题目内容
(2002•聊城)已知点(-2,y1),(-5![](http://thumb.1010pic.com/pic6/res/czsx/web/STSource/20131019105530530257145/SYS201310191055305302571010_ST/0.png)
![](http://thumb.1010pic.com/pic6/res/czsx/web/STSource/20131019105530530257145/SYS201310191055305302571010_ST/1.png)
A.y1>y2>y3
B.y2>y1>y3
C.y2>y3>y1
D.y3>y2>y1
【答案】分析:由二次函数y=2x2+8x+7可知,此函数的对称轴为x=-2,顶点坐标为(-2,-
),二次项系数a=2>0,故此函数的图象开口向上,有最小值,设点(1
,y3)关于x=-2的对称点为A,根据二次函数的性质可知点A′的坐标为(-
,y3),因为二次函数y=2x2+8x+7的图象开口向上,有最小值,在对称轴的左侧为减函数,故看判断y2>y3>y1.
解答:解:∵对称轴为x=-2,顶点坐标为(-2,-
),二次项系数a=2>0
∴此函数的图象开口向上,有最小值,x=-2时y=-![](http://thumb.1010pic.com/pic6/res/czsx/web/STSource/20131019105530530257145/SYS201310191055305302571010_DA/4.png)
设点(1
,y3)关于x=-2的对称点为A,横坐标为a,则
=-2
∴a=-![](http://thumb.1010pic.com/pic6/res/czsx/web/STSource/20131019105530530257145/SYS201310191055305302571010_DA/7.png)
∴点A′的坐标为(-
,y3)
∴x=2时y=-
,故y1最小
∵-5
<-
<-2
∴y2>y3>y1.
故选A.
点评:本题的关键是(1)找到二次函数的对称轴;(2)掌握二次函数y=ax2+bx+c(a≠0)的图象性质.
![](http://thumb.1010pic.com/pic6/res/czsx/web/STSource/20131019105530530257145/SYS201310191055305302571010_DA/0.png)
![](http://thumb.1010pic.com/pic6/res/czsx/web/STSource/20131019105530530257145/SYS201310191055305302571010_DA/1.png)
![](http://thumb.1010pic.com/pic6/res/czsx/web/STSource/20131019105530530257145/SYS201310191055305302571010_DA/2.png)
解答:解:∵对称轴为x=-2,顶点坐标为(-2,-
![](http://thumb.1010pic.com/pic6/res/czsx/web/STSource/20131019105530530257145/SYS201310191055305302571010_DA/3.png)
∴此函数的图象开口向上,有最小值,x=-2时y=-
![](http://thumb.1010pic.com/pic6/res/czsx/web/STSource/20131019105530530257145/SYS201310191055305302571010_DA/4.png)
设点(1
![](http://thumb.1010pic.com/pic6/res/czsx/web/STSource/20131019105530530257145/SYS201310191055305302571010_DA/5.png)
![](http://thumb.1010pic.com/pic6/res/czsx/web/STSource/20131019105530530257145/SYS201310191055305302571010_DA/6.png)
∴a=-
![](http://thumb.1010pic.com/pic6/res/czsx/web/STSource/20131019105530530257145/SYS201310191055305302571010_DA/7.png)
∴点A′的坐标为(-
![](http://thumb.1010pic.com/pic6/res/czsx/web/STSource/20131019105530530257145/SYS201310191055305302571010_DA/8.png)
∴x=2时y=-
![](http://thumb.1010pic.com/pic6/res/czsx/web/STSource/20131019105530530257145/SYS201310191055305302571010_DA/9.png)
∵-5
![](http://thumb.1010pic.com/pic6/res/czsx/web/STSource/20131019105530530257145/SYS201310191055305302571010_DA/10.png)
![](http://thumb.1010pic.com/pic6/res/czsx/web/STSource/20131019105530530257145/SYS201310191055305302571010_DA/11.png)
∴y2>y3>y1.
故选A.
点评:本题的关键是(1)找到二次函数的对称轴;(2)掌握二次函数y=ax2+bx+c(a≠0)的图象性质.
![](http://thumb2018.1010pic.com/images/loading.gif)
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