    Thanks
    64

    Re: Finding asymptotes

    Quote Originally Posted by Hamzawesome View Post
    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  f(x) is defined for all  x\in R it follows that f(x) has no vertical asymptote .

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

    However , f(x) could has an oblique asymptote...
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Member
    Joined
    Jul 2011
    Posts
    196

    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.
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Newbie
    Joined
    Dec 2011
    Posts
    13

    Re: Finding asymptotes

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

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

    However , 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.
    Follow Math Help Forum on Facebook and Google+

  5. #5
    MHF Contributor Siron's Avatar
    Joined
    Jul 2011
    From
    Norway
    Posts
    1,250
    Thanks
    20

    Re: Finding asymptotes

    You're oblique asymptote is not correct, normally it has to be y=x+2, recheck your calculations about \lim_{x\to \infty} f(x)-ax
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Finding asymptotes:
    Posted in the Pre-Calculus Forum
    Replies: 3
    Last Post: February 22nd 2011, 03:52 PM
  2. Finding Asymptotes
    Posted in the Algebra Forum
    Replies: 1
    Last Post: June 2nd 2010, 11:09 PM
  3. Finding asymptotes.
    Posted in the Pre-Calculus Forum
    Replies: 3
    Last Post: May 13th 2010, 09:24 AM
  4. Finding asymptotes
    Posted in the Calculus Forum
    Replies: 1
    Last Post: September 29th 2009, 12:06 AM
  5. Finding Asymptotes
    Posted in the Pre-Calculus Forum
    Replies: 1
    Last Post: July 7th 2009, 10:49 PM

Search Tags


/mathhelpforum @mathhelpforum