解:(1)∵抛物线y=-x
2+bx+c,当x=-
![](http://thumb.1010pic.com/pic5/latex/13.png)
时,y取最大值
![](http://thumb.1010pic.com/pic5/latex/6318.png)
,
∴抛物线的解析式是:y=-(x+
![](http://thumb.1010pic.com/pic5/latex/13.png)
)
2+
![](http://thumb.1010pic.com/pic5/latex/6318.png)
,即y=-x
2-x+6;
当x=0时,y=6,即C点坐标是(0,6),
当y=0时,-x
2-x+6=0,解得:x=2或-3,
![](http://thumb.1010pic.com/pic5/upload/201310/5286c1095c273.png)
即A点坐标是(-3,0),B点坐标是(2,0).
将A(-3,0),C(0,6)代入直线AC的解析式y=kx+m,
得
![](http://thumb.1010pic.com/pic5/latex/253037.png)
,
解得:
![](http://thumb.1010pic.com/pic5/latex/253038.png)
,
则直线的解析式是:y=2x+6;
(2)过点B作BD⊥AC,D为垂足,
∵S
△ABP:S
△BPC=1:3,
∴
![](http://thumb.1010pic.com/pic5/latex/253039.png)
=
![](http://thumb.1010pic.com/pic5/latex/8.png)
,
![](http://thumb.1010pic.com/pic5/upload/201310/5286c1096ae65.png)
∴AP:PC=1:3,
由勾股定理,得AC=
![](http://thumb.1010pic.com/pic5/latex/246948.png)
=3
![](http://thumb.1010pic.com/pic5/latex/559.png)
.
①当点P为线段AC上一点时,过点P作PH⊥x轴,点H为垂足.
∵PH∥OC,
∴
![](http://thumb.1010pic.com/pic5/latex/253040.png)
=
![](http://thumb.1010pic.com/pic5/latex/184915.png)
=
![](http://thumb.1010pic.com/pic5/latex/96.png)
,
∴PH=
![](http://thumb.1010pic.com/pic5/latex/33.png)
,
∴
![](http://thumb.1010pic.com/pic5/latex/33.png)
=2x+6,
∴x=-
![](http://thumb.1010pic.com/pic5/latex/2321.png)
,
∴点P(-
![](http://thumb.1010pic.com/pic5/latex/2321.png)
,
![](http://thumb.1010pic.com/pic5/latex/33.png)
);
②
![](http://thumb.1010pic.com/pic5/upload/201310/5286c1097b1e5.png)
当点P在CA延长线时,作PG⊥x轴,点G为垂足.
∵AP:PC=1:3,
∴AP:AC=1:2.
∵PG∥OC,
∴
![](http://thumb.1010pic.com/pic5/latex/235085.png)
=
![](http://thumb.1010pic.com/pic5/latex/184915.png)
=
![](http://thumb.1010pic.com/pic5/latex/13.png)
,
∴PG=3,
∴-3=2x+6,x=-
![](http://thumb.1010pic.com/pic5/latex/190.png)
,
∴点P(-
![](http://thumb.1010pic.com/pic5/latex/190.png)
,-3).
综上所述,点P的坐标为(-
![](http://thumb.1010pic.com/pic5/latex/2321.png)
,
![](http://thumb.1010pic.com/pic5/latex/33.png)
)或(-
![](http://thumb.1010pic.com/pic5/latex/190.png)
,-3).
(3)设直线y=
![](http://thumb.1010pic.com/pic5/latex/13.png)
x+a与抛物线y=-x
2-x+6的交点为M(x
M,y
M),N(x
N,y
N)(M在N左侧).
则
![](http://thumb.1010pic.com/pic5/latex/253041.png)
,
![](http://thumb.1010pic.com/pic5/latex/253042.png)
为方程组
![](http://thumb.1010pic.com/pic5/latex/253043.png)
的解,
由方程组消去y整理,得:x
2+
![](http://thumb.1010pic.com/pic5/latex/33.png)
x+a-6=0,
∴x
M、x
N是方程x
2+
![](http://thumb.1010pic.com/pic5/latex/33.png)
x+a-6=0的两个根,
∴x
M+x
N=-
![](http://thumb.1010pic.com/pic5/latex/33.png)
,x
M•x
N=a-6,
∴y
M•y
N=(
![](http://thumb.1010pic.com/pic5/latex/13.png)
x
M+a)(
![](http://thumb.1010pic.com/pic5/latex/13.png)
x
N+a)=
![](http://thumb.1010pic.com/pic5/latex/96.png)
x
M•x
N+
![](http://thumb.1010pic.com/pic5/latex/322.png)
(x
M+x
N)+a
2=
![](http://thumb.1010pic.com/pic5/latex/96.png)
(a-6)-
![](http://thumb.1010pic.com/pic5/latex/365.png)
a+a
2.
![](http://thumb.1010pic.com/pic5/upload/201310/5286c1099a232.png)
①存在a的值,使得∠MON=90°.理由如下:
∵∠MON=90°,
∴OM
2+ON
2=MN
2,即
![](http://thumb.1010pic.com/pic5/latex/253044.png)
+
![](http://thumb.1010pic.com/pic5/latex/253045.png)
+
![](http://thumb.1010pic.com/pic5/latex/253046.png)
+
![](http://thumb.1010pic.com/pic5/latex/253047.png)
=(x
M-x
N)
2+(y
M-y
N)
2,
化简得x
M•x
N+y
M•y
N=0,
∴(a-6)+
![](http://thumb.1010pic.com/pic5/latex/96.png)
(a-6)-
![](http://thumb.1010pic.com/pic5/latex/365.png)
a+a
2=0,
整理,得2a
2+a-15=0,
解得a
1=-3,a
2=
![](http://thumb.1010pic.com/pic5/latex/259.png)
,
∴存在a值,使得∠MON=90°,其值为a=-3或a=
![](http://thumb.1010pic.com/pic5/latex/259.png)
;
②∵∠MON>90°,
∴OM
2+ON
2<MN
2,即
![](http://thumb.1010pic.com/pic5/latex/253044.png)
+
![](http://thumb.1010pic.com/pic5/latex/253045.png)
+
![](http://thumb.1010pic.com/pic5/latex/253046.png)
+
![](http://thumb.1010pic.com/pic5/latex/253047.png)
<(x
M-x
N)
2+(y
M-y
N)
2,
化简得x
M•x
N+y
M•y
N<0,
∴(a-6)+
![](http://thumb.1010pic.com/pic5/latex/96.png)
(a-6)-
![](http://thumb.1010pic.com/pic5/latex/365.png)
a+a
2<0,
整理,得2a
2+a-15<0,
解得-3<a<
![](http://thumb.1010pic.com/pic5/latex/259.png)
,
∴当∠MON>90°时,a的取值范围是-3<a<
![](http://thumb.1010pic.com/pic5/latex/259.png)
.
分析:(1)先根据抛物线y=-x
2+bx+c,当x=-
![](http://thumb.1010pic.com/pic5/latex/13.png)
时,y取最大值
![](http://thumb.1010pic.com/pic5/latex/6318.png)
,得到抛物线的顶点坐标为(-
![](http://thumb.1010pic.com/pic5/latex/13.png)
,
![](http://thumb.1010pic.com/pic5/latex/6318.png)
),可写出抛物线的顶点式,再根据抛物线的解析式求出A、C的坐标,然后将A、C的坐标代入
y=kx+m,运用待定系数法即可求出直线的解析式;
(2)根据等高三角形的面积比等于底边比,因此两三角形的面积比实际是AP:PC=1:3,即3AP=PC,可先求出AC的长,然后分情况讨论:
①当P在线段AC上时,过点P作PH⊥x轴,点H为垂足.由PH∥OC,根据平行线分线段成比例定理求出PH的长,进而求出P点的坐标;
②当P在CA的延长线上时,由PG∥OC,根据平行线分线段成比例定理求出PG的长,进而求出P点的坐标;
(3)联立两函数的解析式,设直线y=
![](http://thumb.1010pic.com/pic5/latex/13.png)
x+a与抛物线y=-x
2-x+6的交点为M(x
M,y
M),N(x
N,y
N)(M在N左侧),则x
M、x
N是方程x
2+
![](http://thumb.1010pic.com/pic5/latex/33.png)
x+a-6=0的两个根,由一元二次方程根与系数关系得,x
M+x
N=-
![](http://thumb.1010pic.com/pic5/latex/33.png)
,x
M•x
N=a-6,进而求出y
M•y
N=
![](http://thumb.1010pic.com/pic5/latex/96.png)
(a-6)-
![](http://thumb.1010pic.com/pic5/latex/365.png)
a+a
2.
①由于∠MON=90°,根据勾股定理得出OM
2+ON
2=MN
2,据此列出关于a的方程,解方程即可求出a的值;
②由于∠MON>90°,根据勾股定理得出OM
2+ON
2<MN
2,据此列出关于a的不等式,解不等式即可求出a的范围.
点评:本题考查了二次函数的综合题型,其中涉及到运用待定系数法求函数的解析式,二次函数的性质,三角形的面积,平行线分线段成比例定理,函数与方程的关系,勾股定理,钝角三角形三边的关系等知识,综合性较强,难度较大.运用分类讨论、数形结合及方程思想是解题的关键.