11. Given the quadratic function
$$
y=a x^{2}+b x+c \geqslant 0(a<b) \text {. }
$$

Then the minimum value of $M=\frac{a+2 b+4 c}{b-a}$ is $\qquad$
11. 8.
From the conditions, it is easy to see that $a>0, b^{2}-4 a c \leqslant 0$.
Notice that
$$
\begin{array}{l}
M=\frac{a+2 b+4 c}{b-a}=\frac{a^{2}+2 a b+4 a c}{a(b-a)} \\
\geqslant \frac{a^{2}+2 a b+b^{2}}{a(b-a)} .
\end{array}
$$

Let $t=\frac{b}{a}$. Then $t>1$. Thus,
$$
\begin{array}{l}
M \geqslant \frac{a^{2}+2 a b+b^{2}}{a(b-a)}=\frac{t^{2}+2 t+1}{t-1} \\
=(t-1)+\frac{4}{t-1}+4 \\
\geqslant 2 \sqrt{4}+4=8 .
\end{array}
$$

The equality holds if and only if $t=3, b^{2}=$ $4 a c$, i.e., $b=3 a, c=\frac{9}{4} a$.
Therefore, the minimum value of $M$ is 8.