【答案】
分析:(1)根據(jù)點(diǎn)A、B的坐標(biāo)求出對(duì)稱軸解析式,從而得到頂點(diǎn)坐標(biāo),然后設(shè)頂點(diǎn)式解析式,把點(diǎn)A的坐標(biāo)代入計(jì)算即可得解;
(2)根據(jù)點(diǎn)B、C的坐標(biāo)求出OB、OC的長(zhǎng)度,利用勾股定理求出BC,再求出直線BC的解析式,根據(jù)三角形的面積,當(dāng)平行于BC的直線與拋物線只有一個(gè)交點(diǎn)時(shí)△MBC的面積最大,再根據(jù)平行直線的解析式的k值相等設(shè)出平行線的解析式,然后與拋物線聯(lián)立消掉y得到關(guān)于x的一元二次方程,然后利用根的判別式△=0求出直線的解析式,再根據(jù)等腰直角三角形的性質(zhì)求出點(diǎn)M到BC的距離,然后求解即可;
(3)根據(jù)拋物線的解析式設(shè)點(diǎn)P的坐標(biāo)為(x,x
2-2x-3),根據(jù)拋物線的對(duì)稱性以及點(diǎn)P在點(diǎn)Q的左側(cè),表示出EF=2(1-x),然后根據(jù)正方形的四條邊都相等列式,再分①x<-1時(shí)點(diǎn)P的縱坐標(biāo)是正數(shù),②-1<x<1時(shí),點(diǎn)P的縱坐標(biāo)是負(fù)數(shù)兩種情況去掉絕對(duì)值號(hào),解方程求解即可.
解答:解:(1)∵二次函數(shù)經(jīng)過(guò)點(diǎn)A(-1,0),B(3,0),
∴拋物線的對(duì)稱軸為直線x=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131103202545164126888/SYS201311032025451641268026_DA/0.png)
=1,
∵二次函數(shù)的最小值為-4,
∴頂點(diǎn)坐標(biāo)為(1,-4),
設(shè)頂點(diǎn)式解析式為y=a(x-1)
2-4,
則a(-1-1)
2-4=0,
解得a=1,
所以,二次函數(shù)解析式為y=(x-1)
2-4=x
2-2x-3,即y=x
2-2x-3;
(2)令x=0,則y=-3,
∴點(diǎn)C坐標(biāo)為(0,-3),
∴OB=3,OC=3,
∴△OBC是等腰直角三角形,
根據(jù)勾股定理,BC=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131103202545164126888/SYS201311032025451641268026_DA/1.png)
=3
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131103202545164126888/SYS201311032025451641268026_DA/2.png)
,
不難求出,直線BC的解析式為y=x-3,
根據(jù)三角形的面積,當(dāng)平行于直線BC直線與拋物線只有一個(gè)交點(diǎn)時(shí),點(diǎn)M到BC的距離最大,此時(shí),△MBC的面積最大,
設(shè)過(guò)點(diǎn)M的直線為y=x+e,
聯(lián)立
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131103202545164126888/SYS201311032025451641268026_DA/3.png)
,
整理得,x
2-3x-3-e=0,
△=b
2-4ac=9+4(3+e)=0,
解得e=-
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131103202545164126888/SYS201311032025451641268026_DA/4.png)
,
此時(shí),x
1+x
2=2m=-
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131103202545164126888/SYS201311032025451641268026_DA/5.png)
=3,
解得m=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131103202545164126888/SYS201311032025451641268026_DA/6.png)
,
n=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131103202545164126888/SYS201311032025451641268026_DA/7.png)
-
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131103202545164126888/SYS201311032025451641268026_DA/8.png)
=-
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131103202545164126888/SYS201311032025451641268026_DA/9.png)
,
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131103202545164126888/SYS201311032025451641268026_DA/images10.png)
所以,點(diǎn)M的坐標(biāo)為(
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131103202545164126888/SYS201311032025451641268026_DA/10.png)
,-
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131103202545164126888/SYS201311032025451641268026_DA/11.png)
),
點(diǎn)M到直線BC的距離為|-3-(-
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131103202545164126888/SYS201311032025451641268026_DA/12.png)
)|×
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131103202545164126888/SYS201311032025451641268026_DA/13.png)
=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131103202545164126888/SYS201311032025451641268026_DA/14.png)
,
S
△MBC=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131103202545164126888/SYS201311032025451641268026_DA/15.png)
×3
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131103202545164126888/SYS201311032025451641268026_DA/16.png)
×
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131103202545164126888/SYS201311032025451641268026_DA/17.png)
=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131103202545164126888/SYS201311032025451641268026_DA/18.png)
;
(3)設(shè)點(diǎn)P的坐標(biāo)為(x,x
2-2x-3),
∵點(diǎn)P在點(diǎn)Q的左側(cè),
∴EF=2(1-x),
∵四邊形PQFE為正方形,
∴|x
2-2x-3|=2(1-x),
根據(jù)函數(shù)圖象,①x<-1時(shí),x
2-2x-3=2(1-x),
整理得,x
2=5,
解得x
1=-
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131103202545164126888/SYS201311032025451641268026_DA/19.png)
,x
2=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131103202545164126888/SYS201311032025451641268026_DA/20.png)
(舍去),
x
2-2x-3=(-
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131103202545164126888/SYS201311032025451641268026_DA/21.png)
)
2-2×(-
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131103202545164126888/SYS201311032025451641268026_DA/22.png)
)-3=2
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131103202545164126888/SYS201311032025451641268026_DA/23.png)
+2,
所以,點(diǎn)P的坐標(biāo)為(-
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131103202545164126888/SYS201311032025451641268026_DA/24.png)
,2
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131103202545164126888/SYS201311032025451641268026_DA/25.png)
+2);
②-1<x<1時(shí),-(x
2-2x-3)=2(1-x),
整理得,x
2-4x-1=0,
解得x
1=2-
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131103202545164126888/SYS201311032025451641268026_DA/26.png)
,x
2=2+
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131103202545164126888/SYS201311032025451641268026_DA/27.png)
(舍去),
x
2-2x-3=(2-
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131103202545164126888/SYS201311032025451641268026_DA/28.png)
)
2-2×(2-
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131103202545164126888/SYS201311032025451641268026_DA/29.png)
)-3=2-2
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131103202545164126888/SYS201311032025451641268026_DA/30.png)
,
所以點(diǎn)P的坐標(biāo)為(2-
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131103202545164126888/SYS201311032025451641268026_DA/31.png)
,2-2
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131103202545164126888/SYS201311032025451641268026_DA/32.png)
);
綜上所述,存在點(diǎn)P(-
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131103202545164126888/SYS201311032025451641268026_DA/33.png)
,2
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131103202545164126888/SYS201311032025451641268026_DA/34.png)
+2)或(2-
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131103202545164126888/SYS201311032025451641268026_DA/35.png)
,2-2
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131103202545164126888/SYS201311032025451641268026_DA/36.png)
),使四邊形PQFE為正方形.
點(diǎn)評(píng):本題考查了二次函數(shù)綜合題型,主要考查了待定系數(shù)法求二次函數(shù)解析式,聯(lián)立兩函數(shù)解析式求交點(diǎn)坐標(biāo),等腰直角三角形的性質(zhì),正方形的四條邊都相等的性質(zhì),二次函數(shù)圖象上點(diǎn)的坐標(biāo)特征,綜合性較強(qiáng),難度較大,(1)先求出頂點(diǎn)坐標(biāo),再利用頂點(diǎn)式解析式求解更加簡(jiǎn)便,(2)注意兩平行直線解析式的k值相等的利用,(3)要分點(diǎn)P的縱坐標(biāo)是正數(shù)與負(fù)數(shù)兩種情況討論求解.