解：?$ (1)$?二次函數(shù)的對(duì)稱軸為?$x=-\frac {2a} = 1$?
當(dāng)?$x=1$?時(shí)，?$y=a-\frac {1}{2}$?
∵頂點(diǎn)在一次函數(shù)的圖像上
∴頂點(diǎn)縱坐標(biāo)為?$y=-2×1=-2$?
∴?$a-\frac {1}{2}=-2$?
∴?$a=-\frac {3}{2}$?
?$(2)$?二次函數(shù)表達(dá)式為?$y=\frac {1}{2}x2-x-\frac {3}{2}$?
令?$y=0，$??$\frac {1}{2}x2-x-\frac {3}{2}=0$?
解得?${x}_1=-1，$??${x}_2=3$?
∴?$A(-1，$??$0)，$??$B(3，$??$0)$?
?$(3)$?∵四邊形?$ACBD$?是平行四邊形
∴點(diǎn)?$C、$??$D$?關(guān)于對(duì)角線交點(diǎn)?$(1，$??$0)$?對(duì)稱
又∵點(diǎn)?$D'$?是點(diǎn)?$D$?關(guān)于?$x$?軸的對(duì)稱點(diǎn)
∴點(diǎn)?$C、$??$D'$?關(guān)于拋物線的對(duì)稱軸(直線?$x=1 )$?對(duì)稱
∴點(diǎn)?$D'$?在二次函數(shù)圖像上
