(1)抛物线y=-x2+bx+c过点B(-1,2)和F(2,1),则:2=-1-b+c;1=-4+2b+c.解得:b=2/3,c=11/3.故抛物线解析式为y=-x2+(2/3)x+11/3.(2)①经过t秒时(0<t<1),见图2。DO=t,S矩形DEFG与矩形OABC重合部分面积S1=DO*DE=1*t=t;y=-x2+(2/3)x+11/3的对称轴为X=1/3,设对称轴交X轴于M,则AM=AO+OM=1+1/3=4/3.S⊿AQF=S⊿ABF-S⊿ABQ=AB*AG/2-AB*AM/2=2*(AO+DG-DO)/2-2*(4/3)/2=5/3-t.∴S1+S2=t+(5/3-t)=5/...
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