【答案】
分析:(1)把A、B、O的坐標代入得到方程組,求出方程組的解即可;
(2)根據(jù)對稱軸求出O、B關(guān)于對稱軸對稱,根據(jù)勾股定理求出AB即可;
(3)①若OB∥AP,根據(jù)點A與點P關(guān)于直線x=1對稱,由A(-2,-4),得出P的坐標;②若OA∥BP,設直線OA的表達式為y=kx,設直線BP的表達式為y=2x+m,由B(2,0)求出直線BP的表達式為y=2x-4,得到方程組,求出方程組的解即可;③若AB∥OP,設直線AB的表達式為y=kx+m,求出直線AB,得到方程組求出方程組的解即可;
解答:解:(1)由OB=2,可知B(2,0),
將A(-2,-4),B(2,0),O(0,0)三點坐標代入拋物線y=ax
2+bx+c,
得
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解得:
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∴拋物線的函數(shù)表達式為
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.
答:拋物線的函數(shù)表達式為
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.
(2)由
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,
可得,拋物線的對稱軸為直線x=1,
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且對稱軸x=1是線段OB的垂直平分線,
連接AB交直線x=1于點M,M點即為所求.
∴MO=MB,則MO+MA=MA+MB=AB
作AC⊥x軸,垂足為C,則AC=4,BC=4,∴AB=
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∴MO+MA的最小值為
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.
答:MO+MA的最小值為
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.
(3)①若OB∥AP,此時點A與點P關(guān)于直線x=1對稱,
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由A(-2,-4),得P(4,-4),則得梯形OAPB.
②若OA∥BP,
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設直線OA的表達式為y=kx,由A(-2,-4)得,y=2x.
設直線BP的表達式為y=2x+m,由B(2,0)得,0=4+m,即m=-4,
∴直線BP的表達式為y=2x-4
由
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,解得x
1=-4,x
2=2(不合題意,舍去)
當x=-4時,y=-12,∴點P(-4,-12),則得梯形OAPB.
③若AB∥OP,
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設直線AB的表達式為y=kx+m,則
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,
解得
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,∴AB的表達式為y=x-2.
∵AB∥OP,
∴直線OP的表達式為y=x.
由
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,得 x
2=0,解得x=0,
(不合題意,舍去),此時點P不存在.
綜上所述,存在兩點P(4,-4)或P(-4,-12)
使得以點P與點O、A、B為頂點的四邊形是梯形.
答:在此拋物線上,存在點P,使得以點P與點O、A、B為頂點的四邊形是梯形,點P的坐標是(4,-4)或(-4,-12).
點評:本題主要考查對梯形,解二元二次方程組,解一元二次方程,二次函數(shù)的性質(zhì),用待定系數(shù)法求一次函數(shù)的解析式等知識點的理解和掌握,綜合運用性質(zhì)進行計算是解此題的關(guān)鍵.