009]解:(1)①轴.轴. 四边形为矩形. 轴.轴. 四边形为矩形. 轴.轴. 四边形均为矩形.············ 1分 . . . . . . .······························································································ 2分 ②由(1)知. . .············································································································ 4分 . .································································································ 5分 . .············································································································· 6分 轴. 四边形是平行四边形. .············································································································· 7分 同理. .············································································································· 8分 (2)与仍然相等.························································································· 9分 . . 又. .································· 10分 . . . . . .············································································································ 11分 轴. 四边形是平行四边形. . 同理. .·········································································································· 12分 【查看更多】

 

题目列表(包括答案和解析)

如图,是二次函数y=ax2+bx+c图象的一部分,其对称轴为直线x=1,若其与x轴一交点为A(3,0),则由图象可知,方程ax2+bx+c=0的另一个解是(  )

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已知二次函数y=x2-6x+5.
(1)请写出该函数的对称轴,顶点坐标;
(2)函数图象与x轴交点坐标为
,与y轴的交点坐标为
;
(3)当
时y>0,
时y随x的增大而增大;
(4)写出不等式x2-6x+5<0的解集.

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如图,直线y=k和双曲线y=
k
x
相交于点P,过P点作PA0垂直x轴,垂足A0,由
y=k
y=
k
x
 可解得x=1,即A0横坐标为1.x轴上的点A0、A1、A2、….An的横坐标是连续整数.过点A1、A2、…、An分别作x轴的垂线,与双曲线y=
k
x
(x>0精英家教网)及直线y=k分别交于点B1、B2、…、Bn、C1、C2、….Cn.
(1)求
C1B1
A1B1
的值;
(2)求
C2B2
A2B2
的值;
(3)试猜想
CnBn
AnBn
的值(直接写答案).

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如图是抛物线y=ax2+bx+c的一部分,其对称轴为直线x=1.若抛物线与x轴一个交点为A(3,0),则由图象可知,不等式ax2+bx+c≥0的解集是:
x≤-1或x≥3
x≤-1或x≥3
.

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如图,在平面直角坐标系中,点A、B分别在x轴y轴的正半轴上,线段OA的长是不等式5x-4<3(x+2)的最大整数解,线段OB的长是一元二次方程x2-2x-3=0的一个根,将Rt△ABO沿BE折叠,使AB边落在OB边所在的y轴上,点A与点D重合.
(1)求OA、OB的长;
(2)求直线BE的解析式;
(3)在平面内是否存在点M,使B、O、E、M为顶点的四边形为平行四边形?若存在,请直接写出点M的坐标;若不存在,请说明理由.

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