试题分析:(1)根据拋物线
![](http://thumb.1010pic.com/pic2/upload/papers/20140823/201408230254587631178.png)
经过原点即可求得m的值,再结合二次项系数不为0即可得到结果;
(2)由点B(-2,n)在拋物线
![](http://thumb.1010pic.com/pic2/upload/papers/20140823/20140823025458810561.png)
上可求得n的值,即得B点的坐标,根据平移的规律可得直线l的解析式为
![](http://thumb.1010pic.com/pic2/upload/papers/20140823/20140823025458904542.png)
,由直线l经过B点即可求得结果;
(3)拋物线
![](http://thumb.1010pic.com/pic2/upload/papers/20140823/20140823025458810561.png)
的对称轴为直线x=2,则对称轴与x轴的交点C的坐标为(2,0),直线l与y轴、直线x=2的交点坐标分别为 D(0,-1)、E(2,-5).过点B作BG⊥直线x=2于G,与y轴交于F.则BG=4.在Rt△BGC中,根据勾股定理可求得CB的长,过点E作EH⊥y轴于H.则点H的坐标为 (0,-5).证得△DFB≌△DHE,即可得到点P在直线CD上,即有符合条件的点P是直线CD与该抛物线的交点.设直线CD的解析式为y="kx+a." 将D(0,-1)、C(2,0)代入即可求得直线CD的解析式,从而求得结果.
(1)∵拋物线
![](http://thumb.1010pic.com/pic2/upload/papers/20140823/201408230254587631178.png)
经过原点,
∴m
2-6m+8=0.解得m
1=2,m
2=4.
由题意知m¹4,
∴m=2
∴拋物线的解析式为
![](http://thumb.1010pic.com/pic2/upload/papers/20140823/20140823025458810561.png)
;
(2)∵点B(-2,n)在拋物线
![](http://thumb.1010pic.com/pic2/upload/papers/20140823/20140823025458810561.png)
上,
∴n=3.
∴B点的坐标为(–2,3) .
∵直线l的解析式为
![](http://thumb.1010pic.com/pic2/upload/papers/20140823/20140823025458904542.png)
,直线l经过B点,
∴
![](http://thumb.1010pic.com/pic2/upload/papers/20140823/20140823025459013622.png)
.
∴
![](http://thumb.1010pic.com/pic2/upload/papers/20140823/20140823025459028342.png)
;
(3)∵拋物线
![](http://thumb.1010pic.com/pic2/upload/papers/20140823/20140823025458810561.png)
的对称轴为直线x=2,直线l的解析式为y=-2x-1,
∴拋物线
![](http://thumb.1010pic.com/pic2/upload/papers/20140823/20140823025458810561.png)
的对称轴与x轴的交点C的坐标为(2,0),
直线l与y轴、直线x=2的交点坐标分别为 D(0,-1)、E(2,-5).
过点B作BG⊥直线x=2于G,与y轴交于F.
![](http://thumb.1010pic.com/pic2/upload/papers/20140823/201408230254590915153.png)
则BG=4.
在Rt△BGC中,
![](http://thumb.1010pic.com/pic2/upload/papers/20140823/20140823025459106737.png)
.
∵CE=5,
∴CB=CE.
过点E作EH⊥y轴于H.
则点H的坐标为 (0,-5).
∵点F、D的坐标为F(0,3)、D(0,-1),
∴FD=DH=4,BF=EH=2,∠BFD=∠EHD=90°.
∴△DFB≌△DHE .
∴DB="DE."
∵PB=PE,
∴点P在直线CD上.
∴符合条件的点P是直线CD与该抛物线的交点.
设直线CD的解析式为y="kx+a."
将D(0,-1)、C(2,0)代入,得
![](http://thumb.1010pic.com/pic2/upload/papers/20140823/20140823025459122742.png)
解得
![](http://thumb.1010pic.com/pic2/upload/papers/20140823/20140823025459138765.png)
∴直线CD的解析式为
![](http://thumb.1010pic.com/pic2/upload/papers/20140823/20140823025459153559.png)
.
设点P的坐标为(x,
![](http://thumb.1010pic.com/pic2/upload/papers/20140823/20140823025459169473.png)
),
∴
![](http://thumb.1010pic.com/pic2/upload/papers/20140823/20140823025459200461.png)
=
![](http://thumb.1010pic.com/pic2/upload/papers/20140823/20140823025459169473.png)
.
解得
![](http://thumb.1010pic.com/pic2/upload/papers/20140823/20140823025459216532.png)
,
![](http://thumb.1010pic.com/pic2/upload/papers/20140823/20140823025459231524.png)
.
∴
![](http://thumb.1010pic.com/pic2/upload/papers/20140823/20140823025459247593.png)
,
![](http://thumb.1010pic.com/pic2/upload/papers/20140823/20140823025459262552.png)
.
∴点P的坐标为(
![](http://thumb.1010pic.com/pic2/upload/papers/20140823/20140823025458826404.png)
,
![](http://thumb.1010pic.com/pic2/upload/papers/20140823/20140823025458841476.png)
)或(
![](http://thumb.1010pic.com/pic2/upload/papers/20140823/20140823025458857396.png)
,
![](http://thumb.1010pic.com/pic2/upload/papers/20140823/20140823025458857467.png)
).
点评:此类问题是初中数学的重点和难点,在中考中极为常见,一般以压轴题形式出现,难度较大.