【答案】
分析:(1)先根據(jù)直線AC的解析式求出A、C兩點(diǎn)的坐標(biāo),然后根據(jù)A、B、C三點(diǎn)的坐標(biāo)用待定系數(shù)法即可求出拋物線的解析式.
(2)根據(jù)拋物線的解析式可求出M點(diǎn)的坐標(biāo),由于四邊形OAMC不是規(guī)則的四邊形,因此可過M作x軸的垂線,將四邊形OAMC分成一個直角三角形和一個直角梯形來求解.
(3)①如果DE∥OC,此時點(diǎn)D,E應(yīng)分別在線段OA,CA上,先求出這個區(qū)間t的取值范圍,然后根據(jù)平行線分線段成比例定理,求出此時t的值,然后看t的值是否符合此種情況下t的取值范圍.如果符合則這個t的值就是所求的值,如果不符合,那么就說明不存在這樣的t.
②本題要分三種情況進(jìn)行討論:
當(dāng)E在OC上,D在OA上,即當(dāng)0<t≤1時,此時S=
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OE•OD,由此可得出關(guān)于S,t的函數(shù)關(guān)系式;
當(dāng)E在CA上,D在OA上,即當(dāng)1<t≤2時,此時S=
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OD×E點(diǎn)的縱坐標(biāo).由此可得出關(guān)于S,t的函數(shù)關(guān)系式;
當(dāng)E,D都在CA上時,即當(dāng)2<t<
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相遇時用的時間,此時S=S
△AOE-S
△AOD,由此可得出S,t的函數(shù)關(guān)系式;
綜上所述,可得出不同的t的取值范圍內(nèi),函數(shù)的不同表達(dá)式.
③根據(jù)②的函數(shù)即可得出S的最大值.
解答:解:(1)令y=0,則x=3,
∴A(3,0),C(0,4),
∵二次函數(shù)的圖象過點(diǎn)C(0,4),
∴可設(shè)二次函數(shù)的關(guān)系式為y=ax
2+bx+4.
又∵該函數(shù)圖象過點(diǎn)A(3,0),B(-1,0),
∴
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,
解得a=-
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,b=
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.
∴所求二次函數(shù)的關(guān)系式為y=-
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x
2+
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x+4.
(2)∵y=-
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x
2+
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x+4
=-
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(x-1)
2+
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∴頂點(diǎn)M的坐標(biāo)為(1,
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)
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過點(diǎn)M作MF⊥x軸于F
∴S
四邊形AOCM=S
△AFM+S
梯形FOCM=
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×(3-1)×
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+
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×(4+
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)×1
=10
∴四邊形AOCM的面積為10.
(3)①不存在DE∥OC
∵若DE∥OC,則點(diǎn)D,E應(yīng)分別在線段OA,CA上,此時1<t<2,在Rt△AOC中,AC=5.
設(shè)點(diǎn)E的坐標(biāo)為(x
1,y
1)
∴
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=
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,
∴
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∵DE∥OC,
∴
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∴
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∵t=
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>2,不滿足1<t<2.
∴不存在DE∥OC.
②根據(jù)題意得D,E兩點(diǎn)相遇的時間為
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(秒)
現(xiàn)分情況討論如下:
(�。┊�(dāng)0<t≤1時,S=
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×
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t•4t=3t
2;
(ⅱ)當(dāng)1<t≤2時,設(shè)點(diǎn)E的坐標(biāo)為(x
2,y
2)
∴
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,
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∴
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∴S=
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×
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t×
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=-
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t
2+
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t;
(ⅲ)當(dāng)2<t<
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時,
設(shè)點(diǎn)E的坐標(biāo)為(x
3,y
3),類似ⅱ可得
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設(shè)點(diǎn)D的坐標(biāo)為(x
4,y
4)
∴
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,
∴
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∴S=S
△AOE-S
△AOD
=
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×3×
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-
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×3×
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=-
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t+
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.
③當(dāng)0<t≤1時,S=
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×
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t•4t=3t
2,函數(shù)的最大值是3;
當(dāng)1<t≤2時,S=-
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t
2+
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t.函數(shù)的最大值是:
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,
當(dāng)2<t<
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時,S=-
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t+
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,0<S<
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.
∴S
=
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.
點(diǎn)評:本題著重考查了待定系數(shù)法求二次函數(shù)解析式以及二次函數(shù)的應(yīng)用等知識點(diǎn),綜合性較強(qiáng),考查學(xué)生分類討論,數(shù)形結(jié)合的數(shù)學(xué)思想方法.