Results 1 to 7 of 7

Math Help - asymtote

  1. #1
    Newbie
    Joined
    Oct 2007
    Posts
    13

    asymtote

    i need to draw a graph of

    x^2 + 1 / x

    this is a asymtote type problem

    can anyone help

    thanks
    Follow Math Help Forum on Facebook and Google+

  2. #2
    GAMMA Mathematics
    colby2152's Avatar
    Joined
    Nov 2007
    From
    Alexandria, VA
    Posts
    1,172
    Awards
    1
    Quote Originally Posted by thrusterkel View Post
    i need to draw a graph of

    x^2 + 1 / x

    this is a asymtote type problem

    can anyone help

    thanks
    Vertical asymptotes are where functions are undefined. Where is this function undefined at?

    BTW, the best way to graph such problems is to graph each term seperately and then "add" the graphs.
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Super Member wingless's Avatar
    Joined
    Dec 2007
    From
    Istanbul
    Posts
    585
    Find the asymptote, find the limits when the function approaches -∞ and +∞, find a few points, find whether the function is symmetrical about x-axis or y-axis or the origin... and plot it.
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Newbie
    Joined
    Oct 2007
    Posts
    13
    i have made x = 0 so get the points 0,0 and made y = 0 but get a imaginary number is this right
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Super Member wingless's Avatar
    Joined
    Dec 2007
    From
    Istanbul
    Posts
    585
    Firstly, is it \frac{x^2+1}{x} or x^2 + \frac{1}{x} ?
    Follow Math Help Forum on Facebook and Google+

  6. #6
    Newbie
    Joined
    Oct 2007
    Posts
    13
    <br />
\frac{x^2+1}{x}<br />
    Follow Math Help Forum on Facebook and Google+

  7. #7
    GAMMA Mathematics
    colby2152's Avatar
    Joined
    Nov 2007
    From
    Alexandria, VA
    Posts
    1,172
    Awards
    1
    Quote Originally Posted by thrusterkel View Post
    <br />
\frac{x^2+1}{x}<br />
    In that case, graph it as x+\frac{1}{x}

    For x>1, it follows the path slightly above f(x)=x.

    Likewise, for x<1, it follows the path slightly under f(x)=x.

    When -1 \le x < 0, the graph shoots down to negative infinity as it gets closer to zero.

    When 0 < x \le 1, the graph shoots up to positive infinity as it gets closer to zero.
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. find horizontal asymtote using limits
    Posted in the Calculus Forum
    Replies: 1
    Last Post: October 7th 2009, 09:43 PM

Search Tags


/mathhelpforum @mathhelpforum