已知拋物線y=ax2+bx+c的頂點為(1,0),且經(jīng)過點(0,1).
(1)求該拋物線對應(yīng)的函數(shù)的解析式;
(2)將該拋物線向下平移m(m>0)個單位,設(shè)得到的拋物線的頂點為A,與x軸的兩個交點為B、C,若△ABC為等邊三角形.
①求m的值;
②設(shè)點A關(guān)于x軸的對稱點為點D,在拋物線上是否存在點P,使四邊形CBDP為菱形?若存在,寫出點P的坐標(biāo);若不存在,請說明理由.
【答案】
分析:(1)根據(jù)拋物線的頂點坐標(biāo)及函數(shù)經(jīng)過點(0,1),利用待定系數(shù)法求解即可.
(2)①先寫出平移后的函數(shù)解析式,然后得出A、B、C三點的坐標(biāo),過點A作AH⊥BC于H,根據(jù)△ABC為等邊三角形,可得出關(guān)于m的方程,解出即可;
②求出點D坐標(biāo),分兩種情況進行討論,①PD為對角線,②PD為邊,根據(jù)菱形的性質(zhì)求解即可.
解答:解:(1)由題意可得,
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101191353110572320/SYS201311011913531105723022_DA/0.png)
,
解得
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101191353110572320/SYS201311011913531105723022_DA/1.png)
,
故拋物線對應(yīng)的函數(shù)的解析式為y=x
2-2x+1;
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101191353110572320/SYS201311011913531105723022_DA/images2.png)
(2)①將y=x
2-2x+1向下平移m個單位得:y=x
2-2x+1-m=(x-1)
2-m,
可知A(1,-m),B(1-
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101191353110572320/SYS201311011913531105723022_DA/2.png)
,0),C(1+
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101191353110572320/SYS201311011913531105723022_DA/3.png)
,0),BC=2
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101191353110572320/SYS201311011913531105723022_DA/4.png)
,
過點A作AH⊥BC于H,
∵△ABC為等邊三角形,
∴BH=HC=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101191353110572320/SYS201311011913531105723022_DA/5.png)
BC,∠CAH=30°,
∴AH=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101191353110572320/SYS201311011913531105723022_DA/6.png)
,即
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101191353110572320/SYS201311011913531105723022_DA/7.png)
=m,
由m>0,解得m=3.
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101191353110572320/SYS201311011913531105723022_DA/images9.png)
②在拋物線上存在點P,能使四邊形CBDP為菱形.理由如下:
∵點D與點A關(guān)于x軸對稱,
∴D(1,3),
①當(dāng)DP為對角線時,顯然點P在點A位置上時,符合題意,
故此時點P坐標(biāo)為(1,-3);
②當(dāng)DP為邊時,
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101191353110572320/SYS201311011913531105723022_DA/images10.png)
要使四邊形CBDP為菱形,需DP∥BC,DP=BC.
由點D的坐標(biāo)為(1,3),DP=BC=2
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101191353110572320/SYS201311011913531105723022_DA/8.png)
,可知點P的橫坐標(biāo)為1+2
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101191353110572320/SYS201311011913531105723022_DA/9.png)
,
當(dāng)x=1+2
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101191353110572320/SYS201311011913531105723022_DA/10.png)
時,y=x
2-2x+1-m=x
2-2x-2=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101191353110572320/SYS201311011913531105723022_DA/11.png)
-2(1+2
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101191353110572320/SYS201311011913531105723022_DA/12.png)
)-2=11≠3,
故不存在這樣的點P.
綜上可得,存在使四邊形CBDP為菱形的點P,坐標(biāo)為(1,-3).
點評:此題屬于二次函數(shù)的綜合題,屬于綜合性較強的題目,應(yīng)理清思路,對每一個知識點都應(yīng)熟練掌握并能靈活運用,求出二次函數(shù)的解析式是解此題的關(guān)鍵,應(yīng)熟練掌握三點式和頂點式求拋物線解析式的方法,二次函數(shù)的平移通常指的是圖象的平移,應(yīng)注意總結(jié)平移的規(guī)律.