《二次函数的应用》专题练习

《二次函数的应用》专题练习
《二次函数的应用》专题练习

《二次函数的应用》专题练习

1.某一型号的飞机着陆后滑行的路程s (单位:m )米与时间t (单位:s )之间的函数关系式为:

s =60t -,试问飞机着陆后滑行多远才能停止

2.如图拱桥的形状是抛物线,其函数关系式为231x y -=,当水面离桥顶的高度为3

25

米时,水面的宽度为多少 米

3.如图是抛物线形拱桥,拱顶离水面2m ,水面宽度4m ,水面下降1m ,水面宽度增加多少

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4.如图,已知一抛物线形大门,其地面宽度AB =18m 。一同学站在门内,在离门脚B 点1m 远的D 处,垂直地面 立起一根1.7m 长的木杆,其顶端恰好顶在抛物线形门上C 处。根据这些条件,请你求出该大门的高h 。

?

5.某地要建造一个圆形喷水池,在水池中央垂直于水面安装一个花形柱子OA ,O 恰好在水面中心,安装在柱子顶 端A 处的喷头向外喷水,水流在各个方向上沿形状相同的抛物线路径落下,且在过OA 的任一平面上,抛物线的 形状如图(1)和(2)所示,建立直角坐标系,水流喷出的高度y(米)与水平距离x(米)之间的关系式是

y =-x 2

+2x +

5

4

,请你寻求: (1)柱子OA 的高度为多少米

(2)喷出的水流距水平面的最大高度是多少

(3)若不计其他因素,水池的半径至少要多少米,才能使喷出的水流不至于落在池外。

6.如图,一位运动员在距篮下4米处跳起投篮,球运行的路线是抛物线,当球运行的水平距离为2.5米时,达到 ,

最大高度3.5米,然后准确落入篮圈。已知篮圈中心到地面的距离为3.05米。 (1)建立如图所示的直角坐标系,求抛物线的表达式;

(2)该运动员身高1.8米,在这次跳投中,球在头顶上方0.25米处出手,问:球出手时,他跳离地面的高度是

多少

7.如图,某隧道口的横截面是抛物线形,已知路宽AB 为6米,最高点离地面的距离OC 为5米。以最高点O 为坐

标原点,抛物线的对称轴为y 轴,1米为数轴的单位长度,建立平面直角坐标系,求: (1)以这一部分抛物线为图象的函数解析式,并写出x 的取值范围;

(2)有一辆宽2.8米,高1米的农用货车(货物最高处与地面AB 的距离)能否通过此隧道

(1)0(2)x B

y A

O

x 【 A B

8.一座隧道的截面由抛物线和长方形构成,长方形的长为8m ,宽为2m ,隧道最高点P 位于AB 的中央且距地

面6m ,建立如图所示的坐标系: (1)求抛物线的解析式;

(2)一辆货车高4m ,宽2m ,能否从该隧道内通过,为什么

(3)如果隧道内设双行道,那么这辆货车是否可以顺利通过,为什么

\

9.如图,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,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

10.某服装商销售每件进价为40元的衬衫,市场调查显示,若每件以50元的价格销售,平均每天可销售500件,

价格每提高1元,则平均每天少销售10件。当每件衬衫提价x 元时,可以获得利润y 元。 (1)写出y 与x 之间的函数关系式;

(2)当每件衬衫提价多少元时,可以获得最大利润最大利润是多少

|

P

y B A O 、 x

11.某跳水运动员进行10米跳台跳水训练时,身体(看成一点)在空中运动路线是如图所示坐标系下的经过原点

O 的一条抛物线(图中标出的数据为已知条件)。在跳某个规定动作时,正常情况下该运动员在空中的最高处

距水面3

2

10

m ,入水距池边的距离为4m ,同时运动员在距水面高度为5m 以前,必须完成规定的翻腾动作,并 !

调整好入水的姿势,否则就会出现失误。 (1)求这条抛物线的解析式;

(2)在某次试跳时,测得运动员在空中的运动路线是(1)中的抛物线,且运动员在空中调整好入水姿势时,

距池边的水平距离为5

3

3

m ,问此次跳水会不会失误并通过计算说明理由。

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12.如图,小明在一次高尔夫球训练中,从山坡下P 点打出一球向球洞A 点飞去,球的飞行路线为抛

物线,如果不考虑空气阻力,当球达到最大高度BD 为12米时,球移动的水平距离PD 为9米。

已知山坡PA 与水平方向PC 的夹角为30°,AC ⊥PC 于点C ,P 、A 两点相距38米。请你建立适 当的平面直角坐标系解决下列问题。 ·

(1)求水平距离PC 的长;

(2)求出球的飞行路线所在抛物线的解析式;

(3)判断小明这一杆能否把高尔夫球从P 点直接打入球洞A 点。

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13.某水果商销售每箱进价为40元的苹果,物价部门规定每箱售价不得高于55元。市场调查显示,若每箱以

50元的价格销售,平均每天可销售90箱,价格每提高1元,则平均每天少销售3箱。 (1)求平均每天销售量y (箱)与售价x (元/箱)之间的函数关系; (2)求平均每天销售利润w (元)与销售价x (元/箱)之间的函数关系; !

(3)当每箱苹果的售价为多少元时,可以获得最大利润最大利润是多少

14.为把产品打入国际市场,某企业决定从下面两个投资方案中选择一个进行投资生产。方案一:生产甲产品,每

件产品成本为a 万美元(a 为常数,且3<a <8),每件产品销售价为10万美元,每年最多可生产200件;方 案二:生产乙产品,每件产品成本为8万美元,每件产品销售价为18万美元,每年最多可生产120件。另外,

年销售x 件乙产品...

时需上交2

0.05x 万美元的特别关税。在不考虑其它因素的情况下: (1)分别写出该企业两个投资方案的年利润1y 、2y 与相应生产件数x (x 为正整数)之间的函数关系式,并指 出自变量的取值范围;

(2)分别求出这两个投资方案的最大年利润;

(3)如果你是企业决策者,为了获得最大收益,你会选择哪个投资方案

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15

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《二次函数的应用》专题练习答案

1.解:s =60t -

=-(t 2-40 t )2

=-(t -20)2

+600 (

∵-<0,

∴函数有最大值。 当t =-20时, s 最大值=600,

即飞机着陆后滑行600米才能停止。

2.10米。

3.解:以抛物线的顶点作为原点,水平线作为x 轴,建立直角坐标系,

设抛物线的解析式为2

ax y =, ∵过(2,-2)点,∴21-

=a ,抛物线的解析式为2

2

1x y -=。 当3-=y 时,6±=x ,所以宽度增加(462-)m 。

4.解法一:如图1,建立平面直角坐标系。

;

设抛物线解析式为y =ax 2

+bx 。

由题意知B 、C 两点坐标分别为B(18,0),C(17,。 把B 、C 两点坐标代入抛物线解析式得

解得

∴抛物线的解析式为 y =-+

=-(x -9)2

+。 ∴该大门的高h 为8.1m 。

设抛物线解析式为y=ax2。

由题意得B、C两点坐标分别为B(9,-h),C(8,-h+。

把B、C两点坐标代入y=ax2得

解得。

∴y=-.

∴该大门的高h为8.1m。

说明:此题还可以以AB所在直线为x轴,AB中点为原点,建立直角坐标系,可得抛物线解析式为y=-+。

5.(1)当x=0时,y=

5

4

,故OA的高度为1.25米。

(2)∵y=-x2+2x+

5

4

=-(x-1)2+,

∴顶点是(1,,故喷出的水流距水面的最大高度是2.25米。

(3)解方程-(x-1)2+=0,得

12

15

,

22

x x

=-=。

<

∴B点坐标为

5

,0

2

??

?

??

∴OB=

5

2

故不计其他因素,水池的半径至少要2.5米,才能使喷出的水流不至于落在水池外。

6. (1)设抛物线的表达式为y=ax2+k,

由图知图象过点,,代入求得a=-。

∴抛物线的表达式为y=-+。

(2)设球出手时,他跳离地面的高度为h m,则球出手时,球的高度为h++=(h+ m,

∴h+=-×(-2+,

∴h=(m)。

7.解:(1)设所求函数的解析式为2

ax

y=。

由题意,得函数图象经过点B(3,-5),

.

∴-5=9a。

9

5

-

=

a。

∴所求的二次函数的解析式为2

9

5

x

y-

=。

x的取值范围是3

3≤

-x。

(2)当车宽8.2米时,此时CN为4.1米,对应

45

49

4.1

9

5

2-

=

?

-

=

y,

176

49

O

x

y

A B

C

M N

E

∴农用货车能够通过此隧道。

8.(1)由题意可知抛物线的顶点坐标(4,6),

设抛物线的方程为6)4(2

+-=x a y , 又因为点A (0,2)在抛物线上, 所以有6)40(22

+-=a 。所以a =4

1-

。 \

因此有:6)4(4

1

2+--

=x y 。 (2)令4y =,则有 6)4(4

142

+--=x 。

解得12422422x x =+=-,。

21422x x -=>。

∴货车可以通过。 (3)由(2)可知

211

2222

x x -=>, ∴货车可以通过。

9. 解:(1) M(12,0),P(6,6)。

(2) 设抛物线解析式为:6)6(2+-=x a y 。 ∵抛物线6)6(2

+-=x a y 经过点(0,0),

*

∴6)60(02

+-=a ,即6

1-=a , ∴抛物线解析式为:

x x y x y 26

1

,6)6(6122+-=+--=即 。

(3) 设A(m ,0),则B(12-m ,0),)261,12(2m m m C +-

-,)261

,(2m m m D +-。 ∴“支撑架”总长AD +DC +CB = )261()212()261(2

2m m m m m +-+-++-

=15)3(3

1122312

2+--=++-m m m 。

∵ 此二次函数的图象开口向下。

∴ 当m =3米时,AD +DC +CB 有最大值为15米。

10.设每件衬衫提价x 元时,可以获得利润y 元。根据题意,得 y =(50-40+x)(500-10x )

=-10x 2

+400x +5000, ` =-10(x -20)2

+9000,

因为-10<0,所以,当x =20时,y 的最大值为9000元。

11.解:(1)在给定的直角坐标系中,设抛物线的解析式为y =ax 2

+bx +c 。由题意得,O 、B 两点坐标分别为

(0,0)、(2,-10),顶点纵坐标为

3

2

。则有 ?????

?

?-=++=-=.1024,32

44,02c b a a

b a

c c 解得 ?

???

?????==-=.0,310,625c b a 或 ???????=-=-=.0,2,23c b a 因抛物线对称轴在y 右侧,所以-a

b

2>0,即a 与b 异号,又开口向下,则a <0,b >0, 所以a =-

2

3

,b =-2,c =0不符合题图意,舍去。 故所求抛物线的解析式为y =-625x 2+3

10

x 。

(2)当运动员在空中距池边的水平距离为353m ,即x =353-2=5

8

m 时,

y =(-625)×(58)2+310×58=-3

16

所以此时运动员距水面的高为10-

316=3

14

<5。因此,此次跳水会出现失误。 12.解:(1)依题意得:∠ACP =90°,∠APC =30°,PA =38,

∴AC =34,∴PC =12,

∴PC 的长为12m 。

(2)以P 为原点,PC 所在直线为x 轴建立如图所示的平面直角坐标系,可知:顶点B

(9,12),

抛物线经过原点,

设抛物线的解析式为y =a (x -9)2

+12,

将点P (O )的坐标代入可得:0=a (0-9)2

+12,

求得a =-

27

4, 故抛物线的解析式为:y =-

27

4 (x -9)2

+12。 #

(3)由(1)知点C 的坐标为(12,0),易求得AC =34,

即可得点A 的坐标为(12,34), 当x =12时,y =-

274 (12-9)2

+12=3

32≠34, 故小明不能一杆把高尔夫球从P 点直接打入球洞A 点。

13.(1)y =90-3(x -50),化简得y =-3x +240 (50≤x≤55)

(3)当x =60时,w 有最大值,

又因x <60,所以,当x =55时,w 的最大值为1125元。

即,当每箱苹果的售价为55元时,可获最大利润,为1125元。 14.解:(1)1(10)y a x =- (1≤x≤200,x 为正整数)

22100.05y x x =- (1≤x≤120,x 为正整数)

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(2)①∵3<a <8, ∴10-a >0,即1y 随x 的增大而增大 ,

∴当x =200时,1y 最大值=(10-a )×200=2000-200a (万美元)

②220.05(100)500y x =--+

∵-<0, ∴x=100时, 2y 最大值=500(万美元)

(3)由2000-200a >500,得a <,

∴当3<a <时,选择方案一;

由2000200500a -=,得 7.5a =, ∴当a =时,选择方案一或方案二均可; 由2000200500a -<,得 7.5a >, ∴当<a <8时,选择方案二。

15.解:(1)∵根据表格知道日销售量与时间t 是均匀减少的,

'

∴确定m 与t 是一次函数关系,设函数关系式为:m =kt +b , ∵当t =1,m =94;当t =3,m =90, ∴??

?=+=+90394b k b k , ∴???=-=96

2

b k ,

人教版六年级分数、百分数应用题专项训练及答案

分数、百分数应用题专项训练 1、一桶油第一次取出总数的10%,第二次取出剩下的20%,两次共取出28升。这桶油共有多少升? 2、一桶柴油,第一次用了全桶的20%,第二次用去20千克,第三次用了前两次的和,这时桶里还剩8千克油.问这桶油有多少千克? 3、服装厂一车间人数占全厂的25%,二车间人数比一车间少`1/5`,三车间人数比二车间多`3/10`,三车间是156人,这个服装厂全厂共有多少人? 4、加工一批零件,甲乙二人合作需12天完成;现由甲先工作3天,然后由乙工作2天还剩这批零件的`4/5`没完成. 已知甲每天比乙少加工4个,这批零件共有多少个? 5、某商店同时卖出两件商品,每件各得60元,但其中一件赚20%,另一件亏本20%,问这个商店卖出这两件商品是赚钱还是亏本?赚多少,亏多少? 6、甲、乙两只装有糖水的桶,甲桶有糖水60千克,含糖率4%,乙桶有糖水40千克,含糖率为20%,两桶互相交换多少千克才能使两桶糖水的含糖率相等? 7、现有浓度为10%的盐水20千克,再加入多少千克浓度为30%的盐水,可以得到浓度为22%的盐水? 8、在浓度为40%的酒精溶液中加入5千克水,浓度变为30%,再加入多少千克酒精,浓度变为50%? 9、一批商品,按期望获得 50%的利润来定价。结果只销掉 70%的商品。为尽早销掉剩下的商品,商店决定按定价打折扣销售。这样所获得的全部利润,是原来期望利润的91%,问:打了多少折扣 10、一列火车从甲地开往乙地,如果将车速提高20%,可以比原计划提前1小时到达;如果先以原速度行驶240千米后,再将速度提高25%,则可提前40分钟到达.求甲、乙两地之间的距离及火车原来的速度。

广东高考英语语法填空专题训练+答案

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(完整版)英语完成句子专项练习及答案

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