八年级数学第十四章 一次函数 单元测试 有答案吗?

2025-01-13 07:47:22
推荐回答(2个)
回答1:

第十四章 一次函数测试题
(时间:90分钟 总分120分)
一、相信你一定能填对!(每小题3分,共30分)
1.下列函数中,自变量x的取值范围是x≥2的是( )
A.y= B.y= C.y= D.y= •
2.下面哪个点在函数y= x+1的图象上( )
A.(2,1) B.(-2,1) C.(2,0) D.(-2,0)
3.下列函数中,y是x的正比例函数的是( )
A.y=2x-1 B.y= C.y=2x2 D.y=-2x+1
4.一次函数y=-5x+3的图象经过的象限是( )
A.一、二、三 B.二、三、四
C.一、二、四 D.一、三、四
5.若函数y=(2m+1)x2+(1-2m)x(m为常数)是正比例函数,则m的值为( )
A.m> B.m= C.m< D.m=-
6.若一次函数y=(3-k)x-k的图象经过第二、三、四象限,则k的取值范围是( )
A.k>3 B.07.已知一次函数的图象与直线y=-x+1平行,且过点(8,2),那么此一次函数的解析式为( )
A.y=-x-2 B.y=-x-6 C.y=-x+10 D.y=-x-1
8.汽车开始行驶时,油箱内有油40升,如果每小时耗油5升,则油箱内余油量y(升)与行驶时间t(时)的函数关系用图象表示应为下图中的( )

9.李老师骑自行车上班,最初以某一速度匀速行进,中途由于自行车发生故障,停下修车耽误了几分钟,为了按时到校,李老师加快了速度,仍保持匀速行进,如果准时到校.在课堂上,李老师请学生画出他行进的路程y(千米)与行进时间t(小时)的函数图象的示意图,同学们画出的图象如图所示,你认为正确的是( )

10.一次函数y=kx+b的图象经过点(2,-1)和(0,3),那么这个一次函数的解析式为( )
A.y=-2x+3 B.y=-3x+2 C.y=3x-2 D.y= x-3
二、你能填得又快又对吗?(每小题3分,共30分)
11.已知自变量为x的函数y=mx+2-m是正比例函数,则m=________,该函数的解析式为_________.
12.若点(1,3)在正比例函数y=kx的图象上,则此函数的解析式为________.
13.已知一次函数y=kx+b的图象经过点A(1,3)和B(-1,-1),则此函数的解析式为_________.
14.若解方程x+2=3x-2得x=2,则当x_________时直线y=x+2上的点在直线y=3x-2上相应点的上方.
15.已知一次函数y=-x+a与y=x+b的图象相交于点(m,8),则a+b=_________.
16.若一次函数y=kx+b交于y轴的负半轴,且y的值随x的增大而减少,则k____0,b______0.(填“>”、“<”或“=”)
17.已知直线y=x-3与y=2x+2的交点为(-5,-8),则方程组 的解是________.
18.已知一次函数y=-3x+1的图象经过点(a,1)和点(-2,b),则a=________,b=______.
19.如果直线y=-2x+k与两坐标轴所围成的三角形面积是9,则k的值为_____.
20.如图,一次函数y=kx+b的图象经过A、B两点,与x轴交于点C,则此一次函数的解析式为__________,△AOC的面积为_________.
三、认真解答,一定要细心哟!(共60分)
21.(14分)根据下列条件,确定函数关系式:
(1)y与x成正比,且当x=9时,y=16;
(2)y=kx+b的图象经过点(3,2)和点(-2,1).

22.(12分)一次函数y=kx+b的图象如图所示:
(1)求出该一次函数的表达式;
(2)当x=10时,y的值是多少?
(3)当y=12时,x的值是多少?

23.(12分)一农民带了若干千克自产的土豆进城出售,为了方便,他带了一些零钱备用,按市场价售出一些后,又降价出售.售出土豆千克数与他手中持有的钱数(含备用零钱)的关系如图所示,结合图象回答下列问题:
(1)农民自带的零钱是多少?
(2)降价前他每千克土豆出售的价格是多少?
(3)降价后他按每千克0.4元将剩余土豆售完,这时他手中的钱(含备用零钱)是26元,问他一共带了多少千克土豆?

24.(10分)如图所示的折线ABC表示从甲地向乙地打长途电话所需的电话费y(元)与通话时间t(分钟)之间的函数关系的图象.(1)写出y与t之间的函数关系式.(2)通话2分钟应付通话费多少元?通话7分钟呢?

25.(12分)已知雅美服装厂现有A种布料70米,B种布料52米,现计划用这两种布料生产M、N两种型号的时装共80套.已知做一套M型号的时装需用A种布料1.1米,B种布料0.4米,可获利50元;做一套N型号的时装需用A种布料0.6米,B种布料0.9米,可获利45元.设生产M型号的时装套数为x,用这批布料生产两种型号的时装所获得的总利润为y元.
①求y(元)与x(套)的函数关系式,并求出自变量的取值范围;
②当M型号的时装为多少套时,能使该厂所获利润最大?最大利润是多?

回答2:

老师讲完了就有答案了

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