试题分析:(1)由已知得A(
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824002221893443.png)
,0),B(0,b),则
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824002221909407.png)
=(
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824002221971421.png)
,b),于是
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824002221971421.png)
="2,b=2." ∴k=1,b=2.
(2)由f(x)> g(x),得x+2>x
2-x-6,即(x+2)(x-4)<0, 得-2<x<4,
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824002221878677.png)
=
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824002222049619.png)
=x+2+
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824002222065442.png)
-5
由于x+2>0,则
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824002221878677.png)
≥-3,其中等号当且仅当x+2=1,即x=-1时成立
∴
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824002221878677.png)
的最小值是-3.
点评:(1)向量的坐标就是其终点的坐标减去起点的坐标。(2)注意基本不等式应用的条件:一正二定三相等。本题把式子
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824002222049619.png)
化为x+2+
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824002222065442.png)
-5的形式,从而达到利用基本不等式的条件。