解：????$ (1)$????由題意得，
????$ \begin{cases}{-\dfrac {-2}{2a}=1 }\\{a-2+c=-4} \end{cases}$????
解得????$a=1，$????????c$=-3$????
所以二次函數(shù)的表達(dá)式為????$y=x2- 2x- 3 $????
????$ (2)$????因?yàn)槎魏瘮?shù)????$y=x2-2x-3$????的圖像與????$y$????軸交于點(diǎn)????$C ，$????
與????$x$????軸正半軸交于點(diǎn)????$B$????
所以????$C(0，$????????$-3)，$????????$B(3，$????????$0)$????
設(shè)直線????$AC$????的解析式為????$y= kx+b$????
將????$A(1，$????????$ -4)，$????????$ C(0，$????????$ -3)$????代入，
得????$\begin{cases}{-4=k+b }\\{-3=b} \end{cases}$????
解得????$k=-1，$????????$b=-3$????
所以直線????$AC$????的解析式為????$y= -x- 3$????
同理可得，直線????$AB$????的解析式為????$y= 2x- 6$????
設(shè)點(diǎn)????$P$????坐標(biāo)為????$(t ，$????????$ -t-3) ，$????此時(shí)四邊形????$OPEF$????的面積為????$S$????
因?yàn)????$P(t，$????????$ -t-3) ，$????????$ PE//x$????軸
所以點(diǎn)????$E$????的縱坐標(biāo)為????$-t-3$????
因?yàn)辄c(diǎn)????$E$????在直線????$AB$????上
所以????$-t-3= 2x-6$????
解得，????$ x=\frac {-t+6}{2}$????
所以????$E(\frac {-t+3}{2}，$????????$-t-3)$????
所以????$S=\frac {1}{2}×(t+3)×(\frac {-t+3}{2}+\frac {-t+3}{2}-t)$????
????$ = -t2-\frac {3}{2}t+\frac {9}{2}$????
????$ =-(t+\frac {3}{4})2+\frac {81}{16}$????
所以當(dāng)????$t= -\frac {3}{4}$????時(shí)，四邊形????$OPEF$????的面積取最大值，
最大值為????$\frac {81}{16}.$????此時(shí)點(diǎn)????$P$????的坐標(biāo)為????$(-\frac {3}{4} ，$????????$-\frac {9}{4})$????
