Example 16 (2002 Hunan Province Competition Question) The graph of the quadratic function $f(x)=a x^{2}+b x+c$ is shown in Figure 5-16. Let
$$
\begin{array}{l}
N=|a+b+c|+|2 a-b|, \\
M=|a-b+c|+|2 a+b| . \text { Then ( ). }
\end{array}
$$
A. $M>N$
B. $M=N$
C. $M<N$
D. The relationship between $M$ and $N$ cannot be determined
$a>0, -\frac{b}{2a}>1$.
Thus $b<0$.
Also, $f(0)=c<0$, hence
$M-N=|a-b+c|+|2a+b|-|a+b+c|-|2a-b|=(a-b+c)+(2a+b)-(a+b+c)-(2a-b)=-2(a-c)<0$.
Therefore, $M<N$.