【答案】
分析:(1)已知了拋物線的頂點(diǎn)橫坐標(biāo)為1,即x=-
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131022160015778725481/SYS201310221600157787254023_DA/0.png)
=1,將已知的兩點(diǎn)坐標(biāo)代入拋物線中,聯(lián)立三式即可求出拋物線的解析式.
(2)本題要分兩種情況討論:△BOD∽△BAC或△BDO∽△BAC,解題思路都是通過相似三角形得出的關(guān)于BD、BC、BO、BA的比例關(guān)系式求出BD的長(zhǎng),然后根據(jù)∠OBC=45°的特殊條件用BD的長(zhǎng)求出D點(diǎn)的坐標(biāo).
解答:解:(1)∵二次函數(shù)圖象頂點(diǎn)的橫坐標(biāo)為1,且過點(diǎn)(2,3)和(-3,-12),
∴由
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131022160015778725481/SYS201310221600157787254023_DA/1.png)
,
解得
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131022160015778725481/SYS201310221600157787254023_DA/2.png)
,
∴此二次函數(shù)的表達(dá)式為y=-x
2+2x+3;
(2)假設(shè)存在直線l:y=kx(k≠0)與線段BC交于點(diǎn)D(不與點(diǎn)B,C重合),使得以B,O,D為頂點(diǎn)的三角形與△BAC相似.
在y=-x
2+2x+3中,令y=0,則由-x
2+2x+3=0,
解得x
1=-1,x
2=3
∴A(-1,0),B(3,0)
令x=0,得y=3.
∴C(0,3).
設(shè)過點(diǎn)O的直線l交BC于點(diǎn)D,過點(diǎn)D作DE⊥x軸于點(diǎn)E.
∵點(diǎn)B的坐標(biāo)為(3,0),點(diǎn)C的坐標(biāo)為(0,3),點(diǎn)A的坐標(biāo)為(-1,0).
∴|AB|=4,|OB|=|OC|=3,∠OBC=45°.
∴|BC|=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131022160015778725481/SYS201310221600157787254023_DA/3.png)
=3
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131022160015778725481/SYS201310221600157787254023_DA/4.png)
.
要使△BOD∽△BAC或△BDO∽△BAC,
已有∠B=∠B,則只需
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131022160015778725481/SYS201310221600157787254023_DA/5.png)
,①或
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131022160015778725481/SYS201310221600157787254023_DA/6.png)
②成立.
若是①,則有|BD|=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131022160015778725481/SYS201310221600157787254023_DA/7.png)
=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131022160015778725481/SYS201310221600157787254023_DA/8.png)
=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131022160015778725481/SYS201310221600157787254023_DA/9.png)
.
而∠OBC=45°,
∴|BE|=|DE|.
∴在Rt△BDE中,由勾股定理,
得|BE|
2+|DE|
2=2|BE|
2=|BD|
2=(
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131022160015778725481/SYS201310221600157787254023_DA/10.png)
)
2解得|BE|=|DE|=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131022160015778725481/SYS201310221600157787254023_DA/11.png)
(負(fù)值舍去).
∴|OE|=|OB|-|BE|=3-
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131022160015778725481/SYS201310221600157787254023_DA/12.png)
=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131022160015778725481/SYS201310221600157787254023_DA/13.png)
∴點(diǎn)D的坐標(biāo)為(
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131022160015778725481/SYS201310221600157787254023_DA/14.png)
,
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131022160015778725481/SYS201310221600157787254023_DA/15.png)
)
將點(diǎn)D的坐標(biāo)代入y=kx(k≠0)中,求得k=3,
∴滿足條件的直線l的函數(shù)表達(dá)式為y=3x,
或求出直線AC的函數(shù)表達(dá)式為y=3x+3,則與直線AC平行的直線l的函數(shù)表達(dá)式為y=3x,
此時(shí)易知△BOD∽△BAC,再求出直線BC的函數(shù)表達(dá)式為y=-x+3.聯(lián)立y=3x,y=-x+3求得點(diǎn)D的坐標(biāo)為(
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131022160015778725481/SYS201310221600157787254023_DA/16.png)
,
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131022160015778725481/SYS201310221600157787254023_DA/17.png)
),
若是②,則有|BD|=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131022160015778725481/SYS201310221600157787254023_DA/18.png)
=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131022160015778725481/SYS201310221600157787254023_DA/19.png)
=2
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131022160015778725481/SYS201310221600157787254023_DA/20.png)
,
而∠OBC=45°,
∴|BE|=|DE|,
∴在Rt△BDE中,由勾股定理,
得|BE|
2+|DE|
2=2|BE|
2=|BD|
2=(2
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131022160015778725481/SYS201310221600157787254023_DA/21.png)
)
2解得|BE|=|DE|=2(負(fù)值舍去)
∴|OE|=|OB|-|BE|=3-2=1.
∴點(diǎn)D的坐標(biāo)為(1,2).
將點(diǎn)D的坐標(biāo)代入y=kx(k≠0)中,求得k=2.
∴滿足條件的直線l的函數(shù)表達(dá)式為y=2x.
∴存在直線l:y=3x或y=2x與線段BC交于點(diǎn)D(不與點(diǎn)B,C重合),
使得以B,O,D為頂點(diǎn)的三角形與△BAC相似,且點(diǎn)D的坐標(biāo)分別為(
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131022160015778725481/SYS201310221600157787254023_DA/22.png)
,
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131022160015778725481/SYS201310221600157787254023_DA/23.png)
)或(1,2).
點(diǎn)評(píng):本題是二次函數(shù)綜合題,考查了二次函數(shù)解析式的確定、相似三角形的判定、函數(shù)圖象交點(diǎn)等知識(shí)點(diǎn).綜合性強(qiáng).