5. As shown in Figure 2, the line $x$ $=1$ is the axis of symmetry of the graph of the quadratic function
$$
y=a x^{2}+b x+c
$$

Then we have ( ).
(A) $a+b+c>0$
(B) $b>a+c$
(C) $c>2 b$
(D) $a b c<0$
5.C.

Given $a>0, -\frac{b}{2a}=1$, we have $b<0$, $f(1)=a+b+c=b-a=b+\frac{b}{2}>2b$, hence the answer is (C).