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Math Help - graph

  1. #1
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    graph

    can someone help me find the eqaution of this graph

    it has to be in the formatt of

    y=e
    Attached Thumbnails Attached Thumbnails graph-new-picture-1-.jpg  
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    Last edited by CaptainBlack; April 20th 2006 at 04:55 AM.
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  2. #2
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    Quote Originally Posted by calem_123
    can someone help me find the eqaution of this graph

    it has to be in the formatt of

    y=e
    I don't know about anyone else, but I cant open the spreadsheet.

    RonL
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  3. #3
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    try this

    ok well try this one out, its in word
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    Last edited by calem_123; April 20th 2006 at 04:23 AM.
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  4. #4
    Grand Panjandrum
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    Don't make multiple posts of the same question

    Don't make multiple posts of the same question.

    This is the sub-forum that you will get the best response from. If
    a question is urgent post it here, and don't post duplicates in the
    other fora.

    RonL
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  5. #5
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    ok sorry man

    just need it soon

    thanks
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  6. #6
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    Quote Originally Posted by calem_123
    can someone help me find the eqaution of this graph

    it has to be in the formatt of

    y=e
    I have uploaded a corrected plot.

    Try to avoid using proprietary file formats like Excel and Word, Not everyone
    has these on their machines. Try to stick to something like .JPG, .PNG
    graphics files.

    RonL
    Last edited by CaptainBlack; April 20th 2006 at 05:03 AM.
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  7. #7
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    Quote Originally Posted by calem_123
    can someone help me find the eqaution of this graph

    it has to be in the formatt of

    y=e
    Nobody will be able to fine the equation of the graph.
    They will be able to find an equation which approximates the graph.

    Now your statement about the required format of the equation is not
    clear, I take it to mean that you want an equation in the form:

    <br />
y=e^{f(x)}<br />

    for some function f of x. Is that right?

    RonL
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  8. #8
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    yeah i need an an equation which approximates the graph

    <br />
y=e^{f(x)}<br />

    yes it has to be in that form

    any help wil be greatly appreciated
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  9. #9
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    Using the excell solver to fit an equation of the form:

    <br />
y=K e^{a+bx+cx^2}}<br />

    to a selection of data points from this curve I get:

    <br />
y=25e^{1.098-0.02429x+0.000256x^2}<br />

    but this is not as good a fit as I would like to see.

    RonL
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  10. #10
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    Quote Originally Posted by CaptainBlack
    Using the excell solver to fit an equation of the form:

    <br />
y=K e^{a+bx+cx^2}}<br />

    to a selection of data points from this curve I get:

    <br />
y=25e^{1.098-0.02429x+0.000256x^2}<br />

    but this is not as good a fit as I would like to see.

    RonL
    Oops, the a term is redundant, this is equivalent to:

    <br />
y=74.95e^{-0.02429x+0.000256x^2}<br />

    or absorbing the K instead of the a term:

    <br />
y=e^{3.5343-0.02429x+0.000256x^2}<br />

    (The rational function approximation is better though)

    RonL
    Last edited by CaptainBlack; April 20th 2006 at 07:51 AM.
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  11. #11
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    ok thanks very very much

    i need to know how to find the equation using algabra

    could u explain how to figure it out please, using algabraic methods
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  12. #12
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    Quote Originally Posted by calem_123
    ok thanks very very much

    i need to know how to find the equation using algabra

    could u explain how to figure it out please, using algabraic methods
    Choose three points on the graph (x_1,y_1),\ (x_2,y_2),\ (x_3,y_3), then you have a set of equations:

    y_1=e^{a+bx_1+cx_1^2}
    y_2=e^{a+bx_2+cx_2^2}
    y_3=e^{a+bx_3+cx_3^2},

    now take logs of these equations:

    \log(y_1)=a+bx_1+cx_1^2
    \log(y_2)=a+bx_2+cx_2^2
    \log(y_3)=a+bx_3+cx_3^2.

    This is a set of three linear simultaneous equations in variables a,\ b, and c, which can be solved by the usual methods.

    This method is not advised.

    RonL
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  13. #13
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    ok thanks a lot

    ur a life saver
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  14. #14
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    crap, using that way results in the equation being way off

    thanks though
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  15. #15
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    Quote Originally Posted by calem_123
    crap, using that way results in the equation being way off

    thanks though

    If you have done it right the plot of the equation should pass through
    the points you selected though.

    RonL
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