1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Asymptotes curve (1-2x)/(3x+5)

  1. Jun 15, 2003 #1
    I the curve (1-2x)/(3x+5). I have been asked to find the verticle and horizontal asymptotes. Can anyone help me with a strategy?
     
    Last edited by a moderator: Feb 5, 2013
  2. jcsd
  3. Jun 15, 2003 #2
    Here are some hints.

    Vertical asymptote: check the demominator, see which point doesn't exist.

    horizontal asymptote: take limit of f(x) as x tends to infinity. What is the relation between the limit value you find and the horizontal asymptote?
     
    Last edited: Jun 15, 2003
  4. Jun 15, 2003 #3

    FZ+

    User Avatar

    I dunno but that answer looks a little too complicated...

    y = (1-2x)/(3x+5)
    3xy + 5y = 1 - 2x
    x = (1 - 5y )/(3y + 2)

    Now, find the value y can't have....

    but beware, I have been wrong before (intentionally of course! )
     
  5. Jun 15, 2003 #4
    That's interesting. I never thought of it like that. That suggests to me that the horizontal asymptote of a function should be equivalent to the vertical asymptote of its inverse (if the inverse exists). Let's see. That might be too general.

    Say y(x)= (Ax^n+B)/(Cx^n+D)
    Then y=A/C is its horizontal asymptote.
    It's inverse:
    yCx^n+yD=Ax^n+B
    x^n(yC-A)=B-yD
    x=[(B-yD)/(yC-A)]^(1/n)
    Has as its vertical asymptote y=A/C
    YEAH!
    (For the second graph, x is a function of y - so y=A/C is a vertical line).

    Of course, the converse should also be true.
    I like that. It shows how arbitrary our placement of the axes and the definition of our variables are.
    My algebra skills break down from there. I tried having nonzero coefficients for other powers of x [eg. x^(n-1)] but I'm not sure I can solve for x in that case.
     
  6. Jun 25, 2003 #5
    In this case the largest degrees of the variables are the same (to the first). So for horizontal asymtotes dont u just take the ratio:

    -2x + 1
    3x + 5

    So the horizontal asymtote is (-2/3)

    Right?
     
  7. Jun 25, 2003 #6
    right
    To find horizontal asymptotes (when dealing with rational functions of polynomials), divide both top and bottom of the fraction by the highest power of x and take the limit as x->[oo]
    Essentially, all terms which have an x that is less than the highest power will tend to zero, and that is a shortcut that most people use.

    For example, if the highest power of x is in the denominator, all the terms in the numerator will tend to zero and that asymptote is y=0.

    btw, an asymptote is a line (or some y-value) that the graph approaches, so it should be written as y=k, rather than just k, although I'm sure anyone would know what you mean if you said the horizontal asymptote is -2/3
     
    Last edited: Jun 25, 2003
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook




Similar Discussions: Asymptotes curve (1-2x)/(3x+5)
  1. 3x + 1 - x - 1 = 2 (Replies: 12)

  2. Curved Asymptotes (Replies: 2)

Loading...