10. Given the quadratic function $y=a x^{2}+b x+c$ (where $a$ is a positive integer) whose graph passes through the points $A(-1,4)$ and $B(2,1)$, and intersects the $x$-axis at two distinct points. Then the maximum value of $b+c$ is $\qquad$ .
10. -4 .

From the given, we have $\left\{\begin{array}{l}a-b+c=4, \\ 4 a+2 b+c=1,\end{array}\right.$ solving this yields $\left\{\begin{array}{l}b=-a-1, \\ c=3-2 a .\end{array}\right.$
Since the graph of the quadratic function intersects the $x$-axis at two distinct points, we have,
$$
\begin{array}{l}
\Delta=b^{2}-4 a c>0, \\
(-a-1)^{2}-4 a(3-2 a)>0,
\end{array}
$$

which simplifies to $(9 a-1)(a-1)>0$.
Since $a$ is a positive integer, we have $a>1$, so $a \geqslant 2$.
Since $b+c=-3 a+2 \leqslant-4$, and when $a=2, b=$ $-3, c=-1$, the conditions are satisfied, hence the maximum value of $b+c$ is -4.