b) Using Lagrange's formula for interpolation find the value of \( y \) when \( x=2 \) from the ( \( ) \) following table \[ \begin{array}{lrrrr} x: & -2 & -1 & 0 & 4 \\ y: & -2 & 4 & 1 & 8 \end{array} \]

To find the value of \( y \) when \( x = 2 \) using Lagrange's interpolation formula, we start with the given points: \((-2, -2)\), \((-1, 4)\), \((0, 1)\), and \((4, 8)\). The formula is given by: \[ P(x) = \sum_{i=0}^{n} y_i L_i(x) \] where \(L_i(x)\) is the Lagrange basis polynomial defined as: \[ L_i(x) = \prod_{j=0, j \neq i}^{n} \frac{x - x_j}{x_i - x_j} \] For our interpolation, we have \(n = 3\) (since there are 4 points). Let's calculate: 1. **For \(L_0(x)\)** (\(y = -2\) when \(x = -2\)): \[ L_0(x) = \frac{(x + 1)(x)(x - 4)}{(-2 + 1)(-2 - 0)(-2 - 4)} = \frac{(x + 1)x(x - 4)}{(-1)(-2)(-6)} = -\frac{(x + 1)x(x - 4)}{12} \] 2. **For \(L_1(x)\)** (\(y = 4\) when \(x = -1\)): \[ L_1(x) = \frac{(x + 2)(x)(x - 4)}{(-1 + 2)(-1 - 0)(-1 - 4)} = \frac{(x + 2)x(x - 4)}{(1)(-1)(-5)} = \frac{(x + 2)x(x - 4)}{5} \] 3. **For \(L_2(x)\)** (\(y = 1\) when \(x = 0\)): \[ L_2(x) = \frac{(x + 2)(x + 1)(x - 4)}{(0 + 2)(0 + 1)(0 - 4)} = \frac{(x + 2)(x + 1)(x - 4)}{(2)(1)(-4)} = -\frac{(x + 2)(x + 1)(x - 4)}{8} \] 4. **For \(L_3(x)\)** (\(y = 8\) when \(x = 4\)): \[ L_3(x) = \frac{(x + 2)(x + 1)(x)}{(4 + 2)(4 + 1)(4 - 0)} = \frac{(x + 2)(x + 1)(x)}{(6)(5)(4)} = \frac{(x + 2)(x + 1)(x)}{120} \] Now, plug \(x = 2\) into the Lagrange polynomial: \[ P(2) = (-2) L_0(2) + 4 L_1(2) + 1 L_2(2) + 8 L_3(2) \] Calculating each \(L\): - \(L_0(2) = -\frac{(2 + 1)(2)(2 - 4)}{12} = -\frac{3 \cdot 2 \cdot (-2)}{12} = 1\) - \(L_1(2) = \frac{(2 + 2)(2)(2 - 4)}{5} = \frac{4 \cdot 2 \cdot (-2)}{5} = -\frac{16}{5}\) - \(L_2(2) = -\frac{(2 + 2)(2 + 1)(2 - 4)}{8} = -\frac{4 \cdot 3 \cdot (-2)}{8} = 3\) - \(L_3(2) = \frac{(2 + 2)(2 + 1)(2)}{120} = \frac{4 \cdot 3 \cdot 2}{120} = \frac{24}{120} = \frac{1}{5}\) Substituting back: \[ P(2) = (-2)(1) + 4(-\frac{16}{5}) + 1(3) + 8(\frac{1}{5}) \] \[ = -2 - \frac{