Newtons form \ \

The coordinates are (-4, 3), (0, 0), (2, -6), (5, 5) \ \

Write the n = 4 points (-4, 3), (0, 0), (2, -6), (5, 5) we can write \ \

P(x) = C0  + C1 (x + 4) + C2 (x + 4)(x) +  C3 (x + 4)(x)(x - 2) ---> (1) \ \

This polynomial must go through the four points, so \ \

C0 = 3 \ \

C0 + C1(0 + 4) = 0 \ \

C0 + C1(2 + 4) + C2(2 + 4)(2) = -6 \ \

C0 + C1(5 + 4)+C2(5 + 4)(5) + C3(5 + 4)(5)(5 - 2) = 5 \ \

C0 = 3 \ \

 4C1 = -3 \ \

C1 = -3/4 \ \

C0 + 6 C1 +12 C2 = -6 \ \

3 + 6(-3/4) + 12(C2) = -6 \ \

12(C2) = -6 -3 + 9/2 \ \

12C2 = -9 + 4.5 \ \

12C2 = -4.5 \ \

C2 = -4.5/12 \ \

C2 = - 0.375 \ \

C0 + C1(5 + 4)+C2(5 + 4)(5) + C3(5 + 4)(5)(5 - 2) = 5 \ \

3 + 9C1 + 45C2 + 165C3 = 5 \ \

3 +9(-3/4) + 45(-0.375) + 165C3 = 5 \ \

3 - 6.75 - 16.875 + 165C3 = 5 \ \

-20.625 + 165C3 = 5 \ \

165C3 = 5 + 20.625 \ \

C3 = 25.625/165 \ \

C3 = 0.15 \ \

MW = b \ \

W = [C0 C1 C2 C3] \ \

Substitute the C0, C1, C2, C3 values in equation (1) \ \

P(x) = C0  + C1 (x + 4) + C2 (x + 4)(x) +  C3 (x + 4)(x )(x - 2) \ \

P(x) = 3 + (-3/4) (x + 4) + (-0.375) (x^2 + 4x) +(0.15) (x^3 + 2 x^2 - 8x) \ \

P(x) = 3 - 0.75 x - 3 -0.375 x^2 - 1.5x + 0.15x^3 + 0.3x^2 - 1.2x \ \

P(x) = 0.15x ^3 - 0.075x ^2 - 3.45x + 3 \ \

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