1. For all real numbers $x$, the value of all quadratic functions $f(x)=a x^{2}+b x+c(a<b)$ is always non-negative. Then the minimum value of $M=\frac{a+b+c}{b-a}$ is ( ).
(A) $\frac{1}{2}$
(B) $\frac{1}{3}$
(C) 2
(D) 3
-1. (D).
From the given conditions, we know $a>0$, and $b^{2}-4 a c \leqslant 0$, which means $c \geqslant \frac{b^{2}}{4 a}$. Let $b-a=t(t>0)$, then
$$
\begin{aligned}
M & \geqslant \frac{a+(t+a)+\frac{(a+t)^{2}}{4 a}}{t} \\
& =\frac{(3 a+t)^{2}}{4 a t} \geqslant \frac{(2 \sqrt{3 a t})^{2}}{4 a t}=3 .
\end{aligned}
$$

The equality holds if and only if $3 a=t, c=\frac{b^{2}}{4 a}$, that is, $b=c=4 a$. Therefore, $M_{\text {min }}=3$.