解:(1)令x=0則y=3,
∴點(diǎn)A(0,3),
設(shè)直線AB的解析式為y=kx+b(k≠0),
則
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,
解得
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,
∴直線AB的解析式為y=-
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x+3;
(2)令y=0,則-
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x+3=0,
解得x=4,
∴點(diǎn)B(4,0),
點(diǎn)B關(guān)于y軸的對稱點(diǎn)D的坐標(biāo)為(-4,0),
∴BD=4-(-4)=4+4=8,
由勾股定理得,AB=
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=
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=5,
設(shè)點(diǎn)D到直線AB的距離為h,
則sin∠ABO=
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=
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,
即
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=
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,
解得h=4.8,
即點(diǎn)D到直線AB的距離是4.8;
(3)對稱軸為直線x=
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,
當(dāng)
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≤-1,即m≤-2時(shí),x=-1時(shí)二次函數(shù)的最小值為-3,
(-1)
2-m•(-1)+3=-3,
解得m=-7;
當(dāng)-1<
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<1,即-2<m<2時(shí),x=
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時(shí)二次函數(shù)有最小值為-3,
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=-3,
解得m=±2
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,都不滿足-2<m<2,舍去;
當(dāng)
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≥1即m≥時(shí),x=1時(shí)二次函數(shù)的最小值為-3,
(-1)
2-m•1+3=-3,
解得m=7,
綜上所述,實(shí)數(shù)m的值為7或-7.
分析:(1)令x=0求出y的值得到點(diǎn)A的坐標(biāo),然后設(shè)直線AB的解析式為y=kx+b(k≠0),利用待定系數(shù)法求一次函數(shù)解析式;
(2)令y=0求出點(diǎn)B的坐標(biāo),然后根據(jù)關(guān)于y軸對稱的點(diǎn)的橫坐標(biāo)互為相反數(shù)求出點(diǎn)D的坐標(biāo),然后根據(jù)∠ABO的正弦值列式計(jì)算即可得解;
(3)表示出拋物線的對稱軸,然后根據(jù)對稱軸的位置,分別根據(jù)二次函數(shù)的增減性和最值問題列式計(jì)算即可得解.
點(diǎn)評:本題是二次函數(shù)綜合題型,主要利用了待定系數(shù)法求一次函數(shù)解析式,翻折的性質(zhì),勾股定理,銳角三角函數(shù)的應(yīng)用,二次函數(shù)的最值問題,難點(diǎn)在于(3)根據(jù)對稱軸的位置情況分情況討論.