Question

Values of \(y\) for different values of \(x\) are tabulated below.

\(x\)	\(-2\)	\(1\)	\(2\)
\(y\)	\(28\)	\(4\)	\(16\)

If a second-degree interpolating polynomial \(P_2(x)\) is used to represent \(y\), the value of \(P_2(0)\) is (rounded off to the nearest integer).

Hint:

Build the unique quadratic through the three points (Lagrange or Newton's divided differences) and evaluate it at \(x=0\).

Answer 1

Correct Answer: 2

Step 1: Build a divided-difference table instead of using Lagrange's formula directly.
Order the nodes as $x_0=-2, y_0=28$; $x_1=1, y_1=4$; $x_2=2, y_2=16$. Newton's divided difference method builds the interpolating polynomial from successive differences of the data instead of from Lagrange weight functions.

Step 2: First-order divided differences.
\[ f[x_0,x_1] = \frac{y_1-y_0}{x_1-x_0} = \frac{4-28}{1-(-2)} = \frac{-24}{3} = -8 \]
\[ f[x_1,x_2] = \frac{y_2-y_1}{x_2-x_1} = \frac{16-4}{2-1} = \frac{12}{1} = 12 \]

Step 3: Second-order divided difference.
\[ f[x_0,x_1,x_2] = \frac{f[x_1,x_2]-f[x_0,x_1]}{x_2-x_0} = \frac{12-(-8)}{2-(-2)} = \frac{20}{4} = 5 \]

Step 4: Write Newton's divided-difference interpolating polynomial.
\[ P_2(x) = y_0 + f[x_0,x_1](x-x_0) + f[x_0,x_1,x_2](x-x_0)(x-x_1) \]
\[ P_2(x) = 28 + (-8)(x-(-2)) + (5)(x-(-2))(x-1) \]
\[ P_2(x) = 28 - 8(x+2) + 5(x+2)(x-1) \]

Step 5: Evaluate at $x=0$.
\[ P_2(0) = 28 - 8(0+2) + 5(0+2)(0-1) \]
\[ P_2(0) = 28 - 8(2) + 5(2)(-1) \]
\[ P_2(0) = 28 - 16 - 10 \]
\[ P_2(0) = 2 \]
This matches the value found through Lagrange's method, confirming the interpolated value is correct, since both methods must give the same unique quadratic through three points.

Final Answer:
\[ \boxed{P_2(0) = 2} \]