【答案】
分析:(1)根據(jù)直線y=-2x+4求出點A、C的坐標(biāo),再利用待定系數(shù)法求二次函數(shù)解析式解答即可;
(2)根據(jù)拋物線解析式求出點P的坐標(biāo),過點P作PD⊥y軸于D,根據(jù)點P、C的坐標(biāo)求出PD、CD,然后根據(jù)S
△APC=S
梯形APOD-S
△AOC-S
△PCD,列式求出△APC的面積,再根據(jù)拋物線解析式求出點B的坐標(biāo),從而得到AB的長度,然后利用三角形的面積公式求出△ABQ的點Q的縱坐標(biāo)的值,然后代入拋物線求解即可得到點Q的坐標(biāo);
(3)根據(jù)點E在x軸上,根據(jù)點M在直線y=-2x+4上,設(shè)點M的坐標(biāo)為(a,-2a+4),然后分①∠EMF=90°時,利用點M到坐標(biāo)軸的距離相等列式求解即可;②∠MFE=90°時,根據(jù)等腰直角三角形的性質(zhì),點M的橫坐標(biāo)的長度等于縱坐標(biāo)長度的一半,然后列式進行計算即可得解.
解答:解:(1)令x=0,則y=4,
令y=0,則-2x+4=0,解得x=2,
所以,點A(2,0),C(0,4),
∵拋物線y=-2x
2+bx+c經(jīng)過點A、C,
∴
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131103202710184444158/SYS201311032027101844441025_DA/0.png)
,
解得
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131103202710184444158/SYS201311032027101844441025_DA/1.png)
,
∴拋物線的解析式為:y=-2x
2+2x+4;
(2)∵y=-2x
2+2x+4=-2(x-
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131103202710184444158/SYS201311032027101844441025_DA/2.png)
)
2+
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131103202710184444158/SYS201311032027101844441025_DA/3.png)
,
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131103202710184444158/SYS201311032027101844441025_DA/images4.png)
∴點P的坐標(biāo)為(
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131103202710184444158/SYS201311032027101844441025_DA/4.png)
,
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131103202710184444158/SYS201311032027101844441025_DA/5.png)
),
如圖,過點P作PD⊥y軸于D,
又∵C(0,4),
∴PD=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131103202710184444158/SYS201311032027101844441025_DA/6.png)
,CD=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131103202710184444158/SYS201311032027101844441025_DA/7.png)
-4=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131103202710184444158/SYS201311032027101844441025_DA/8.png)
,
∴S
△APC=S
梯形APOD-S
△AOC-S
△PCD,
=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131103202710184444158/SYS201311032027101844441025_DA/9.png)
×(
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131103202710184444158/SYS201311032027101844441025_DA/10.png)
+2)×
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131103202710184444158/SYS201311032027101844441025_DA/11.png)
-
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131103202710184444158/SYS201311032027101844441025_DA/12.png)
×
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131103202710184444158/SYS201311032027101844441025_DA/13.png)
×
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131103202710184444158/SYS201311032027101844441025_DA/14.png)
-
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131103202710184444158/SYS201311032027101844441025_DA/15.png)
×2×4,
=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131103202710184444158/SYS201311032027101844441025_DA/16.png)
-
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131103202710184444158/SYS201311032027101844441025_DA/17.png)
-4,
=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131103202710184444158/SYS201311032027101844441025_DA/18.png)
,
令y=0,則-2x
2+2x+4=0,
解得x
1=-1,x
2=2,
∴點B的坐標(biāo)為(-1,0),
∴AB=2-(-1)=3,
設(shè)△ABQ的邊AB上的高為h,
∵△ABQ的面積等于△APC面積的4倍,
∴
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131103202710184444158/SYS201311032027101844441025_DA/19.png)
×3h=4×
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131103202710184444158/SYS201311032027101844441025_DA/20.png)
,
解得h=4,
∵4<
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131103202710184444158/SYS201311032027101844441025_DA/21.png)
,
∴點Q可以在x軸的上方也可以在x軸的下方,
即點Q的縱坐標(biāo)為4或-4,
當(dāng)點Q的縱坐標(biāo)為4時,-2x
2+2x+4=4,
解得x
1=0,x
2=1,
此時,點Q的坐標(biāo)為(0,4)或(1,4),
當(dāng)點Q的縱坐標(biāo)為-4時,-2x
2+2x+4=-4,
解得x
1=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131103202710184444158/SYS201311032027101844441025_DA/22.png)
,x
2=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131103202710184444158/SYS201311032027101844441025_DA/23.png)
,
此時點Q的坐標(biāo)為(
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131103202710184444158/SYS201311032027101844441025_DA/24.png)
,-4)或(
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131103202710184444158/SYS201311032027101844441025_DA/25.png)
,-4),
綜上所述,存在點Q(0,4)或(1,4)或(
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131103202710184444158/SYS201311032027101844441025_DA/26.png)
,-4)或(
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131103202710184444158/SYS201311032027101844441025_DA/27.png)
,-4);
(3)存在.
理由如下:如圖,∵點M在直線y=-2x+4上,
∴設(shè)點M的坐標(biāo)為(a,-2a+4),
①∠EMF=90°時,∵△MEF是等腰直角三角形,
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131103202710184444158/SYS201311032027101844441025_DA/images29.png)
∴|a|=|-2a+4|,
即a=-2a+4或a=-(-2a+4),
解得a=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131103202710184444158/SYS201311032027101844441025_DA/28.png)
或a=4,
∴點F坐標(biāo)為(0,
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131103202710184444158/SYS201311032027101844441025_DA/29.png)
)時,點M的坐標(biāo)為(
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131103202710184444158/SYS201311032027101844441025_DA/30.png)
,
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131103202710184444158/SYS201311032027101844441025_DA/31.png)
),
點F坐標(biāo)為(0,-4)時,點M的坐標(biāo)為(4,-4);
②∠MFE=90°時,∵△MEF是等腰直角三角形,
∴|a|=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131103202710184444158/SYS201311032027101844441025_DA/32.png)
|-2a+4|,
即a=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131103202710184444158/SYS201311032027101844441025_DA/33.png)
(-2a+4),
解得a=1,
-2a+4=2×1=2,
此時,點F坐標(biāo)為(0,1),點M的坐標(biāo)為(1,2),
或a=-
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131103202710184444158/SYS201311032027101844441025_DA/34.png)
(-2a+4),
此時無解,
綜上所述,點F坐標(biāo)為(0,
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131103202710184444158/SYS201311032027101844441025_DA/35.png)
)時,點M的坐標(biāo)為(
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131103202710184444158/SYS201311032027101844441025_DA/36.png)
,
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131103202710184444158/SYS201311032027101844441025_DA/37.png)
),
點F坐標(biāo)為(0,-4)時,點M的坐標(biāo)為(4,-4);
點F坐標(biāo)為(0,1),點M的坐標(biāo)為(1,2).
點評:本題是二次函數(shù)綜合題型,主要考查了直線與坐標(biāo)軸的交點的求解,待定系數(shù)法求二次函數(shù)解析式,二次函數(shù)頂點坐標(biāo)的求解,二次函數(shù)圖象上點的坐標(biāo)特征,三角形的面積,等腰直角三角形的性質(zhì),綜合性較強,難度較大,(3)要注意分情況討論.