解:(1)設(shè)函數(shù)解析式為
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,
解出
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,
∴
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;
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(2)求出點P的坐標(biāo)為(3,2),
由梯形中位線定理得,AC+OD=3×2=6,m+n=6,
∴n=6-m(0≤m≤6);
(3)方法一:①當(dāng)△ACE∽△ODP時(如圖1),∠ACO=∠ODP,
∵AB∥x軸,∴∠ACO=∠COD
∴∠COD=∠ODP,OC=CD,又CF⊥OD,∴AC=OF=
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OD,
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∴m=
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(6-m)解得:m=2
②當(dāng)△ACE∽△OPD時(如圖2),∠ACO=∠OPD,∵∠ACO=∠COD
∴∠COD=∠OPD,可得△OPD∽△COD,可得OD
2=DP•DC,
即OD
2=
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CD
2=(6-m)
2=
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(
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)
2,解得:m=
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方法二:得出AE=
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1當(dāng)△ACE∽△ODP時,可求出m=2
②當(dāng)△ACE∽△OPD時,可求出m=
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.
分析:(1)已知拋物線的頂點縱坐標(biāo)以及對稱軸,根據(jù)待定系數(shù)法即可求得二次函數(shù)的解析式;
(2)首先求得A點的坐標(biāo),P的縱坐標(biāo)是A的縱坐標(biāo)的一半,即可求得P的縱坐標(biāo),代入二次函數(shù)解析式即可求得P的坐標(biāo);
(3)分△ACE∽△ODP和△ACE∽△OPD,兩種情況,根據(jù)相似三角形的對應(yīng)邊的比相等,即可求得m的值.
點評:本題考查了二次函數(shù)解析式的確定、相似三角形的性質(zhì)等知識點.(3)題中,要根據(jù)相似三角形對應(yīng)邊和對應(yīng)角的不同分類討論,不要漏解.