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