【答案】
分析:(1)首先過點(diǎn)A作AF⊥x軸于點(diǎn)F,由點(diǎn)A的坐標(biāo)為(2,
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101191512123846220/SYS201311011915121238462025_DA/0.png)
),∠ABC=30°,利用直角三角形的性質(zhì),即可求得點(diǎn)B與C的坐標(biāo),然后利用待定系數(shù)法即可求得二次函數(shù)的解析式;
(2)由(1),可求得拋物線的對(duì)稱軸,又由點(diǎn)B、C關(guān)于直線x=1對(duì)稱,求△AEC的周長的最小值,即為求AE+EC+AC的最小值,由對(duì)稱性知,AE+EC的最小值為AB的長,即當(dāng)點(diǎn)E運(yùn)動(dòng)到AB與拋物線對(duì)稱軸x=1的交點(diǎn)處時(shí),△AEC的周長最小,繼而可求得答案;
(3)首先連接結(jié)PO,設(shè)點(diǎn)P的坐標(biāo)為(t,-
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101191512123846220/SYS201311011915121238462025_DA/1.png)
),過點(diǎn)P分別向 x軸,y軸作垂線,垂足分別為N、G,由S
四邊形PDBC=S
△POC+S
△POD+S
△BOD,即可求得答案.
解答:解:(1)過點(diǎn)A作AF⊥x軸于點(diǎn)F,在Rt△AFB中,
∵∠ABC=30°,點(diǎn)A的坐標(biāo)為(2,
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101191512123846220/SYS201311011915121238462025_DA/2.png)
),
∴OF=2,AF=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101191512123846220/SYS201311011915121238462025_DA/3.png)
,∠ACF=60°,
∴BF=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101191512123846220/SYS201311011915121238462025_DA/4.png)
=3,
∴OB=BF-OF=3-2=1,
∴點(diǎn)B的點(diǎn)標(biāo)為(-1,0),
在Rt△AFC中,由∠ACF=60°,
∴FC=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101191512123846220/SYS201311011915121238462025_DA/5.png)
=1,
∴點(diǎn)C的坐標(biāo)為(3,0),
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101191512123846220/SYS201311011915121238462025_DA/images6.png)
將A、B、C三點(diǎn)坐標(biāo)分別代入y=ax
2+bx+c得:
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101191512123846220/SYS201311011915121238462025_DA/6.png)
,
解得:
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101191512123846220/SYS201311011915121238462025_DA/7.png)
,
∴該拋線的解析式為:y=-
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101191512123846220/SYS201311011915121238462025_DA/8.png)
x
2+
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101191512123846220/SYS201311011915121238462025_DA/9.png)
x+
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101191512123846220/SYS201311011915121238462025_DA/10.png)
…(4分)
(2)∵y=-
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101191512123846220/SYS201311011915121238462025_DA/11.png)
x
2+
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101191512123846220/SYS201311011915121238462025_DA/12.png)
x+
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101191512123846220/SYS201311011915121238462025_DA/13.png)
=-
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101191512123846220/SYS201311011915121238462025_DA/14.png)
(x-1)
2+
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101191512123846220/SYS201311011915121238462025_DA/15.png)
,
∴拋物線的對(duì)稱軸為x=1,
∴點(diǎn)B、C關(guān)于直線x=1對(duì)稱,
求△AEC的周長的最小值,即為求AE+EC+AC的最小值,
由對(duì)稱性知,AE+EC的最小值為AB的長,即當(dāng)點(diǎn)E運(yùn)動(dòng)到AB與拋物線對(duì)稱軸x=1的交點(diǎn)處時(shí),△AEC的周長最小,
由B(-1,0),A(2,
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101191512123846220/SYS201311011915121238462025_DA/16.png)
)可得AB所在直線的解析式為:y=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101191512123846220/SYS201311011915121238462025_DA/17.png)
x+
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101191512123846220/SYS201311011915121238462025_DA/18.png)
,…(7分)
當(dāng)x=1時(shí),y=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101191512123846220/SYS201311011915121238462025_DA/19.png)
,
故點(diǎn)E的坐標(biāo)為(1,
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101191512123846220/SYS201311011915121238462025_DA/20.png)
),
此時(shí),△AEC的周長最小,最小值為AB+AC=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101191512123846220/SYS201311011915121238462025_DA/21.png)
+2…(8分)
(3)連接結(jié)PO,設(shè)點(diǎn)P的坐標(biāo)為(t,-
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101191512123846220/SYS201311011915121238462025_DA/22.png)
)其中O<t<3,
過點(diǎn)P分別向 x軸,y軸作垂線,垂足分別為N、G,
由(1)知,點(diǎn)D的坐標(biāo)為(0,
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101191512123846220/SYS201311011915121238462025_DA/23.png)
)…(9分)
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101191512123846220/SYS201311011915121238462025_DA/images25.png)
則S
四邊形PDBC=S
△POC+S
△POD+S
△BOD=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101191512123846220/SYS201311011915121238462025_DA/24.png)
×OC×PN+
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101191512123846220/SYS201311011915121238462025_DA/25.png)
×OD×PG+
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101191512123846220/SYS201311011915121238462025_DA/26.png)
×OB×OD
=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101191512123846220/SYS201311011915121238462025_DA/27.png)
×3×(-
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101191512123846220/SYS201311011915121238462025_DA/28.png)
)+
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101191512123846220/SYS201311011915121238462025_DA/29.png)
×
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101191512123846220/SYS201311011915121238462025_DA/30.png)
×t+
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101191512123846220/SYS201311011915121238462025_DA/31.png)
×1×
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101191512123846220/SYS201311011915121238462025_DA/32.png)
=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101191512123846220/SYS201311011915121238462025_DA/33.png)
…(11分)
故當(dāng)
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101191512123846220/SYS201311011915121238462025_DA/34.png)
時(shí),四邊形PDBC的面積最大,最大面積為
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101191512123846220/SYS201311011915121238462025_DA/35.png)
,
此時(shí)點(diǎn)P的坐標(biāo)為(
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101191512123846220/SYS201311011915121238462025_DA/36.png)
,
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101191512123846220/SYS201311011915121238462025_DA/37.png)
).…(12分)
點(diǎn)評(píng):此題考查了待定系數(shù)法求函數(shù)的解析式、三角形周長最小值問題以及四邊形面積最小值問題.此題綜合性很強(qiáng),難度很大,解題的關(guān)鍵是注意數(shù)形結(jié)合思想與方程思想的應(yīng)用,注意準(zhǔn)確作出輔助線.