Quote:
Originally Posted by theNerd
does this solution fit if we used 2nd degree polynomials instead of cubic?
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You cannot fit a 2
nd degree polynomial to 2 arbitrary points in the XY plane if the tangents at both points are specified.
A 2
nd degree polynomial has only 3 adjustable parameters, and there are 4 equations to be satisfied:
Y
1 = f(X
1)
Y
2 = f(X
2)
Tangent1 = f
'(X
1)
Tangent2 = f
'(X
2)