# Curve Fitting

• Nov 23rd 2010, 02:24 PM
dwsmith
Curve Fitting
If I have a set of data points, how can solve the curvelinear best fit equation for the Laffer Curve, -ax^2+bx=0?

Dustin.
• Nov 23rd 2010, 02:43 PM
pickslides
I'm not sure exactly with you're asking Dustbin

you have the data points and you want to find a model i.e. \$\displaystyle \displaystyle f(x)=-ax^2+bx \$ for these points or you want to solve \$\displaystyle \displaystyle -ax^2+bx=0\$ once you have the model?
• Nov 23rd 2010, 02:45 PM
dwsmith
I have a set of data points tax and rev/pop. I want to run a curvelinear regression to obtain the function and find the rev maximizing rate.
• Nov 23rd 2010, 02:58 PM
pickslides
Have you tried to fit a least squares on your line?

If not maybe find a line of best fit on \$\displaystyle \displaystyle f(x)=-ax^2+bx\$ using two points from your data set.
• Nov 23rd 2010, 03:01 PM
dwsmith
How do I do that for a quadratic?
• Nov 23rd 2010, 03:10 PM
pickslides
Which one, the least squares or a simple line of best fit?
• Nov 23rd 2010, 03:12 PM
dwsmith
Doesn't least squares generate the polynomial of best fit?
• Nov 23rd 2010, 03:22 PM
pickslides
It does, but it may give you one in the form of \$\displaystyle y=ax^2+bx+c\$ instead of a laffer curve i.e \$\displaystyle y=ax^2+bx\$

Therefore if tyou force the model to be \$\displaystyle y=ax^2+bx\$ and solve for a and b using two reliable points you may get a better fit.

Regression is a dark beast.
• Nov 23rd 2010, 03:27 PM
dwsmith
The Laffer Curve is forced through the origin.
• Nov 23rd 2010, 03:31 PM
pickslides
I am aware of this hence why we don't want a curve that has a nonzero y-intercept.

Maybe you can post your dataset?
• Nov 23rd 2010, 03:39 PM
dwsmith
Alabama:
Rate: 5, 5, 5, 5, 5, 5, 5, 5, 5, 6.5, 6.5, 6.5, 6.5, 6.5
Rev/Pop: 76.44097206, 60.17941474, 50.72157557, 51.45070669, 54.28532149,50.89583831, 51.39448729, 41.88461272, 64.81533834, 47.43814622, 55.25741448, 72.27369959, 97.92370105, 85.43364431
• Nov 23rd 2010, 03:55 PM
pickslides
Sorry Dustbin, I have plotted that dataset and as you probably know yourself it is horrible. Is there any reason you want to fit a parabola to that?