Question

Given the following data: x: (-2, 1, 2), y: (28, 4, 16). Let \(P_2(x)\) be the quadratic interpolating polynomial passing through the above three points. Find the value of \(P_2(0)\).

Hint:

When asked to find the value of an interpolating polynomial at \(x=0\), you are essentially looking for the constant term (\(c\) in \(ax^2+bx+c\)). Setting up and solving the system of equations for \(c\) is often the most direct method.

Answer 1

Step 1: Understanding the Question:
We need to find the value of a quadratic polynomial at \(x=0\). The polynomial is defined by three points it passes through: \((-2, 28)\), \((1, 4)\), and \((2, 16)\).
Step 2: Key Formula or Approach:
We can use Lagrange's interpolation formula or set up a system of equations for a general quadratic \(P_2(x) = ax^2 + bx + c\). Since we need to find \(P_2(0)\), which is simply the constant term \(c\), solving the system of equations is a direct approach.
Step 3: Detailed Calculation using System of Equations:
Let the polynomial be \(P_2(x) = ax^2 + bx + c\). We substitute the three given points into this equation:

1. For point \((-2, 28)\): \(a(-2)^2 + b(-2) + c = 28 \implies 4a - 2b + c = 28\)
2. For point \((1, 4)\): \(a(1)^2 + b(1) + c = 4 \implies a + b + c = 4\)
3. For point \((2, 16)\): \(a(2)^2 + b(2) + c = 16 \implies 4a + 2b + c = 16\)

We now have a system of three linear equations. We need to solve for \(c\). Subtract Equation (1) from Equation (3): \[ (4a + 2b + c) - (4a - 2b + c) = 16 - 28 \] \[ 4b = -12 \implies b = -3 \] Now substitute \(b=-3\) into Equation (2): \[ a + (-3) + c = 4 \implies a + c = 7 \implies a = 7 - c \] Substitute \(b=-3\) and \(a = 7-c\) into Equation (3): \[ 4(7-c) + 2(-3) + c = 16 \] \[ 28 - 4c - 6 + c = 16 \] \[ 22 - 3c = 16 \] \[ 3c = 22 - 16 \] \[ 3c = 6 \implies c = 2 \] Step 4: Final Answer:
The value of the polynomial at \(x=0\) is \(P_2(0) = a(0)^2 + b(0) + c = c\). Therefore, \(P_2(0) = 2\).