【答案】
分析:(1)由于A、B、C三點(diǎn)的坐標(biāo)已知,代入函數(shù)解析式中利用待定系數(shù)法就可以確定函數(shù)的解析式;
(2)若點(diǎn)D為線段OA的一個(gè)三等分點(diǎn),那么根據(jù)已知條件可以確定D的坐標(biāo)為(0,1)或,(0,2),而C的坐標(biāo)已知,利用待定系數(shù)法就可以確定直線CD的解析式;
(3)如圖,由題意,可得M(0,

),點(diǎn)M關(guān)于x軸的對(duì)稱點(diǎn)為M′(0,-

),點(diǎn)A關(guān)于拋物線對(duì)稱軸x=3的對(duì)稱點(diǎn)為A'(6,3),連接A'M',根據(jù)軸對(duì)稱性及兩點(diǎn)間線段最短可知,A'M'的長(zhǎng)就是所求點(diǎn)P運(yùn)動(dòng)的最短總路徑的長(zhǎng),根據(jù)待定系數(shù)法可求出直線A'M'的解析式為y=
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x-
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,從而求出E、F兩點(diǎn)的坐標(biāo),再根據(jù)勾股定理可以求出A'M'=
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,也就求出了最短總路徑的長(zhǎng).
解答:
解:(1)根據(jù)題意,c=3,
所以
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解得
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所以拋物線解析式為y=

x
2-

x+3.
(2)依題意可得OA的三等分點(diǎn)分別為(0,1),(0,2).
設(shè)直線CD的解析式為y=kx+b.
當(dāng)點(diǎn)D的坐標(biāo)為(0,1)時(shí),直線CD的解析式為y=-

x+1;(3分)
當(dāng)點(diǎn)D的坐標(biāo)為(0,2)時(shí),直線CD的解析式為y=-

x+2.(4分)
(3)如圖,由題意,可得M(0,
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).
點(diǎn)M關(guān)于x軸的對(duì)稱
點(diǎn)為M′(0,-

),
點(diǎn)A關(guān)于拋物線對(duì)稱軸x=3的對(duì)稱點(diǎn)為A'(6,3).
連接A'M'.
根據(jù)軸對(duì)稱性及兩點(diǎn)間線段最短可知,A'M'的長(zhǎng)就是所求點(diǎn)P運(yùn)動(dòng)的最短總路徑的長(zhǎng).(5分)
所以A'M'與x軸的交點(diǎn)為所求E點(diǎn),與直線x=3的交點(diǎn)為所求F點(diǎn).
可求得直線A'M'的解析式為y=
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x-

.
可得E點(diǎn)坐標(biāo)為(2,0),F(xiàn)點(diǎn)坐標(biāo)為(3,

).(7分)
由勾股定理可求出
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.
所以點(diǎn)P運(yùn)動(dòng)的最短總路徑(ME+EF+FA)的長(zhǎng)為
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.(8分)
點(diǎn)評(píng):本題著重考查了待定系數(shù)法求二次函數(shù)解析式,一次函數(shù)的解析式,圖形的對(duì)稱變換,求最短線段之和等重要知識(shí)點(diǎn),綜合性強(qiáng),能力要求極高.考查學(xué)生分類討論,數(shù)形結(jié)合的數(shù)學(xué)思想方法.