1. Given the function $f(x)=a x^{2}+b x+c(a \neq 0, a$, $b$, $c$ are all constants), the graph of function $f_{1}(x)$ is symmetric to the graph of function $f(x)$ about the $y$-axis, and the graph of function $f_{2}(x)$ is symmetric to the graph of function $f_{1}(x)$ about the line $y=1$. Then the analytical expression of function $f_{2}(x)$ is $\qquad$.
$$
-1 \cdot f_{2}(x)=-a x^{2}+b x+2-c \text {. }
$$

Let point $P(x, y)$ be on the graph of the function $f_{2}(x)$. Then the point $P^{\prime}(x, 2-y)$, which is symmetric to $P$ with respect to the line $y=1$, is on the graph of the function $f_{1}(x)$. Furthermore, the point $P^{\prime \prime}(-x, 2-y)$, which is symmetric to $P^{\prime}$ with respect to the $y$-axis, is on the graph of the function $f(x)$.
Thus, $2-y=a x^{2}-b x+c$, which means
$$
y=-a x^{2}+b x+2-c \text {. }
$$

Therefore, $f_{2}(x)=-a x^{2}+b x+2-c$.