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I have 4 data points, from which I want to calculate a hyperbola. It seems that the Excel trendline feature can't do it for me, so how do I find the relationship?

The points are: (x,y)

(3, 0.008) (6, 0,006) (10, 0.003) (13, 0.002)


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Probably you wanted to fit a hyperbola, not a "hyperbolic function". Just reciprocate your x-coordinates and proceed with linear regression as usual. – J. M. May 10 '11 at 9:20
Can you explain that further? I don't really understand what you mean. – MathsStudent May 10 '11 at 9:37
Do the usual linear fit on the points $\left(\frac1{x},y\right)$... – J. M. May 10 '11 at 9:45
With this data, a straight line is in fact a far closer fit in terms of the sum of squares of residuals. – Henry May 10 '11 at 12:27
up vote 3 down vote accepted

A hyperbola takes the form $y = k \frac{1}{x}$. This may be difficult to deal with. So instead, let's consider the reciprocals of our x values as J.M. suggested. For example, instead of looking at $(2.5, 0.007713)$, we consider $(\frac{1}{2.5}, 0.007713)$. Then since we have flipped all of our x values, we are looking to fit something of the form $y = k \dfrac{1}{ \frac{1}{x} } = k x$. This can be accomplished by doing any standard linear regression technique.

This is just an extension of J.M.'s comment.

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Great, thanks!! – MathsStudent May 10 '11 at 10:56

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