Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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)

Thanks!

share|improve this question
    
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
add comment

1 Answer

up vote 2 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.

share|improve this answer
    
Great, thanks!! –  MathsStudent May 10 '11 at 10:56
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.