已知:關(guān)于x的二次函數(shù)y=-x2+(m+2)x-m.
(1)求證:不論m為任何實(shí)數(shù),二次函數(shù)的圖象的頂點(diǎn)P總是在x軸的上方;
(2)設(shè)二次函數(shù)圖象與y軸交于A,過(guò)點(diǎn)A作x軸的平行線與圖象交于另外一點(diǎn)B.若頂點(diǎn)P在第一象限,當(dāng)m為何值時(shí),△PAB是等邊三角形.
【答案】
分析:(1)只要求出頂點(diǎn)的縱坐標(biāo)為正,就能確定頂點(diǎn)P總是在x軸的上方,根據(jù)頂點(diǎn)的縱坐標(biāo)公式求解;
(2)根據(jù)圖形可以看出,對(duì)稱軸把等邊三角形分成兩個(gè)全等的30°的直角三角形,根據(jù)點(diǎn)的坐標(biāo)與線段的關(guān)系可以求解.
解答:(1)證明:二次函數(shù)y=-x
2+(m+2)x-m中,a=-1,b=m+2,c=-m,
∴頂點(diǎn)P的縱坐標(biāo)為
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131103104005990364168/SYS201311031040059903641022_DA/0.png)
=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131103104005990364168/SYS201311031040059903641022_DA/1.png)
=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131103104005990364168/SYS201311031040059903641022_DA/2.png)
>0,
∴頂點(diǎn)P總在x軸上方;
(2)解:二次函數(shù)y=-x
2+(m+2)x-m與y軸交于點(diǎn)A(0,-m),
頂點(diǎn)P(
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131103104005990364168/SYS201311031040059903641022_DA/3.png)
,
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131103104005990364168/SYS201311031040059903641022_DA/4.png)
),
過(guò)P作PC⊥AB于C,則C(
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131103104005990364168/SYS201311031040059903641022_DA/5.png)
,-m),
因?yàn)辄c(diǎn)P在第一象限,所以
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131103104005990364168/SYS201311031040059903641022_DA/6.png)
>0,
AC=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131103104005990364168/SYS201311031040059903641022_DA/7.png)
,PC=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131103104005990364168/SYS201311031040059903641022_DA/8.png)
,
∵△PAB是等邊三角形,
∴∠PAC=60°,
由tan∠PAC=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131103104005990364168/SYS201311031040059903641022_DA/9.png)
得
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131103104005990364168/SYS201311031040059903641022_DA/10.png)
=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131103104005990364168/SYS201311031040059903641022_DA/11.png)
(
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131103104005990364168/SYS201311031040059903641022_DA/12.png)
),
整理得:(m+2)
2=2
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131103104005990364168/SYS201311031040059903641022_DA/13.png)
(m+2),
∴m+2=2
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131103104005990364168/SYS201311031040059903641022_DA/14.png)
∴m=2
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131103104005990364168/SYS201311031040059903641022_DA/15.png)
-2,
即m=2
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131103104005990364168/SYS201311031040059903641022_DA/16.png)
-2時(shí),△PAB是等邊三角形.
點(diǎn)評(píng):解答此題的關(guān)鍵是求出對(duì)稱軸,頂點(diǎn)縱坐標(biāo),然后由圖象解答,鍛煉了學(xué)生數(shù)形結(jié)合的思想方法.