【答案】
分析:(1)根據(jù)圖表可以得到,拋物線經(jīng)過的四點(diǎn)的坐標(biāo),根據(jù)待定系數(shù)法,設(shè)y=ax
2+bx+c把其中三點(diǎn)的坐標(biāo),就可以解得函數(shù)的解析式.進(jìn)而就可以求出A、B、C的坐標(biāo).
(2)易證△ADG∽△AOC,AD=2-m,根據(jù)相似三角形的對應(yīng)邊的比相等,就可以用m表示出DG的長,再根據(jù)△BEF∽△BOC,就可以表示出BE,就可以得到OE,因而ED就可以表示出來.因而S與m的函數(shù)關(guān)系就可以得到.
(3)當(dāng)矩形DEFG的面積S取最大值時(shí),就是函數(shù)的值是最大值時(shí),根據(jù)二次函數(shù)的性質(zhì)就可以求出相應(yīng)的m的值.則矩形的四個(gè)頂點(diǎn)的坐標(biāo)就可以求出,根據(jù)待定系數(shù)法就可以求出直線DF的解析式.就可以求出直線DF與拋物線的交點(diǎn)的坐標(biāo),根據(jù)FM=k•DF,就可以表示出M的坐標(biāo),把M的坐標(biāo)代入函數(shù)就可以得到一個(gè)關(guān)于k的方程,求出k的值,判斷是否滿足函數(shù)的解析式.
解答:解:(1)解法一:設(shè)y=ax
2+bx+c(a≠0),
任取x,y的三組值代入,求出解析式y(tǒng)=
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x
2+x-4,
令y=0,求出x
1=-4,x
2=2;
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令x=0,得y=-4,
∴A、B、C三點(diǎn)的坐標(biāo)分別是A(2,0),B(-4,0),C(0,-4).
解法二:由拋物線P過點(diǎn)(1,-
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),(-3,-
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)可知,
拋物線P的對稱軸方程為x=-1,
又∵拋物線P過(2,0)、(-2,-4),
∴由拋物線的對稱性可知,
點(diǎn)A、B、C的坐標(biāo)分別為A(2,0),B(-4,0),C(0,-4).
(2)由題意,
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=
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,而AO=2,OC=4,AD=2-m,故DG=4-2m,
又
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=
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,EF=DG,得BE=4-2m,
∴DE=3m,
∴S
DEFG=DG•DE=(4-2m)3m=12m-6m
2(0<m<2).
(3)∵S
DEFG=12m-6m
2(0<m<2),
∴m=1時(shí),矩形的面積最大,且最大面積是6.
當(dāng)矩形面積最大時(shí),其頂點(diǎn)為D(1,0),G(1,-2),F(xiàn)(-2,-2),E(-2,0),
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設(shè)直線DF的解析式為y=kx+b,易知,k=
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,b=-
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,
∴y=
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x-
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,
又可求得拋物線P的解析式為:y=
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x
2+x-4,
令
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x-
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=
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x
2+x-4,可求出x=
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.
設(shè)射線DF與拋物線P相交于點(diǎn)N,則N的橫坐標(biāo)為
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,過N作x軸的垂線交x軸于H,
有
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=
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=
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=
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,
點(diǎn)M不在拋物線P上,即點(diǎn)M不與N重合時(shí),此時(shí)k的取值范圍是
k≠
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且k>0.
點(diǎn)評:本題主要考查了待定系數(shù)法求函數(shù)的解析式,并且本題還考查了函數(shù)交點(diǎn)坐標(biāo)的求法.就是求函數(shù)的解析式組成的方程組.