Question

A third degree polynomial passes through (0,1),(1,-1),(2,-1),(3,2). The value at
x=4 is

Answer 1

Answer:

InterpolatingPolynomial finds the lowest degree polynomial fitting the points. For three points this is a second degree polynomial.

f[x_] = InterpolatingPolynomial[pts, x] (* (1 + x - xi) (1 - x + xi) *)

Step-by-step explanation:

To find a higher degree polynomial add additional points at arbitrary unique locations.

poly[degree_Integer?(# > 1 &)][x_] := InterpolatingPolynomial[ Join[pts, {xa[#],

ya[#]} & /@ Range[degree - 2]], x] // FullSimplify

Verifying that poly of degree 2 is identical to f

f[x] === poly[2][x] (* True *)

The third degree polynomial is

poly[3][x] (* (1 + x - xi) (1 + (x - xi) (-1 + ((-1 + x - xi) (1 + ( 1 + ya[1]/(-1 - xi + xa[1]))/(-xi + xa[1])))/(1 - xi + xa[1]))) *)

Verifying that this polynomial passes through the original points

pts === ({#, poly[3][#]} & /@ {xi - 1, xi, xi + 1}) (* True *)

I think so

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