【答案】
分析:(1)先求出點(diǎn)A、B的坐標(biāo),再根據(jù)旋轉(zhuǎn)的性質(zhì)求出點(diǎn)C、D的坐標(biāo),然后利用待定系數(shù)法求拋物線解析式即可;
(2)根據(jù)拋物線解析式求出對(duì)稱軸解析式,然后求出點(diǎn)P的坐標(biāo),過(guò)點(diǎn)P作PQ⊥x軸,則PQ∥y軸,根據(jù)兩直線平行,內(nèi)錯(cuò)角相等可得∠OPQ=∠POC,然后利用點(diǎn)P的坐標(biāo),根據(jù)銳角的正切值的定義列式計(jì)算即可得解;
(3)根據(jù)點(diǎn)M在x軸上,且△ABM與△APD相似可知,點(diǎn)M一定在點(diǎn)A的右側(cè),然后求出AP、AB、AD的長(zhǎng)度,因?yàn)閷?duì)應(yīng)邊不明確,所以分①AP和AB是對(duì)應(yīng)邊,②AP和AM是對(duì)應(yīng)邊,然后根據(jù)相似三角形對(duì)應(yīng)邊成比例列式求出AM的長(zhǎng)度,再根據(jù)點(diǎn)A的坐標(biāo)求解即可.
解答:解:(1)當(dāng)y=0時(shí),
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131022165051882888256/SYS201310221650518828882023_DA/0.png)
x+1=0,解得x=-2,
當(dāng)x=0時(shí),y=1,
所以A(-2,0),B(0,1),
∵△AOB順時(shí)針旋轉(zhuǎn)90°得到△COD,
∴C(0,2),D(1,0),
∵拋物線y=ax
2+bx+c過(guò)點(diǎn)A、D、C,
∴
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131022165051882888256/SYS201310221650518828882023_DA/1.png)
,
解得
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131022165051882888256/SYS201310221650518828882023_DA/2.png)
,
∴拋物線解析式為y=-x
2-x+2;
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131022165051882888256/SYS201310221650518828882023_DA/images3.png)
(2)根據(jù)(1),拋物線對(duì)稱軸為x=-
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131022165051882888256/SYS201310221650518828882023_DA/3.png)
=-
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131022165051882888256/SYS201310221650518828882023_DA/4.png)
=-
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131022165051882888256/SYS201310221650518828882023_DA/5.png)
,
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131022165051882888256/SYS201310221650518828882023_DA/6.png)
×(-
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131022165051882888256/SYS201310221650518828882023_DA/7.png)
)+1=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131022165051882888256/SYS201310221650518828882023_DA/8.png)
,
∴點(diǎn)P的坐標(biāo)為(-
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131022165051882888256/SYS201310221650518828882023_DA/9.png)
,
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131022165051882888256/SYS201310221650518828882023_DA/10.png)
),
過(guò)點(diǎn)P作PQ⊥x軸于Q,則PQ∥y軸,
∴∠POC=∠OPQ,
∵tan∠OPQ=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131022165051882888256/SYS201310221650518828882023_DA/11.png)
=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131022165051882888256/SYS201310221650518828882023_DA/12.png)
,
∴tan∠POC=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131022165051882888256/SYS201310221650518828882023_DA/13.png)
;
(3)∵點(diǎn)M在x軸上,且△ABM與△APD相似,
∴點(diǎn)M必在點(diǎn)A的右側(cè),
AP=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131022165051882888256/SYS201310221650518828882023_DA/14.png)
=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131022165051882888256/SYS201310221650518828882023_DA/15.png)
,
AB=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131022165051882888256/SYS201310221650518828882023_DA/16.png)
=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131022165051882888256/SYS201310221650518828882023_DA/17.png)
,
AD=1-(-2)=1+2=3,
∵∠A=∠A,
∴①AP和AB是對(duì)應(yīng)邊時(shí),
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131022165051882888256/SYS201310221650518828882023_DA/18.png)
=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131022165051882888256/SYS201310221650518828882023_DA/19.png)
,
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131022165051882888256/SYS201310221650518828882023_DA/images21.png)
即
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131022165051882888256/SYS201310221650518828882023_DA/20.png)
=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131022165051882888256/SYS201310221650518828882023_DA/21.png)
,
解得AM=4,
設(shè)點(diǎn)M坐標(biāo)為(x,0),
則x-(-2)=4,
解得x=2,
所以點(diǎn)M的坐標(biāo)為(2,0),
②AP和AM是對(duì)應(yīng)邊時(shí),
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131022165051882888256/SYS201310221650518828882023_DA/22.png)
=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131022165051882888256/SYS201310221650518828882023_DA/23.png)
,
即
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131022165051882888256/SYS201310221650518828882023_DA/24.png)
=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131022165051882888256/SYS201310221650518828882023_DA/25.png)
,
解得AM=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131022165051882888256/SYS201310221650518828882023_DA/26.png)
,
設(shè)點(diǎn)M坐標(biāo)為(x,0),
則x-(-2)=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131022165051882888256/SYS201310221650518828882023_DA/27.png)
,
解得x=-
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131022165051882888256/SYS201310221650518828882023_DA/28.png)
,
所以點(diǎn)M的坐標(biāo)為(-
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131022165051882888256/SYS201310221650518828882023_DA/29.png)
,0),
綜上所述,存在點(diǎn)M(2,0)或(-
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131022165051882888256/SYS201310221650518828882023_DA/30.png)
,0),使△ABM與△APD相似.
點(diǎn)評(píng):本題是對(duì)二次函數(shù)的綜合考查,有旋轉(zhuǎn)變換的性質(zhì),待定系數(shù)法求函數(shù)解析式,銳角三角形函數(shù),兩點(diǎn)間的距離公式,相似三角形對(duì)應(yīng)邊成比例,綜合性較強(qiáng),求出二次函數(shù)解析式是解題的關(guān)鍵.