(2003•内蒙古)已知关于x的二次函数y=-x2+(2m+3)x+4-m2的图象与x轴交于A、B两点,点A在点B的左边,与y轴的交点C在原点的上方,若A、B两点到原点的距离AO、OB满足4(OB-AO)=3AO•OB.
(1)求这个二次函数的解析式;
(2)求这个二次函数图象的顶点M的坐标,并画出函数图象的略图;
(3)求△AMC的面积.
【答案】
分析:(1)本题可根据韦达定理和题中给出的OA、OB的关系式来求m的值,以此来得出抛物线的解析式;
(2)根据(1)得出的抛物线的解析式可用配方法或公式法求出M的坐标;
(3)由于三角形ACM的面积无法直接求出,设AM与y轴的交点为D,可将其分割成三角形ADC和CDM两部分来求.可先求出直线AM的解析式,得出D的坐标后再求三角形AMC的面积.
解答:![](http://thumb.1010pic.com/pic6/res/czsx/web/STSource/20131019105641310711237/SYS201310191056413107112028_DA/images0.png)
解:(1)∵抛物线与y轴的交点在原点上方,且抛物线开口向下
∴A、B必在原点两侧.
∵点A在点B的左边,因此A在x轴的负半轴,B在x轴的正半轴.
设A(x
1,0),B(x
2,0),那么OA=-x
1,OB=x
2.
则有:x
1+x
2=2m+3,x
1x
2=m
2-4.
∵4(OB-AO)=3AO•OB,即4(x
2+x
1)=-3x
1x
2;
4(2m+3)=-3(m
2-4),
解得m=0,m=-
![](http://thumb.1010pic.com/pic6/res/czsx/web/STSource/20131019105641310711237/SYS201310191056413107112028_DA/0.png)
,
∵抛物线与y轴的交点C在y轴正半轴
∴4-m
2>0,即-2<m<2,
∴m=0.
∴抛物线的解析式为y=-x
2+3x+4;
(2)由(1)知:y=-x
2+3x+4=-(x-
![](http://thumb.1010pic.com/pic6/res/czsx/web/STSource/20131019105641310711237/SYS201310191056413107112028_DA/1.png)
)
2+
![](http://thumb.1010pic.com/pic6/res/czsx/web/STSource/20131019105641310711237/SYS201310191056413107112028_DA/2.png)
,
∴M(
![](http://thumb.1010pic.com/pic6/res/czsx/web/STSource/20131019105641310711237/SYS201310191056413107112028_DA/3.png)
,
![](http://thumb.1010pic.com/pic6/res/czsx/web/STSource/20131019105641310711237/SYS201310191056413107112028_DA/4.png)
);
(3)设直线AM与y轴的交点为D.
易知A(-1,0),M(
![](http://thumb.1010pic.com/pic6/res/czsx/web/STSource/20131019105641310711237/SYS201310191056413107112028_DA/5.png)
,
![](http://thumb.1010pic.com/pic6/res/czsx/web/STSource/20131019105641310711237/SYS201310191056413107112028_DA/6.png)
),
∴直线AM的解析式为y=
![](http://thumb.1010pic.com/pic6/res/czsx/web/STSource/20131019105641310711237/SYS201310191056413107112028_DA/7.png)
x+
![](http://thumb.1010pic.com/pic6/res/czsx/web/STSource/20131019105641310711237/SYS201310191056413107112028_DA/8.png)
.
∴D(
![](http://thumb.1010pic.com/pic6/res/czsx/web/STSource/20131019105641310711237/SYS201310191056413107112028_DA/9.png)
,
![](http://thumb.1010pic.com/pic6/res/czsx/web/STSource/20131019105641310711237/SYS201310191056413107112028_DA/10.png)
),
∴CD=OC-OD=4-
![](http://thumb.1010pic.com/pic6/res/czsx/web/STSource/20131019105641310711237/SYS201310191056413107112028_DA/11.png)
=
![](http://thumb.1010pic.com/pic6/res/czsx/web/STSource/20131019105641310711237/SYS201310191056413107112028_DA/12.png)
,
∴S
△ACM=S
△ACD+S
△CDM=
![](http://thumb.1010pic.com/pic6/res/czsx/web/STSource/20131019105641310711237/SYS201310191056413107112028_DA/13.png)
×
![](http://thumb.1010pic.com/pic6/res/czsx/web/STSource/20131019105641310711237/SYS201310191056413107112028_DA/14.png)
×1+
![](http://thumb.1010pic.com/pic6/res/czsx/web/STSource/20131019105641310711237/SYS201310191056413107112028_DA/15.png)
×
![](http://thumb.1010pic.com/pic6/res/czsx/web/STSource/20131019105641310711237/SYS201310191056413107112028_DA/16.png)
×
![](http://thumb.1010pic.com/pic6/res/czsx/web/STSource/20131019105641310711237/SYS201310191056413107112028_DA/17.png)
=
![](http://thumb.1010pic.com/pic6/res/czsx/web/STSource/20131019105641310711237/SYS201310191056413107112028_DA/18.png)
.
点评:本题考查了二次函数与一元二次方程的关系、一元二次方程根与系数的关系、函数解析式的确定、图形面积的求法、函数图象交点等知识.