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Homework Help: Building Functions from Asymptotes

  1. Sep 29, 2008 #1
    1. The problem statement, all variables and given/known data
    Use the given information about the asymptotes of a function to write an equation for a rational function. You may leave your function in "factored" form.


    2. Relevant equations

    1. Vertical Asymptote at x=3 and x=-2

    2. Vertical Asymptote at x=5, Horizontal Asymptote at y=3

    3. Hole at (2,3), Horizontal Asymptote at y=0

    4. Horizontal Asymptote at y=0, Vertical Asymptote at x=6, Goes through the point (3,-2)

    5. Vertical Asymptote at x=2, Oblique Asymptote with Equation y=3x-5

    3. The attempt at a solution

    I know the basics for finding horizontal, vertical, and oblique asymptotes when you have the equation. I'm just having a hard time going backwards.

    1. f(x)=1/(x-3)(x+2)

    2. f(x)=3x/(x-5)

    3. I know the denominator has to be a higher degree than the numerator, bu the "hole" in the graph through me for a loop. No idea where to start.

    4. I'm having a similar problem, I don't know how to find a point that goes through the asymptote.

    5. f(x)=3x2-11x+10/(x-2)

    I got that by setting the denominator equal to zero to find '(x-2)'. I then multiplied the oblique asymptote 'y=3x-5' by '(x-2)' to get the numerator. I think i did something wrong though.

    Any help would be greatly appreciated!

    Thanks
    Lizardjuice7
    1. The problem statement, all variables and given/known data



    2. Relevant equations



    3. The attempt at a solution
     
  2. jcsd
  3. Sep 29, 2008 #2

    Dick

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    Science Advisor
    Homework Helper

    None these problems necessarily determines a unique function, you just have to find one that works. 1 and 2 look good. A 'hole in the graph' is referring to a function like (x-2)/(x-2). It's equal to 1 everywhere except at x=2, where it's undefined. That's the 'hole in the graph'. For 4 the point (-3,2) really supposed to be on the asymptote or on the graph? It's not on either asymptote. For 5 the equation should be (3x-5) times something that goes to 1 as x->infinity, right? And that something should have a vertical asymptote at x=2.
     
    Last edited: Sep 29, 2008
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