【答案】
分析:(1)首先根據(jù)矩形的性質(zhì)以及A、C點(diǎn)的坐標(biāo)確定點(diǎn)B的坐標(biāo),再利用待定系數(shù)法確定該拋物線的解析式.
(2)設(shè)拋物線的對(duì)稱軸與x軸的交點(diǎn)為D,若矩形的頂點(diǎn)恰好落在拋物線對(duì)稱軸上時(shí),該頂點(diǎn)、O、D正好構(gòu)成一個(gè)直角三角形,由勾股定理即可確定這個(gè)頂點(diǎn)的坐標(biāo).
(3)觀察圖示可知:當(dāng)點(diǎn)E運(yùn)動(dòng)到y(tǒng)軸負(fù)半軸上時(shí),CE最長(zhǎng),找出了這個(gè)關(guān)鍵位置,解答問(wèn)題就簡(jiǎn)單多了.
解答:![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101191401970408980/SYS201311011914019704089024_DA/images0.png)
解:(1)∵矩形OABC,A(2
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101191401970408980/SYS201311011914019704089024_DA/0.png)
,0),C(0,2),∴B(2
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101191401970408980/SYS201311011914019704089024_DA/1.png)
,2).
∴拋物線的對(duì)稱軸為x=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101191401970408980/SYS201311011914019704089024_DA/2.png)
.∴b=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101191401970408980/SYS201311011914019704089024_DA/3.png)
.
∴二次函數(shù)的解析式為:y=-x
2+2
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101191401970408980/SYS201311011914019704089024_DA/4.png)
x+2.
(2)①當(dāng)頂點(diǎn)A落在對(duì)稱軸上時(shí),設(shè)點(diǎn)A的對(duì)應(yīng)點(diǎn)為點(diǎn)A′,連接OA′,
設(shè)對(duì)稱軸x=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101191401970408980/SYS201311011914019704089024_DA/5.png)
與x軸交于點(diǎn)D,∴OD=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101191401970408980/SYS201311011914019704089024_DA/6.png)
.
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101191401970408980/SYS201311011914019704089024_DA/images8.png)
∴OA′=OA=2
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101191401970408980/SYS201311011914019704089024_DA/7.png)
.
在Rt△OA′D中,根據(jù)勾股定理A′D=3.
∴A′(
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101191401970408980/SYS201311011914019704089024_DA/8.png)
,-3).
②當(dāng)頂點(diǎn)落C對(duì)稱軸上時(shí)(如圖),設(shè)點(diǎn)C的對(duì)應(yīng)點(diǎn)為點(diǎn)C′,連接OC′,
在Rt△OC′D中,根據(jù)勾股定理C′D=1.
∴C′(
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101191401970408980/SYS201311011914019704089024_DA/9.png)
,1).
③當(dāng)頂點(diǎn)落B對(duì)稱軸上時(shí),同理①可求出點(diǎn)B′的坐標(biāo)是(
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101191401970408980/SYS201311011914019704089024_DA/10.png)
,-3);
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101191401970408980/SYS201311011914019704089024_DA/images13.png)
(3)如右圖,設(shè)AC、OB的交點(diǎn)為E;
在Rt△OAB中,OA=2
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101191401970408980/SYS201311011914019704089024_DA/11.png)
,AB=2,∴∠BOA=30°,OE=AB=2;
在OE旋轉(zhuǎn)過(guò)程中,可將點(diǎn)E的軌跡看作是以O(shè)為圓心,以O(shè)E為半徑的圓(旋轉(zhuǎn)角度:0°~180°);
由圖可看出,當(dāng)點(diǎn)E運(yùn)動(dòng)到y(tǒng)軸負(fù)半軸上時(shí)(即點(diǎn)E′的位置),CE最長(zhǎng);
此時(shí),旋轉(zhuǎn)的角度:∠EOE′=∠BOA+90°=30°+90°=120°;
CE的最長(zhǎng)值:CE′=OC+OE′=2+2=4;
故填:120°,4.
點(diǎn)評(píng):該題主要考查了函數(shù)解析式的確定、矩形的性質(zhì)、圖形的旋轉(zhuǎn)以及勾股定理的應(yīng)用等綜合知識(shí);題目的難度不大,需要注意數(shù)形結(jié)合思想的應(yīng)用.