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天津耀华嘉诚国际中学九年级上册期末精选试卷检测题

天津耀华嘉诚国际中学九年级上册期末精选试卷检测题
天津耀华嘉诚国际中学九年级上册期末精选试卷检测题

天津耀华嘉诚国际中学九年级上册期末精选试卷检测题

一、初三数学 一元二次方程易错题压轴题(难)

1.如图,在四边形ABCD 中,9054ABC BCD AB BC cm CD cm ∠=∠=?===,,点

P 从点C 出发以1/cm s 的速度沿CB 向点B 匀速移动,点M 从点A 出发以15/cm s 的速

度沿AB 向点B 匀速移动,点N 从点D 出发以/acm s 的速度沿DC 向点C 匀速移动.点

P M N 、、同时出发,当其中一个点到达终点时,其他两个点也随之停止运动,设移动时

间为ts . (1)如图①,

①当a 为何值时,点P B M 、、为顶点的三角形与PCN △全等?并求出相应的t 的值; ②连接AP BD 、交于点E ,当AP BD ⊥时,求出t 的值; (2)如图②,连接AN MD 、交于点F .当38

83

a t ==

,时,证明:ADF CDF S S ??=.

【答案】(1)① 2.5t =, 1.1a =或2t =,0.5a =;②1t =;(2)见解析 【解析】 【分析】

(1)①当PBM PCN ?△△时或当MBP PCN ?△△时,分别列出方程即可解决问题; ②当AP BD ⊥时,由ABP BCD ?△△,推出BP CD =,列出方程即可解决问题; (2)如图②中,连接AC 交MD 于O 只要证明AOM COD ?△△,推出OA OC =,可得ADO CDO S S ??=,AFO CFO S S ??=,推出ADO AFO CDO CFO S S S S ????-=-,即ADF CDF S S ??=;

【详解】

解:(1)①90ABC BCD ∠=∠=?,

∴当PBM PCN ?△△时,有BM NC =,即5t t -=①

5 1.54t at -=-②

由①②可得 1.1a =, 2.5t =.

当MBP PCN ?△△时,有BM PC =,BP NC =,即5 1.5t t -=③ 54t at -=-④,

由③④可得0.5a =,2t =.

综上所述,当 1.1a =, 2.5t =或0.5a =,2t =时,以P 、B 、M 为顶点的三角形与

PCN △全等; ②AP BD ⊥,

90BEP ∴∠=?,

90APB CBD ∴∠+∠=?,

90ABC ∠=?,

90APB BAP ∴∠+∠=?, BAP CBD ∴∠=∠,

在ABP △和BCD 中,

BAP CBD AB BC

ABC BCD ∠=∠??

=??∠=∠?

, ()ABP BCD ASA ∴?△△,

BP CD ∴=, 即54t -=, 1t ∴=;

(2)当38a =,8

3

t =时,1DN at ==,而4CD =,

DN CD ∴<,

∴点N 在点C 、D 之间, 1.54AM t ==,4CD =, AM CD ∴=,

如图②中,连接AC 交MD 于O , 90ABC BCD ∠=∠=?, 180ABC BCD ∴∠+∠=?, //AB BC ∴,

AMD CDM ∴∠=∠,BAC DCA ∠=∠, 在AOM 和COD △中, AMD CDM AM CD

BAC DCA ∠=∠??

=??∠=∠?

, ()AOM COD ASA ∴?△△,

OA OC ∴=,

ADO CDO S S ??∴=,AFO CFO S S ??=, ADO AFO CDO CFO S S S S ????∴-=-, ADF CDF S S ??∴=.

【点睛】

本题考查三角形综合题、全等三角形的判定和性质、等高模型等知识,解题的关键是灵活运用所学知识解决问题,学会用分类讨论的思想思考问题,属于中考压轴题.

2.已知关于x 的一元二次方程kx 2﹣2(k +1)x +k ﹣1=0有两个不相等的实数根x 1,x 2. (1)求k 的取值范围; (2)是否存在实数k ,使12

11

x x -=1成立?若存在,请求出k 的值;若不存在,请说明理由.

【答案】(1)k >﹣1

3

且k ≠0;(2)存在,7213,k =±详见解析 【解析】 【分析】

(1)根据一元二次方程的根的判别式,建立关于k 的不等式,求得k 的取值范围. (2)利用根与系数的关系,根据

21

1212

11,x x x x x x --=即可求出k 的值,看是否满足(1)中k 的取值范围,从而确定k 的值是否存在. 【详解】

解:(1)由题意知,k ≠0且△=b 2﹣4ac >0 ∴b 2﹣4ac =[﹣2(k +1)]2﹣4k (k ﹣1)>0, 即4k 2+8k +4﹣4k 2+4k >0, ∴12k >﹣4 解得:k >1

3

-且k ≠0

(2)存在,且7213.k =±理由如下:

∵12122(1)1

,,k k x x x x k k

+-+=

= 又有21

1212

111,x x x x x x --

== 2112,x x x x ∴-=

2222

2121122,x x x x x x ∴-+=

22121212()4(),x x x x x x ∴+-=

22

22441(

)(),k k k k k k

+--∴-= 22(22)(44)(1),k k k k ∴+--=- 21430,k k ∴--= 1,14,3,a b c ==-=-

24208,b ac ∴?=-= 14413

7213.2

k ±∴=

=± k >13

-且k ≠0,

172130.21,3-≈--> 1

7213.3

+->

∴满足条件的k 值存在,且7213.k =± . 【点睛】

本题考查的是一元二次方程根的判别式,一元二次方程根与系数的关系,掌握以上知识是解题的关键.

3.图1是李晨在一次课外活动中所做的问题研究:他用硬纸片做了两个三角形,分别为△ABC 和△DEF ,其中∠B=90°,∠A=45°,BC=

,∠F=90°,∠EDF=30°, EF=2.将△DEF

的斜边DE 与△ABC 的斜边AC 重合在一起,并将△DEF 沿AC 方向移动.在移动过程中,D 、E 两点始终在AC 边上(移动开始时点D 与点A 重合). (1)请回答李晨的问题:若CD=10,则AD= ;

(2)如图2,李晨同学连接FC ,编制了如下问题,请你回答: ①∠FCD 的最大度数为 ; ②当FC ∥AB 时,AD= ;

③当以线段AD 、FC 、BC 的长度为三边长的三角形是直角三角形,且FC 为斜边时,AD= ; ④△FCD 的面积s 的取值范围是 .

【答案】(1)2;(2)① 60°;②;③;④

.

【解析】

试题分析:(1)根据等腰直角三角形的性质,求出AC的长,即可得到AD的长.

(2)①当点E与点C重合时,∠FCD的角度最大,据此求解即可.

②过点F作FH⊥AC于点H,应用等腰直角三角形的判定和性质,含30度角直角三角形的性质求解即可.

③过点F作FH⊥AC于点H,AD=x,应用含30度角直角三角形的性质把FC用x来表示,根据勾股定理列式求解.

④设AD=x,把△FCD的面积s表示为x的函数,根据x的取值范围来确定s的取值范围.试题解析:(1)∵∠B=90°,∠A=45°,BC=,∴AC=12.

∵CD=10,∴AD=2.

(2)①∵∠F=90°,∠EDF=30°,∴∠DEF=60°.

∵当点E与点C重合时,∠FCD的角度最大,∴∠FCD的最大度数=∠DEF="60°."

② 如图,过点F作FH⊥AC于点H,

∵∠EDF=30°, EF=2,∴DF=. ∴DH=3,FH=.

∵FC∥AB,∠A=45°,∴∠FCH="45°." ∴HC=. ∴DC=DH+HC=.

∵AC=12,∴AD=.

③如图,过点F作FH⊥AC于点H,设AD=x,

由②知DH=3,FH=,则HC=.

在Rt△CFH中,根据勾股定理,得.

∵以线段AD、FC、BC的长度为三边长的三角形是直角三角形,且FC为斜边,

∴,即,解得.

④设AD=x,易知,即.

而,

当时,;当时,.

∴△FCD的面积s的取值范围是.

考点:1.面动平移问题;2.等腰直角三角形的判定和性质;3.平行的性质;4.含30度角直角三角形的性质;5.勾股定理;6.由实际问题列函数关系式;7.求函数值.

4.某建材销售公司在2019年第一季度销售,A B 两种品牌的建材共126件,A 种品牌的建材售价为每件6000元,B 种品牌的建材售价为每件9000元.

(1)若该销售公司在第一季度售完两种建材后总销售额不低于96.6万元,求至多销售A 种品牌的建材多少件?

(2)该销售公司决定在2019年第二季度调整价格,将A 种品牌的建材在上一个季度的基础上下调%a ,B 种品牌的建材在上一个季度的基础上上涨%a ;同时,与(1)问中最低销售额的销售量相比,A 种品牌的建材的销售量增加了1

%2

a ,B 种品牌的建材的销售量减少了2

%5a ,结果2019年第二季度的销售额比(1)问中最低销售额增加2%23

a ,求a 的值.

【答案】(1)至多销售A 品牌的建材56件;(2)a 的值是30. 【解析】 【分析】

(1)设销售A 品牌的建材x 件,根据售完两种建材后总销售额不低于96.6万元,列不等式求解;

(2)根据题意列出方程求解即可. 【详解】

(1)设销售A 品牌的建材x 件.

根据题意,得()60009000126966000x x +-≥, 解这个不等式,得56x ≤, 答:至多销售A 品牌的建材56件.

(2)在(1)中销售额最低时,B 品牌的建材70件, 根据题意,得

()()()12260001%561%90001%701%6000569000701%2523a a a a a ??????-?+++?-=?+?+ ? ? ?

??????

令%a y =,整理这个方程,得2

1030y y -=,

解这个方程,得1230,10

y y ==

, ∴10a =(舍去),230a =,

即a的值是30.【点睛】

本题考查了一元二次方程和一元一次不等式的应用,解答本题的关键是读懂题意,设出未知数,找出合适的等量关系和不等关系,列方程组和不等式求解.

5.使得函数值为零的自变量的值称为函数的零点.例如,对于函数1

y x

=-,令y=0,可得x=1,我们就说1是函数1

y x

=-的零点.

己知函数222(3)

y x mx m

=--+(m m为常数).

(1)当m=0时,求该函数的零点;

(2)证明:无论m取何值,该函数总有两个零点;

(3)设函数的两个零点分别为1x和2x,且

12

111

4

x x

+=-,此时函数图象与x轴的交点分别为A、B(点A在点B左侧),点M在直线10

y x

=-上,当MA+MB最小时,求直线AM 的函数解析式.

【答案】(1)当m=0时,该函数的零点为6和6

-.

(2)见解析,

(3)AM的解析式为

1

1

2

y x

=--.

【解析】

【分析】

(1)根据题中给出的函数的零点的定义,将m=0代入y=x2-2mx-2(m+3),然后令y=0即可解得函数的零点;

(2)令y=0,函数变为一元二次方程,要想证明方程有两个解,只需证明△>0即可;(3)根据题中条件求出函数解析式进而求得A、B两点坐标,个、作点B关于直线y=x-10的对称点B′,连接AB′,求出点B′的坐标即可求得当MA+MB最小时,直线AM的函数解析式

【详解】

(1)当m=0时,该函数的零点为6和6

-.

(2)令y=0,得△=

∴无论m取何值,方程总有两个不相等的实数根.

即无论m取何值,该函数总有两个零点.

(3)依题意有,

解得

∴函数的解析式为.

令y=0,解得

∴A(

),B(4,0)

作点B 关于直线10y x =-的对称点B’,连结AB’, 则AB’与直线10y x =-的交点就是满足条件的M 点.

易求得直线10y x =-与x 轴、y 轴的交点分别为C (10,0),D (0,10). 连结CB’,则∠BCD=45° ∴BC=CB’=6,∠B’CD=∠BCD=45° ∴∠BCB’=90° 即B’(106-,)

设直线AB’的解析式为y kx b =+,则

20{106k b k b -+=+=-,解得112

k b =-=-, ∴直线AB’的解析式为1

12

y x =--, 即AM 的解析式为1

12

y x =-

-.

二、初三数学 二次函数易错题压轴题(难)

6.二次函数22(0)63

m m y x x m m =-+>的图象交y 轴于点A ,顶点为P ,直线PA 与x 轴交于点B .

(1)当m =1时,求顶点P 的坐标; (2)若点Q (a ,b )在二次函数22(0)63

m m

y x x m m =-+>的图象上,且0b m ->,试求a 的取值范围;

(3)在第一象限内,以AB 为边作正方形ABCD . ①求点D 的坐标(用含m 的代数式表示);

②若该二次函数的图象与正方形ABCD 的边CD 有公共点,请直接写出符合条件的整数m 的值.

【答案】(1)P (2,

1

3

);(2)a 的取值范围为:a <0或a >4;(3)①D (m ,m +3); ②2,3,4. 【解析】 【分析】

(1)把m =1代入二次函数22(0)63

m m y x x m m =-+>解析式中,进而求顶点P 的坐标即可;

(2)把点Q (a ,b )代入二次函数22(0)63

m m

y x x m m =

-+>解析式中,根据0b m ->得到关于a 的一元二次不等式即一元一次不等式组,解出a 的取值范围即可;

(3)①过点D 作DE ⊥x 轴于点E ,过点A 作AF ⊥DE 于点F ,求出二次函数与y 轴的交点A 的坐标,得到OA 的长,再根据待定系数法求出直线AP 的解析式,进而求出与x 轴的交点B 的坐标,得到OB 的长;通过证明△ADF ≌△ABO ,得到AF=OA=m ,DF=OB=3,DE=DF+EF= DF+OA=m+3,求出点D 的坐标;

②因为二次函数的图象与正方形ABCD 的边CD 有公共点,由①同理可得:C (m+3,3),分当x 等于点D 的横坐标时与当x 等于点C 的横坐标两种情况,进行讨论m 可能取的整数值即可. 【详解】

解:(1)当m =1时,二次函数为212

163

y x x =-+, ∴顶点P 的坐标为(2,

1

3

); (2)∵点Q (a ,b )在二次函数22(0)63

m m y x x m m =-+>的图象上, ∴2263

m m

b a a m =

-+, 即:2263

m m

b m a a -=

-

∵0b

m ->,

2263m m a a ->0, ∵m >0,

∴2263

a a ->0, 解得:a <0或a >4,

∴a 的取值范围为:a <0或a >4;

(3)①如下图,过点D 作DE ⊥x 轴于点E ,过点A 作AF ⊥DE 于点F ,

∵二次函数的解析式为2263

m m

y x x m =-+, ∴顶点P (2,

3

m

), 当x=0时,y=m , ∴点A (0,m ), ∴OA=m ;

设直线AP 的解析式为y=kx+b(k≠0), 把点A (0,m ),点P (2,

3

m

)代入,得: 23

m b m

k b =??

?=+??, 解得:3m k b m

?

=-???=?,

∴直线AP 的解析式为y=3

m

-x+m , 当y=0时,x=3, ∴点B (3,0); ∴OB=3;

∵四边形ABCD 是正方形, ∴AD=AB ,∠DAF+∠FAB=90°, 且∠OAB+∠FAB =90°, ∴∠DAF=∠OAB , 在△ADF 和△ABO 中,

DAF OAB AFD AOB AD AB ∠=∠??

∠=∠??=?

, ∴△ADF ≌△ABO (AAS ),

∴AF=OA=m ,DF=OB=3,DE=DF+EF= DF+OA=m+3, ∴点D 的坐标为:(m ,m+3); ②由①同理可得:C (m+3,3),

∵二次函数的图象与正方形ABCD 的边CD 有公共点,

∴当x =m 时,3y m ≤+,可得3

2

2363

m m

m m -+≤+,化简得:32418m m -≤.

∵0m >,∴2

184m m m -≤

,∴2

18(2)4m m

--≤, 显然:m =1,2,3,4是上述不等式的解,

当5m ≥时,2

(2)45m --≥,18 3.6m ≤,此时,218(2)4m m

-->, ∴符合条件的正整数m =1,2,3,4;

当x = m +3时,y ≥3,可得2

(3)2(3)

363

m m m m m ++-+≥,

∵0m >,∴2

1823m m m ++≥

,即2

18(1)2m m

++≥, 显然:m =1不是上述不等式的解,

当2m ≥时,2

(1)211m ++≥,189m ≤,此时,218(1)2m m

++>恒成立, ∴符合条件的正整数m =2,3,4;

综上:符合条件的整数m 的值为2,3,4. 【点睛】

本题考查二次函数与几何问题的综合运用,熟练掌握二次函数的图象和性质、一次函数的图象和性质、正方形的性质是解题的关键.

7.如图,在平面直角坐标系中,抛物线y =﹣12

x 2

+bx +c 与x 轴交于B ,C 两点,与y 轴交于点A ,直线y =﹣

1

2

x +2经过A ,C 两点,抛物线的对称轴与x 轴交于点D ,直线MN 与

对称轴交于点G,与抛物线交于M,N两点(点N在对称轴右侧),且MN∥x轴,MN=7.

(1)求此抛物线的解析式.

(2)求点N的坐标.

(3)过点A的直线与抛物线交于点F,当tan∠FAC=

1

2

时,求点F的坐标.

(4)过点D作直线AC的垂线,交AC于点H,交y轴于点K,连接CN,△AHK沿射线AC 以每秒1个单位长度的速度移动,移动过程中△AHK与四边形DGNC产生重叠,设重叠面积为S,移动时间为t(0≤t5S与t的函数关系式.

【答案】(1)y=﹣

1

2

x2+

3

2

x+2;(2)点N的坐标为(5,-3);(3)点F的坐标为:(3,2)或(

17

3

,﹣

50

9

);(4)

2

535

,0

45

3593535

,(

4

35935

5)

4

t t

S t

t

???

≤≤

? ?

?

???

=-<≤

+<≤

【解析】

【分析】

(1)点A、C的坐标分别为(0,2)、(4,0),将点A、C坐标代入抛物线表达式即可求解;

(2)抛物线的对称轴为:x=

3

2

,点N的横坐标为:

37

5

22

+=,即可求解;

(3)分点F在直线AC下方、点F在直线AC的上方两种情况,分别求解即可;

(4)分0≤t

3535

<t

3535<t5

【详解】

解:(1)直线y=﹣

1

2

x+2经过A,C两点,则点A、C的坐标分别为(0,2)、(4,0),

则c=2,抛物线表达式为:y=﹣

1

2

x2+bx+2,

将点C坐标代入上式并解得:b=3 2

故抛物线的表达式为:y=﹣

1

2

x2+

3

2

x+2…①;

(2)抛物线的对称轴为:x=

3

2

点N的横坐标为:

37

5

22

+=,

故点N的坐标为(5,-3);

(3)∵tan∠ACO=

21

42

AO

CO

===tan∠FAC=

1

2

即∠ACO=∠FAC,

①当点F在直线AC下方时,

设直线AF交x轴于点R,

∵∠ACO=∠FAC,则AR=CR,

设点R(r,0),则r2+4=(r﹣4)2,解得:r=

3

2

即点R的坐标为:(

3

2

,0),

将点R、A的坐标代入一次函数表达式:y=mx+n得:

2

3

2

n

m n

=

?

?

?

+=

??

,解得:

4

3

2

m

n

?

=-

?

?

?=

?

故直线AR的表达式为:y=﹣

4

3

x+2…②,

联立①②并解得:x=

17

3

,故点F(

17

3

,﹣

50

9

);

②当点F在直线AC的上方时,

∵∠ACO=∠F′AC,∴AF′∥x轴,

则点F ′(3,

2);

综上,点F 的坐标为:(3,2)或(

173,﹣509

); (4)如图2,设∠ACO =α,则tanα=1

2

AO CO =,则sinα=5,cosα=5;

①当0≤t ≤

35

时(左侧图), 设△AHK 移动到△A ′H ′K ′的位置时,直线H ′K ′分别交x 轴于点T 、交抛物线对称轴于点S ,

则∠DST =∠ACO =α,过点T 作TL ⊥KH , 则LT =HH ′=t ,∠LTD =∠ACO =α,

则DT ='5

2co 5

c s os L HH T t αα===,DS =tan DT α

, S =S △DST =12?DT ×DS =2

54

t ; 35<t 35

时(右侧图),

同理可得:

S =''DGS T S 梯形=12

?DG ×(GS ′+DT ′)=12?3+(52t +52t ﹣32)=35924

-; 35

<t 53594

+; 综上,S =2535,023593535,(245435935(5)10

44t t t t t t ??≤≤? ???

??

?-<≤??

?+<≤??.

【点睛】

本题考查的是二次函数综合运用,涉及到一次函数、图形平移、图形的面积计算等,其中(3)、(4),要注意分类求解,避免遗漏.

8.如图,直线3y

x

与x 轴、y 轴分别交于点A ,C ,经过A ,C 两点的抛物线

2y ax bx c =++与x 轴的负半轴的另一交点为B ,且tan 3CBO ∠=

(1)求该抛物线的解析式及抛物线顶点D 的坐标;

(2)点P 是射线BD 上一点,问是否存在以点P ,A ,B 为顶点的三角形,与ABC 相似,若存在,请求出点P 的坐标;若不存在,请说明理由

【答案】(1)2

43y x x =++,顶点(2,1)D --;(2)存在,52,33P ??

--

???

或(4,3)-- 【解析】 【分析】

(1)利用直线解析式求出点A 、C 的坐标,从而得到OA 、OC ,再根据tan ∠CBO=3求出OB ,从而得到点B 的坐标,然后利用待定系数法求出二次函数解析式,整理成顶点式形式,然后写出点D 的坐标;

(2)根据点A 、B 的坐标求出AB ,判断出△AOC 是等腰直角三角形,根据等腰直角三角形的性质求出AC ,∠BAC=45°,再根据点B 、D 的坐标求出∠ABD=45°,然后分①AB 和BP 是对应边时,△ABC 和△BPA 相似,利用相似三角形对应边成比例列式求出BP ,过点P 作PE ⊥x 轴于E ,求出BE 、PE ,再求出OE 的长度,然后写出点P 的坐标即可;②AB 和BA 是对应边时,△ABC 和△BAP 相似,利用相似三角形对应边成比例列式求出BP ,过点P 作PE ⊥x 轴于E ,求出BE 、PE ,再求出OE 的长度,然后写出点P 的坐标即可. 【详解】

解:(1)令y=0,则x+3=0, 解得x=-3, 令x=0,则y=3,

∴点A (-3,0),C (0,3), ∴OA=OC=3, ∵tan ∠CBO=3OC

OB

=, ∴OB=1, ∴点B (-1,0),

把点A、B、C的坐标代入抛物线解析式得,

930

3

a b c

a b c

c

-+=

?

?

-+=

?

?=

?

,解得:

1

4

3

a

b

c

=

?

?

=

?

?=

?

∴该抛物线的解析式为:243

y x x

=++,∵y=x2+4x+3=(x+2)2-1,

∴顶点(2,1)

D--;

(2)∵A(-3,0),B(-1,0),

∴AB=-1-(-3)=2,

∵OA=OC,∠AOC=90°,

∴△AOC是等腰直角三角形,

∴AC=2OA=32,∠BAC=45°,

∵B(-1,0),D(-2,-1),

∴∠ABD=45°,

①AB和BP是对应边时,△ABC∽△BPA,

∴AB AC

BP BA

=,

232

2

BP

=,

解得BP=

22

3

过点P作PE⊥x轴于E,

则BE=PE=

2

3

×

2

2

=

2

3

∴OE=1+2

3

=

5

3

∴点P的坐标为(-5

3

,-

2

3

);

②AB 和BA 是对应边时,△ABC ∽△BAP , ∴AB AC

BA BP =,

22=

解得BP= 过点P 作PE ⊥x 轴于E ,

则BE=PE=2

=3, ∴OE=1+3=4,

∴点P 的坐标为(-4,-3);

综合上述,当52,33P ??-- ???

或(4,3)--时,以点P ,A ,B 为顶点的三角形与ABC ?相似; 【点睛】

本题是二次函数综合题型,主要利用了直线与坐标轴交点的求解,待定系数法求二次函数解析式,等腰直角三角形的判定与性质,相似三角形的判定与性质,难点在于(2)要分情况讨论.

9.定义:在平面直角坐标系中,O 为坐标原点,设点P 的坐标为(x ,y ),当x <0时,点P 的变换点P′的坐标为(﹣x ,y );当x≥0时,点P 的变换点P′的坐标为(﹣y ,x ). (1)若点A (2,1)的变换点A′在反比例函数y=

k

x

的图象上,则k= ; (2)若点B (2,4)和它的变换点B'在直线y=ax+b 上,则这条直线对应的函数关系式为 ,∠BOB′的大小是 度.

(3)点P 在抛物线y=x 2﹣2x ﹣3的图象上,以线段P P′为对角线作正方形PMP'N ,设点P 的横坐标为m ,当正方形PMP′N 的对角线垂直于x 轴时,求m 的取值范围.

(4)抛物线y=(x ﹣2)2+n 与x 轴交于点C ,D (点C 在点D 的左侧),顶点为E ,点P 在该抛物线上.若点P 的变换点P′在抛物线的对称轴上,且四边形ECP′D 是菱形,求n 的值.

【答案】(1) -2;(2) y=13x+103,90;(3) m <0,或;(4) n=﹣8,n=﹣2,n=﹣3. 【解析】 【分析】

(1)先求出A 的变换点A ′,然后把A ′代入反比例函数即可得到结论; (2)确定点B ′的坐标,把问题转化为方程组解决;

(3)分三种情形讨论:①当m <0时;②当m ≥0,PP '⊥x 轴时;③当m ≥0,MN ⊥x 轴

时.

(4)利用菱形的性质,得到点E 与点P '关于x 轴对称,从而得到点P '的坐标为(2,﹣n ).分两种情况讨论:①当点P 在y 轴左侧时,点P 的坐标为(﹣2,﹣n ),代入抛物线解析式,求解即可;②当点P 在y 轴右侧时,点P 的坐标为(﹣n ,﹣2).代入抛物线解析式,求解即可. 【详解】

(1)∵A (2,1)的变换点为A ′(-1,2),把A ′(-1,2)代入y =k

x

中,得到k =-2. 故答案为:-2.

(2)点B (2,4)的变换点B ′(﹣4,2),把(2,4),(﹣4,2)代入y =ax +b 中.

得到:2442a b a b +=??-+=?,解得:13

10

3a b ?=????=??

,∴11033y x =+.

∵OB 2=2224+=20,OB ′2=2224+=20,BB ′2=22

(42)(24)--+-=40,∴OB 2+OB ′2=BB ′2,

∴∠BOB ′=90°. 故答案为:y =

13x +10

3

,90. (3)①当m <0时,点P 与点P '关于y 轴对称,此时MN 垂直于x 轴,所以m <0. ②当m ≥0,PP '⊥x 轴时,则点P '的坐标为(m ,m ),点P 的坐标为(m ,﹣m ). 将点P (m ,﹣m )代入y =x 2﹣2x ﹣3,得:﹣m =m 2﹣2m ﹣3.

解得:12m m ==

(不合题意,舍去).

所以12

m +=

③当m ≥0,MN ⊥x 轴时,则PP '∥x 轴,点P 的坐标为(m ,m ). 将点P (m ,m )代入y =x 2﹣2x ﹣3,得:m =m 2﹣2m ﹣3.

解得:123322

m m ==

(不合题意,舍去).

所以m =

. 综上所述:m 的取值范围是m <0,m

或m

=32

. (4)∵四边形ECP 'D 是菱形,∴点E 与点P '关于x 轴对称. ∵点E 的坐标为(2,n ),∴点P '的坐标为(2,﹣n ). ①当点P 在y 轴左侧时,点P 的坐标为(﹣2,﹣n ). 代入y =(x ﹣2)2+n ,得:﹣n =(﹣2﹣2)2+n ,解得:n =﹣8. ②当点P 在y 轴右侧时,点P 的坐标为(﹣n ,﹣2).

代入y=(x﹣2)2+n,得:﹣2=(﹣n﹣2)2+n.解得:n1=﹣2,n2=﹣3.

综上所述:n的值是n=﹣8,n=﹣2,n=﹣3.

【点睛】

本题是二次函数综合题、一次函数的应用、待定系数法、变换点的定义等知识,解题的关键是理解题意,学会用分类讨论的射线思考问题,学会用方程的思想思考问题,属于中考压轴题.

10.在平面直角坐标系中,二次函数y=ax2+bx+2的图象与x轴交于A(﹣3,0),B(1,0)两点,与y轴交于点C.

(1)求这个二次函数的关系解析式;

(2)求直线AC的函数解析式;

(3)点P是直线AC上方的抛物线上一动点,是否存在点P,使△ACP的面积最大?若存在,求出点P的坐标;若不存在,说明理由;

【答案】(1)y=﹣

2

3

x2﹣

4

3

x+2;(2)

2

2

3

y x

=+;(3)存在,(

35

,

22

-)

【解析】

【分析】

(1)直接用待定系数法即可解答;

(2)先确定C点坐标,设直线AC的函数解析式y=kx+b,最后用待定系数法求解即可;(3)连接PO,作PM⊥x轴于M,PN⊥y轴于N,然后求出△ACP面积的表达式,最后利用二次函数的性质求最值即可.

【详解】

解:(1)∵抛物线y=ax2+bx+2过点A(﹣3,0),B(1,0),

0932

02

a b

a b

=-+

?

?

=++

?

解得

2

3

4

3

a

b

?

=-

??

?

?=-

??

∴二次函数的关系解析式为y=﹣

2

3

x2﹣

4

3

x+2;

2)∵当x=0时,y=2, ∴C (0,2)

设直线AC 的解析式为y kx b =+,把A 、C 两点代入得

0=32k b b -+??

=? 解得232

k b ?

=

???=? ∴直线AC 的函数解析式为2

23

y x =+; (3)存在.

如图: 连接PO ,作PM⊥x 轴于M ,PN⊥y 轴于N

设点P 坐标为(m ,n ),则n=224

233

m m --+),PN=-m ,AO=3

当x=0时,y=22

400233

-?-?+=2, ∴点C 的坐标为(0,2),OC=2 ∵PAC

PAO

PCO

ACO

S

S

S

S

=+-

21241

1322()322332

2m m m ??=

??--++??--?? ??? =23m m -- ∵a=-1<0

∴函数S △PAC =-m 2

-3m 有最大值

∴b 当m=()33

212

-=--?-

∴当m=32

-时,S △PAC 有最大值n=222423435

223332322m m ??--+=-?-?+= ???

∴当△ACP 的面积最大时,P 的坐标为(35

,22

-). 【点睛】

本题是二次函数压轴题,综合考查了二次函数的图象与性质、待定系数法、二次函数极值等知识点,根据题意表示出△PAC 的面积是解答本题的关键.

三、初三数学 旋转易错题压轴题(难)

人教版九年级英语期末测试卷及答案

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