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解:(1)將點A與B的坐標代入拋物線的解析式得:
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,
解得:
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,
∴拋物線的解析式為:y=-x
2-2x+3;
(2)∵拋物線的解析式為:y=-x
2-2x+3,
∴點C的坐標為(0,3),
設(shè)點E的坐標為(x,y),過點E作EF∥AB交y軸于F,
∴EF=-x,OB=3,OC=3,OF=-x
2-2x+3,CF=3-(-x
2-2x+3)=x
2+2x,∴S
△BEC=S
梯形OBEF+S
△EFC-S
△BOC
=
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(EF+OB)•OF+
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EF•CF-
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OB•OC
=
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×(-x+3)×(-x
2-2x+3)+
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×(-x)×(x
2+2x)-
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×3×3
=-
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(x+
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)
2+
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,
∴當(dāng)x=-
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時,△BCE的面積最大,最大面積為
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;
∴y=-x
2-2x+3=
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,
∴點E的坐標為(-
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,
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);
(3)存在.
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如果AP=BP,則點P在AB的垂直平分線上,即是拋物線的頂點,
∵y=-x
2-2x+3=-(x+1)
2+4,
∴此時P點的坐標為(-1,4);
如果AB=BP,則如圖①:
如果AB=AP,則如圖②:
∴存在使得△ABP為等腰三角形的P點3個;
有一點的坐標為(-1,4).
分析:(1)由拋物線y=ax
2+bx+3(a≠0)與x軸交于點A(1,0)和點B(-3,0),利用待定系數(shù)法,將點A與B的坐標代入拋物線的解析式即可求得a與b的值,則可得此拋物線的解析式;
(2)根據(jù)已知可求得點C的坐標,然后作輔助線:EF∥AB,設(shè)點E的坐標為(x,y),由S
△BEC=S
梯形OBEF+S
△EFC-S
△BOC即可求得關(guān)于x的二次函數(shù),配方即可求得x的值,代入解析式,求得y的值;
(3)分別從AP=BP與AB=BP與AB=AP去分析,可得到存在符合條件的點有3個,其中最好求得是P在頂點時的坐標,配方求解即可.
點評:此題考查了待定系數(shù)法求二次函數(shù)的解析式,三角形的面積最大值問題以及求拋物線上的點的問題.此題綜合性很強,注意數(shù)形結(jié)合與方程思想的應(yīng)用.