【答案】
分析:(1)將m=2代入原式,得到二次函數(shù)的頂點(diǎn)式,據(jù)此即可求出B點(diǎn)的坐標(biāo);
(2)延長(zhǎng)EA,交y軸于點(diǎn)F,證出△AFC≌△AED,進(jìn)而證出△ABF∽△DAE,利用相似三角形的性質(zhì),求出DE=4;
(3)①根據(jù)點(diǎn)A和點(diǎn)B的坐標(biāo),得到x=2m,y=-
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101193257762512377/SYS201311011932577625123023_DA/0.png)
m
2+m+4,將m=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101193257762512377/SYS201311011932577625123023_DA/1.png)
代入y=-
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101193257762512377/SYS201311011932577625123023_DA/2.png)
m
2+m+4,即可求出二次函數(shù)的表達(dá)式;
②作PQ⊥DE于點(diǎn)Q,則△DPQ≌△BAF,然后分(如圖1)和(圖2)兩種情況解答.
解答:解:(1)當(dāng)m=2時(shí),y=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101193257762512377/SYS201311011932577625123023_DA/3.png)
(x-2)
2+1,
把x=0代入y=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101193257762512377/SYS201311011932577625123023_DA/4.png)
(x-2)
2+1,得:y=2,
∴點(diǎn)B的坐標(biāo)為(0,2).
(2)延長(zhǎng)EA,交y軸于點(diǎn)F,
∵AD=AC,∠AFC=∠AED=90°,∠CAF=∠DAE,
∴△AFC≌△AED,
∴AF=AE,
∵點(diǎn)A(m,-
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101193257762512377/SYS201311011932577625123023_DA/5.png)
m
2+m),點(diǎn)B(0,m),
∴AF=AE=|m|,BF=m-(-
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101193257762512377/SYS201311011932577625123023_DA/6.png)
m
2+m)=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101193257762512377/SYS201311011932577625123023_DA/7.png)
m
2,
∵∠ABF=90°-∠BAF=∠DAE,∠AFB=∠DEA=90°,
∴△ABF∽△DAE,
∴
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101193257762512377/SYS201311011932577625123023_DA/8.png)
=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101193257762512377/SYS201311011932577625123023_DA/9.png)
,即:
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101193257762512377/SYS201311011932577625123023_DA/10.png)
=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101193257762512377/SYS201311011932577625123023_DA/11.png)
,
∴DE=4.
(3)①∵點(diǎn)A的坐標(biāo)為(m,-
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101193257762512377/SYS201311011932577625123023_DA/12.png)
m
2+m),
∴點(diǎn)D的坐標(biāo)為(2m,-
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101193257762512377/SYS201311011932577625123023_DA/13.png)
m
2+m+4),
∴x=2m,y=-
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101193257762512377/SYS201311011932577625123023_DA/14.png)
m
2+m+4,
∴y=-
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101193257762512377/SYS201311011932577625123023_DA/15.png)
•
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101193257762512377/SYS201311011932577625123023_DA/16.png)
+
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101193257762512377/SYS201311011932577625123023_DA/17.png)
+4,
∴所求函數(shù)的解析式為:y=-
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101193257762512377/SYS201311011932577625123023_DA/18.png)
x
2+
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101193257762512377/SYS201311011932577625123023_DA/19.png)
x+4,
②作PQ⊥DE于點(diǎn)Q,則△DPQ≌△BAF,
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101193257762512377/SYS201311011932577625123023_DA/images20.png)
(Ⅰ)當(dāng)四邊形ABDP為平行四邊形時(shí)(如圖1),
點(diǎn)P的橫坐標(biāo)為3m,
點(diǎn)P的縱坐標(biāo)為:(-
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101193257762512377/SYS201311011932577625123023_DA/20.png)
m
2+m+4)-(
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101193257762512377/SYS201311011932577625123023_DA/21.png)
m
2)=-
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101193257762512377/SYS201311011932577625123023_DA/22.png)
m
2+m+4,
把P(3m,-
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101193257762512377/SYS201311011932577625123023_DA/23.png)
m
2+m+4)的坐標(biāo)代入y=-
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101193257762512377/SYS201311011932577625123023_DA/24.png)
x
2+
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101193257762512377/SYS201311011932577625123023_DA/25.png)
x+4得:
-
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101193257762512377/SYS201311011932577625123023_DA/26.png)
m
2+m+4=-
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101193257762512377/SYS201311011932577625123023_DA/27.png)
×(3m)
2+
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101193257762512377/SYS201311011932577625123023_DA/28.png)
×(3m)+4,
解得:m=0(此時(shí)A,B,D,P在同一直線上,舍去)或m=8.
(Ⅱ)當(dāng)四邊形ABPD為平行四邊形時(shí)(如圖2),
點(diǎn)P的橫坐標(biāo)為m,
點(diǎn)P的縱坐標(biāo)為:(-
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101193257762512377/SYS201311011932577625123023_DA/29.png)
m
2+m+4)+(
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101193257762512377/SYS201311011932577625123023_DA/30.png)
m
2)=m+4,
把P(m,m+4)的坐標(biāo)代入y=-
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101193257762512377/SYS201311011932577625123023_DA/31.png)
x
2+
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101193257762512377/SYS201311011932577625123023_DA/32.png)
x+4得:
m+4=-
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101193257762512377/SYS201311011932577625123023_DA/33.png)
m
2+
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101193257762512377/SYS201311011932577625123023_DA/34.png)
m+4,
解得:m=0(此時(shí)A,B,D,P在同一直線上,舍去)或m=-8,
綜上所述:m的值為8或-8.
點(diǎn)評(píng):本題是二次函數(shù)綜合題,涉及四邊形的知識(shí),同時(shí)也是存在性問題,解答時(shí)要注意數(shù)形結(jié)合及分類討論.