91Ó°ÊÓ

The general formula for Lagrange interpolation is given by: \[P(x) = \sum_{i=0}^{n} L_i(x) \cdot y_i\] where \[L_i(x) = \prod_{j=0, j\neq i}^{n} \frac{x - x_j}{x_i - x_j}\] and \(P(x)\) is the polynomial of smallest degree that matches the given points. Now, let's find the interpolated polynomial for each set of points. (a) Given points are \((-2,-6),(-1,5),(1,3)\):

For \(i=0\), we have \(x_0=-2\) and \(y_0=-6\). Substitute into the formula for \(L_0(x)\): \[L_0(x) = \frac{x - x_1}{x_0 - x_1} \cdot \frac{x - x_2}{x_0 - x_2} = \frac{x - (-1)}{-2 - (-1)} \cdot \frac{x - 1}{-2 - 1} = -(x + 1)(x - 1)\]

For \(i=1\), we have \(x_1=-1\) and \(y_1=5\). Substitute into the formula for \(L_1(x)\): \[L_1(x) = \frac{x - x_0}{x_1 - x_0} \cdot \frac{x - x_2}{x_1 - x_2} = \frac{x - (-2)}{-1 - (-2)} \cdot \frac{x - 1}{-1 - 1} = (x + 2)(x - 1)\]

For \(i=2\), we have \(x_2=1\) and \(y_2=3\). Substitute into the formula for \(L_2(x)\): \[L_2(x) = \frac{x - x_0}{x_2 - x_0} \cdot \frac{x - x_1}{x_2 - x_1} = \frac{x - (-2)}{1 - (-2)} \cdot \frac{x - (-1)}{1 - (-1)} = \frac{1}{3}(x + 2)(x + 1)\]

Now, we can combine the terms using the general formula for Lagrange interpolation: \[ P(x) = y_0 \cdot L_0(x) + y_1 \cdot L_1(x) + y_2 \cdot L_2(x) \] \[ P(x) = -6\cdot[-(x + 1)(x - 1)]+ 5\cdot[(x + 2)(x - 1)] + 3\cdot\left[\frac{1}{3}(x + 2)(x + 1)\right] \] Simplify the expression to obtain the polynomial of smallest degree: \[ P(x) = -6x^2 + x + 5x^2 - 5x - 10 + x^2 + x \] \[ P(x) = x^2 -3x - 10 \] So, for set (a), the polynomial of smallest degree that passes through all the points is \(P(x) = x^2 -3x - 10\). (b) Given points are \((-4,24),(1,9),(3,3)\): Following the same methodology as above, we obtain: \(P(x) = -2x^2 + x + 8\) (c) Given points are \((-2,3),(-1,-6),(1,0),(3,-2)\): Following the same methodology as above, we obtain: \(P(x) = \frac{1}{12} x^3 - \frac{5}{4}x\) (d) Given points are \((-3,-30),(-2,7),(0,15),(1,10)\): Following the same methodology as above, we obtain: \(P(x) = -2x^3 - 3x^2 + 12x + 15\)