【答案】
分析:(Ⅰ)首先利用拋物線經(jīng)過(guò)O(0,0),B(1,1)兩點(diǎn),且解析式的二次項(xiàng)系數(shù)為-
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求出拋物線解析式,再利用a=1求出拋物線的頂點(diǎn)坐標(biāo)即可;
(Ⅱ)利用當(dāng)y=0時(shí),有
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,求出x的值,進(jìn)而得出點(diǎn)N的坐標(biāo),再利用若點(diǎn)M在點(diǎn)B右側(cè),此時(shí)a>1,BM=a-1;若點(diǎn)M在點(diǎn)B左側(cè),此時(shí)0<a<1,BM=1-a得出答案即可;
(Ⅲ)利用平移后的拋物線只有一個(gè)不動(dòng)點(diǎn),故此方程有兩個(gè)相等的實(shí)數(shù)根,得出判別式△=(a-2h)
2-4(h
2-ak)=0,進(jìn)而求出k與h,a的關(guān)系即可得出頂點(diǎn)(h,k)在直線
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上.
解答:解:設(shè)該拋物線的解析式為
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,
∵拋物線經(jīng)過(guò)(0,0)、(1,1)兩點(diǎn),
∴
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,
解得
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.
∴該拋物線的解析式為
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(Ⅰ)當(dāng)a=1時(shí),該拋物線的解析式為y=-x
2+2x,
y=-x
2+2x=-(x
2-2x+1)+1=-(x-1)
2+1.
該拋物線的頂點(diǎn)坐標(biāo)為(1,1);
(Ⅱ)∵點(diǎn)N在x軸上,∴點(diǎn)N的縱坐標(biāo)為0.
當(dāng)y=0時(shí),有
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,
解得x
1=0,x
2=a+1.
∵點(diǎn)N異于原點(diǎn),∴點(diǎn)N的坐標(biāo)為(a+1,0).
∴ON=a+1,
∵點(diǎn)M在射線AB上,∴點(diǎn)M的縱坐標(biāo)為1.
當(dāng)y=1時(shí),有
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,
整理得出
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,
解得x
1=1,x
2=a.
點(diǎn)M的坐標(biāo)為(1,1)或(a,1).
當(dāng)點(diǎn)M的坐標(biāo)為(1,1)時(shí),M與B重合,
此時(shí)a=1,BM=0,ON=2.ON+BM與ON-BM的值都是常數(shù)2.
當(dāng)點(diǎn)M的坐標(biāo)為(a,1)時(shí),
若點(diǎn)M在點(diǎn)B右側(cè),此時(shí)a>1,BM=a-1.
∴ON+BM=(a+1)+(a-1)=2a,ON-BM=(a+1)-(a-1)=2.
若點(diǎn)M在點(diǎn)B左側(cè),此時(shí)0<a<1,BM=1-a.
∴ON+BM=(a+1)+(1-a)=2,ON-BM=(a+1)-(1-a)=2a.
∴當(dāng)0<a≤1時(shí),ON+BM的值是常數(shù)2,
當(dāng)a≥1時(shí),ON-BM的值是常數(shù)2.
(Ⅲ)設(shè)平移后的拋物線的解析式為
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,
由不動(dòng)點(diǎn)的定義,得方程:
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,
即t
2+(a-2h)t+h
2-ak=0.
∵平移后的拋物線只有一個(gè)不動(dòng)點(diǎn),∴此方程有兩個(gè)相等的實(shí)數(shù)根.
∴判別式△=(a-2h)
2-4(h
2-ak)=0,
有a-4h+4k=0,即
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.
∴頂點(diǎn)(h,k)在直線
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上.
點(diǎn)評(píng):此題主要考查了二次函數(shù)的綜合應(yīng)用以及根的判別式的性質(zhì)等知識(shí),利用分類討論的思想得出M與B的不同位置關(guān)系得出答案是解題關(guān)鍵.