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: Set up the interpolation data.
We are given three points: \((x_0,y_0)=(-2,28)\), \((x_1,y_1)=(1,4)\), \((x_2,y_2)=(2,16)\). Since there are three points, a unique second-degree (quadratic) polynomial \(P_2(x)\) passes through all of them. We will use Lagrange's interpolation formula to build \(P_2(x)\) directly and then evaluate it at \(x=0\).

Step 2: Recall Lagrange's interpolation formula.
\[ P_2(x) = y_0 L_0(x) + y_1 L_1(x) + y_2 L_2(x) \]
where each \(L_i(x)\) equals 1 at \(x=x_i\) and 0 at the other nodes:
\[ L_0(x) = \frac{(x-x_1)(x-x_2)}{(x_0-x_1)(x_0-x_2)}, \quad L_1(x) = \frac{(x-x_0)(x-x_2)}{(x_1-x_0)(x_1-x_2)}, \quad L_2(x) = \frac{(x-x_0)(x-x_1)}{(x_2-x_0)(x_2-x_1)} \]

Step 3: Evaluate each \(L_i\) directly at \(x=0\), since we only need \(P_2(0)\).
\[ L_0(0) = \frac{(0-1)(0-2)}{(-2-1)(-2-2)} = \frac{(-1)(-2)}{(-3)(-4)} = \frac{2}{12} = \frac{1}{6} \]
\[ L_1(0) = \frac{(0-(-2))(0-2)}{(1-(-2))(1-2)} = \frac{(2)(-2)}{(3)(-1)} = \frac{-4}{-3} = \frac{4}{3} \]
\[ L_2(0) = \frac{(0-(-2))(0-1)}{(2-(-2))(2-1)} = \frac{(2)(-1)}{(4)(1)} = \frac{-2}{4} = -\frac{1}{2} \]

Step 4: Combine using the given \(y\) values.
\[ P_2(0) = 28 \times \frac{1}{6} + 4 \times \frac{4}{3} + 16 \times \left(-\frac{1}{2}\right) \]
\[ P_2(0) = \frac{28}{6} + \frac{16}{3} - 8 = \frac{14}{3} + \frac{16}{3} - 8 = \frac{30}{3} - 8 = 10 - 8 \]
\[ P_2(0) = 2 \]

Step 5: Sanity check.
The value 2 lies between the given \(y\) values in a way consistent with the data dropping sharply from \(x=-2\) to \(x=1\) before rising again to \(x=2\), so a moderate value like 2 at \(x=0\) (between the low point near \(x=1\) and the higher point at \(x=-2\)) is reasonable.

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