解:(1)直線AE中,y=mx+n,則E(0,n);
∵正方形ABCD,
∴AB=BC,則tan∠CAB=1,
∴OA=OE=n,即A(-n,0);
△AOE中,AO=n,OE=n,
則S
△AOE=
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OA•OE=
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,又S
正方形ABCD=AB
2,
∵S
△AOE=2S
正方形ABCD,
∴
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n
2=2AB
2,即AB=
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n,
故OB=OA-AB=n-
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n=
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n,即B(-
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n,0);
∴A(-n,0),B(-
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n,0).
∵G是拋物線的頂點,且A(-n,0),B(-
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n,0),
∴G點的橫坐標(biāo)為-
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n;
易知G是線段AC的中點,故BC=AB=2y
G,
∴G點的縱坐標(biāo)為
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n;
即G(-
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n,
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n);
設(shè)拋物線的解析式為y=a(x+
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n)
2+
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n,將A(-n,0)代入上式,得:
a×
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n
2+
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n=0,即a=-
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;
∴y=-
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(x+
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n)
2+
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n=-
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x
2-6x-2n;
故abc=(-
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)×(-6)×(-2n)=-48.
(2)根據(jù)(1)得到的拋物線解析式,易知F(0,-2n);
∵E(0,n),A(-n,0),G(-
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n,
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n),
∴S
△AEF=
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EF•OA=
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,S
△EGF=
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EF•|x
G|=
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n
2,
∴S
△AGF=S
△AEF-S
△EGF=
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n
2-
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n
2=
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n
2,又n>
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,
故S
△AGF的范圍為:S
△AGF>
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.
分析:(1)根據(jù)直線AE的解析式可得到點E的坐標(biāo),根據(jù)正方形ABCD的邊長相等得到AB=BC,即AO=OE,由此可求得點A的坐標(biāo);易求得△AOE的面積,即可得到正方形ABCD的面積,由于AB=BC,可用AB表示出正方形ABCD的面積,進而可得到AB的值(含n的表達式),由此可確定點B的坐標(biāo).由于點G是拋物線的頂點,即在拋物線的對稱軸上,根據(jù)A、B的坐標(biāo),可求得點G的橫坐標(biāo),而G點在直線AE上,那么G點的縱坐標(biāo)應(yīng)該是AB的
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(由于AB=BC=2y
G),由此可確定點G的坐標(biāo);可將拋物線設(shè)為頂點坐標(biāo)式,將A或B的坐標(biāo)代入其中,即可求出含n的拋物線解析式,進而可求出abc的值;
(2)△AGF的面積無法直接求出,分析圖形后可知△AGF的面積為△AEF、△EGF的面積差,這兩個三角形的頂點的坐標(biāo)都已求出,即可得到△AGF的面積表達式(含n的式子),根據(jù)已知的n的取值范圍,即可求得△AGF的面積范圍.
點評:此題是二次函數(shù)的綜合題,涉及到函數(shù)圖象與坐標(biāo)軸交點坐標(biāo)的求法、函數(shù)解析式的確定、圖形面積的求法等重要知識,由于本題中大部分?jǐn)?shù)據(jù)都是字母,乍看之下無從下手,但是只要將字母當(dāng)做已知數(shù)來對待,即可按照常規(guī)思路解決問題.