What is the polynomial of smallest degree that passes through $(-2, 2), (-1, 1), (0, 2),(1,-1)$, and $(2, 10)$?
To find the polynomial of smallest degree that passes through the given points \((-2, 2)\), \((-1, 1)\), \((0, 2)\), \((1, -1)\), and \((2, 10)\), we can use Lagrange's interpolation formula. The Lagrange interpolation polynomial is given by:

\[ f(x) = \sum_{i=0}^{n} y_i \prod_{\substack{0 \le j \le n \\ j \ne i}} \frac{x - x_j}{x_i - x_j} \]

Here, \( n = 4 \) because we have 5 points. Let's denote the points as \((x_0, y_0), (x_1, y_1), (x_2, y_2), (x_3, y_3), (x_4, y_4)\) where:
\[
(x_0, y_0) = (-2, 2), \quad (x_1, y_1) = (-1, 1), \quad (x_2, y_2) = (0, 2), \quad (x_3, y_3) = (1, -1), \quad (x_4, y_4) = (2, 10)
\]

Using the Lagrange interpolation formula, we can write:

\[ f(x) = \sum_{i=0}^{4} y_i \prod_{\substack{0 \le j \le 4 \\ j \ne i}} \frac{x - x_j}{x_i - x_j} \]

Let's compute each term separately:

1. For \(i = 0\):
\[ L_0(x) = \prod_{\substack{0 \le j \le 4 \\ j \ne 0}} \frac{x - x_j}{x_0 - x_j} = \frac{(x + 1)(x)(x - 1)(x - 2)}{(-2 + 1)(-2)(-2 - 1)(-2 - 2)} = \frac{(x + 1)x(x - 1)(x - 2)}{24} \]
\[ y_0 L_0(x) = 2 \cdot \frac{(x + 1)x(x - 1)(x - 2)}{24} = \frac{(x + 1)x(x - 1)(x - 2)}{12} \]

2. For \(i = 1\):
\[ L_1(x) = \prod_{\substack{0 \le j \le 4 \\ j \ne 1}} \frac{x - x_j}{x_1 - x_j} = \frac{(x + 2)(x)(x - 1)(x - 2)}{(-1 + 2)(-1)(-1 - 1)(-1 - 2)} = \frac{(x + 2)x(x - 1)(x - 2)}{6} \]
\[ y_1 L_1(x) = 1 \cdot \frac{(x + 2)x(x - 1)(x - 2)}{6} = \frac{(x + 2)x(x - 1)(x - 2)}{6} \]

3. For \(i = 2\):
\[ L_2(x) = \prod_{\substack{0 \le j \le 4 \\ j \ne 2}} \frac{x - x_j}{x_2 - x_j} = \frac{(x + 2)(x + 1)(x - 1)(x - 2)}{(0 + 2)(0 + 1)(0 - 1)(0 - 2)} = \frac{(x + 2)(x + 1)(x - 1)(x - 2)}{2} \]
\[ y_2 L_2(x) = 2 \cdot \frac{(x + 2)(x + 1)(x - 1)(x - 2)}{2} = (x + 2)(x + 1)(x - 1)(x - 2) \]

4. For \(i = 3\):
\[ L_3(x) = \prod_{\substack{0 \le j \le 4 \\ j \ne 3}} \frac{x - x_j}{x_3 - x_j} = \frac{(x + 2)(x + 1)(x)(x - 2)}{(1 + 2)(1 + 1)(1)(1 - 2)} = \frac{(x + 2)(x + 1)x(x - 2)}{-6} \]
\[ y_3 L_3(x) = -1 \cdot \frac{(x + 2)(x + 1)x(x - 2)}{-6} = \frac{(x + 2)(x + 1)x(x - 2)}{6} \]

5. For \(i = 4\):
\[ L_4(x) = \prod_{\substack{0 \le j \le 4 \\ j \ne 4}} \frac{x - x_j}{x_4 - x_j} = \frac{(x + 2)(x + 1)(x)(x - 1)}{(2 + 2)(2 + 1)(2)(2 - 1)} = \frac{(x + 2)(x + 1)x(x - 1)}{24} \]
\[ y_4 L_4(x) = 10 \cdot \frac{(x + 2)(x + 1)x(x - 1)}{24} = \frac{10(x + 2)(x + 1)x(x - 1)}{24} = \frac{5(x + 2)(x + 1)x(x - 1)}{12} \]

Now, summing all these terms, we get:

\[ f(x) = \frac{(x + 1)x(x - 1)(x - 2)}{12} + \frac{(x + 2)x(x - 1)(x - 2)}{6} + (x + 2)(x + 1)(x - 1)(x - 2) + \frac{(x + 2)(x + 1)x(x - 2)}{6} + \frac{5(x + 2)(x + 1)x(x - 1)}{12} \]

Expanding and simplifying this polynomial, we get:

\[ f(x) = x^4 + x^3 - 3x^2 - 2x + 2 \]

The final answer is \(\boxed{x^4 + x^3 - 3x^2 - 2x + 2}\)