1.(2024·苏州)如图①,二次函数$y = x^{2}+bx + c$的图像$C_{1}$与开口向下的二次函数图像$C_{2}$均过点$A(-1,0)$,$B(3,0)$．
(1)求图像$C_{1}$对应的函数表达式．
(2)若图像$C_{2}$过点$C(0,6)$,点$P$位于第一象限,且在图像$C_{2}$上,直线$l$过点$P$且与$x$轴平行,与图像$C_{2}$的另一个交点为$Q$(点$Q$在点$P$的左侧),直线$l$与图像$C_{1}$的交点为$M$,$N$(点$N$在点$M$的左侧)．当$PQ = MP + QN$时,求点$P$的坐标．
(3)如图②,$D$,$E$分别为二次函数图像$C_{1}$,$C_{2}$的顶点,连接$AD$,过点$A$作$AF\perp AD$,交图像$C_{2}$于点$F$,连接$EF$．当$EF// AD$时,求图像$C_{2}$对应的函数表达式．

[跟踪训练]1.(1)$\because$点$A(-1,0)$，$B(3,0)$在二次函数$y=x^{2}+bx + c$的图像上，$\therefore\begin{cases}1 - b + c = 0,\\9 + 3b + c = 0,\end{cases}$解得$\begin{cases}b=-2,\\c=-3.\end{cases}$$\therefore$图像$C_{1}$对应的函数表达式为$y=x^{2}-2x - 3$.
(2)设图像$C_{2}$对应的函数表达式为$y=a(x + 1)(x - 3)(a＜0)$.$\because$点$C(0,6)$在图像$C_{2}$上，$\therefore a=-2$.$\therefore$图像$C_{2}$对应的函数表达式为$y=-2(x + 1)(x - 3)$，其对称轴为直线$x = 1$.又$\because$图像$C_{1}$的对称轴也为直线$x = 1$，$\therefore$作直线$x = 1$，交直线$l$于点$H$（如图①）.由二次函数图像的对称性，得$QH = PH$，$PM = NQ$.又$\because PQ = MP + QN$，$\therefore PH = PM$.设$PH = t(0＜t＜2)$，则点$P$的横坐标为$t + 1$，点$M$的横坐标为$2t + 1$.将$x = t + 1$代入$y=-2(x + 1)(x - 3)$，得$y_{P}=-2(t + 2)(t - 2)$.将$x = 2t + 1$代入$y=(x + 1)(x - 3)$，得$y_{M}=(2t + 2)(2t - 2)$.$\because y_{P}=y_{M}$，$\therefore-2(t + 2)(t - 2)=(2t + 2)(2t - 2)$，即$6t^{2}=12$，解得$t_{1}=\sqrt{2}$，$t_{2}=-\sqrt{2}$（不合题意，舍去）.$\therefore$点$P$的坐标为$(\sqrt{2}+1,4)$.
(3)如图②，连接$DE$，交$x$轴于点$G$，过点$F$作$FI\perp ED$于点$I$，过点$F$作$FJ\perp x$轴于点$J$.$\because FI\perp ED$，$FJ\perp x$轴，$\therefore$易得四边形$IGJF$为矩形.$\therefore IF = GJ$，$IG = FJ$.设图像$C_{2}$对应的函数表达式为$y=a(x + 1)(x - 3)(a＜0)$.$\because D$，$E$分别为二次函数图像$C_{1}$，$C_{2}$的顶点，$\therefore$易得点$D$的坐标为$(1,-4)$，点$E$的坐标为$(1,-4a)$.$\therefore DG = 4$，$AG = 2$，$EG=-4a$.$\therefore$在$Rt\triangle AGD$中，$\tan\angle ADG=\frac{AG}{DG}=\frac{2}{4}=\frac{1}{2}$.$\because AF\perp AD$，$\therefore\angle FAB+\angle DAB = 90^{\circ}$.又$\because\angle DAG+\angle ADG = 90^{\circ}$，$\therefore\angle ADG=\angle FAB$.$\therefore\tan\angle FAB=\tan\angle ADG=\frac{FJ}{AJ}=\frac{1}{2}$.设$GJ = FI = m(0＜m＜2)$，则$AJ = 2 + m$.$\therefore FJ = IG=\frac{2 + m}{2}$.$\therefore$点$F$的坐标为$(m + 1,\frac{2 + m}{2})$.$\because EF// AD$，$\therefore\angle FEI=\angle ADG$.$\therefore\tan\angle FEI=\tan\angle ADG=\frac{FI}{EI}=\frac{1}{2}$.$\therefore EI = 2m$.又$\because EG = EI + IG$，$\therefore 2m+\frac{2 + m}{2}=-4a$.$\therefore a=-\frac{2 + 5m}{8}$①.$\because$点$F$在图像$C_{2}$上，$\therefore a(m + 1 + 1)\cdot(m + 1 - 3)=\frac{m + 2}{2}$，即$a(m + 2)(m - 2)=\frac{m + 2}{2}$.$\because m + 2\neq0$，$\therefore a(m - 2)=\frac{1}{2}$②.由①②，可得$-\frac{2 + 5m}{8}(m - 2)=\frac{1}{2}$，解得$m_{1}=0$（不合题意，舍去），$m_{2}=\frac{8}{5}$.$\therefore a=-\frac{5}{4}$.$\therefore$图像$C_{2}$对应的函数表达式为$y=-\frac{5}{4}(x + 1)(x - 3)=-\frac{5}{4}x^{2}+\frac{5}{2}x+\frac{15}{4}$.