解:(1)由函數(shù)圖象經(jīng)過(guò)原點(diǎn)得,函數(shù)解析式為y=ax
2+bx(a≠0),
又∵函數(shù)的頂點(diǎn)坐標(biāo)為(3,-
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),
∴
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,
解得:
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,
故函數(shù)解析式為:y=
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x
2-
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x,
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由二次函數(shù)圖象的對(duì)稱性可得點(diǎn)A的坐標(biāo)為(6,0);
(2)∵S
△POA=2S
△AOB,
∴點(diǎn)P到OA的距離是點(diǎn)B到OA距離的2倍,即點(diǎn)P的縱坐標(biāo)為2
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,
代入函數(shù)解析式得:2
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=
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x
2-
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x,
解得:x
1=3+3
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,x
2=3-3
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,
即滿足條件的點(diǎn)P有兩個(gè),其坐標(biāo)為:P
1(3+3
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,2
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),P
2(3-3
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,2
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).
分析:(1)根據(jù)函數(shù)經(jīng)過(guò)原點(diǎn),可得c=0,然后根據(jù)函數(shù)的對(duì)稱軸,及函數(shù)圖象經(jīng)過(guò)點(diǎn)(3,-
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)可得出函數(shù)解析式,根據(jù)二次函數(shù)的對(duì)稱性可直接得出點(diǎn)A的坐標(biāo).
(2)根據(jù)題意可得點(diǎn)P到OA的距離是點(diǎn)B到OA距離的2倍,即點(diǎn)P的縱坐標(biāo)為2
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,代入函數(shù)解析式可得出點(diǎn)P的橫坐標(biāo).
點(diǎn)評(píng):此題考查了二次函數(shù)的綜合題目,涉及了待定系數(shù)法求函數(shù)解析式,三角形的面積及一元二次方程的解,綜合性較強(qiáng),需要我們仔細(xì)分析,分步解答.