11. Given the quadratic function $y=a x^{2}+b x+c(b$ is an integer) whose graph does not lie below the $x$-axis at any point, the graph intersects the $y$-axis at point $C$, and the vertex is at point $E$. The axis of symmetry is to the right of the line $x=c-\frac{1}{12}$. Let $t=\frac{a+2 b+12 c}{a}$.
(1) Find the minimum value of $t$;
(2) When $t$ takes its minimum value, and $C E>\frac{5}{24}$, find the equation of the quadratic function.
11. (1) According to the problem, we have
$$
a>0, b^{2}-4 a c \leqslant 0 \Rightarrow c \geqslant \frac{b^{2}}{4 a} \text {. }
$$

Therefore, $t=\frac{a+2 b+12 c}{a} \geqslant \frac{a+2 b+\frac{3 b^{2}}{a}}{a}$
$$
=1+2\left(\frac{b}{a}\right)+3\left(\frac{b}{a}\right)^{2}=3\left(\frac{b}{a}+\frac{1}{3}\right)^{2}+\frac{2}{3} \text {. }
$$

When $a=-3 b$, and $c=-\frac{b}{12}$, $t$ reaches its minimum value $\frac{2}{3}$.
(2) From the graph of the quadratic function, the axis of symmetry is to the right of the line $x=c-\frac{1}{12}$, so $-\frac{b}{2 a}>c-\frac{1}{12}$.
From (1), we know that when $t$ reaches its minimum value,
$a=-3 b$, and $c=-\frac{b}{12}$.
Substituting into $-\frac{b}{2 a}>c-\frac{1}{12}$, we get
$\frac{1}{6}>-\frac{1}{12} b-\frac{1}{12} \Rightarrow b>-3$.
Also, $b=-\frac{1}{3} a\frac{5}{24}$ (which fits the problem).
In summary, the quadratic function is
$y=6 x^{2}-2 x+\frac{1}{6}$.