【答案】
分析:(1)求出方程ax
2+2ax-3a=0(a≠0),即可得到A點(diǎn)坐標(biāo)和B點(diǎn)坐標(biāo);把A的坐標(biāo)代入直線l即可判斷A是否在直線上;
(2)根據(jù)點(diǎn)H、B關(guān)于過A點(diǎn)的直線l:
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對(duì)稱,得出AH=AB=4,過頂點(diǎn)H作HC⊥AB交AB于C點(diǎn),求出AC和HC的長,得出頂點(diǎn)H的坐標(biāo),代入二次函數(shù)解析式,求出a,即可得到二次函數(shù)解析式;
(3)解方程組
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,即可求出K的坐標(biāo),根據(jù)點(diǎn)H、B關(guān)于直線AK對(duì)稱,得出HN+MN的最小值是MB,過點(diǎn)K作直線AH的對(duì)稱點(diǎn)Q,連接QK,交直線AH于E,得到BM+MK的最小值是BQ,即BQ的長是HN+NM+MK的最小值,由勾股定理得QB=8,即可得出答案.
解答:解:(1)依題意,得ax
2+2ax-3a=0(a≠0),
兩邊都除以a得:
即x
2+2x-3=0,
解得x
1=-3,x
2=1,
∵B點(diǎn)在A點(diǎn)右側(cè),
∴A點(diǎn)坐標(biāo)為(-3,0),B點(diǎn)坐標(biāo)為(1,0),
答:A、B兩點(diǎn)坐標(biāo)分別是(-3,0),(1,0).
證明:∵直線l:
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,
當(dāng)x=-3時(shí),
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,
∴點(diǎn)A在直線l上.
(2)∵點(diǎn)H、B關(guān)于過A點(diǎn)的直線l:

對(duì)稱,
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∴AH=AB=4,
過頂點(diǎn)H作HC⊥AB交AB于C點(diǎn),
則
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,
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,
∴頂點(diǎn)
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,
代入二次函數(shù)解析式,解得
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,
∴二次函數(shù)解析式為
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,
答:二次函數(shù)解析式為
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.
(3)直線AH的解析式為
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,
直線BK的解析式為
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,
由
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,
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解得
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,
即
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,
則BK=4,
∵點(diǎn)H、B關(guān)于直線AK對(duì)稱,K(3,2
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),
∴HN+MN的最小值是MB,

,
過K作KD⊥x軸于D,作點(diǎn)K關(guān)于直線AH的對(duì)稱點(diǎn)Q,連接QK,交直線AH于E,

,
則QM=MK,
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,AE⊥QK,
∴根據(jù)兩點(diǎn)之間線段最短得出BM+MK的最小值是BQ,即BQ的長是HN+NM+MK的最小值,
∵BK∥AH,
∴∠BKQ=∠HEQ=90°,
由勾股定理得QB=
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=

=8,
∴HN+NM+MK的最小值為8,
答:HN+NM+MK和的最小值是8.
點(diǎn)評(píng):本題主要考查對(duì)勾股定理,解二元一次方程組,二次函數(shù)與一元二次方程,二次函數(shù)與X軸的交點(diǎn),用待定系數(shù)法求二次函數(shù)的解析式等知識(shí)點(diǎn)的理解和掌握,綜合運(yùn)用這些性質(zhì)進(jìn)行計(jì)算是解此題的關(guān)鍵,此題是一個(gè)綜合性比較強(qiáng)的題目,有一定的難度.