6. Given the quadratic function $y=a x^{2}+b x+c(a \neq 0)$, where $a, b, c$ simultaneously satisfy the following conditions:
(1) $x-2$ is a factor of the left side of the equation $a x^{2}+b x-c=0$;
(2) The remainder when $a x^{2}+b x+c$ is divided by $x+2$ is -4;
(3) The difference between $c x^{2}+b x+c$ and -6 can be divided by $x+1$.
Then the quadratic function $y=a x^{2}+b x+c$ has ( ).
(A) a maximum value of $\frac{25}{4}$
(B) a maximum value of $-\frac{25}{4}$
(C) a minimum value of $\frac{25}{4}$
(D) a minimum value of $-\frac{25}{4}$
6. D

According to the problem, using the root value method and the remainder theorem, we can obtain the system of equations
$$
\begin{array}{l}
\left\{\begin{array} { l } 
{ 4 a + 2 b + c = 0 , } \\
{ 4 a - 2 b + c + 4 = 0 , } \\
{ a - b + c + 6 = 0 . }
\end{array} \text { Solving, we get } \left\{\begin{array}{l}
a=1, \\
b=1, \\
c=-6 .
\end{array}\right.\right. \\
\therefore y=x^{2}+x-6 \text {. } \\
\because a=1>0, \therefore y \text { has a minimum value. } \\
\text { Therefore, } y_{\text {min }}=\frac{4 a c-b^{2}}{4 a}=-\frac{25}{4} \text {. } \\
\end{array}
$$