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Function of the graph.

  1. Nov 3, 2005 #1
    y = x+(m/(ax^n+b)+c?

    I'm trying to find the equation for a graph in which the equation changes for different points.
    The graph can be made accurate upto 1500, but after that it becomes inacurate for my numbers, using this equation;
    As you can see, it's sort of like modifying x by a percentage, that decreases as x gets bigger, an equation which changes as x does. At 0 the modifer is 180% or +80%, at 50 it's 177% (+77%), the modifer drops 3% each 50 along the x axis. At 500 its 150%, at 1000 its 120% all the way down to 0% at 3000, The problem with this is that when ever I put this into a calculator it doesn't go 3000+(0% of 3000) which WOULD = 3000, instead it somehow calculates 3000 x 0% = 0.
    The idea is that as x increases y increases but at a slower rate, like an exponential function (if X > 0). Problem is I can't find any exponential function to fix it (My calculator is gay, and even though the numbers are fairly round I can't manually find one that fits it, every time I get close and add a +c to the equation it goes spastic), it was suggested to me that maybe it is a, [tex] x+(\frac{m}{ax^n+b})+c[/tex] type equation which would complicate things.
    Maybe I am looking to hard, maybe I'm just an idiot but after a few hours of thinking about it my head is startinig to hurt and I'm getting lazy so any help will be usefull.
    I might have to edit this if the tex is stuffed up, first time using it lol.
    Edit: Okay I think I got the tex stuff down. Hehe.
    Last edited: Nov 3, 2005
  2. jcsd
  3. Nov 7, 2005 #2
    Bump, 60 views no reply. Is my question too difficult or just silly?
  4. Nov 7, 2005 #3


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    Staff Emeritus
    Science Advisor
    Gold Member

    You have not provided any usefull information. It is not easy, or even possible, to always find a relationship which describes and arbitrary data set.

    Tell us more about the data? Where did it come from? What is the physical situation?

    While I admire your efforts, they do not help us help you.
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