Given $n+1$ points in the plane, Lagrangian interpolation gives the unique polynomial of degree less or equal to $n$.
From the examples I've seen so far, interpolation polynomials all have degree exactly $n$ but I don't know if there might be a set of points for which the interpolation gives a polynomial of degree strictly less than $n$.
Many thanks for your answers.
You get to choose the order of the polynomial you want to use. If you use degree $n$, you get an exact interpolating polynomial that will go through every point. Problem is, it might oscillate quite a bit, because essentially you're overfitting a model to your data (at least, quite often that is the case). If the data doesn't warrant such a high-order polynomial, consider using linear regression to fit a lower-order polynomial (this will no longer be a Lagrange interpolating polynomial, but it might be much more useful) to your data.
You can definitely get polynomials of degree less than $n$: if some of your data points are essentially "redundant". So, for example, if you choose the points $(-2,4), (-1,1), (0,0), (1,1),$ and $(2,4)$, they're all on $y=x^2$.
