how can i calculate the polynomial that has the following asymptotes

how can i calculate the polynomial that has the tangent lines (1) y = x where x = 1, and (2) y = 1 where x = 365

I realize this may not be the proper forum but I figured somebody here could answer this in jiffy.

Also, I am not looking for an algorithm to answer this. I'd just like like to see the process.

Tha开发者_JS百科nks.

I guess I should have mentioned that i'm writing an algorithm for scaling the y-axis of flotr graph

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The specification of the curve can be expressed as four constraints:

y(1)   = 1,     y'(1)   = 1      => tangent is (y=x) when x=1
y(365) = 1,     y'(365) = 0      => tangent is (y=1) when x=365

We therefore need a family of curves with at least four degrees of freedom to match these constraints; the simplest type of polynomial is a cubic,

y  = a*x^3 + b*x^2 + c*x + d
y' = 3*a*x^2 + 2*b*x + c

and the constraints give the following equations for the parameters:

a + b + c + d = 1
3*a + 2*b + c = 1
48627125*a + 133225*b + 365*c + d = 1
399675*a + 730*b + c = 0

I'm too old and too lazy to solve these myself, so I googled a linear equation solver to give the answer:

a = 1/132496, b = -731/132496, c = 133955/132496, d  = -729/132496

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I will post this type of question in mathoverflow.net next time. thanks

my solution in javascript was to adapt the equation of a circle:

        var radius = Math.pow((2*Math.pow(365, 2)), 1/2);
        var t = 365; //offset
        this.tMax = (Math.pow(Math.pow(r, 2) - Math.pow(x, 2), 1/2) - t) * (t / (r - t)) + 1;

the above equation has the above specified asymptotes. it is part of a step polynomial for scaling an axis for a flotr graph.

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well, you are missing data (you need another point to determine the polynomial)

a*(x-1)^2+b*(x-1)+c=y-1
a*(x-365)^2+b*(x-365)+c=y-1

you can solve the exact answer for b but A depends on C (or vv)

and your question is off topic anyways, and you need to revise your algebra