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Determine Equation by Graph

  1. Dec 9, 2004 #1
    OK I am given the graph of a reciprocal function the asymptotes are x=1 and x=-1
    Quadrants ____1____|____2_____
    3 | 4

    Ok in the first quadrant there is a > looking curve going through the invariant point 1 and in the second quadrant there is a < curve going through the invariant point 1. then there is a parabola with a max of -1 inbetween the asymptotes.

    The question says determine the equation of the graph, so I need to find f(x) and then to get the reciprocal just put 1/(f(X)). I sketched f(x) but how do I find out the eqn? :confused:
  2. jcsd
  3. Dec 9, 2004 #2


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    Does it look like the [tex]f(x)=\frac{1}{4*x^2-1}[/tex] in your other thread?

    If so, try a function with a similar form: [tex]f(x)=\frac{a}{bx^2-c}[/tex]
    and find a,b and c by plugging in coordinates from the graph.
  4. Dec 11, 2004 #3
    no i graphed the equation in my other post it does not look similar to this one. I still have no idea of how to determine the equation, all I know is to find the equation of f(x) first then find 1/f(x). :cry:

    Well this is what I just tried since the vertex of f(x) is 0,-1 I subbed this into the equations y=a(x+h)^2-K and then I took a point from the graph (the invariant point, 1,1) and then solved for a... a=2 so my equation for f(x) became 2x^(2)-1 that means the reciprocal must be 1/2x^(2)-1
    Im not sure if this is the correct way, I tried graphing the equation using my ti 83 graphing calculator, but the reciprocal function looks different from the one on my hand out, so i cant check If I am right? What do I do? does this seem right? :cry: please HELP ME !!!
    Last edited: Dec 11, 2004
  5. Dec 14, 2004 #4
    Please Help Me Please

    :cry: No one is helping me out, Im stuck well I tried and for my graph I got the equation 1/(2x^(2)-1) . HELP PLEASE
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