解:(1)∵拋物線過A(3,0),B(6,0),
∴
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,
解得:
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,
∴所求拋物線的函數(shù)表達(dá)式是y=
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x
2-x+2.
(2)①∵當(dāng)x=0時(shí),y=2,
∴點(diǎn)C的坐標(biāo)為(0,2).
設(shè)直線BC的函數(shù)表達(dá)式是y=kx+b.
則有
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,
解得:
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.
∴直線BC的函數(shù)表達(dá)式是y=-
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x+2.
∵0<x<6,點(diǎn)P、Q的橫坐標(biāo)相同,
∴PQ=y
Q-y
P=(-
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x+2)-(
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x
2-x+2)
=-
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x
2+
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x
=-
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(x-3)
2+1
∴當(dāng)x=3時(shí),線段PQ的長度取得最大值.最大值是1.
②解:當(dāng)∠OAQ=90°時(shí),點(diǎn)P與點(diǎn)A重合,
∴P(3,0)
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當(dāng)∠QOA=90°時(shí),點(diǎn)P與點(diǎn)C重合,
∴x=0(不合題意)
當(dāng)∠OQA=90°時(shí),
設(shè)PQ與x軸交于點(diǎn)D.
∵∠ODQ+∠ADQ=90°,∠QAD+∠AQD=90°,
∴∠OQD=∠QAD.
又∵∠ODQ=∠QDA=90°,
∴△ODQ∽△QDA.
∴
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,即DQ
2=OD•DA.
∴(-
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x+2)
2=x(3-x),
10x
2-39x+36=0,
∴x
1=
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,x
2=
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,
∴y
1=
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×(
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)
2-
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+2=
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;
y
2=
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×(
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)
2-
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+2=
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;
∴P(
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,
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)或P(
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,
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).
∴所求的點(diǎn)P的坐標(biāo)是P(3,0)或P(
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,
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)或P(
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,
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).
分析:(1)已知了A,B的坐標(biāo),可用待定系數(shù)法求出函數(shù)的解析式.
(2)①Q(mào)P其實(shí)就是一次函數(shù)與二次函數(shù)的差,二次函數(shù)的解析式在(1)中已經(jīng)求出,而一次函數(shù)可根據(jù)B,C的坐標(biāo),用待定系數(shù)法求出.那么讓一次函數(shù)的解析式減去二次函數(shù)的解析式,得出的新的函數(shù)就是關(guān)于PQ,x的函數(shù)關(guān)系式,那么可根據(jù)函數(shù)的性質(zhì)求出PQ的最大值以及相對(duì)應(yīng)的x的取值.
(3)分三種情況進(jìn)行討論:
當(dāng)∠QOA=90°時(shí),Q與C重合,顯然不合題意.因此這種情況不成立;
當(dāng)∠OAQ=90°時(shí),P與A重合,因此P的坐標(biāo)就是A的坐標(biāo);
當(dāng)∠OQA=90°時(shí),如果設(shè)QP與x軸的交點(diǎn)為D,那么根據(jù)射影定理可得出DQ
2=OD•DA.由此可得出關(guān)于x的方程即可求出x的值,然后將x代入二次函數(shù)式中即可得出P的坐標(biāo).
點(diǎn)評(píng):本題主要考查了二次函數(shù)的綜合應(yīng)用,用數(shù)形結(jié)合的思想來求解是解題的基本思路.