解：?$(1)$?∵點(diǎn)?$A$?，?$B$?在函數(shù)?$y=\frac {1}{4}x2$?的圖象上
?$A$?，?$B $?的橫坐標(biāo)分別為?$-2$?，?$4$?
∴?$A$?的坐標(biāo)為?$(-2$?，?$1)$?，?$B$?的坐標(biāo)為?$(4$?，?$4)$?
設(shè)直線?$AB$?的函數(shù)表達(dá)式為?$y=kx+b$?
∴?$\begin {cases}{-2k+b=1}\\{4k+b=4}\end {cases}$?，解得?$\begin {cases}{k=\frac {1}{2}}\\{b=2}\end {cases}$?
∴直線?$AB$?的函數(shù)表達(dá)式為?$y=\frac 12x+2$?
?$(2)$?在函數(shù)?$y=\frac {1}{2}x+2$?中，令?$x=0$?，得?$y=2$?
∴?$C$?的坐標(biāo)為?$(0$?，?$2)$?
∴?$OC=2$?
∴?$S_{△AOB}=S_{△AOC}+S_{△BOC}$?
?$=\frac {1}{2}×2×2+\frac {1}{2}×2×4=6$?
