I see you have attached a .xlsx file.
xy = [-0.020830515 13256.65
x = xy(:,1);
y = xy(:,2);
I can plot them. However, as I do, I see they do NOT look like a quadratic polynomial. That is simply not a quadratic I see plotted.
I then used the basic fitting tools to plot a quadratic polynomial fit. As you see, I was correct. So asking for polyfit to produce THE quadratic polynomial exact fit is something that simply makes no sense. Sorry, but a basic quadratic will not fit those points exactly. It simply does not have the correct shape to do so. How you generated the points isan unknown to us. But it was not a basic quadratic polynomial, and trying to force it into that mold will always fail.
If I had to make a wild guess, I might wonder if it is possible that you have a hyperbolic relationship, which perhaps you have confused with a quadratic. After all, they are both conic sections.
ft = fittype('a + b./(1 + c*x)')
ft(a,b,c,x) = a + b./(1 + c*x)
mdl = fit(x,y,ft)
Warning: Start point not provided, choosing random start point.
> In curvefit.attention.Warning/throw (line 30)
In fit>iFit (line 299)
In fit (line 108)
mdl(x) = a + b./(1 + c*x)
Coefficients (with 95
a = 9416 (9265, 9568)
b = 17.07 (13.39, 20.74)
c = 47.78 (47.73, 47.82)
Note that this model is not in fact exact. However, it is far closer to a good fit than a quadratic polynomial.