【答案】
分析:(1)把(0,3)代入函數(shù)解析式y(tǒng)=ax
2+bx+c中,易求c;
(2)當(dāng)a=-1時,函數(shù)解析式是y=-x
2+bx+3,設(shè)D點坐標(biāo)是(e,3),E點坐標(biāo)是(6,f),分別把D、E的坐標(biāo)代入y=-x
2+bx+3中,易求e=b以及f=-33+6b,結(jié)合三角形面積公式,易得S=-3b
2+18b,求關(guān)于b的二次函數(shù)的最大值即可;
(3)設(shè)M的坐標(biāo)是(g,3),N的坐標(biāo)是(6,h),根據(jù)圖乙知直線OF與BC的交點坐標(biāo)(6,2),進而求直線OF的解析式是y=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101193054086935959/SYS201311011930540869359024_DA/0.png)
x,而OF又是MN的中垂線,那么MN的中點就在直線OF上,于是可得g=3h+3①,g
2+9=36+h
2②,解關(guān)于g、h的二元二次方程組,易求g、h(負(fù)數(shù)舍去),進而可得M、N的坐標(biāo),再把M、N的坐標(biāo)代入y=ax
2+bx+3中,得到關(guān)于a、b的二元一次方程組,解可求a、b,進而可得二次函數(shù)解析式.
解答:解:(1)把(0,3)代入函數(shù)解析式y(tǒng)=ax
2+bx+c中,得
c=3;
(2)若a=-1,且拋物線與矩形有且只有三個交點A、D、E,
則D、E分別在線段AB、BC上,或分別在AB、OC上,
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101193054086935959/SYS201311011930540869359024_DA/images1.png)
若D、E分別在線段AB、BC上,
在y=-x
2+bx+3中,令y=3,得x
2-bx=0,解得:x=0或x=b,故D(b,3),
令x=6,得:y=6b-33,故E(6,6b-33),
∵0≤6b-33<3,
∴
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101193054086935959/SYS201311011930540869359024_DA/1.png)
≤b<6,
又∵AD=|b|=b,EB=|3-(6b-33)|=36-6b,
△ADE的面積S=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101193054086935959/SYS201311011930540869359024_DA/2.png)
AD•BE=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101193054086935959/SYS201311011930540869359024_DA/3.png)
b(36-6b)=-3b
2+18b=-3(b-3)
2+27,
則當(dāng)b=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101193054086935959/SYS201311011930540869359024_DA/4.png)
時,S有最大值
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101193054086935959/SYS201311011930540869359024_DA/5.png)
.
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101193054086935959/SYS201311011930540869359024_DA/images7.png)
若D、E分別在AB、OC上,
△ADE的面積S=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101193054086935959/SYS201311011930540869359024_DA/6.png)
AD•BE=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101193054086935959/SYS201311011930540869359024_DA/7.png)
b•3=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101193054086935959/SYS201311011930540869359024_DA/8.png)
b,
∵拋物線的對稱軸為:x=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101193054086935959/SYS201311011930540869359024_DA/9.png)
,
當(dāng)過點C時,拋物線為:y=-x
2+
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101193054086935959/SYS201311011930540869359024_DA/10.png)
x+3,
∴0<
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101193054086935959/SYS201311011930540869359024_DA/11.png)
≤
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101193054086935959/SYS201311011930540869359024_DA/12.png)
,
∴當(dāng)b=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101193054086935959/SYS201311011930540869359024_DA/13.png)
時,S有最大值
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101193054086935959/SYS201311011930540869359024_DA/14.png)
.
(3)當(dāng)點M、N分別在AB、OC上時,過M作MG⊥OC于點G,連接OM,
∴MG=OA=3,∠2+∠MNO=90°,
∵OF垂直平分MN,
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101193054086935959/SYS201311011930540869359024_DA/images17.png)
∴OM=ON,∠1+∠MNO=90°,
∴∠1=∠2,
∴tan∠1=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101193054086935959/SYS201311011930540869359024_DA/15.png)
=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101193054086935959/SYS201311011930540869359024_DA/16.png)
,tan∠2=tan∠1=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101193054086935959/SYS201311011930540869359024_DA/17.png)
,
∴GN=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101193054086935959/SYS201311011930540869359024_DA/18.png)
GM=1,設(shè)N(n,0),則G(n-1,0)
∴M(n-1,3)
∴AM=n-1,ON=n=OM,
在直角△AOM中,OM
2=OA
2+AM
2,
∴n
2=3
2+(n-1)
2,解得:n=5,
∴M(4,3),N(5,0),
把M、N代入二次函數(shù)的解析式得:
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101193054086935959/SYS201311011930540869359024_DA/19.png)
解得:
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101193054086935959/SYS201311011930540869359024_DA/20.png)
,
則函數(shù)的解析式是:y=-
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101193054086935959/SYS201311011930540869359024_DA/21.png)
x
2+
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101193054086935959/SYS201311011930540869359024_DA/22.png)
x+3;
如右圖,
當(dāng)點M、N分別在AB、BC邊上時,
設(shè)M的坐標(biāo)是(g,3),N的坐標(biāo)是(6,h),
直線OF與BC交點的橫坐標(biāo)是6,縱坐標(biāo)是3-1=2,
把(6,2)代入函數(shù)y=kx中,得k=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101193054086935959/SYS201311011930540869359024_DA/23.png)
,
故直線OF的解析式是y=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101193054086935959/SYS201311011930540869359024_DA/24.png)
x,
∵OF垂直平分MN,
∴點(
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101193054086935959/SYS201311011930540869359024_DA/25.png)
,
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101193054086935959/SYS201311011930540869359024_DA/26.png)
)在直線y=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101193054086935959/SYS201311011930540869359024_DA/27.png)
x上,OM=ON,
∴
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101193054086935959/SYS201311011930540869359024_DA/28.png)
•
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101193054086935959/SYS201311011930540869359024_DA/29.png)
=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101193054086935959/SYS201311011930540869359024_DA/30.png)
,g
2+9=36+h
2,
即g=3h+3①,g
2+9=36+h
2,②
解關(guān)于①②的方程組,得
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101193054086935959/SYS201311011930540869359024_DA/31.png)
或
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101193054086935959/SYS201311011930540869359024_DA/32.png)
(負(fù)數(shù)不合題意,舍去),
把(
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101193054086935959/SYS201311011930540869359024_DA/33.png)
,3)、(6,
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101193054086935959/SYS201311011930540869359024_DA/34.png)
)代入二次函數(shù)y=ax
2+bx+3中,得
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101193054086935959/SYS201311011930540869359024_DA/35.png)
,
解得
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101193054086935959/SYS201311011930540869359024_DA/36.png)
.
故所求二次函數(shù)解析式是y=-
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101193054086935959/SYS201311011930540869359024_DA/37.png)
x
2+
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101193054086935959/SYS201311011930540869359024_DA/38.png)
x+3.
則二次函數(shù)解析式是y=-
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101193054086935959/SYS201311011930540869359024_DA/39.png)
x
2+
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101193054086935959/SYS201311011930540869359024_DA/40.png)
x+3或y=-
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101193054086935959/SYS201311011930540869359024_DA/41.png)
x
2+
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101193054086935959/SYS201311011930540869359024_DA/42.png)
x+3.
點評:本題考查了二次函數(shù)綜合題,解題的關(guān)鍵是理解題意,并能畫出草圖,利用線段垂直平分線的性質(zhì)、解方程組、兩點之間的距離公式來解決問題.