(1)设c1的解析式为y=ax2+bx+c,由图象可知:c1过A(-1,0),B(0,3),C(2,3)三点.
解得: ∴抛物线c1的解析式为y=-x2+2x+3.![](http://img.shitiku.com.cn/uploads/allimg/20191020/20191020110414-91283.png)
(2)∵y=-x2+2x+3=-(x-1)2+4. ∴抛物线c1的顶点D的坐标为(1,4); 过D作DF⊥x轴于F,由图象可知:OA=1,OB=3,OF=1,DF=4; 令y=0,则-x2+2x+3=0, 解得x1=-1,x2=3 ∴OE=3,则FE=2. S△ABO=OA•OB=×1×3=; S△DFE=DF•FE=×4×2=4; S梯形BOFD=(BO+DF)•OF=. ∴S四边形ABDE=S△AOB+S梯形BOFD+S△DFE=9(平方单位).
(3)如图,过B作BK⊥DF于K,则BK=OF=1.![](http://img.shitiku.com.cn/uploads/allimg/20191020/20191020110415-81518.png) DK=DF-OB=4-3=1. ∴BD==, 又DE==2; AB=,BE=3; 在△ABO和△BDE中, AO=1,BO=3,AB=; BD=,BE=3,DE=2. ∵=== ∴△AOB∽△DBE.
(4),,,,,,. |