Author 
Comment/Response 
Bill Simpson

11/20/12 7:03pm
Mathematica has never shown the steps to calculate a result, often even the algorithm used is undocumented.
Sometimes you can write a simple calculation yourself and learn something in the process.
From http://mathworld.wolfram.com/LagrangeInterpolatingPolynomial.html P(x) a Lagrange polynomial is a sum of product/product.
If I haven't made any mistakes then this is one way of calculating that
In[1]:= p[points_]:=Sum[Product[If[i==j,1,xpoints[[i,1]]],{i,1,Length[points]}]/ Product[If[i==j,1,points[[i,1]]points[[j,1]]],i,1,Length[points]}] *points[[j,2]], {j,1,Length[points]}];
poly=p[{{1,1},{2,4},{3,2}}]
Out[2]=
1/2(3+x)(2+x)4(3+x)(1+x)+(2+x)(1+x)
In[3]:= {x,poly}/.{{x>1},{x>2},{x>3}}
Out[3]= {{1,1},{2,4},{3,2}}
And so the same input x values create the same output y values.
You can also Expand that polynomial to get it into a more usual form if you like.
Please check this carefully on different sets of data until you are fairly sure I didn't make any misatkes. You can also compare that definition of p with the example on the web page and try to understand how the two are related.
URL: , 
