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Thread: Determining Asymptotes from Field Data

  1. #1
    Mar 2009

    Determining Asymptotes from Field Data

    Not too sure if this is possible, but I have a set of data and was wondering if there was a mathematically method of calculating its vertical asymptote.

    Known Information:

    Data is from a reciprocal function, although it is "field" data so does not all strictly satisfy an equation.

    x: 1.5, 2, 3, 4, 5, 6, 7, 8, 10, 15
    y: 56, 20, 10.2, 7.6, 6, 5, 4.3, 3.8, 3.1, 2.5

    The horizontal asymptote is at y = 1

    Asked Question:
    "Using what you have discovered about the function y=(A/(x+B))+C, find a reciprocal function that could be used as a model for this data. Fully justify all decisions regarding the values of A, B and C based on your knowledge of transformations"

    I don't believe there is enough information to calculate the function of the above data. If I had the vertical asymptote, I would be able to use a pair of coordinates and the two asymptotes to find A. Right?

    Even so, since the data is "field data", isnt the only way to "determine" a function by using hyperbolic regression (line of best fit)?

    In brief;
    Is there an algebraic method to determine the vertical asymptote?
    If not, can I simply use (as the question asks) transformations of A and C to determine B?
    More likely, how would I be able (as accurately as possible) use hyperbolic or linear regression to determine the vertical asymptote? (Help with this would be much appreciated, regression scares me :S)

    Thanks in advance. I know this question is long and a little messy but its really holding me back from finishing my work.
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  2. #2
    MHF Contributor

    Apr 2005
    If you assume that the function of the form given, [tex]y= A/(x+ B)+ C, that contains 3 unknown parameters, A, B, and C. You need 3 equations to determine those three values. Rather than saying "there is not enough information", I would say "there is too much information"!

    Put any three of those pairs of points into that formula to get three equations to solve for A, B, and C. If the function were, in fact, of that form, it wouldn't matter which of them you used- you would get the same thing.

    For example, using just the first three pairs, x= 1.5, y= 56 gives 56= A/(1.5- B)+ C; x= 2, y= 20 gives 20= A/(2- B)+ C; x= 3, y= 10.2 gives 10.2= A/(3- B)+ C. Subtrating the second equation from the first eliminates C: 36= A/(1.5-B)- A/(2- B)= (2A- AB- (1.5A- AB))/((1.5-B)(2-B))= .5A/((1.5-B)(2-B)) or .5A= 36(1.5-B)(2-B). Subtracting the third equation from the first gives 47.8= A/(1.5-B)- A/(3-B)= (3A-AB- (1.5A- AB)/((1.5-B)(3-B))= 1.5A((1.5-B)(3- B)) or 1.5A= 47.8(1.5-B)(3-B). We can easily eliminate A from those two equations, leaving a quadratic equation in B.
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  3. #3
    Forum Admin topsquark's Avatar
    Jan 2006
    Wellsville, NY
    It almost looks like you need a set of regression equations to fit the data to a given curve... (Too bad I can't figure out how to do it with the given equation.)

    I'm not sure which is more valid, the data or the comments, but if we really do have a function y = A/(x - B) + C and have an asymptote at y = 1, then C = 1. Then all you need is A and B and can use HallsofIvy's method.

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