已知二次函數y=ax2+bx+c的最大值為12.5,且不等式ax2+bx+c>0的解集為-2<x<3
(1)求a、b、c的值;
(2)求函數圖象頂點的坐標.
【答案】
分析:(1)設二次函數y=ax
2+bx+c=a(x-e)(x-f),根據二次函數與一元二次不等式的關系得到y(tǒng)=ax
2+bx+c=a(x+2)(x-3)=ax
2-ax-6a,根據二次函數y=ax
2+bx+c的最大值公式即可求出a的值,即可求出b、c的值;
(2)把二次函數y=-2x
2+2x+12通過配方得出頂點式-2
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131103195819424970072/SYS201311031958194249700018_DA/0.png)
+
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131103195819424970072/SYS201311031958194249700018_DA/1.png)
,即可求出頂點坐標.
解答:(1)解:設二次函數y=ax
2+bx+c=a(x-e)(x-f),
∵不等式ax
2+bx+c>0的解集為-2<x<3,
∴y=ax
2+bx+c=a(x+2)(x-3)=ax
2-ax-6a,
∵二次函數y=ax
2+bx+c的最大值為12.5,
∴
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131103195819424970072/SYS201311031958194249700018_DA/2.png)
=12.5,
解得:a=-2,
∴y=-2x
2+2x+12,
∴a=-2,b=2,c=12,
答:a=-2,b=2,c=12.
(2)解:y=-2x
2+2x+12=-2(x
2-x+
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131103195819424970072/SYS201311031958194249700018_DA/3.png)
)+12+
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131103195819424970072/SYS201311031958194249700018_DA/4.png)
,
=-2
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131103195819424970072/SYS201311031958194249700018_DA/5.png)
+
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131103195819424970072/SYS201311031958194249700018_DA/6.png)
,
∴函數圖象頂點的坐標是(
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131103195819424970072/SYS201311031958194249700018_DA/7.png)
,
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131103195819424970072/SYS201311031958194249700018_DA/8.png)
),
答:函數圖象頂點的坐標是(
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131103195819424970072/SYS201311031958194249700018_DA/9.png)
,
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131103195819424970072/SYS201311031958194249700018_DA/10.png)
).
點評:本題主要考查對二次函數的三種形式,解一元一次方程,二次函數與不等式的關系,二次函數的圖象上的點的坐標特征,二次函數的最值,二次函數與X軸的交點坐標等知識點的理解和掌握,理解二次函數與一元二次不等式、一元二次方程的關系是解此題的關鍵,題型較好,難度適中.