【答案】
分析:(1)在Rt△AOC中求出AC的長度,然后求出sin∠CAO的值,過點(diǎn)B作BF⊥x軸于點(diǎn)F,由∠BCF=∠CAO,可求出BF,繼而得出FC,從而求得點(diǎn)B的坐標(biāo),利用待定系數(shù)法可求出一次函數(shù)和反比例函數(shù)的關(guān)系式;
(2)不等式的含義為:當(dāng)x<0時(shí),求出一次函數(shù)值y=kx+b小于反比例函數(shù)y=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101193535181321708/SYS201311011935351813217024_DA/0.png)
的x的取值范圍,結(jié)合圖形即可直接寫出答案.
(3)根據(jù)軸對(duì)稱的性質(zhì),找到點(diǎn)A關(guān)于x的對(duì)稱點(diǎn)A',連接BA',則BA'與x軸的交點(diǎn)即為點(diǎn)M的位置,求出直線BA'的解析式,可得出點(diǎn)M的坐標(biāo),根據(jù)B、A'的坐標(biāo)可求出AM+BM的最小值.
解答:解:(1)過點(diǎn)B作BF⊥x軸于點(diǎn)F,
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101193535181321708/SYS201311011935351813217024_DA/images1.png)
在Rt△AOC中,AC=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101193535181321708/SYS201311011935351813217024_DA/1.png)
=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101193535181321708/SYS201311011935351813217024_DA/2.png)
,則sin∠CAO=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101193535181321708/SYS201311011935351813217024_DA/3.png)
=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101193535181321708/SYS201311011935351813217024_DA/4.png)
,
∵∠BCA=90°,
∴∠BCF+∠ACO=90°,
又∵∠CAO+∠ACO=90°,
∴∠BCF=∠CAO,
∴sin∠BCF=sin∠CAO=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101193535181321708/SYS201311011935351813217024_DA/5.png)
=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101193535181321708/SYS201311011935351813217024_DA/6.png)
,
∴BF=1,
∴CF=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101193535181321708/SYS201311011935351813217024_DA/7.png)
=2,
∴點(diǎn)B的坐標(biāo)為(-3,1),
將點(diǎn)B的坐標(biāo)代入反比例函數(shù)解析式可得:1=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101193535181321708/SYS201311011935351813217024_DA/8.png)
,
解得:k=-3,
故可得反比例函數(shù)解析式為y=-
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101193535181321708/SYS201311011935351813217024_DA/9.png)
;
將點(diǎn)B、C的坐標(biāo)代入一次函數(shù)解析式可得:
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101193535181321708/SYS201311011935351813217024_DA/10.png)
,
解得:
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101193535181321708/SYS201311011935351813217024_DA/11.png)
.
故可得一次函數(shù)解析式為y=-
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101193535181321708/SYS201311011935351813217024_DA/12.png)
x-
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101193535181321708/SYS201311011935351813217024_DA/13.png)
.
(2)結(jié)合點(diǎn)B的坐標(biāo)及圖象,可得當(dāng)x<0時(shí),kx+b-
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101193535181321708/SYS201311011935351813217024_DA/14.png)
<0的解集為:-3<x<0;
(3)作點(diǎn)A關(guān)于x軸的對(duì)稱點(diǎn)A′,連接 B A′與x軸 的交點(diǎn)即為點(diǎn)M,
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101193535181321708/SYS201311011935351813217024_DA/images16.png)
設(shè)直線BA'的解析式為y=ax+b,將點(diǎn)A'及點(diǎn)B的坐標(biāo)代入可得:
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101193535181321708/SYS201311011935351813217024_DA/15.png)
,
解得:
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101193535181321708/SYS201311011935351813217024_DA/16.png)
.
故直線BA'的解析式為y=-x-2,
令y=0,可得-x-2=0,
解得:x=-2,
故點(diǎn)M 的坐標(biāo)為(-2,0),
AM+BM=BM+MA′=BA′=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101193535181321708/SYS201311011935351813217024_DA/17.png)
=3
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101193535181321708/SYS201311011935351813217024_DA/18.png)
.
綜上可得:點(diǎn)M的坐標(biāo)為(-2,0),AM+BM的最小值為3
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101193535181321708/SYS201311011935351813217024_DA/19.png)
.
點(diǎn)評(píng):本題考查了反比例函數(shù)的綜合應(yīng)用,涉及了待定系數(shù)法求函數(shù)解析式、軸對(duì)稱求最短路徑及一次函數(shù)與反比例函數(shù)的交點(diǎn)問題,綜合考察的知識(shí)點(diǎn)較多,注意培養(yǎng)自己解綜合題的能力,將所學(xué)知識(shí)融會(huì)貫通.