5. Given the quadratic function $f(x)=a x^{2}+b x+c$ whose graph is shown in Figure 2, let
$$
\begin{array}{l}
p=|a-b+c|+|2 a+b|, \\
q=|a+b+c|+|2 a-b| .
\end{array}
$$

Then ().
(A) $p>q$
(B) $p=q$
(C) $p<q$
(D) The relationship between $p$ and $q$ cannot be determined
5.C.

From the graph, we know $f(0)=0=c, f(1)=a+b+c>0$. Thus, $a+b>0, f(-1)=a-b+c=a-b$. Therefore,
$$
\left.\frac{b}{2 a}0\right\}.
$$
And $2 a-b=a+(a-b)<0$, so
$$
\begin{array}{l}
p=-(a-b)+2 a+b=a+2 b, \\
q=a+b-(2 a-b)=2 b-a, \\
p-q=(a+2 b)-(2 b-a)=2 a<0 .
\end{array}
$$