解:(1)依題意,得
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,
解得,
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,
拋物線的解析式為y=-x
2-2x+3,
頂點(diǎn)坐標(biāo)為(-1,4);
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(2)如圖,∵AB=4,OC=3,
∴CD
1=CD
2=AB=4,
D的坐標(biāo)為D
1(-4,3),D
2(4,3),
∵D
3E=OC=3,AE=OB,可得E點(diǎn)坐標(biāo)為(-2,0),
∴D
3(-2,-3);
(3)拋物線y=-x
2-2x+3與y軸的交點(diǎn)C的坐標(biāo)為(0,3),
設(shè)點(diǎn)Q的坐標(biāo)為(-1,m),
①若∠QAC=90°,如圖1,設(shè)拋物線的對(duì)稱軸與x軸的交點(diǎn)
為E,則E(-1,0),則AE=2,EQ=-m,
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由△AEQ∽△COA,得
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,
∴
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,
∴m=-2,
∴點(diǎn)Q的坐標(biāo)為(-1,-2);
②若∠QCA=90°,如圖2,作QF⊥y軸于點(diǎn)F,則QF=1,F(xiàn)C=m-3,
由△QFC∽△COA,得
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,
∴
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,
∴m=4,
∴點(diǎn)Q的坐標(biāo)為(-1,4);
③若∠CQA=90°,如圖3,設(shè)AC的中點(diǎn)為O
1,則O
1的坐標(biāo)為
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,作O
1G⊥l于點(diǎn)G,則QG=
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,O
1G=
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,
由勾股定理得,
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,
∵
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,
∴
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,
解得,
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,
∴點(diǎn)Q的坐標(biāo)為
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,
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;
綜上所述,使△ACQ為直角三角形,點(diǎn)Q的坐標(biāo)為
(-1,-2)、(-1,4)、
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或
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.
分析:(1)將A(-3,0)、B(1,0)分別代入y=ax
2+bx+3,組成關(guān)于a、b的方程組,解方程組即可求出a、b的值,從而得到二次函數(shù)解析式;
(2)根據(jù)題意畫(huà)出圖形,根據(jù)平行四邊形的性質(zhì)及AB的長(zhǎng)為4,OC=3,即可輕松得出點(diǎn)D的坐標(biāo);
(3)拋物線y=-x
2-2x+3與y軸的交點(diǎn)C的坐標(biāo)為(0,3),設(shè)點(diǎn)Q的坐標(biāo)為(-1,m),然后分三種情況討論①若∠QAC=90°,△AEQ∽△COA,利用相似三角形的性質(zhì)解答;②若∠QCA=90°,由△QFC∽△COA,利用相似三角形的性質(zhì)解答;③若∠CQA=90°,作O
1G⊥l于點(diǎn)G,則QG=
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,O
1G=
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,
由勾股定理得到關(guān)于m的方程,解方程求出m的值.
點(diǎn)評(píng):本題考查了二次函數(shù)綜合題,涉及求函數(shù)解析式、平行四邊形的性質(zhì)、相似三角形的判定與性質(zhì)、勾股定理,要注意分類討論的作用.