解:(1)令y
1=0,得△=(-2t)
2-4(2t-1)=4t
2-8t+4=4(t-1)
2,
∵t>1,∴△=4(t-1)
2>0,
∴無論t取何值,方程x
2-2tx+(2t-1)=0總有兩個不相等的實數(shù)根,
∴無論t取何值,拋物線C
1與y軸總有兩個交點.
(2)解方程x
2-2tx+(2t-1)=0得,x
1=1,x
2=2t-1,
∵t>1,∴2t-1>1.得A(1,0),B(2t-1,0),
∵D(m,n),E(m+2,n),∴DE=AB=2,
即2t-1-1=2,解得t=2.
∴二次函數(shù)為
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,
顯然將拋物線C
1向上平移1個單位可得拋物線C
2:
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,
故n=1.
(3)由(2)得拋物線C
2:
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,D(1,1),E(3,1),
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翻折后,頂點F(2,0)的對應點為F'(2,2),
如圖,當直線
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經(jīng)過點D(1,1)時,記為l
1,
此時
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,圖形G與l
1只有一個公共點;
當直線
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經(jīng)過點E(3,1)時,記為l
2,此時
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,圖形G與l
2有三個公共點;
當b<3時,由圖象可知,只有當直線l:
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位于l
1與l
2之間時,圖形G與直線l有且只有兩個公共點,
∴符合題意的b的取值范圍是
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.
分析:(1)求出b
2-4ac的值,根據(jù)根的判別式為正數(shù)即可得到答案;
(2)首先用含有t的字母表示出點A與點B的坐標,然后根據(jù)點D和點E的坐標得到DE=AB=2,從而求得t值,配方后利用平移規(guī)律得到平移個數(shù)即可;
(3)分三種情況討論后即可求得變量t的取值范圍.
點評:本題主要考查對二次函數(shù)圖象上點的坐標特征,根與系數(shù)的關(guān)系,解一元二次方程,平移的性質(zhì)等知識點的理解和掌握,能根據(jù)性質(zhì)進行推理是解此題的關(guān)鍵.