**#2** by **Thommo** » Apr 11, 2017 4:55 pm

There are always infinitely many curves that link any finite set of points. Any set of 2 points will lie on exactly one polynomial of degree 1 and infinitely many polynomials of higher degree. Any set of 3 points will lie on exactly one polynomial of degree 2 and infinitely many polynomials of higher degree. And so on.

You can easily find these solutions by solving simultaneous equations for a polynomial of suitable degree.

There will also be infinitely many other solutions for various curve types and sets of points, including many smooth curves or functions (e.g. fitting logarithmic or exponential curves to sets of two points is often easy).

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However, I've written that because it's interesting, I'm not sure it's what you want.

You've plotted a few examples of a family of curves, which have the function f(x)=(x+k)(x+2k)(x+2.5k).

Expanding that out: f(x) = x3 + 5.5kx2 + 9.5k2x + 5k3

Differentiate to find the turning points: f'(x) = 3x2 + 11kx + 9.5k2

Find the solutions by putting that into the quadratic formula with a = 3, b = 11k, c = 9.5k2 gives two solutions of x = (-11k + SQRT(7)k)/6 and (-11k - SQRT(7)k)/6 we can note that you've simply plotted the turning points stemming from the second solution by inspection.

This allows us to relate x and k by the formula x = (-11 - SQRT(7))k/6 or x = -2.274k (taking k to 3 d.p). Or equivalently k = -x/2.274.

Finally we can substitute that back into f(x) to get the curve of the solution family as g(x) = (0.560x)(0.121x)(-0.099x) = -0.006706415x3.

To answer the follow-on question we just do the same for the other solution family x = (-11k + SQRT(7)k)/6 = -1.392k, giving k = 0.718x. Substituting that back in to f(x) gives h(x) = (0.282x)(-0.436x)(-0.795x) = 0.09782757x3.

Last edited by

Thommo on Apr 11, 2017 5:29 pm, edited 2 times in total.