Let $p_2(x)$ be the interpolating polynomial for the data $(0 , 0) , (0.5 , y) , (1,3)$ from Lagrange formula. The coefficient of $x^2$ in $p_2(x)$ is $-2$ , Find the value of $y$ .
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$\begingroup$We are given the three points:
$$(x_0, y_0) = (0, 0), (x_1, y_1) = \left( \dfrac 12, y \right), (x_2, y_2) = (1, 3)$$
The quadratic polynomial passing through these points can be expressed as:
$$P_2(x) = y_0 L_0(x) + y_1 L_1(x) + y_2 L_2(x)$$
where:
- $L_0(x) = \dfrac{(x − x_1)(x − x_2)}{(x_0 − x_1)(x_0 − x_2)} = 1 - 3x + 2x^2$
- $L_1(x) = \dfrac{(x − x_0)(x − x_2)}{(x_1 − x_0)(x_1 − x_2)} = 4x - 4 x^2$
- $L_2(x) = \dfrac{(x − x_0)(x − x_1)}{(x_2 − x_0)(x_2 − x_1)} = -x + 2x^2$
This yields:
$$P_2(x) = y_0 L_0(x) + y_1 L_1(x) + y_2 L_2(x) = 0 + y(4x - 4x^2) + 3(-x+2x^2)$$
Thus
$$P_2(x) = 4 x y - 4x^2 y - 3x + 6x^2$$
We are also told the coefficient of the $x^2$ term in $P_2(x) = -2$ and asked to find $y$.
Equating the $x^2$ terms, we have:
$$-4x^2 y + 6x^2 = -2x^2 \implies -4y + 6 = - 2 \implies y = 2$$
Therefore:
$$P_2(x) = -2x^2 + 5x$$
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