【答案】
分析:(1)根據(jù)函數(shù)經(jīng)過原點(diǎn),可得c=0,然后根據(jù)函數(shù)的對稱軸,及函數(shù)圖象經(jīng)過點(diǎn)(3,-
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101193738848240573/SYS201311011937388482405022_DA/0.png)
)可得出函數(shù)解析式,根據(jù)二次函數(shù)的對稱性可直接得出點(diǎn)A的坐標(biāo).
(2)點(diǎn)Q不與點(diǎn)B重合.先求出∠BOA的度數(shù),然后可確定∠Q
1OA=的度數(shù),繼而利用解直角三角形的知識求出x,得出Q
1的坐標(biāo),利用二次函數(shù)圖象函數(shù)的對稱性可得出Q
2的坐標(biāo).
(3)將M(0,
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101193738848240573/SYS201311011937388482405022_DA/1.png)
)、Q
1(9,3
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101193738848240573/SYS201311011937388482405022_DA/2.png)
)代入y=kx+b,得直線QR的解析式為
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101193738848240573/SYS201311011937388482405022_DA/3.png)
,求與拋物線的交點(diǎn)R:P點(diǎn)在直線QR下方且在拋物線上,故設(shè)P(x,
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101193738848240573/SYS201311011937388482405022_DA/4.png)
),
如圖,過P作直線平行于y軸,交QR于點(diǎn)K,則K(x,
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101193738848240573/SYS201311011937388482405022_DA/5.png)
),則S
△PQR=S
△QPK+S
△RPK=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101193738848240573/SYS201311011937388482405022_DA/6.png)
PK(9-x+x+1)=-
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101193738848240573/SYS201311011937388482405022_DA/7.png)
,所以根據(jù)求二次函數(shù)最值的方法知當(dāng)x=4時(shí),S
△PQR最大=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101193738848240573/SYS201311011937388482405022_DA/8.png)
,則易求點(diǎn)P的坐標(biāo).同理求得P
2(0,0),S
△PQR最大=3
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101193738848240573/SYS201311011937388482405022_DA/9.png)
.
解答:解:(1)由函數(shù)圖象經(jīng)過原點(diǎn)得,函數(shù)解析式為y=ax
2+bx(a≠0),
又∵函數(shù)的頂點(diǎn)坐標(biāo)為B(3,
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101193738848240573/SYS201311011937388482405022_DA/10.png)
),
∴
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101193738848240573/SYS201311011937388482405022_DA/11.png)
,
解得,
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101193738848240573/SYS201311011937388482405022_DA/12.png)
∴該函數(shù)解析式為:
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101193738848240573/SYS201311011937388482405022_DA/13.png)
,
∴由二次函數(shù)圖象的對稱性可知,點(diǎn)A與原點(diǎn)關(guān)于x=3對稱,
∴點(diǎn)A的坐標(biāo)為(6,0);
綜上所述,拋物線的解析式為
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101193738848240573/SYS201311011937388482405022_DA/14.png)
,點(diǎn)A的坐標(biāo)為(6,0);
(2)過B作BC⊥x軸于點(diǎn)C,Rt△OCB中,tan∠OBC=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101193738848240573/SYS201311011937388482405022_DA/15.png)
,
∴∠OBC=60°,
∴∠OBA=120°,△AOB是頂角為120°的等腰三角形,當(dāng)點(diǎn)Q在x軸下方時(shí),必與點(diǎn)B重合(舍去全等情況),
∴當(dāng)Q在x軸上方時(shí),過Q作QD⊥x軸,
∵△QAO∽△AOB,
∴必有OA=AQ=6,且∠OAQ=120°,
∴∠QAD=60°,
∴AD=3,QD=3
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101193738848240573/SYS201311011937388482405022_DA/16.png)
,
∴Q(9,3
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101193738848240573/SYS201311011937388482405022_DA/17.png)
).
∵Q(9,3
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101193738848240573/SYS201311011937388482405022_DA/18.png)
)滿足
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101193738848240573/SYS201311011937388482405022_DA/19.png)
,
∴Q在拋物線上,
根據(jù)對稱性Q
2(
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101193738848240573/SYS201311011937388482405022_DA/20.png)
也滿足條件,
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101193738848240573/SYS201311011937388482405022_DA/images21.png)
∴符合條件的Q點(diǎn)有兩個:Q
1(9,3
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101193738848240573/SYS201311011937388482405022_DA/21.png)
)、Q
2(
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101193738848240573/SYS201311011937388482405022_DA/22.png)
;
(3)設(shè)直線QR的解析式為y=kx+b(k≠0).
將M(0,
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101193738848240573/SYS201311011937388482405022_DA/23.png)
)、Q
1(9,3
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101193738848240573/SYS201311011937388482405022_DA/24.png)
)代入y=kx+b,得直線QR的解析式為
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101193738848240573/SYS201311011937388482405022_DA/25.png)
,
令
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101193738848240573/SYS201311011937388482405022_DA/26.png)
=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101193738848240573/SYS201311011937388482405022_DA/27.png)
解得x
1=-1,x
2=9(即Q點(diǎn)舍去),
∴R(-1,
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101193738848240573/SYS201311011937388482405022_DA/28.png)
),
∵P點(diǎn)在直線QR下方且在拋物線上,故設(shè)P(x,
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101193738848240573/SYS201311011937388482405022_DA/29.png)
).
如圖,過P作直線平行于y軸,交QR于點(diǎn)K,則K(x,
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101193738848240573/SYS201311011937388482405022_DA/30.png)
)
則S
△PQR=S
△QPK+S
△RPK=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101193738848240573/SYS201311011937388482405022_DA/31.png)
PK(9-x+x+1)=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101193738848240573/SYS201311011937388482405022_DA/32.png)
[
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101193738848240573/SYS201311011937388482405022_DA/33.png)
-(
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101193738848240573/SYS201311011937388482405022_DA/34.png)
)]×10
=-
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101193738848240573/SYS201311011937388482405022_DA/35.png)
當(dāng)x=4時(shí),S
△PQR最大=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101193738848240573/SYS201311011937388482405022_DA/36.png)
,
∴點(diǎn)P的坐標(biāo)為(4,
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101193738848240573/SYS201311011937388482405022_DA/37.png)
).
同理過Q
2(
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101193738848240573/SYS201311011937388482405022_DA/38.png)
、M的直線交拋物線R
2,在Q
2R
2下方拋物線取點(diǎn)P
2,
解得P
2(0,0),S
△PQR最大=3
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101193738848240573/SYS201311011937388482405022_DA/39.png)
.
點(diǎn)評:此題屬于二次函數(shù)的綜合題目,涉及了待定系數(shù)法求二次函數(shù)解析式,相似三角形的判定與性質(zhì),三角形的面積及一元二次方程的解,綜合性較強(qiáng),需要我們仔細(xì)分析,分步解答.