解：?$(1) $?∵ 二次函數(shù)?$y=a(x-1)^2+h$?的圖像與?$x$?軸交于點?$A(-2，$??$0)，$?
與?$y$?軸交于點?$C(0，$??$4)$?
∴?$\begin{cases}{a(-2-1)^2+h=0}\\{a(0-1)^2+h=4}\end{cases}，$?解得?$ \begin{cases}{a=-\dfrac 12}\\{h=\dfrac 92}\end{cases}$?
∴ 該二次函數(shù)的表達式為?$y=-\frac {1}{2} (x-1)^2+\frac {9}{2}=- \frac {1}{2} x^2+x+4 $?
?$(2)$?令?$y=0，$?則?$- \frac {1}{2} x^2+x+4=0$?
解得?$x_{1}=-2，$??$x_{2}=4$?
∴ 點?$B$?的坐標(biāo)為?$(4，$??$0)$?
∵?$ E$?是?$BC$?的中點
∴點?$E$?的坐標(biāo)為?$(2，$??$2)$?
設(shè)直線?$AE$?相應(yīng)的函數(shù)表達式為?$y=mx+n$?
則?$\begin{cases}{-2m+n=0}\\{2m+n=2}\end{cases}，$? 解得?$\begin{cases}{m=\dfrac {1}{2}}\\{n=1}\end{cases}$?
∴ 直線?$AE$?相應(yīng)的函數(shù)表達式為?$y=\frac {1}{2} x+1$?
聯(lián)立方程組?$ \begin{cases}{y=\dfrac {1}{2} x+1}\\{y=-\dfrac 12x^2+x+4}\end{cases}，$?解得?$\begin{cases}{x=3}\\{y=\dfrac 52}\end{cases}，$?或?$\begin{cases}{x=-2}\\{y=0}\end{cases}$?
∴點?$D$?的坐標(biāo)為?$(3，$??$\frac {5}{2} )$?
