2. The graph of the quadratic function $f(x) = a x^{2} + b x + c$ is shown in Figure 1. Let
$$
\begin{array}{c}
N=|a+b+c| \\
+|2 a-b|, \\
M=|a-b+c| \\
+|2 a+b| \text {. Then ( ). }
\end{array}
$$
(A) $M>N$
(B) $M=N$
(C) $M<N$
(D) The relationship between $M$ and $N$ cannot be determined
$$
\begin{array}{l}
\text { 2. (C). } \\
\text { As shown in Figure 1, } f(1)=a+b+c0, a>0,-\frac{b}{2 a}>1 \text {. }
\end{array}
$$
2. (C).

As shown in Figure 1, $f(1) = a + b + c0, a > 0, -\frac{b}{2a} > 1$.
Thus, $b < 0$.
Also, $f(0) = c0$, hence
$M - N = |a - b + c| + |2a + b| - |a + b + c| - |2a - b| = (a - b + c) + (a + b + c) - (2a + b) - (2a - b) = -2(a - c) < 0$.
$\therefore M < N$.