2015年中考数学压轴题精选（二次函数）（16题） - 附详细解答和评分标准

1、（10广东茂名25题）（本题满分10分）

如图，在平面直角坐标系中，抛物线y=－

22x+bx+c经3y 过A（0，－4）、B（x1，0）、 C（x2，0）三点，且x2-x1=5． （1）求b、c的值；（4分）

（2）在抛物线上求一点D，使得四边形BDCE是以BC为对 角线的菱形；（3分）

（3）在抛物线上是否存在一点P，使得四边形BPOH是以OB为对角线的菱形？若存在，求出点P的坐标，并判断这个菱形是否为正方形？若不存在，请说明理由．（3分）

解：

解：（1）解法一： ∵抛物线y=－

B C O x

A （第25题图）

22， x+bx+c经过点A（0，－4）

322x+bx+c=0的两个根， 3 ∴c=－4 ??1分

又由题意可知，x1、x2是方程－∴x1+x2=

33b， x1x2=－c=6 ·········································································· 2分 222由已知得（x2-x1）=25 又（x2-x∴

1）=（x2+x1）－4x122x2=

92b－24 492b－24=25 414解得b=± ···················································································································· 3分

314当b=时，抛物线的对称轴在y轴的右边，不合题意，舍去．

3∴b=－

14． ··················································································································· 4分 3解法二：∵x1、x2是方程－

即方程2x－3b222x+bx+c=0的两个根， 3x+12=0的两个根．

1

∴x=

3b?9b2?96， ·················································································· 2分

49b2?96∴x2－x1==5，

2 解得 b=±14 ········································································································ 3分 3 （以下与解法一相同．）

（2）∵四边形BDCE是以BC为对角线的菱形，根据菱形的性质，点D必在抛物线的

对称轴上， ··········································································································· 5分

22142725···································· 6分 x－x－4=－（x+）2+ ·

33326725 ∴抛物线的顶点（－，）即为所求的点D． ·········································· 7分

26 又∵y=－

（3）∵四边形BPOH是以OB为对角线的菱形，点B的坐标为（－6，0），

根据菱形的性质，点P必是直线x=-3与

2214································································· 8分 x-x-4的交点， ·

332142 ∴当x=－3时，y=－×（－3）－×（－3）－4=4，

33抛物线y=－

∴在抛物线上存在一点P（－3，4），使得四边形BPOH为菱形． ··················· 9分 四边形BPOH不能成为正方形，因为如果四边形BPOH为正方形，点P的坐标

只能是（－3，3），但这一点不在抛物线上． ························································· 10分 2、（08广东肇庆25题）（本小题满分10分）

已知点A（a，y1）、B（2a，y2）、C（3a，y3）都在抛物线y?5x2?12x上. （1）求抛物线与x轴的交点坐标； （2）当a=1时，求△ABC的面积；

（3）是否存在含有y1、y2、y3，且与a无关的等式？如果存在，试给出一个，并加以证明；如果不存在，说明理由.

解：（1）由5x?12x=0， ····················································································· （1分）

212． ························································································· （2分） 512∴抛物线与x轴的交点坐标为（0，0）、（?，0）． ·········································· （3分）

5得x1?0，x2??（2）当a=1时，得A（1，17）、B（2，44）、C（3，81）， ································· （4分）

2

分别过点A、B、C作x轴的垂线，垂足分别为D、E、F，则有

······················································· （5分） S?ABC=S梯形ADFC -S梯形ADEB -S梯形BEFC · =

(17?81)?2(17?44)?1(44?81)?1-- ······································· （6分）

222=5（个单位面积） ·············································································· （7分）

（3）如：y3?3(y2?y1)． ················································································ （8分）

事实上，y3?5?(3a)2?12?(3a) =45a2+36a． 3（y2?y1）=3[5×（2a）2+12×2a-（5a2+12a）] =45a2+36a．·············· （9分） ∴y3?3(y2?y1)． ··························································································· （10分） 3、（08辽宁沈阳26题）（本题14分）26．如图所示，在平面直角坐标系中，矩形ABOC的边BO在x轴的负半轴上，边OC在y轴的正半轴上，且AB?1，OB?3，矩形ABOC绕点O按顺时针方向旋转60后得到矩形EFOD．点A的对应点为点E，点B的对应点为点F，点C的对应点为点D，抛物线y?ax2?bx?c过点A，E，D．

（1）判断点E是否在y轴上，并说明理由； （2）求抛物线的函数表达式；

（3）在x轴的上方是否存在点P，点Q，使以点O，B，P，Q为顶点的平行四边形的面积是矩形ABOC面积的2倍，且点P在抛物线上，若存在，请求出点P，点Q的坐标；若不存在，请说明理由．

解：（1）点E在y轴上 ···································································································· 1分 理由如下：

连接AO，如图所示，在Rt△ABO中，

B A F C D O 第26题图

x

y E AB?1，BO?3，?AO?2

?sin?AOB?1，??AOB?30 2由题意可知：?AOE?60

??BOE??AOB??AOE?30?60?90

点B在x轴上，?点E在y轴上． ··············································································· 3分

3

（2）过点D作DM?x轴于点M

OD?1，?DOM?30

?在Rt△DOM中，DM?点D在第一象限，

13，OM? 22?31? ·································································································· 5分 ?点D的坐标为???2，?2??由（1）知EO?AO?2，点E在y轴的正半轴上

2) ?点E的坐标为(0，··································································································· 6分 ?点A的坐标为(?31)， ·抛物线y?ax2?bx?c经过点E，

?c?2

?31?2D，由题意，将A(?31)代入y?ax?bx?2中得 ，，???22???8??3a?3b?2?1a???9?? 解得 ?3?31b?2??a??b??53?422?9?853x?2····························································· 9分 ?所求抛物线表达式为：y??x2?99（3）存在符合条件的点P，点Q． ················································································ 10分 理由如下：

矩形ABOC的面积?ABBO?3 ?以O，B，P，Q为顶点的平行四边形面积为23．

由题意可知OB为此平行四边形一边， 又

OB?3

?OB边上的高为2 ··········································································································· 11分

2) 依题意设点P的坐标为(m，点P在抛物线y??

8253x?x?2上 994

853??m2?m?2?2

99解得，m1?0，m2??53 8?P2)，P2??1(0，?53?2??8，?

??以O，B，P，Q为顶点的四边形是平行四边形，

?PQ∥OB，PQ?OB?3， 2)时， ?当点P1的坐标为(0，点Q的坐标分别为Q1(?3，2)，Q2(3，2)；

A B F y E C D x ?53?2?当点P2的坐标为???8，?时，

??点Q的坐标分别为Q3??

O M ?133??33?，2Q，2?，． ···················································· 14分 ?4?????8???8?4、（08辽宁12市26题）（本题14分）26．如图16，在平面直角坐标系中，直线y??3x?3与x轴交于点A，与y轴交于点C，抛物线y?ax?223x?c(a?0)3y 经过A，B，C三点．

（1）求过A，B，C三点抛物线的解析式并求出顶点F的坐标；

（2）在抛物线上是否存在点P，使△ABP为直角三角形，若存在，直接写出P点坐标；若不存在，请说明理由； （3）试探究在直线AC上是否存在一点M，使得△MBF的周长最小，若存在，求出M点的坐标；若不存在，请说明理由．

解：（1）直线y??3x?3与x轴交于点A，与y轴交于点C．

A C O F B x

图16 ?A(?1，0)，C(0，································································································ 1分 ?3) ·

点A，C都在抛物线上，

5

3?0???b

23b?

233?BC的解析式为y??x? ····················································································· 2分

4232?y??x?3?x1??1???4（2）由?，得?9

y1??y??3x?3??4??42?x2?2 ························································· 4分 ??y2?09??0) ?C??1，?，B(2，4??9 ······································································································· 5分 4199?S△ABC??4?? ································································································· 6分

242（3）过点N作NP?MB于点P EO?MB ?NP∥EO

?△BNP∽△BEO ········································································································ 7分 BNNP?? ···················································································································· 8分 BEEO?AB?4，CD?由直线y??33?3?x?可得：E?0，? 42?2?35，则BE? 22?在△BEO中，BO?2，EO??62tNP，?NP?t ······························································································· 9分 ?5352216?S?t(4?t)

25312S??t2?t(0?t?4) ··························································································· 10分

55312S??(t?2)2? ····································································································· 11分

5512此抛物线开口向下，?当t?2时，S最大?

512····························· 12分 ?当点M运动2秒时，△MNB的面积达到最大，最大为． ·5

11

8、（08新疆自治区24题）（10分）某工厂要赶制一批抗震救灾用的大型活动板房．如图，板房一面的形状是由矩形和抛物线的一部分组成，矩形长为12m，抛物线拱高为5.6m． （1）在如图所示的平面直角坐标系中，求抛物线的表达式． （2）现需在抛物线AOB的区域内安装几扇窗户，窗户的底边在AB上，每扇窗户宽1.5m，高1.6m，相邻窗户之间的间距均为0.8m，左右两边窗户的窗角所在的点到抛物线的水平距离至少为0.8m．请计算最多可安装几扇这样的窗户？

解：（1）设抛物线的表达式为y?ax2 ···················· 1分

?5.6)在抛物线的图象上． 点B(6，∴?5.6?36a

7 ··································································· 3分 4572x ·∴抛物线的表达式为y??··················································································· 4分 45a??（2）设窗户上边所在直线交抛物线于C、D两点，D点坐标为（k，t）

已知窗户高1.6m，∴t??5.6?(?1.6)??4 ································································ 5分

?4?

?72k 45·················································································· 6分 k1≈5.07，k2≈?5.07（舍去） ·

∴CD?5.07?2≈10.14（m） ····················································································· 7分

又设最多可安装n扇窗户

12

∴1.5n?0.8(n?1)≤10.14 ····························································································· 9分

n≤4.06．

答：最多可安装4扇窗户． ··························································································· 10分 （本题不要求学生画出4个表示窗户的小矩形）

9、（08广东梅州23题）23．本题满分11分．

如图11所示，在梯形ABCD中，已知AB∥CD， AD⊥DB，AD=DC=CB，AB=4．以AB所在直线为x轴，过D且垂直于AB的直线为y轴建立平面直角坐标系．

（1）求∠DAB的度数及A、D、C三点的坐标；

（2）求过A、D、C三点的抛物线的解析式及其对称轴L． （3）若P是抛物线的对称轴L上的点，那么使?PDB为等腰三角形的点P有几个?（不必求点P的坐标，只需说明理由）

解： （1） ?DC∥AB，AD=DC=CB，

····················································································· 0.5分 ? ∠CDB=∠CBD=∠DBA， ·

∠DAB=∠CBA， ?∠DAB=2∠DBA， ·············· 1分

∠DAB+∠DBA=90， ?∠DAB=60， ··········· 1.5分 ∠DBA=30，?AB=4， ?DC=AD=2， ·········· 2分 Rt?AOD，OA=1，OD=3， ····························· 2.5分 ，D（0， 3），C（2， 3）． · 4分 ?A（-1，0）

（2）根据抛物线和等腰梯形的对称性知，满足条件的抛物

线必过点A（－1，0），B（3，0）， 故可设所求为 y=a （x+1）（ x-3） ··································································· 6分 将点D（0，

???3）的坐标代入上式得， a=?3． 3所求抛物线的解析式为 y=?3(x?1)(x?3). ·············································· 7分 3其对称轴L为直线x=1． ····························································································· 8分 （3） ?PDB为等腰三角形，有以下三种情况：

13

①因直线L与DB不平行，DB的垂直平分线与L仅有一个交点P1，P1D=P1B， ?P1DB为等腰三角形； ························································································· 9分 ②因为以D为圆心，DB为半径的圆与直线L有两个交点P2、P3，DB=DP2，DB=DP3， ?P2DB， ?P3DB为等腰三角形；

③与②同理，L上也有两个点P4、P5，使得 BD=BP4，BD=BP5． ······················· 10分 由于以上各点互不重合，所以在直线L上，使?PDB为等腰三角形的点P有5个． 10、（08广东中山22题）将两块大小一样含30°角的直角三角板，叠放在一起，使得它们的斜边

AB重合，直角边不重合，已知AB=8，BC=AD=4，AC与BD相交于点E，连结CD． (1)填空：如图9，AC= ，BD= ；四边形ABCD是 梯形. (2)请写出图9中所有的相似三角形（不含全等三角形）.

(3)如图10，若以AB所在直线为x轴，过点A垂直于AB的直线为y轴建立如图10的

平面直角坐标系，保持ΔABD不动，将ΔABC向x轴的正方向平移到ΔFGH的位置，FH与BD相交于点P，设AF=t，ΔFBP面积为S，求S与t之间的函数关系式，并写出t的取值值范围.

解：（1）43，43，??????????1分

等腰；??????????2分

（2）共有9对相似三角形.（写对3－5对得1分，写对6－8对得2分，写对9对得3分）

①△DCE、△ABE与△ACD或△BDC两两相似，分别是：△DCE∽△ABE，△DCE∽△ACD，△DCE∽△BDC，△ABE∽△ACD，△ABE∽△BDC；(有5对)

②△ABD∽△EAD，△ABD∽△EBC；(有2对) ③△BAC∽△EAD，△BAC∽△EBC；(有2对)

所以，一共有9对相似三角形.????????????????5分

y

（3）由题意知，FP∥AE， ∴ ∠1＝∠PFB，

又∵ ∠1＝∠2＝30°，

DCH ∴ ∠PFB＝∠2＝30°,

∴ FP＝BP.??????????6分 过点P作PK⊥FB于点K，则FK?BK?∵ AF＝t，AB＝8，

Ay D E A 图9

B A F 图10

C

D C E P B G x H 1FB. 21FEP2K 图10BGx 14

∴ FB＝8－t，BK?1(8?t). 2在Rt△BPK中，PK?BK?tan?2?13(8?t)tan30??(8?t). ????????7分 26∴ △FBP的面积S?113?FB?PK??(8?t)?(8?t)， 226∴ S与t之间的函数关系式为： S?332416(t?8)2，或S?t?t?3. ?????????????8分 121233t的取值范围为：0?t?8. ??????????????????????9分 11、（08湖北十堰25题）已知抛物线y??ax2?2ax?b与x轴的一个交点为A(-1,0)，与y轴的正半轴交于点C．

⑴直接写出抛物线的对称轴，及抛物线与x轴的另一个交点B的坐标； ⑵当点C在以AB为直径的⊙P上时，求抛物线的解析式；

⑶坐标平面内是否存在点M，使得以点M和⑵中抛物线上的三点A、B、C为顶点的四边形是平行四边形？若存在，请求出点M的坐标；若不存在，请说明理由．

解：⑴对称轴是直线：x?1，点B的坐标是(3,0)． ??2分

说明：每写对1个给1分，“直线”两字没写不扣分．

⑵如图，连接PC，∵点A、B的坐标分别是A(-1,0)、B (3,0)，

11∴AB＝4．∴PC?AB??4?2．

22在Rt△POC中，∵OP＝PA－OA＝2－1＝1， ∴OC?PC2?PO2?22?12?3．

∴b＝3． ????????????3分 当x??1，y?0时，?a?2a?3?0，

15

∴a?3． ????????????4分 33223 ??????5分 x?x?3．33∴y??⑶存在．???????????6分

理由：如图，连接AC、BC．设点M的坐标为M(x,y)．

①当以AC或BC为对角线时，点M在x轴上方，此时CM∥AB，且CM＝AB． 由⑵知，AB＝4，∴|x|＝4，y?OC?3．

∴x＝±4．∴点M的坐标为M(4,3)或(?4,3)．?9分

说明：少求一个点的坐标扣1分．

②当以AB为对角线时，点M在x轴下方． 过M作MN⊥AB于N，则∠MNB＝∠AOC＝90°．

∵四边形AMBC是平行四边形，∴AC＝MB，且AC∥MB．

∴∠CAO＝∠MBN．∴△AOC≌△BNM．∴BN＝AO＝1，MN＝CO＝3． ∵OB＝3，∴0N＝3－1＝2．

∴点M的坐标为M(2,?3)． ???????????12分

说明：求点M的坐标时，用解直角三角形的方法或用先求直线解析式，

然后求交点M的坐标的方法均可，请参照给分．

综上所述，坐标平面内存在点M，使得以点A、B、C、M为顶点的四边形是平行四边形．其坐标为M1(4,3),M2(?4,3),M3(2,?3)．

说明：①综上所述不写不扣分；②如果开头“存在”二字没写，但最后解答全部正确，

不扣分。

16

12、（08四川达州23题）如图，将△AOB置于平面直角坐标系中，其中点O为坐标原点，点A的坐标为(3，0)，?ABO?60．

（1）若△AOB的外接圆与y轴交于点D，求D点坐标．

（2）若点C的坐标为(?1，0)，试猜想过D，C的直线与△AOB的外接圆的位置关系，并加以说明．

（3）二次函数的图象经过点O和A且顶点在圆上， 求此函数的解析式．

C

0

解：（1）连结AD，则∠ADO＝∠B＝60

0

在Rt△ADO中，∠ADO＝60 所以OD＝OA÷3＝3÷3＝3 所以D点的坐标是（0，3）

（2）猜想是CD与圆相切

∵ ∠AOD是直角，所以AD是圆的直径

又∵ Tan∠CDO=CO/OD=1/3=3, ∠CDO＝30

0

y B D O F E A x F D y B ∴∠CDA=∠CDO+∠ADO=Rt∠ 即CD⊥AD ∴ CD切外接圆于点D

（3）依题意可设二次函数的解析式为 ：

y=α（x－0）(x－3)

由此得顶点坐标的横坐标为：x=?E C O A x 3a3=; 2a210

∠B＝30 2即顶点在OA的垂直平分线上，作OA的垂直平分线EF，则得∠EFA＝

3333 可得一个顶点坐标为（，3）

222313） 同理可得另一个顶点坐标为（，?22得到EF＝3EA＝

分别将两顶点代入y=α（x－0）(x－3)可解得α的值分别为?

2323， 3917

则得到二次函数的解析式是y=?2323x(x－3)或y= x(x－3) 3913、（08湖北仙桃等4市25题）如图，直角梯形OABC中，AB∥OC,O为坐标原点，点A在y轴正半轴上，点C在x轴正半轴上，点B坐标为（2，23），∠BCO= 60°，

OH?BC于点H.动点P从点H出发，沿线段HO向点O运动，动点Q从点O出发，沿

线段OA向点A运动，两点同时出发，速度都为每秒1个单位长度.设点P运动的时间为t秒.

（1） 求OH的长；

（2） 若?OPQ的面积为S（平方单位）. 求S与t之间的函数关系式.并求t为何

值时，?OPQ的面积最大，最大值是多少？

（3） 设PQ与OB交于点M.①当△OPM为等腰三角形时，求（2）中S的值. ②探究线段OM长度的最大值是多少，直接写出结论. y

B A

H Q M

P

O C 解：（1）∵AB∥OC

x ∴ ?OAB??AOC?90 在Rt?OAB中，AB?2 ,AO?23

0 ∴OB?4， ?ABO?60

0y A B H P ∴?BOC?60 而?BCO?60

∴?BOC为等边三角形 ∴OH?OBcos30?4?00Q M 3?23?（3分） 2（2）∵OP?OH?PH?23?t

0O C x t3t yp?OPsin300?3?

22113t) ∴S??OQ?xp??t?(3?222323t?t （0?t?23）??????????（6分） =?42∴xp?OPcos30?3?0 18

333 (t?3)2?4433∴当t?3时，S最大????????????????（7分） 4y （3）①若?OPM为等腰三角形，则：

B A （i）若OM?PM，?MPO??MOP??POC ∴PQ∥OC

tH ∴OQ?yp 即t?3? MQ 2P 23解得：t? O C3 323232323此时S??????????????（8分） ?()???43233y 0（ii）若OP?OM，?OPM??OMP?75

B A ∴?OQP?450

过P点作PE?OA，垂足为E，则有： EQ?EP H Q 即S??即t?(3?x 13t)?3?t 22M解得：t?2 此时S??E O P Cx 33?22??2?3?3??????????????（9分） 42（iii）若OP?PM，?POM??PMO??AOB

∴PQ∥OA

此时Q在AB上，不满足题意.?????????????????（10分）

3 ②线段OM长的最大值为????????????????????(12分)

214、（08甘肃兰州28题）（本题满分12分）如图19-1，OABC是一张放在平面直角坐标系

中的矩形纸片，O为原点，点A在x轴的正半轴上，点C在y轴的正半轴上，OA?5，

OC?4．

（1）在OC边上取一点D，将纸片沿AD翻折，使点O落在BC边上的点E处，求D，E两点的坐标；

（2）如图19-2，若AE上有一动点P（不与A，E重合）自A点沿AE方向向E点匀速运动，运动的速度为每秒1个单位长度，设运动的时间为t秒（0?t?5），过P点作ED的平行线交AD于点M，过点M作AE的平行线交DE于点N．求四边形PMNE的面积S与时间t之间的函数关系式；当t取何值时，S有最大值？最大值是多少？

（3）在（2）的条件下，当t为何值时，以A，M，E为顶点的三角形为等腰三角形，并求出相应的时刻点M的坐标．

y y E E C C B B N D D P M x x O O A A 19 图19-1 图19-2

（本题满分12分） 解：（1）依题意可知，折痕AD是四边形OAED的对称轴， ?在Rt△ABE中，AE?AO?5，AB?4．

?BE?AE2?AB2?52?42?3．?CE?2．

?E点坐标为（2，4）． ········································································································· 2分

在Rt△DCE中，DC?CE?DE， 又

222DE?OD．

?(4?OD)2?22?OD2 ． 解得：CD?5． 2?5??D点坐标为?0，? ············································································································· 3分

?2?PM∥ED，?△APM∽△AED．

PMAP5??，又知AP?t，ED?，AE?5 EDAE2t5t?PM???， 又PE?5?t．

522而显然四边形PMNE为矩形．

t15?S矩形PMNE?PMPE??(5?t)??t2?t ······························································· 5分

222（2）如图①

?S四边形PMNE51?5?25???t???，又0??5

22?2?82525时，S矩形PMNE有最大值． ··············································································· 6分 28（3）（i）若以AE为等腰三角形的底，则ME?MA（如图①） 在Rt△AED中，ME?MA，PM?AE，?P为AE的中点，

15?t?AP?AE?．

22y 又PM∥ED，?M为AD的中点． E C B 过点M作MF?OA，垂足为F，则MF是△OAD的中位线， N P D 1515?MF?OD?，OF?OA?，

M 2422?当t??当t?55??时，?0??5?，△AME为等腰三角形． 22??O F 图① A x 此时M点坐标为?，?． ··································································································· 8分

?55??24? 20

（ii）若以AE为等腰三角形的腰，则AM?AE?5（如图②）

5?5?2225． 在Rt△AOD中，AD?OD?AO????5?22??过点M作MF?OA，垂足为F．

PM∥ED，?△APM∽△AED．

2y C N D M O F 图② A x E P B APAM?． AEAD1AMAE5?5?t?AP???25，?PM?t?5．

52AD52??MF?MP?5，OF?OA?AF?OA?AP?5?25，

（0?25?5），此时M点坐标为(5?25，5)． ···························· 11分 ?当t?25时，综合（i）（ii）可知，t?5或t?25时，以A，M，E为顶点的三角形为等腰三角形，相2应M点的坐标为?，?或(5?25，5)． ······································································ 12分 15、（08天津市卷26题）（本小题10分） 已知抛物线y?3ax2?2bx?c，

（Ⅰ）若a?b?1，c??1，求该抛物线与x轴公共点的坐标；

（Ⅱ）若a?b?1，且当?1?x?1时，抛物线与x轴有且只有一个公共点，求c的取值范围； x2?1时，（Ⅲ）若a?b?c?0，且x1?0时，对应的y1?0；对应的y2?0，试判断当0?x?1?55??24?时，抛物线与x轴是否有公共点？若有，请证明你的结论；若没有，阐述理由．

解（Ⅰ）当a?b?1，c??1时，抛物线为y?3x2?2x?1， 方程3x2?2x?1?0的两个根为x1??1，x2?1． 3∴该抛物线与x轴公共点的坐标是??1··················································· 2分 0?． ·，0?和?，（Ⅱ）当a?b?1时，抛物线为y?3x2?2x?c，且与x轴有公共点．

?1

?3??

1对于方程3x2?2x?c?0，判别式??4?12c≥0，有c≤． ··········································· 3分

3①当c?111时，由方程3x2?2x??0，解得x1?x2??． 333此时抛物线为y?3x2?2x?

?1?1与x轴只有一个公共点??，··································· 4分 0?． ·3?3?21

②当c?1时， 3x1??1时，y1?3?2?c?1?c， x2?1时，y2?3?2?c?5?c．

1由已知?1?x?1时，该抛物线与x轴有且只有一个公共点，考虑其对称轴为x??，

3应有??y1≤0，?1?c≤0， 即?

y?0.?2?5?c?0.1或?5?c≤?1． ······················································································· 6分 3解得?5?c≤?1． 综上，c?（Ⅲ）对于二次函数y?3ax2?2bx?c，

由已知x1?0时，y1?c?0；x2?1时，y2?3a?2b?c?0， 又a?b?c?0，∴3a?2b?c?(a?b?c)?2a?b?2a?b． 于是2a?b?0．而b??a?c，∴2a?a?c?0，即a?c?0．

∴a?c?0． ····················································································································· 7分 ∵关于x的一元二次方程3ax2?2bx?c?0的判别式

??4b2?12ac?4(a?c)2?12ac?4[(a?c)2?ac]?0，

∴抛物线y?3ax2?2bx?c与x轴有两个公共点，顶点在x轴下方． ································ 8分 又该抛物线的对称轴x??b， 3a由a?b?c?0，c?0，2a?b?0， 得?2a?b??a， ∴

y 1b2???． 33a3O 1 x 又由已知x1?0时，y1?0；x2?1时，y2?0，观察图象，

可知在0?x?1范围内，该抛物线与x轴有两个公共点． ················································ 10分 16、（08江苏镇江28题）（本小题满分8分）探索研究 如图，在直角坐标系xOy中，点P为函数y?12x在第一象限内的图象上的任一点，点A41)，直线l过B(0，?1)且与x轴平行，过P作y轴的平行线分别交x轴，l于的坐标为(0，y P A O C x

22

C，Q，连结AQ交x轴于H，直线PH交y轴于R．

（1）求证：H点为线段AQ的中点； （2）求证：①四边形APQR为平行四边形；

②平行四边形APQR为菱形；

（3）除P点外，直线PH与抛物线y?

（1）法一：由题可知AO?CQ?1．

12x有无其它公共点？并说明理由． 4?AOH??QCH?90，?AHO??QHC，

?△AOH≌△QCH． ································································································ （1分） ?OH?CH，即H为AQ的中点． ··········································································· （2分）

法二：

A(0，1)，B(0，?1)，?OA?OB． ······························································· （1分）

又BQ∥x轴，?HA?HQ． ····················································································· （2分） （2）①由（1）可知AH?QH，?AHR??QHP，

AR∥PQ，??RAH??PQH，

?△RAH≌△PQH． ································································································· （3分） ?AR?PQ，

又AR∥PQ，?四边形APQR为平行四边形． ························································ （4分）

②设P?m，m?，

??142??1PQ∥y轴，则Q(m，?1)，则PQ?1?m2．

4过P作PG?y轴，垂足为G，在Rt△APG中，

1?1??1?AP?AG?PG??m2?1??m2??m2?1??m2?1?PQ．

4?4??4?2222····················································································· （6分） ?平行四边形APQR为菱形． ·

23

（3）设直线PR为y?kx?b，由OH?CH，得H??m??1?，2?，P?m，m2?代入得： ?2??4?m?m?k?b?0，k?，??m12?2?2PRy?x?m． ·? 直线为························· （7分） ???1241?km?b?m2.?b??m2.???4?4设直线PR与抛物线的公共点为?x，x2?，代入直线PR关系式得：

??14??12m11?1?x?x?m2?0，(x?m)2?0，解得x?m．得公共点为?m，m2?． 4244?4?所以直线PH与抛物线y?

12x只有一个公共点P． ················································ （8分） 4 24

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