【答案】
分析:(1)由直線y=-x+4交x軸的正半軸于點(diǎn)A,交y軸于點(diǎn)B,即可求得A,B的坐標(biāo),又由拋物線上有不同的兩點(diǎn)E(k+3,-k
2+1)和F(-k-1,-k
2+1)的縱坐標(biāo)相等,即可求得此拋物線的對(duì)稱軸,利用待定系數(shù)法即可求得解析式;
(2)分別從當(dāng)點(diǎn)P(x,x)在直線AB上時(shí)與當(dāng)點(diǎn)Q(
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101190913580278072/SYS201311011909135802780023_DA/0.png)
,
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101190913580278072/SYS201311011909135802780023_DA/1.png)
)在直線AB上時(shí)分析,即可求得x的取值范圍;
(3)首先求得當(dāng)點(diǎn)E(x,
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101190913580278072/SYS201311011909135802780023_DA/2.png)
)在直線AB上時(shí)x的值,再分別從當(dāng)2≤x<
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101190913580278072/SYS201311011909135802780023_DA/3.png)
時(shí)與當(dāng)
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101190913580278072/SYS201311011909135802780023_DA/4.png)
≤x≤4時(shí)去分析,注意三角形的面積求解方法與二次函數(shù)最大值的求解方法的應(yīng)用.
解答:解:(1)當(dāng)x=0時(shí),y=4,即B(0,4),
當(dāng)y=0時(shí),x=4,即A(4,0),
∵拋物線上有不同的兩點(diǎn)E(k+3,-k
2+1)和F(-k-1,-k
2+1)的縱坐標(biāo)相等,
∴點(diǎn)E和點(diǎn)F關(guān)于拋物線對(duì)稱軸對(duì)稱,
∴對(duì)稱軸x=-
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101190913580278072/SYS201311011909135802780023_DA/5.png)
=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101190913580278072/SYS201311011909135802780023_DA/6.png)
=1,
把點(diǎn)A,點(diǎn)B代入拋物線解析式中求得a=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101190913580278072/SYS201311011909135802780023_DA/7.png)
,b=1,c=4,
∴拋物線解析式為y=-
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101190913580278072/SYS201311011909135802780023_DA/8.png)
x
2+x+4;
(2)當(dāng)點(diǎn)P(x,x)在直線AB上時(shí),x=-x+4,
解得x=2,
當(dāng)點(diǎn)Q(
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101190913580278072/SYS201311011909135802780023_DA/9.png)
,
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101190913580278072/SYS201311011909135802780023_DA/10.png)
)在直線AB上時(shí),
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101190913580278072/SYS201311011909135802780023_DA/11.png)
=-
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101190913580278072/SYS201311011909135802780023_DA/12.png)
+4,
解得x=4.
所以,若正方形PEQF與直線AB有公共點(diǎn),則2≤x≤4.
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101190913580278072/SYS201311011909135802780023_DA/images13.png)
(3)當(dāng)點(diǎn)E(x,
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101190913580278072/SYS201311011909135802780023_DA/13.png)
)在直線AB上時(shí),(此時(shí)點(diǎn)F也在直線AB上)
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101190913580278072/SYS201311011909135802780023_DA/14.png)
=-x+4,
解得x=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101190913580278072/SYS201311011909135802780023_DA/15.png)
.
①當(dāng)2≤x<
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101190913580278072/SYS201311011909135802780023_DA/16.png)
時(shí),直線AB分別與PE、PF有交點(diǎn),設(shè)交點(diǎn)分別為C、D,
此時(shí),PC=x-(-x+4)=2x-4,
又PD=PC,
所以S
△PCD=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101190913580278072/SYS201311011909135802780023_DA/17.png)
PC
2=2(x-2)
2,
從而:S=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101190913580278072/SYS201311011909135802780023_DA/18.png)
x
2-2(x-2)
2=-
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101190913580278072/SYS201311011909135802780023_DA/19.png)
x
2+8x-8=-
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101190913580278072/SYS201311011909135802780023_DA/20.png)
(x-
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101190913580278072/SYS201311011909135802780023_DA/21.png)
)
2+
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101190913580278072/SYS201311011909135802780023_DA/22.png)
.
∵2≤
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101190913580278072/SYS201311011909135802780023_DA/23.png)
<
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101190913580278072/SYS201311011909135802780023_DA/24.png)
,
∴當(dāng)x=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101190913580278072/SYS201311011909135802780023_DA/25.png)
時(shí),S
max=
![](http://img.manfen5.com/res/czsx/web/STSource/20131101190913580278072/SYS201311011909135802780023_DA/26.png)
.
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101190913580278072/SYS201311011909135802780023_DA/images28.png)
②當(dāng)
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101190913580278072/SYS201311011909135802780023_DA/27.png)
≤x≤4時(shí),直線AB分別與QE、QF有交點(diǎn),設(shè)交點(diǎn)分別為M、N,
此時(shí),QN=(-
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101190913580278072/SYS201311011909135802780023_DA/28.png)
+4)-
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101190913580278072/SYS201311011909135802780023_DA/29.png)
=-x+4,
又QM=QN,
∴S
△QMN=
![](http://img.manfen5.com/res/czsx/web/STSource/20131101190913580278072/SYS201311011909135802780023_DA/30.png)
QN
2=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101190913580278072/SYS201311011909135802780023_DA/31.png)
(x-4)
2,
即S=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101190913580278072/SYS201311011909135802780023_DA/32.png)
(x-4)
2.
其中當(dāng)x=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101190913580278072/SYS201311011909135802780023_DA/33.png)
時(shí),S
max=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101190913580278072/SYS201311011909135802780023_DA/34.png)
.
綜合①②得,當(dāng)x=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101190913580278072/SYS201311011909135802780023_DA/35.png)
時(shí),S
max=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101190913580278072/SYS201311011909135802780023_DA/36.png)
.
點(diǎn)評(píng):此題考查了待定系數(shù)法求二次函數(shù)的解析式,函數(shù)自變量的取值范圍的確定、二次函數(shù)最大值的確定以及三角形面積的求解等知識(shí).此題綜合性很強(qiáng),注意數(shù)形結(jié)合思想與分類討論思想的應(yīng)用.