Let $P_2$ be the space of polynomials of degree at most 2. An inner product on this space is defined by the formula: $(p, q) = \frac{1}{8} \int_{-1}^{1} p(x) q(x) dx$ We need to find a polynomial $h(x) = h_0 + h_1 x + h_2 x^2 \in P_2$ such that $(p, h) = p(2)$ for every $p(x) \in P_2$. We need to find the coefficients $h_0, h_1, h_2$.

1. Problem Description

Let $P_2$ be the space of polynomials of degree at most

2. An inner product on this space is defined by the formula:

$(p, q) = \frac{1}{8} \int_{-1}^{1} p(x) q(x) dx$

We need to find a polynomial $h(x) = h_0 + h_1 x + h_2 x^2 \in P_2$ such that $(p, h) = p(2)$ for every $p(x) \in P_2$. We need to find the coefficients $h_0, h_1, h_2$.

2. Solution Steps

Since $(p, h) = p(2)$ for every $p \in P_2$, we can choose some specific polynomials $p(x)$ to obtain equations for $h_0, h_1, h_2$. We'll choose $p(x) = 1, p(x) = x, p(x) = x^2$.

Case 1: $p(x) = 1$.

Then $p(2) = 1$. Also, $(1, h) = \frac{1}{8} \int_{-1}^{1} (h_0 + h_1 x + h_2 x^2) dx = \frac{1}{8} [h_0 x + \frac{h_1 x^2}{2} + \frac{h_2 x^3}{3}]_{-1}^{1} = \frac{1}{8} (2h_0 + \frac{2h_2}{3}) = \frac{h_0}{4} + \frac{h_2}{12}$.

So, $\frac{h_0}{4} + \frac{h_2}{12} = 1$. This implies $3h_0 + h_2 = 12$.

Case 2: $p(x) = x$.

Then $p(2) = 2$. Also, $(x, h) = \frac{1}{8} \int_{-1}^{1} x(h_0 + h_1 x + h_2 x^2) dx = \frac{1}{8} \int_{-1}^{1} (h_0 x + h_1 x^2 + h_2 x^3) dx = \frac{1}{8} [\frac{h_0 x^2}{2} + \frac{h_1 x^3}{3} + \frac{h_2 x^4}{4}]_{-1}^{1} = \frac{1}{8} (\frac{2h_1}{3}) = \frac{h_1}{12}$.

So, $\frac{h_1}{12} = 2$. This implies $h_1 = 24$.

Case 3: $p(x) = x^2$.

Then $p(2) = 4$. Also, $(x^2, h) = \frac{1}{8} \int_{-1}^{1} x^2 (h_0 + h_1 x + h_2 x^2) dx = \frac{1}{8} \int_{-1}^{1} (h_0 x^2 + h_1 x^3 + h_2 x^4) dx = \frac{1}{8} [\frac{h_0 x^3}{3} + \frac{h_1 x^4}{4} + \frac{h_2 x^5}{5}]_{-1}^{1} = \frac{1}{8} (\frac{2h_0}{3} + \frac{2h_2}{5}) = \frac{h_0}{12} + \frac{h_2}{20}$.

So, $\frac{h_0}{12} + \frac{h_2}{20} = 4$. This implies $5h_0 + 3h_2 = 240$.

Now we have a system of equations:

$3h_0 + h_2 = 12$

$h_1 = 24$

$5h_0 + 3h_2 = 240$

Multiply the first equation by 3: $9h_0 + 3h_2 = 36$.

Subtract this from the third equation: $(5h_0 + 3h_2) - (9h_0 + 3h_2) = 240 - 36$, so $-4h_0 = 204$.

Thus, $h_0 = -51$.

Substitute $h_0 = -51$ into $3h_0 + h_2 = 12$: $3(-51) + h_2 = 12$, so $-153 + h_2 = 12$.

Thus, $h_2 = 165$.

Therefore, $h(x) = -51 + 24x + 165x^2$.

3. Final Answer

$h_0 = -51$, $h_1 = 24$, $h_2 = 165$

$h(x) = -51 + 24x + 165x^2$