【答案】
分析:(1)根據(jù)二次函數(shù)的判別式,可以判斷函數(shù)的圖象與x軸交點(diǎn)情況;
(2)把A點(diǎn)坐標(biāo)為(-1,0)代入函數(shù)解析式,求出m的值,令y=0,求出一元二次方程的解即可;
(3)根據(jù)二次函數(shù)的性質(zhì)判斷其增減性;
解答:解:(1)對(duì)于關(guān)于x的二次函數(shù)y=x
2-mx+
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131103101115306695490/SYS201311031011153066954019_DA/0.png)
,
由于△=(-m)
2-4×1×
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131103101115306695490/SYS201311031011153066954019_DA/1.png)
=-m
2-2<0,
所以此函數(shù)的圖象與x軸沒(méi)有交點(diǎn);
對(duì)于關(guān)于x的二次函數(shù)y=x
2-mx-
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131103101115306695490/SYS201311031011153066954019_DA/2.png)
,
由于△=(-m)
2-4×1×(-
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131103101115306695490/SYS201311031011153066954019_DA/3.png)
)=3m
2+4>0
所以此函數(shù)的圖象與x軸有兩個(gè)不同的交點(diǎn).
故圖象經(jīng)過(guò)A、B兩點(diǎn)的二次函數(shù)為y=x
2-mx-
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131103101115306695490/SYS201311031011153066954019_DA/4.png)
;
(2)將A(-1,0)代入y=x
2-mx-
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131103101115306695490/SYS201311031011153066954019_DA/5.png)
,得1+m-
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131103101115306695490/SYS201311031011153066954019_DA/6.png)
=0.
整理,得-m
2+2m=0.
解之,得m=0,或m=2.
當(dāng)m=0時(shí),y=x
2-1.
令y=0,得x
2-1=0.
解這個(gè)方程,得x
1=-1,x
2=1,
此時(shí),B點(diǎn)的坐標(biāo)是B(1,0);
當(dāng)m=2時(shí),y=x
2-2x-3.
令y=0,得x
2-2x-3=0.
解這個(gè)方程,得x
1=-1,x
2=3,
此時(shí),B點(diǎn)的坐標(biāo)是B(3,0).
(3)當(dāng)m=0時(shí),二次函數(shù)為y=x
2-1,此函數(shù)的圖象開(kāi)口向上,對(duì)稱軸為直線x=0,
所以當(dāng)x<0時(shí),函數(shù)值y隨x的增大而減�。�
當(dāng)m=2時(shí),二次函數(shù)為y=x
2-2x-3=(x-1)
2-4,此函數(shù)的圖象開(kāi)口向上,
對(duì)稱軸為直線x=1,所以當(dāng)x<1時(shí),函數(shù)值y隨x的增大而減�。�
點(diǎn)評(píng):主要考查了二次函數(shù)的與x軸交點(diǎn)的求法,以及二次函數(shù)的增減性.