各地中考数学压轴题预测.pdf

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1、学习好资料欢迎下载yxBADCNG(M)DBCO(A)IyxBADCNMDBCGO(A)IyxNMDBCO(A)2010年各地中考数学压轴题预测1(2010 年江苏省泰州市济川实验初中中考模拟题)如图 1，把一个边长为22的正方形ABCD 放在平面直角坐标系中，点A在坐标原点，点C在 y 轴的正半轴上，经过B、C、D三点的抛物线c1交 x 轴于点 M、N(M在 N的左边).(1)求抛物线 c1的解析式及点M、N的坐标；(2)如图 2，另一个边长为22的正方形/DCBA的中心 G在点 M上，/B、/D在 x 轴的负半轴上(/D在/B的左边)，点/A在第三象限，当点 G沿着抛物线c1从点 M移到点

2、 N，正方形随之移动，移动中/DB始终与 x 轴平行.直接写出点/A、/B移动路线形成的抛物线/)(cA、/)(cB的函数关系式；如图 3，当正方形/DCBA第一次移动到与正方形ABCD有一边在同一直线上时，求点 G的坐标答案：(1)y=21x2+4,M(22,0),N(22,0)yA=21x2+2,yB=21(x2)2+4G(113，313)2.（09 年福建厦门,11 分)已知二次函数yx2xc精品资料-欢迎下载-欢迎下载 名师归纳-第 1 页，共 14 页 -学习好资料欢迎下载(1)若点 A(1，a)、B(2，2n1)在二次函数yx2xc 的图象上，求此二次函数的最小值；(2)若点 D(

3、x1，y1)、E(x2，y2)、P(m，n)(mn)在二次函数yx2xc 的图象上，且D、E 两点关于坐标原点成中心对称，连接 OP当 2 2OP22时，试判断直线DE 与抛物线 yx2xc38的交点个数，并说明理由解析：(1)解：法 1：由题意得n2c，2n12c.1 分解得n1，c 1.2 分法 2：抛物线 yx2xc 的对称轴是x12，且12(1)212，A、B 两点关于对称轴对称.n2n1 1 分n1，c 1.2 分 有 yx2 x1 3 分(x12)254.二次函数yx2x1 的最小值是54.4 分(2)解：点 P(m，m)(m0)，PO2m.2 22m22.2m12.5 分法 1：

4、点 P(m，m)(m0)在二次函数yx2xc 的图象上，mm2mc，即 c m22m.开口向下，且对称轴m1，当 2m 12时，有1c0.6 分法 2：2m12，1m12.1(m1)22.点 P(m，m)(m0)在二次函数yx2xc 的图象上，mm2mc，即 1c(m1)2.11c2.1c0.6 分 点 D、E 关于原点成中心对称，法 1：x2 x1，y2 y1.y1x12x1c，y1x12x1c.2y1 2x1，y1 x1.精品资料-欢迎下载-欢迎下载 名师归纳-第 2 页，共 14 页 -文档编码:CI9V2L10T10M5 HC7X2Z10X4T5 ZK6T9A5W4D10文档编码:CI

5、9V2L10T10M5 HC7X2Z10X4T5 ZK6T9A5W4D10文档编码:CI9V2L10T10M5 HC7X2Z10X4T5 ZK6T9A5W4D10文档编码:CI9V2L10T10M5 HC7X2Z10X4T5 ZK6T9A5W4D10文档编码:CI9V2L10T10M5 HC7X2Z10X4T5 ZK6T9A5W4D10文档编码:CI9V2L10T10M5 HC7X2Z10X4T5 ZK6T9A5W4D10文档编码:CI9V2L10T10M5 HC7X2Z10X4T5 ZK6T9A5W4D10文档编码:CI9V2L10T10M5 HC7X2Z10X4T5 ZK6T9A5W4D1

6、0文档编码:CI9V2L10T10M5 HC7X2Z10X4T5 ZK6T9A5W4D10文档编码:CI9V2L10T10M5 HC7X2Z10X4T5 ZK6T9A5W4D10文档编码:CI9V2L10T10M5 HC7X2Z10X4T5 ZK6T9A5W4D10文档编码:CI9V2L10T10M5 HC7X2Z10X4T5 ZK6T9A5W4D10文档编码:CI9V2L10T10M5 HC7X2Z10X4T5 ZK6T9A5W4D10文档编码:CI9V2L10T10M5 HC7X2Z10X4T5 ZK6T9A5W4D10文档编码:CI9V2L10T10M5 HC7X2Z10X4T5 ZK6

7、T9A5W4D10文档编码:CI9V2L10T10M5 HC7X2Z10X4T5 ZK6T9A5W4D10文档编码:CI9V2L10T10M5 HC7X2Z10X4T5 ZK6T9A5W4D10文档编码:CI9V2L10T10M5 HC7X2Z10X4T5 ZK6T9A5W4D10文档编码:CI9V2L10T10M5 HC7X2Z10X4T5 ZK6T9A5W4D10文档编码:CI9V2L10T10M5 HC7X2Z10X4T5 ZK6T9A5W4D10文档编码:CI9V2L10T10M5 HC7X2Z10X4T5 ZK6T9A5W4D10文档编码:CI9V2L10T10M5 HC7X2Z10

8、X4T5 ZK6T9A5W4D10文档编码:CI9V2L10T10M5 HC7X2Z10X4T5 ZK6T9A5W4D10文档编码:CI9V2L10T10M5 HC7X2Z10X4T5 ZK6T9A5W4D10文档编码:CI9V2L10T10M5 HC7X2Z10X4T5 ZK6T9A5W4D10文档编码:CI9V2L10T10M5 HC7X2Z10X4T5 ZK6T9A5W4D10文档编码:CI9V2L10T10M5 HC7X2Z10X4T5 ZK6T9A5W4D10文档编码:CI9V2L10T10M5 HC7X2Z10X4T5 ZK6T9A5W4D10文档编码:CI9V2L10T10M5

9、HC7X2Z10X4T5 ZK6T9A5W4D10文档编码:CI9V2L10T10M5 HC7X2Z10X4T5 ZK6T9A5W4D10文档编码:CI9V2L10T10M5 HC7X2Z10X4T5 ZK6T9A5W4D10文档编码:CI9V2L10T10M5 HC7X2Z10X4T5 ZK6T9A5W4D10文档编码:CI9V2L10T10M5 HC7X2Z10X4T5 ZK6T9A5W4D10文档编码:CI9V2L10T10M5 HC7X2Z10X4T5 ZK6T9A5W4D10文档编码:CI9V2L10T10M5 HC7X2Z10X4T5 ZK6T9A5W4D10文档编码:CI9V2L

10、10T10M5 HC7X2Z10X4T5 ZK6T9A5W4D10文档编码:CI9V2L10T10M5 HC7X2Z10X4T5 ZK6T9A5W4D10文档编码:CI9V2L10T10M5 HC7X2Z10X4T5 ZK6T9A5W4D10文档编码:CI9V2L10T10M5 HC7X2Z10X4T5 ZK6T9A5W4D10文档编码:CI9V2L10T10M5 HC7X2Z10X4T5 ZK6T9A5W4D10文档编码:CI9V2L10T10M5 HC7X2Z10X4T5 ZK6T9A5W4D10文档编码:CI9V2L10T10M5 HC7X2Z10X4T5 ZK6T9A5W4D10文档编

11、码:CI9V2L10T10M5 HC7X2Z10X4T5 ZK6T9A5W4D10文档编码:CI9V2L10T10M5 HC7X2Z10X4T5 ZK6T9A5W4D10文档编码:CI9V2L10T10M5 HC7X2Z10X4T5 ZK6T9A5W4D10文档编码:CI9V2L10T10M5 HC7X2Z10X4T5 ZK6T9A5W4D10文档编码:CI9V2L10T10M5 HC7X2Z10X4T5 ZK6T9A5W4D10文档编码:CI9V2L10T10M5 HC7X2Z10X4T5 ZK6T9A5W4D10学习好资料欢迎下载设直线 DE：ykx.有 x1kx1.由题意，存在x1x2.

12、存在 x1，使 x10.7 分k 1.直线 DE：y x.8 分法 2：设直线 DE：ykx.则根据题意有kxx2xc，即 x2(k1)xc0.1c0，(k1)24c0.方程 x2(k1)xc0 有实数根.7 分x1x20，k10.k 1.直线 DE：y x.8 分若y x，yx2xc38.则有x2c380.即 x2c38.当 c380 时，即 c38时，方程x2c38有相同的实数根，即直线 y x 与抛物线 yx2xc38有唯一交点.9 分当c380 时，即 c38时，即 1c38时，方程 x2c38有两个不同实数根，即直线 y x 与抛物线 yx2xc38有两个不同的交点.10 分当c38

13、0 时，即 c38时，即38c0 时，方程 x2c38没有实数根，即直线 y x 与抛物线 yx2xc38没有交点.11 分3如图，已知平面直角坐标系xOy中，点A(2,m)，B(-3,n)为两动点，其中m 1，连结OAOB，精品资料-欢迎下载-欢迎下载 名师归纳-第 3 页，共 14 页 -文档编码:CI9V2L10T10M5 HC7X2Z10X4T5 ZK6T9A5W4D10文档编码:CI9V2L10T10M5 HC7X2Z10X4T5 ZK6T9A5W4D10文档编码:CI9V2L10T10M5 HC7X2Z10X4T5 ZK6T9A5W4D10文档编码:CI9V2L10T10M5 HC

14、7X2Z10X4T5 ZK6T9A5W4D10文档编码:CI9V2L10T10M5 HC7X2Z10X4T5 ZK6T9A5W4D10文档编码:CI9V2L10T10M5 HC7X2Z10X4T5 ZK6T9A5W4D10文档编码:CI9V2L10T10M5 HC7X2Z10X4T5 ZK6T9A5W4D10文档编码:CI9V2L10T10M5 HC7X2Z10X4T5 ZK6T9A5W4D10文档编码:CI9V2L10T10M5 HC7X2Z10X4T5 ZK6T9A5W4D10文档编码:CI9V2L10T10M5 HC7X2Z10X4T5 ZK6T9A5W4D10文档编码:CI9V2L10

15、T10M5 HC7X2Z10X4T5 ZK6T9A5W4D10文档编码:CI9V2L10T10M5 HC7X2Z10X4T5 ZK6T9A5W4D10文档编码:CI9V2L10T10M5 HC7X2Z10X4T5 ZK6T9A5W4D10文档编码:CI9V2L10T10M5 HC7X2Z10X4T5 ZK6T9A5W4D10文档编码:CI9V2L10T10M5 HC7X2Z10X4T5 ZK6T9A5W4D10文档编码:CI9V2L10T10M5 HC7X2Z10X4T5 ZK6T9A5W4D10文档编码:CI9V2L10T10M5 HC7X2Z10X4T5 ZK6T9A5W4D10文档编码:

16、CI9V2L10T10M5 HC7X2Z10X4T5 ZK6T9A5W4D10文档编码:CI9V2L10T10M5 HC7X2Z10X4T5 ZK6T9A5W4D10文档编码:CI9V2L10T10M5 HC7X2Z10X4T5 ZK6T9A5W4D10文档编码:CI9V2L10T10M5 HC7X2Z10X4T5 ZK6T9A5W4D10文档编码:CI9V2L10T10M5 HC7X2Z10X4T5 ZK6T9A5W4D10文档编码:CI9V2L10T10M5 HC7X2Z10X4T5 ZK6T9A5W4D10文档编码:CI9V2L10T10M5 HC7X2Z10X4T5 ZK6T9A5W4

17、D10文档编码:CI9V2L10T10M5 HC7X2Z10X4T5 ZK6T9A5W4D10文档编码:CI9V2L10T10M5 HC7X2Z10X4T5 ZK6T9A5W4D10文档编码:CI9V2L10T10M5 HC7X2Z10X4T5 ZK6T9A5W4D10文档编码:CI9V2L10T10M5 HC7X2Z10X4T5 ZK6T9A5W4D10文档编码:CI9V2L10T10M5 HC7X2Z10X4T5 ZK6T9A5W4D10文档编码:CI9V2L10T10M5 HC7X2Z10X4T5 ZK6T9A5W4D10文档编码:CI9V2L10T10M5 HC7X2Z10X4T5 Z

18、K6T9A5W4D10文档编码:CI9V2L10T10M5 HC7X2Z10X4T5 ZK6T9A5W4D10文档编码:CI9V2L10T10M5 HC7X2Z10X4T5 ZK6T9A5W4D10文档编码:CI9V2L10T10M5 HC7X2Z10X4T5 ZK6T9A5W4D10文档编码:CI9V2L10T10M5 HC7X2Z10X4T5 ZK6T9A5W4D10文档编码:CI9V2L10T10M5 HC7X2Z10X4T5 ZK6T9A5W4D10文档编码:CI9V2L10T10M5 HC7X2Z10X4T5 ZK6T9A5W4D10文档编码:CI9V2L10T10M5 HC7X2Z

19、10X4T5 ZK6T9A5W4D10文档编码:CI9V2L10T10M5 HC7X2Z10X4T5 ZK6T9A5W4D10文档编码:CI9V2L10T10M5 HC7X2Z10X4T5 ZK6T9A5W4D10文档编码:CI9V2L10T10M5 HC7X2Z10X4T5 ZK6T9A5W4D10文档编码:CI9V2L10T10M5 HC7X2Z10X4T5 ZK6T9A5W4D10文档编码:CI9V2L10T10M5 HC7X2Z10X4T5 ZK6T9A5W4D10文档编码:CI9V2L10T10M5 HC7X2Z10X4T5 ZK6T9A5W4D10文档编码:CI9V2L10T10M

20、5 HC7X2Z10X4T5 ZK6T9A5W4D10文档编码:CI9V2L10T10M5 HC7X2Z10X4T5 ZK6T9A5W4D10文档编码:CI9V2L10T10M5 HC7X2Z10X4T5 ZK6T9A5W4D10文档编码:CI9V2L10T10M5 HC7X2Z10X4T5 ZK6T9A5W4D10学习好资料欢迎下载OAOB，作BCx轴于C点，ADx轴于D点（1）求证：mn=6；（2）当10AOBS时，抛物线经过AB，两点且以y轴为对称轴，求抛物线对应的二次函数的关系式；（3）在（2）的条件下，设直线AB交y轴于点F，过点F作直线l交抛物线于PQ，两点，问是否存在直线l，使S

21、POF:SQOF=1:2？若存在，求出直线l对应的函数关系式；若不存在，请说明理由解:（1）AB，点坐标分别为(2,m)，(-3,n)，BC=n,OC=3,OD=2,AD=m，又OAOB，易证CBODOA，OABODACODOCB,mn32,mn=6 4（2）由（1）得，BOmOA3，又10AOBS，1102OB OA，即,20OABO602mBO，又92222nOCBCOB,60)9(2nm,又mn=6,2032mnm=6(舍去不合题意，m32),n=1A坐标为(2 6)B，坐标为(31)，易得抛物线解析式为210yx 8（3）直线AB为4yx，且与y轴交于(0 4)F，点，4,OF假设存在

22、直线l交抛物线于PQ，两点，且使SPOF:SQOF=1:2，如图所示，则有PF:FQ=1:2，作PMy轴于M点，QNy轴于N点，P在抛物线210yx上，设P坐标为2(10)tt，则 FM=641022tt，易证PMFQNF，21QFPFFNMFQNPM,QN=2PM=-2t,NF=2MF=1222t，822tONQ点坐标为)82,2(2tt,Q点在抛物线210yx上，1048222tt，解得)3(,3舍去tt，P坐标为)7,3(（，Q坐标为)2,32(，精品资料-欢迎下载-欢迎下载 名师归纳-第 4 页，共 14 页 -文档编码:CI9V2L10T10M5 HC7X2Z10X4T5 ZK6T9

23、A5W4D10文档编码:CI9V2L10T10M5 HC7X2Z10X4T5 ZK6T9A5W4D10文档编码:CI9V2L10T10M5 HC7X2Z10X4T5 ZK6T9A5W4D10文档编码:CI9V2L10T10M5 HC7X2Z10X4T5 ZK6T9A5W4D10文档编码:CI9V2L10T10M5 HC7X2Z10X4T5 ZK6T9A5W4D10文档编码:CI9V2L10T10M5 HC7X2Z10X4T5 ZK6T9A5W4D10文档编码:CI9V2L10T10M5 HC7X2Z10X4T5 ZK6T9A5W4D10文档编码:CI9V2L10T10M5 HC7X2Z10X4

24、T5 ZK6T9A5W4D10文档编码:CI9V2L10T10M5 HC7X2Z10X4T5 ZK6T9A5W4D10文档编码:CI9V2L10T10M5 HC7X2Z10X4T5 ZK6T9A5W4D10文档编码:CI9V2L10T10M5 HC7X2Z10X4T5 ZK6T9A5W4D10文档编码:CI9V2L10T10M5 HC7X2Z10X4T5 ZK6T9A5W4D10文档编码:CI9V2L10T10M5 HC7X2Z10X4T5 ZK6T9A5W4D10文档编码:CI9V2L10T10M5 HC7X2Z10X4T5 ZK6T9A5W4D10文档编码:CI9V2L10T10M5 HC

25、7X2Z10X4T5 ZK6T9A5W4D10文档编码:CI9V2L10T10M5 HC7X2Z10X4T5 ZK6T9A5W4D10文档编码:CI9V2L10T10M5 HC7X2Z10X4T5 ZK6T9A5W4D10文档编码:CI9V2L10T10M5 HC7X2Z10X4T5 ZK6T9A5W4D10文档编码:CI9V2L10T10M5 HC7X2Z10X4T5 ZK6T9A5W4D10文档编码:CI9V2L10T10M5 HC7X2Z10X4T5 ZK6T9A5W4D10文档编码:CI9V2L10T10M5 HC7X2Z10X4T5 ZK6T9A5W4D10文档编码:CI9V2L10

26、T10M5 HC7X2Z10X4T5 ZK6T9A5W4D10文档编码:CI9V2L10T10M5 HC7X2Z10X4T5 ZK6T9A5W4D10文档编码:CI9V2L10T10M5 HC7X2Z10X4T5 ZK6T9A5W4D10文档编码:CI9V2L10T10M5 HC7X2Z10X4T5 ZK6T9A5W4D10文档编码:CI9V2L10T10M5 HC7X2Z10X4T5 ZK6T9A5W4D10文档编码:CI9V2L10T10M5 HC7X2Z10X4T5 ZK6T9A5W4D10文档编码:CI9V2L10T10M5 HC7X2Z10X4T5 ZK6T9A5W4D10文档编码:

27、CI9V2L10T10M5 HC7X2Z10X4T5 ZK6T9A5W4D10文档编码:CI9V2L10T10M5 HC7X2Z10X4T5 ZK6T9A5W4D10文档编码:CI9V2L10T10M5 HC7X2Z10X4T5 ZK6T9A5W4D10文档编码:CI9V2L10T10M5 HC7X2Z10X4T5 ZK6T9A5W4D10文档编码:CI9V2L10T10M5 HC7X2Z10X4T5 ZK6T9A5W4D10文档编码:CI9V2L10T10M5 HC7X2Z10X4T5 ZK6T9A5W4D10文档编码:CI9V2L10T10M5 HC7X2Z10X4T5 ZK6T9A5W4

28、D10文档编码:CI9V2L10T10M5 HC7X2Z10X4T5 ZK6T9A5W4D10文档编码:CI9V2L10T10M5 HC7X2Z10X4T5 ZK6T9A5W4D10文档编码:CI9V2L10T10M5 HC7X2Z10X4T5 ZK6T9A5W4D10文档编码:CI9V2L10T10M5 HC7X2Z10X4T5 ZK6T9A5W4D10文档编码:CI9V2L10T10M5 HC7X2Z10X4T5 ZK6T9A5W4D10文档编码:CI9V2L10T10M5 HC7X2Z10X4T5 ZK6T9A5W4D10文档编码:CI9V2L10T10M5 HC7X2Z10X4T5 Z

29、K6T9A5W4D10文档编码:CI9V2L10T10M5 HC7X2Z10X4T5 ZK6T9A5W4D10文档编码:CI9V2L10T10M5 HC7X2Z10X4T5 ZK6T9A5W4D10文档编码:CI9V2L10T10M5 HC7X2Z10X4T5 ZK6T9A5W4D10文档编码:CI9V2L10T10M5 HC7X2Z10X4T5 ZK6T9A5W4D10文档编码:CI9V2L10T10M5 HC7X2Z10X4T5 ZK6T9A5W4D10文档编码:CI9V2L10T10M5 HC7X2Z10X4T5 ZK6T9A5W4D10学习好资料欢迎下载易得直线PQ为43xy根据抛物线

30、的对称性可得直线PQ的另解为43xy14【说明：本卷由乌牛中学谢理福老师（665650）命题，实验中学俞志莉老师（660558）、城西中学全岳强审阅。各题可能有不同的正确解法，可参考上述步骤相应给分，各阅卷老师在确认答案正确无误后才可开始评卷 】4.已知关于x的一元二次方程22410 xxk有实数根，k为正整数.（1）求k的值；（2）当此方程有两个非零的整数根时，将关于x的二次函数2241yxxk的图象向下平移8 个单位，求平移后的图象的解析式；（3）在（2）的条件下，将平移后的二次函数的图象在x轴下方的部分沿x轴翻折，图象的其余部分保持不变，得到一个新的图象.请你结合这个新的图象回答：当直线

31、12yxb bk与此图象有两个公共点时，b的取值范围.5（福建漳州满分14 分）如图 1，已知：抛物线212yxbxc与x轴交于AB、两点，与y轴交于点C，经过BC、两点的直线是122yx，连结AC（1）BC、两点坐标分别为B（_，_）、C（_，_），抛物线的函数关系式为_；（2）判断ABC的形状，并说明理由；（3）若ABC内部能否截出面积最大的矩形DEFC（顶点DEF、G在ABC各边上）？若能，求出在AB边上的矩形顶点的坐标；若不能，请说明理由6.（福建漳州本题满分9 分）如图 17，某公路隧道横截面为抛物线，其最大高度为6 米，底部宽度OM为 12 米.现以 O点为原点，C A O B x

32、 y C A O B x y 图 1 图 2(备用)（第 26 题）精品资料-欢迎下载-欢迎下载 名师归纳-第 5 页，共 14 页 -文档编码:CI9V2L10T10M5 HC7X2Z10X4T5 ZK6T9A5W4D10文档编码:CI9V2L10T10M5 HC7X2Z10X4T5 ZK6T9A5W4D10文档编码:CI9V2L10T10M5 HC7X2Z10X4T5 ZK6T9A5W4D10文档编码:CI9V2L10T10M5 HC7X2Z10X4T5 ZK6T9A5W4D10文档编码:CI9V2L10T10M5 HC7X2Z10X4T5 ZK6T9A5W4D10文档编码:CI9V2L1

33、0T10M5 HC7X2Z10X4T5 ZK6T9A5W4D10文档编码:CI9V2L10T10M5 HC7X2Z10X4T5 ZK6T9A5W4D10文档编码:CI9V2L10T10M5 HC7X2Z10X4T5 ZK6T9A5W4D10文档编码:CI9V2L10T10M5 HC7X2Z10X4T5 ZK6T9A5W4D10文档编码:CI9V2L10T10M5 HC7X2Z10X4T5 ZK6T9A5W4D10文档编码:CI9V2L10T10M5 HC7X2Z10X4T5 ZK6T9A5W4D10文档编码:CI9V2L10T10M5 HC7X2Z10X4T5 ZK6T9A5W4D10文档编码

34、:CI9V2L10T10M5 HC7X2Z10X4T5 ZK6T9A5W4D10文档编码:CI9V2L10T10M5 HC7X2Z10X4T5 ZK6T9A5W4D10文档编码:CI9V2L10T10M5 HC7X2Z10X4T5 ZK6T9A5W4D10文档编码:CI9V2L10T10M5 HC7X2Z10X4T5 ZK6T9A5W4D10文档编码:CI9V2L10T10M5 HC7X2Z10X4T5 ZK6T9A5W4D10文档编码:CI9V2L10T10M5 HC7X2Z10X4T5 ZK6T9A5W4D10文档编码:CI9V2L10T10M5 HC7X2Z10X4T5 ZK6T9A5W

35、4D10文档编码:CI9V2L10T10M5 HC7X2Z10X4T5 ZK6T9A5W4D10文档编码:CI9V2L10T10M5 HC7X2Z10X4T5 ZK6T9A5W4D10文档编码:CI9V2L10T10M5 HC7X2Z10X4T5 ZK6T9A5W4D10文档编码:CI9V2L10T10M5 HC7X2Z10X4T5 ZK6T9A5W4D10文档编码:CI9V2L10T10M5 HC7X2Z10X4T5 ZK6T9A5W4D10文档编码:CI9V2L10T10M5 HC7X2Z10X4T5 ZK6T9A5W4D10文档编码:CI9V2L10T10M5 HC7X2Z10X4T5

36、ZK6T9A5W4D10文档编码:CI9V2L10T10M5 HC7X2Z10X4T5 ZK6T9A5W4D10文档编码:CI9V2L10T10M5 HC7X2Z10X4T5 ZK6T9A5W4D10文档编码:CI9V2L10T10M5 HC7X2Z10X4T5 ZK6T9A5W4D10文档编码:CI9V2L10T10M5 HC7X2Z10X4T5 ZK6T9A5W4D10文档编码:CI9V2L10T10M5 HC7X2Z10X4T5 ZK6T9A5W4D10文档编码:CI9V2L10T10M5 HC7X2Z10X4T5 ZK6T9A5W4D10文档编码:CI9V2L10T10M5 HC7X2

37、Z10X4T5 ZK6T9A5W4D10文档编码:CI9V2L10T10M5 HC7X2Z10X4T5 ZK6T9A5W4D10文档编码:CI9V2L10T10M5 HC7X2Z10X4T5 ZK6T9A5W4D10文档编码:CI9V2L10T10M5 HC7X2Z10X4T5 ZK6T9A5W4D10文档编码:CI9V2L10T10M5 HC7X2Z10X4T5 ZK6T9A5W4D10文档编码:CI9V2L10T10M5 HC7X2Z10X4T5 ZK6T9A5W4D10文档编码:CI9V2L10T10M5 HC7X2Z10X4T5 ZK6T9A5W4D10文档编码:CI9V2L10T10

38、M5 HC7X2Z10X4T5 ZK6T9A5W4D10文档编码:CI9V2L10T10M5 HC7X2Z10X4T5 ZK6T9A5W4D10文档编码:CI9V2L10T10M5 HC7X2Z10X4T5 ZK6T9A5W4D10文档编码:CI9V2L10T10M5 HC7X2Z10X4T5 ZK6T9A5W4D10文档编码:CI9V2L10T10M5 HC7X2Z10X4T5 ZK6T9A5W4D10文档编码:CI9V2L10T10M5 HC7X2Z10X4T5 ZK6T9A5W4D10文档编码:CI9V2L10T10M5 HC7X2Z10X4T5 ZK6T9A5W4D10文档编码:CI9

39、V2L10T10M5 HC7X2Z10X4T5 ZK6T9A5W4D10文档编码:CI9V2L10T10M5 HC7X2Z10X4T5 ZK6T9A5W4D10学习好资料欢迎下载OM 所在直线为x 轴建立直角坐标系.(1)直接写出点M及抛物线顶点P的坐标；(2)求这条抛物线的解析式；(3)若要搭建一个矩形“支撑架”AD-DC-CB，使 C、D点在抛物线上，A、B点在地面 OM上，则这个“支撑架”总长的最大值是多少？7(广西贺州本题满分10 分)如图，抛物线2124yxx的顶点为A，与 y 轴交于点B（1）求点 A、点 B的坐标（2）若点 P是 x 轴上任意一点，求证：PAPBAB（3）当PBP

40、A最大时，求点P的坐标8、（贵州安顺本题满分12 分）如图，已知抛物线与x交于 A(1，0)、E(3，0)两点，与y轴交于点B(0，3)。（1）求抛物线的解析式；（2）设抛物线顶点为D，求四边形AEDB的面积；（3）AOB与 DBE是否相似？如果相似，请给以证明；如果不相似，请说明理由。9.（海南省13 分）如图12，已知抛物线经过坐标原点O和 x 轴上另一点E，顶点 M的坐标为 (2,4)；矩B O A x y 第 28 题图精品资料-欢迎下载-欢迎下载 名师归纳-第 6 页，共 14 页 -文档编码:CI9V2L10T10M5 HC7X2Z10X4T5 ZK6T9A5W4D10文档编码:C

41、I9V2L10T10M5 HC7X2Z10X4T5 ZK6T9A5W4D10文档编码:CI9V2L10T10M5 HC7X2Z10X4T5 ZK6T9A5W4D10文档编码:CI9V2L10T10M5 HC7X2Z10X4T5 ZK6T9A5W4D10文档编码:CI9V2L10T10M5 HC7X2Z10X4T5 ZK6T9A5W4D10文档编码:CI9V2L10T10M5 HC7X2Z10X4T5 ZK6T9A5W4D10文档编码:CI9V2L10T10M5 HC7X2Z10X4T5 ZK6T9A5W4D10文档编码:CI9V2L10T10M5 HC7X2Z10X4T5 ZK6T9A5W4D

42、10文档编码:CI9V2L10T10M5 HC7X2Z10X4T5 ZK6T9A5W4D10文档编码:CI9V2L10T10M5 HC7X2Z10X4T5 ZK6T9A5W4D10文档编码:CI9V2L10T10M5 HC7X2Z10X4T5 ZK6T9A5W4D10文档编码:CI9V2L10T10M5 HC7X2Z10X4T5 ZK6T9A5W4D10文档编码:CI9V2L10T10M5 HC7X2Z10X4T5 ZK6T9A5W4D10文档编码:CI9V2L10T10M5 HC7X2Z10X4T5 ZK6T9A5W4D10文档编码:CI9V2L10T10M5 HC7X2Z10X4T5 ZK

43、6T9A5W4D10文档编码:CI9V2L10T10M5 HC7X2Z10X4T5 ZK6T9A5W4D10文档编码:CI9V2L10T10M5 HC7X2Z10X4T5 ZK6T9A5W4D10文档编码:CI9V2L10T10M5 HC7X2Z10X4T5 ZK6T9A5W4D10文档编码:CI9V2L10T10M5 HC7X2Z10X4T5 ZK6T9A5W4D10文档编码:CI9V2L10T10M5 HC7X2Z10X4T5 ZK6T9A5W4D10文档编码:CI9V2L10T10M5 HC7X2Z10X4T5 ZK6T9A5W4D10文档编码:CI9V2L10T10M5 HC7X2Z1

44、0X4T5 ZK6T9A5W4D10文档编码:CI9V2L10T10M5 HC7X2Z10X4T5 ZK6T9A5W4D10文档编码:CI9V2L10T10M5 HC7X2Z10X4T5 ZK6T9A5W4D10文档编码:CI9V2L10T10M5 HC7X2Z10X4T5 ZK6T9A5W4D10文档编码:CI9V2L10T10M5 HC7X2Z10X4T5 ZK6T9A5W4D10文档编码:CI9V2L10T10M5 HC7X2Z10X4T5 ZK6T9A5W4D10文档编码:CI9V2L10T10M5 HC7X2Z10X4T5 ZK6T9A5W4D10文档编码:CI9V2L10T10M5

45、 HC7X2Z10X4T5 ZK6T9A5W4D10文档编码:CI9V2L10T10M5 HC7X2Z10X4T5 ZK6T9A5W4D10文档编码:CI9V2L10T10M5 HC7X2Z10X4T5 ZK6T9A5W4D10文档编码:CI9V2L10T10M5 HC7X2Z10X4T5 ZK6T9A5W4D10文档编码:CI9V2L10T10M5 HC7X2Z10X4T5 ZK6T9A5W4D10文档编码:CI9V2L10T10M5 HC7X2Z10X4T5 ZK6T9A5W4D10文档编码:CI9V2L10T10M5 HC7X2Z10X4T5 ZK6T9A5W4D10文档编码:CI9V2

46、L10T10M5 HC7X2Z10X4T5 ZK6T9A5W4D10文档编码:CI9V2L10T10M5 HC7X2Z10X4T5 ZK6T9A5W4D10文档编码:CI9V2L10T10M5 HC7X2Z10X4T5 ZK6T9A5W4D10文档编码:CI9V2L10T10M5 HC7X2Z10X4T5 ZK6T9A5W4D10文档编码:CI9V2L10T10M5 HC7X2Z10X4T5 ZK6T9A5W4D10文档编码:CI9V2L10T10M5 HC7X2Z10X4T5 ZK6T9A5W4D10文档编码:CI9V2L10T10M5 HC7X2Z10X4T5 ZK6T9A5W4D10文档

47、编码:CI9V2L10T10M5 HC7X2Z10X4T5 ZK6T9A5W4D10文档编码:CI9V2L10T10M5 HC7X2Z10X4T5 ZK6T9A5W4D10文档编码:CI9V2L10T10M5 HC7X2Z10X4T5 ZK6T9A5W4D10文档编码:CI9V2L10T10M5 HC7X2Z10X4T5 ZK6T9A5W4D10文档编码:CI9V2L10T10M5 HC7X2Z10X4T5 ZK6T9A5W4D10文档编码:CI9V2L10T10M5 HC7X2Z10X4T5 ZK6T9A5W4D10学习好资料欢迎下载形 ABCD 的顶点 A与点 O重合，AD、AB分别在 x

48、 轴、y 轴上，且AD=2，AB=3.（1）求该抛物线所对应的函数关系式；（2）将矩形 ABCD 以每秒 1 个单位长度的速度从图12 所示的位置沿x 轴的正方向匀速平行移动，同时一动点 P也以相同的速度从点 A出发向 B匀速移动，设它们运动的时间为t 秒（0t 3），直线 AB 与 该抛物线的交点为N（如图 13 所示）.当 t=25时，判断点P是否在直线ME上，并说明理由；设以 P、N、C、D为顶点的多边形面积为S，试问 S是否存在最大值？若存在，求出这个最大值；若不存在，请说明理由24.（1）因所求抛物线的顶点M 的坐标为(2,4)，故可设其关系式为224ya x(1 分)又抛物线经过O

49、(0,0)，于是得20240a，(2 分)解得a=-1(3 分)所求函数关系式为224yx，即24yxx.(4 分)（2）点 P 不在直线 ME 上.(5 分)根据抛物线的对称性可知E 点的坐标为(4,0)，又 M 的坐标为(2,4)，设直线 ME 的关系式为y=kx+b.于是得4204bkbk，解得82bk所以直线ME 的关系式为y=-2x+8.(6 分)由已知条件易得，当t25时，OA=AP25，25,25P(7 分)P 点的坐标不满足直线ME 的关系式 y=-2x+8.当 t25时，点 P 不在直线 ME 上.(8 分)S存在最大值.理由如下：(9 分)点 A 在 x 轴的非负半轴上，且

50、N 在抛物线上，OA=AP=t.点 P，N 的坐标分别为(t,t)、(t,-t2+4t)AN=-t2+4t(0t3),图 13B C O A D E M y x P N 图 12B C O(A)D E M y x 精品资料-欢迎下载-欢迎下载 名师归纳-第 7 页，共 14 页 -文档编码:CI9V2L10T10M5 HC7X2Z10X4T5 ZK6T9A5W4D10文档编码:CI9V2L10T10M5 HC7X2Z10X4T5 ZK6T9A5W4D10文档编码:CI9V2L10T10M5 HC7X2Z10X4T5 ZK6T9A5W4D10文档编码:CI9V2L10T10M5 HC7X2Z10