1. Let there be a quadratic function $f(x)=a x^{2}+b x+c, f\left(x_{1}\right)=f\left(x_{2}\right)\left(x_{1} \neq x_{2}\right)$, then the value of $f\left(x_{1}+x_{2}\right)$ is equal to
A. $c$
B. $a$
C. $b$
D. $a+b$
1. A

From $f\left(x_{1}\right)=f\left(x_{2}\right)$, we have $a x_{1}^{2}+b x_{1}+c=a x_{2}^{2}+b x_{2}+c$, which means $\left(x_{1}-x_{2}\right)\left[a\left(x_{1}+x_{2}\right)+b\right]=0, x_{1}+x_{2}=-\frac{b}{a}$, thus $f\left(-\frac{b}{a}\right)=a\left(-\frac{b}{a}\right)^{2}+b\left(-\frac{b}{a}\right)+c=c$.