Values of y for different values of x are tabulated below.

If a second-degree interpolating polynomial P₂(x) is ________ used to represent y, the value of P₂(0) is (rounded off to the nearest integer).

The correct option is 2.

Step-by-Step Explanation:

1. Analysis of the Data Table from the Image:
By analyzing the table shown in the image, we can extract the given data points for the variable $x$ and its corresponding function values $y$:
• For $x_0=-2$, the corresponding value is $y_0=28$.
• For $x_1=1$, the corresponding value is $y_1=4$.
• For $x_2=2$, the corresponding value is $y_2=16$.

We need to find the value of $P_2\left(0\right)$ using a second-degree interpolating polynomial $P_2\left(x\right)$ that represents these points.

Let's solve this using two different approaches to understand the solution thoroughly.

Method 1: Lagrange Interpolation Formula
The formula for a second-degree Lagrange interpolating polynomial with three points is given by:

$$P_2\left(x\right)=y_0\frac{(x-x_1)(x-x_2)}{(x_0-x_1)(x_0-x_2)}+y_1\frac{(x-x_0)(x-x_2)}{(x_1-x_0)(x_1-x_2)}+y_2\frac{(x-x_0)(x-x_1)}{(x_2-x_0)(x_2-x_1)}$$

To find $P_2\left(0\right)$, we substitute $x=0$ along with the values of the points into the expression:

$$P_2\left(0\right)=28\frac{(0-1)(0-2)}{(-2-1)(-2-2)}+4\frac{(0-(-2\left)\right)(0-2)}{(1-(-2\left)\right)(1-2)}+16\frac{(0-(-2\left)\right)(0-1)}{(2-(-2\left)\right)(2-1)}$$

Now, simplify each term individually:

• First term:
$$28·\frac{(-1)(-2)}{(-3)(-4)}=28·\frac{2}{12}=28·\frac{1}{6}=\frac{14}{3}$$

• Second term:
$$4·\frac{\left(2\right)(-2)}{\left(3\right)(-1)}=4·\frac{-4}{-3}=\frac{16}{3}$$

• Third term:
$$16·\frac{\left(2\right)(-1)}{\left(4\right)\left(1\right)}=16·\frac{-2}{4}=-8$$

Summing these terms together gives:

$$P_2\left(0\right)=\frac{14}{3}+\frac{16}{3}-8$$

$$P_2\left(0\right)=\frac{30}{3}-8=10-8=2$$

Method 2: Solving a System of Quadratic Equations
Alternatively, let the quadratic polynomial be represented in the general form:

$$P_2\left(x\right)=a{x}^2+bx+c$$

Substitute the three data points into this general equation:

1. For $x=-2$, $y=28$:
$$4a-2b+c=28$$

2. For $x=1$, $y=4$:
$$a+b+c=4$$

3. For $x=2$, $y=16$:
$$4a+2b+c=16$$

Subtracting equation (1) from equation (3):

$$(4a+2b+c)-(4a-2b+c)=16-28$$

$$4b=-12\Rightarrow b=-3$$

Substitute $b=-3$ back into equation (2) and equation (3):

$$a-3+c=4\Rightarrow a+c=7$$

$$4a-6+c=16\Rightarrow 4a+c=22$$

Subtracting these two new equations:

$$(4a+c)-(a+c)=22-7$$

$$3a=15\Rightarrow a=5$$

Substitute $a=5$ into $a+c=7$:

$$5+c=7\Rightarrow c=2$$

Thus, the interpolating polynomial is:

$$P_2\left(x\right)=5x^2-3x+2$$

At $x=0$:

$$P_2\left(0\right)=5{\left(0\right)}^2-3\left(0\right)+2=2$$

Both methods yield the exact value of 2.