【答案】
分析:(1)先根據(jù)直線的解析式求出A點(diǎn)的坐標(biāo),然后根據(jù)M點(diǎn)的坐標(biāo),用頂點(diǎn)式二次函數(shù)通式來(lái)設(shè)拋物線的解析式,將A點(diǎn)的坐標(biāo)代入拋物線中即可求出二次函數(shù)的解析式.
(2)PQ的長(zhǎng),實(shí)際是直線AB的函數(shù)值與拋物線的函數(shù)值的差.據(jù)此可得出l,x的函數(shù)關(guān)系式.
(3)要想使PQMA為梯形,只有一種情況,即MQ∥AP,可根據(jù)直線AB的斜率和M點(diǎn)的坐標(biāo)求出直線MQ的解析式,聯(lián)立拋物線的解析式即可求出Q點(diǎn)的坐標(biāo),將Q的橫坐標(biāo)代入直線AB中即可求出P點(diǎn)的坐標(biāo),得出然后可根據(jù)A,M,Q,P的坐標(biāo)求出AP,MQ,AM的長(zhǎng),進(jìn)而可求出梯形AMQP的面積(可設(shè)直線AB與x軸的交點(diǎn)為N,利用∠ANO=45°來(lái)求個(gè)各邊的長(zhǎng)).
解答:解:(1)依題意,設(shè)二次函數(shù)的解析式為y=a(x-2)
2,
由于直線y=x+2與y軸交于(0,2),
∴x=0,y=2
滿足y=a(x-2)
2,于是求得a=
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,
二次函數(shù)的解析式為y=
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(x-2)
2;
(2)設(shè)P點(diǎn)坐標(biāo)為:P(x,y),則Q點(diǎn)坐標(biāo)為(x,
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x
2-2x+2)
依題意得,PQ=l=(x+2)-
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(x-2)
2=-
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+3x,
由
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,
求得點(diǎn)B的坐標(biāo)為(6,8),
∴0<x<6;
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(3)由(2)知P的橫坐標(biāo)為0<x<6時(shí),必有對(duì)應(yīng)的點(diǎn)Q在拋物線上;
反之,Q的橫坐標(biāo)為0<x<6時(shí),在線段AB上必有一點(diǎn)P與之對(duì)應(yīng).
假設(shè)存在符合條件的點(diǎn)P,由題意得AM與PQ不會(huì)平行,
因此梯形的兩底只能是AP與MQ,
∵過(guò)點(diǎn)M(2,0)且平行AB的直線方程為y=x-2,
由
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,
消去y得:x
2-6x+8=0,即(x-2)(x-4)=0,
解得x=2或x=4,
∵當(dāng)x=2時(shí),P點(diǎn)、Q點(diǎn)、M點(diǎn) 三點(diǎn)共線,與A點(diǎn)只能構(gòu)成三角形,而不能構(gòu)成梯形;
∴x=2這個(gè)解舍去.
∴過(guò)M點(diǎn)的直線與拋物線的另一交點(diǎn)為(4,2),
∵此交點(diǎn)橫坐標(biāo)4,落在0<x<6范圍內(nèi),
∴Q的坐標(biāo)為(4,2)時(shí),P(4,6)符合條件,
即存在符合條件的點(diǎn)P,其坐標(biāo)為(4,6),
設(shè)直線AB與x軸交于N,由條件可知,△ANM是等腰直角三角形,即AM=AN=2
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,
AP=PN-AN=6
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-2
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=4
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,MQ=2
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,
AM為梯形PQMA的高,
故S
梯形PQMA=
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(2
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+4
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)•2
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=12.
點(diǎn)評(píng):本題考查了二次函數(shù)解析式的確定、圖形的面積求法、函數(shù)圖象交點(diǎn)、梯形的判定等知識(shí)及綜合應(yīng)用知識(shí)、解決問(wèn)題的能力.