(I assume you intended to write 1/y, and not I/y, where the variable I is not defined.)
y = [1.6E5 2.5E5 4.1E5 8E5 1E6 2E6 7E6 2E7];
x = [17 20 27 59 62 81 89 95];
Often it is the case, when you cannot fit a model to your data, it means you model does not have the correct shape. This model would be a nice STRAIGHT line, when we plot it as as 1/y versus x.
If we plot 1/y versus x, we would expect to see that behavior in the data. And here, it looks like we have not at all a straight line.
plot(x,1./y,'o')
I might also point out that since your data is scaled to go from 0-100 in x, so the model you wanted to write was probably
I/y = x/a + (100-x)/b.
How can we estimate a and b in that model? Easily enough. We could even use polyfit. There is no need to use a custom model in fit. But you can. Next, note that I'll fit the model as
1/y = x*a + (100-x)*b
and then invert a and b. fit manages to make it converge better if I do so.
mdl = fittype('x*a + (100 - x)*b','indep','x')
fittedmdl = fit(x(:),1./y(:),mdl,'start',[0.001 0.001]) % note that fit wants COLUMN vectors of data.
plot(fittedmdl,x,1./y)
a = 1./fittedmdl.a
b = 1./fittedmdl.b
Again though, I don't think your model fits the data very well.
