【答案】
分析:(1)根據(jù)二次函數(shù)的對(duì)稱性,利用點(diǎn)B的坐標(biāo)與對(duì)稱軸求解;
(2)利用待定系數(shù)法求二次函數(shù)解析式列式計(jì)算即可得解;
(3)假設(shè)存在,根據(jù)拋物線解析式設(shè)點(diǎn)P的坐標(biāo)為(x,-
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101192017056886384/SYS201311011920170568863023_DA/0.png)
x
2+6x-9),過點(diǎn)C作CE⊥AB于點(diǎn)E,過點(diǎn)P作PF⊥x軸于點(diǎn)F,則S
四邊形ABPC=S
△ACE+S
梯形CEFP+S
△BPF,再根據(jù)三角形的面積公式與梯形的面積公式列式整理,然后根據(jù)二次函數(shù)的最值問題解答;
(4)根據(jù)A、B的坐標(biāo)求出AB的長(zhǎng)度,根據(jù)勾股定理求出BC的值,再分①BN=MN時(shí),過點(diǎn)N作ND⊥BM于點(diǎn)D,然后利用∠ABC的余弦列式計(jì)算即可得解,②BN=BM時(shí),用t表示出BM、BN,列出方程計(jì)算即可得解,③BM=MN時(shí),過點(diǎn)M作MH⊥BN于點(diǎn)H,然后利用∠ABC的余弦列式計(jì)算即可得解.
解答:解:(1)∵B點(diǎn)坐標(biāo)為(6,0),拋物線對(duì)稱軸為直線x=4,
4×2-6=2,
∴點(diǎn)A的坐標(biāo)為(2,0);
(2)設(shè)拋物線解析式為y=ax
2+bx+c,
∵A(2,0),B(6,0),C(3,
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101192017056886384/SYS201311011920170568863023_DA/1.png)
),
∴
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101192017056886384/SYS201311011920170568863023_DA/2.png)
,
解得
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101192017056886384/SYS201311011920170568863023_DA/3.png)
,
∴拋物線解析式為y=-
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101192017056886384/SYS201311011920170568863023_DA/4.png)
x
2+6x-9;
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101192017056886384/SYS201311011920170568863023_DA/images5.png)
(3)存在.理由如下:
如圖,設(shè)存在點(diǎn)P(x,-
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101192017056886384/SYS201311011920170568863023_DA/5.png)
x
2+6x-9),使得四邊形ABPC的面積最大,
過點(diǎn)C作CE⊥AB于點(diǎn)E,過點(diǎn)P作PF⊥x軸于點(diǎn)F,
∵A(2,0),B(6,0),C(3,
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101192017056886384/SYS201311011920170568863023_DA/6.png)
),
∴S
四邊形ABPC=S
△ACE+S
梯形CEFP+S
△BPF=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101192017056886384/SYS201311011920170568863023_DA/7.png)
×(3-2)×
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101192017056886384/SYS201311011920170568863023_DA/8.png)
+
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101192017056886384/SYS201311011920170568863023_DA/9.png)
(
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101192017056886384/SYS201311011920170568863023_DA/10.png)
-
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101192017056886384/SYS201311011920170568863023_DA/11.png)
x
2+6x-9)×(x-3)+
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101192017056886384/SYS201311011920170568863023_DA/12.png)
×(6-x)×(-
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101192017056886384/SYS201311011920170568863023_DA/13.png)
x
2+6x-9)
=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101192017056886384/SYS201311011920170568863023_DA/14.png)
+
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101192017056886384/SYS201311011920170568863023_DA/15.png)
(x-3)+
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101192017056886384/SYS201311011920170568863023_DA/16.png)
(-
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101192017056886384/SYS201311011920170568863023_DA/17.png)
x
2+6x-9)×(x-3)+
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101192017056886384/SYS201311011920170568863023_DA/18.png)
×(6-x)×(-
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101192017056886384/SYS201311011920170568863023_DA/19.png)
x
2+6x-9)
=-
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101192017056886384/SYS201311011920170568863023_DA/20.png)
(x
2-9x+14)
=-
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101192017056886384/SYS201311011920170568863023_DA/21.png)
(x-
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101192017056886384/SYS201311011920170568863023_DA/22.png)
)
2+
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101192017056886384/SYS201311011920170568863023_DA/23.png)
,
∵3<
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101192017056886384/SYS201311011920170568863023_DA/24.png)
<6,
∴當(dāng)x=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101192017056886384/SYS201311011920170568863023_DA/25.png)
時(shí),四邊形ABPC的面積有最大值,最大值為
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101192017056886384/SYS201311011920170568863023_DA/26.png)
,
此時(shí),-
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101192017056886384/SYS201311011920170568863023_DA/27.png)
x
2+6x-9=-
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101192017056886384/SYS201311011920170568863023_DA/28.png)
×(
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101192017056886384/SYS201311011920170568863023_DA/29.png)
)
2+6×
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101192017056886384/SYS201311011920170568863023_DA/30.png)
-9=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101192017056886384/SYS201311011920170568863023_DA/31.png)
,
∴點(diǎn)P的坐標(biāo)為(
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101192017056886384/SYS201311011920170568863023_DA/32.png)
,
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101192017056886384/SYS201311011920170568863023_DA/33.png)
);
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101192017056886384/SYS201311011920170568863023_DA/images35.png)
(4)∵A(2,0),B(6,0),
∴AB=6-2=4,
∵B(6,0),C(3,
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101192017056886384/SYS201311011920170568863023_DA/34.png)
),
∴BC=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101192017056886384/SYS201311011920170568863023_DA/35.png)
=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101192017056886384/SYS201311011920170568863023_DA/36.png)
.
①BN=MN時(shí),如圖,過點(diǎn)N作ND⊥BM于點(diǎn)D,則BD=MD=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101192017056886384/SYS201311011920170568863023_DA/37.png)
(4-t),
cos∠ABC=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101192017056886384/SYS201311011920170568863023_DA/38.png)
=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101192017056886384/SYS201311011920170568863023_DA/39.png)
,
解得t=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101192017056886384/SYS201311011920170568863023_DA/40.png)
,
②BN=BM時(shí),如圖,BM=4-t,BN=2t,
所以,4-t=2t,
解得t=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101192017056886384/SYS201311011920170568863023_DA/41.png)
,
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101192017056886384/SYS201311011920170568863023_DA/images44.png)
③BM=MN時(shí),如圖,過點(diǎn)M作MH⊥BN于點(diǎn)H,
則BH=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101192017056886384/SYS201311011920170568863023_DA/42.png)
BN=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101192017056886384/SYS201311011920170568863023_DA/43.png)
×2t=t,
BM=4-t,
cos∠ABC=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101192017056886384/SYS201311011920170568863023_DA/44.png)
=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101192017056886384/SYS201311011920170568863023_DA/45.png)
,
解得t=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101192017056886384/SYS201311011920170568863023_DA/46.png)
,
綜上所述,當(dāng)t為
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101192017056886384/SYS201311011920170568863023_DA/47.png)
或
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101192017056886384/SYS201311011920170568863023_DA/48.png)
或
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101192017056886384/SYS201311011920170568863023_DA/49.png)
秒時(shí),△MNB為等腰三角形.
點(diǎn)評(píng):本題綜合考查了二次函數(shù),主要利用了二次函數(shù)的對(duì)稱性,待定系數(shù)法求二次函數(shù)解析式,不規(guī)則圖形的面積的求解,二次函數(shù)的最值問題,以及等腰三角形的性質(zhì),(3)運(yùn)算量比較大,計(jì)算時(shí)要認(rèn)真仔細(xì),(4)要根據(jù)等腰三角形腰的不同分情況討論.