【答案】
分析:(1)根據(jù)圖象求出點(diǎn)A、B關(guān)于y軸的對(duì)稱(chēng)點(diǎn),然后利用待定系數(shù)法求二次函數(shù)解析式解答即可;
(2)分①AO、CO為一邊時(shí),矩形的長(zhǎng)與寬分別是CO、AO,然后根據(jù)矩形的周長(zhǎng)公式列式計(jì)算即可得解,②AC為一邊時(shí),先根據(jù)勾股定理求出AC的長(zhǎng)度,再利用三角形的面積求出點(diǎn)O到AC的長(zhǎng)度,即為矩形的寬,然后根據(jù)矩形的周長(zhǎng)公式列式計(jì)算即可得解;
(3)作點(diǎn)C關(guān)于直線BM的對(duì)稱(chēng)點(diǎn)C′,過(guò)C′作C′N(xiāo)⊥x軸交BM于點(diǎn)P,此時(shí)PC+PN最小,然后對(duì)稱(chēng)性求出拋物線C
1的解析式,再求出點(diǎn)M的坐標(biāo),然后利用待定系數(shù)法求直線解析式求出BM的解析式,再根據(jù)互相相垂直的直線的解析式的k值互為負(fù)倒數(shù)求出直線CC′的解析式,與直線BM的解析式聯(lián)立求出交點(diǎn)坐標(biāo),然后根據(jù)中點(diǎn)坐標(biāo)公式求出點(diǎn)C′的縱坐標(biāo),絕對(duì)值即為PC+PN的最小值.
解答:解:(1)根據(jù)圖形,點(diǎn)A、B關(guān)于y軸的對(duì)稱(chēng)點(diǎn)分別為(1,0)(-2,0),點(diǎn)C的坐標(biāo)為(0,-2),
設(shè)拋物線C
2的解析式為y=ax
2+bx+c,
則
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101191416941866852/SYS201311011914169418668024_DA/0.png)
,
解得
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101191416941866852/SYS201311011914169418668024_DA/1.png)
,
所以,拋物線C
2的解析式為y=x
2+x-2;
(2)①AO、CO為一邊時(shí),都是以CO、AO為長(zhǎng)與寬的矩形,
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101191416941866852/SYS201311011914169418668024_DA/images2.png)
∵A(-1,0)C(0,-2),
∴AO=1,CO=2,
∴周長(zhǎng)為:2(1+2)=2×3=6,
②AC為一邊時(shí),根據(jù)勾股定理,AC=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101191416941866852/SYS201311011914169418668024_DA/2.png)
=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101191416941866852/SYS201311011914169418668024_DA/3.png)
=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101191416941866852/SYS201311011914169418668024_DA/4.png)
,
根據(jù)三角形的面積,設(shè)點(diǎn)O到AC的距離為h,則
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101191416941866852/SYS201311011914169418668024_DA/5.png)
×
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101191416941866852/SYS201311011914169418668024_DA/6.png)
•h=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101191416941866852/SYS201311011914169418668024_DA/7.png)
×1×2,
解得h=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101191416941866852/SYS201311011914169418668024_DA/8.png)
,
所以,周長(zhǎng)為2(
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101191416941866852/SYS201311011914169418668024_DA/9.png)
+
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101191416941866852/SYS201311011914169418668024_DA/10.png)
)=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101191416941866852/SYS201311011914169418668024_DA/11.png)
;
(3)根據(jù)軸對(duì)稱(chēng)與最短距離問(wèn)題,作點(diǎn)C關(guān)于直線BM的對(duì)稱(chēng)點(diǎn)C′,過(guò)C′作C′N(xiāo)⊥x軸交BM于點(diǎn)P,此時(shí)PC+PN最小,
根據(jù)對(duì)稱(chēng)性,拋物線C
1的解析式為y=x
2-x-2=(x-
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101191416941866852/SYS201311011914169418668024_DA/12.png)
)
2-
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101191416941866852/SYS201311011914169418668024_DA/13.png)
,
所以,頂點(diǎn)M的坐標(biāo)為(
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101191416941866852/SYS201311011914169418668024_DA/14.png)
,-
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101191416941866852/SYS201311011914169418668024_DA/15.png)
),
設(shè)直線BM的解析式為y=kx+b,
則
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101191416941866852/SYS201311011914169418668024_DA/16.png)
,
解得
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101191416941866852/SYS201311011914169418668024_DA/17.png)
,
所以,直線BM的解析式為y=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101191416941866852/SYS201311011914169418668024_DA/18.png)
x-3,
∵直線CC′與直線BM垂直,且經(jīng)過(guò)點(diǎn)C(0,-2),
∴直線CC′的解析式為y=-
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101191416941866852/SYS201311011914169418668024_DA/19.png)
x-2,
聯(lián)立
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101191416941866852/SYS201311011914169418668024_DA/20.png)
,
解得
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101191416941866852/SYS201311011914169418668024_DA/21.png)
,
∴交點(diǎn)坐標(biāo),即CC′的中點(diǎn)坐標(biāo)為(
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101191416941866852/SYS201311011914169418668024_DA/22.png)
,-
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101191416941866852/SYS201311011914169418668024_DA/23.png)
),
根據(jù)中點(diǎn)坐標(biāo),C′的縱坐標(biāo)為2×(-
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101191416941866852/SYS201311011914169418668024_DA/24.png)
)-(-2)=-
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101191416941866852/SYS201311011914169418668024_DA/25.png)
+2=-
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101191416941866852/SYS201311011914169418668024_DA/26.png)
,
∵|-
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101191416941866852/SYS201311011914169418668024_DA/27.png)
|=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101191416941866852/SYS201311011914169418668024_DA/28.png)
,
∴PC+PN的最小值為
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101191416941866852/SYS201311011914169418668024_DA/29.png)
.
點(diǎn)評(píng):本題是對(duì)二次函數(shù)的綜合考查,主要利用了待定系數(shù)法求函數(shù)解析式,矩形的性質(zhì),利用軸對(duì)稱(chēng)求最短距離,以及中點(diǎn)坐標(biāo)公式,(3)中利用互相垂直的直線的解析式的k值互為負(fù)倒數(shù)求出CC′的直線是解題的關(guān)鍵,也是本題的難點(diǎn).