4. A third-degree polynomial passes through the points (0,-12), (1,0) ,(3,6) and (4,12).
a). Determine this polynomial using Lagrange's interpolation formula
Lagrange's interpolation formula states that for a polynomial of degree n that passes through n+1 distinct points, the polynomial can be written as:
P(x) = Σ [y_i * l_i(x)]
where P(x) is the polynomial, y_i are the y-values of the points, l_i(x) are the Lagrange basis polynomials.
The Lagrange basis polynomials are given by:
l_i(x) = Π [ (x - x_j) / (x_i - x_j) ] for i ≠ j
where i and j range from 0 to n.
Let's use Lagrange's interpolation formula to determine the polynomial that passes through the points (0,-12), (1,0), (3,6), and (4,12).
First, let's define the Lagrange basis polynomials:
l_0(x) = ((x - 1) * (x - 3) * (x - 4)) / ((0 - 1) * (0 - 3) * (0 - 4)) = (x^3 - 8x^2 + 19x - 12) / 6
l_1(x) = ((x - 0) * (x - 3) * (x - 4)) / ((1 - 0) * (1 - 3) * (1 - 4)) = -(x^3 - 7x^2 + 12x) / 2
l_2(x) = ((x - 0) * (x - 1) * (x - 4)) / ((3 - 0) * (3 - 1) * (3 - 4)) = (x^3 - 5x^2 - 4x) / 2
l_3(x) = ((x - 0) * (x - 1) * (x - 3)) / ((4 - 0) * (4 - 1) * (4 - 3)) = -(x^3 - 4x^2 - 3x) / 6
Now, let's substitute these basis polynomials and the y-values of the points into the Lagrange's interpolation formula:
P(x) = (-12 * l_0(x)) + (0 * l_1(x)) + (6 * l_2(x)) + (12 * l_3(x))
P(x) = (-12 * ((x^3 - 8x^2 + 19x - 12) / 6)) + (0 * (-(x^3 - 7x^2 + 12x) / 2)) + (6 * ((x^3 - 5x^2 - 4x) / 2)) + (12 * (-(x^3 - 4x^2 - 3x) / 6))
Simplifying the expression:
P(x) = -2x^3 + 9x^2 - 10x - 12
Therefore, the polynomial that passes through the points (0,-12), (1,0), (3,6), and (4,12) is:
P(x) = -2x^3 + 9x^2 - 10x - 12