题目内容
在直角坐标系中有三点A(0,-1),B(1,3)C(2,6).已知直线y=ax+b上横坐标为0,1,2的点分别为D,E,F,试求a,b的值使AD2+BE2+CF2达到最小值 .
【答案】分析:先求出D(0,b),E(1,a+b),F(2,2a+b),根据坐标可列出AD、BE、CF的表达式.
解答:解:由题意可得:D(0,b),E(1,a+b),F(2,2a+b),
∴AD2+BE2+CF2=(b+1)2+(a+b-3)2+(2a+b-6)2,
=(b+1)2+[(a-3)+b]2+[2(a-3)+b]2,
=3b2+2b+1+5(a-3)2+6(a-3)b,
=5[a-3+(
)]2+
b2+2b+1,
=5[a-3+(
)]2+
(b+
)2+
,
∴a-3+
=0,b+
=0.
解得a=
,b=-
时,有最小值为
.
故答案为:
.
点评:此题考查了函数图象上点的坐标特征,将AD2+BE2+CF2转化为完全平方式,再根据非负数的性质求出最值是常用的方法.
解答:解:由题意可得:D(0,b),E(1,a+b),F(2,2a+b),
∴AD2+BE2+CF2=(b+1)2+(a+b-3)2+(2a+b-6)2,
=(b+1)2+[(a-3)+b]2+[2(a-3)+b]2,
=3b2+2b+1+5(a-3)2+6(a-3)b,
=5[a-3+(
![](http://thumb.1010pic.com/pic6/res/czsx/web/STSource/20131103195546676183172/SYS201311031955466761831008_DA/0.png)
![](http://thumb.1010pic.com/pic6/res/czsx/web/STSource/20131103195546676183172/SYS201311031955466761831008_DA/1.png)
=5[a-3+(
![](http://thumb.1010pic.com/pic6/res/czsx/web/STSource/20131103195546676183172/SYS201311031955466761831008_DA/2.png)
![](http://thumb.1010pic.com/pic6/res/czsx/web/STSource/20131103195546676183172/SYS201311031955466761831008_DA/3.png)
![](http://thumb.1010pic.com/pic6/res/czsx/web/STSource/20131103195546676183172/SYS201311031955466761831008_DA/4.png)
![](http://thumb.1010pic.com/pic6/res/czsx/web/STSource/20131103195546676183172/SYS201311031955466761831008_DA/5.png)
∴a-3+
![](http://thumb.1010pic.com/pic6/res/czsx/web/STSource/20131103195546676183172/SYS201311031955466761831008_DA/6.png)
![](http://thumb.1010pic.com/pic6/res/czsx/web/STSource/20131103195546676183172/SYS201311031955466761831008_DA/7.png)
解得a=
![](http://thumb.1010pic.com/pic6/res/czsx/web/STSource/20131103195546676183172/SYS201311031955466761831008_DA/8.png)
![](http://thumb.1010pic.com/pic6/res/czsx/web/STSource/20131103195546676183172/SYS201311031955466761831008_DA/9.png)
![](http://thumb.1010pic.com/pic6/res/czsx/web/STSource/20131103195546676183172/SYS201311031955466761831008_DA/10.png)
故答案为:
![](http://thumb.1010pic.com/pic6/res/czsx/web/STSource/20131103195546676183172/SYS201311031955466761831008_DA/11.png)
点评:此题考查了函数图象上点的坐标特征,将AD2+BE2+CF2转化为完全平方式,再根据非负数的性质求出最值是常用的方法.
![](http://thumb2018.1010pic.com/images/loading.gif)
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