2021年2021年八年级数学一次函数练习题.pdf

精品好资料欢迎下载人教课标版八年级（上）数学检测试卷第十一章一次函数 B 卷（考试时间为 90分钟，满分 100 分）一二三总分一、填空题（每题2 分，共 20 分）1.在圆的周长公式C=2 r 中，变量是 _，常量是 _.2.在函数2xy中，自变量x的取值范围是 _.3.函数221xy中,当 x=_时,函数的值等于2.4.一次函数的图象经过点（-2，3）与（1，-1），它的解析式是_ _.5.将直线 y3x 向下平移5 个单位，得到直线；将直线 y-x-5 向上平移5个单位，得到直线.6.东方超市鲜鸡蛋每个0.4 元，那么所付款y 元与买鲜鸡蛋个数x（个）之间的函数关系式是 _7.平行四边形相邻的两边长为x、y，周长是30，则 y 与 x 的函数关系式是_8.出租车收费按路程计算，3km 内（包括3km）收费 8 元;超过 3km 每增加 1km 加收元，则路程 x3km 时，车费y（元）与x(km)之间的函数关系式是_9.已知点 P（3a 1，a+3）是第二象限内坐标为整数的点，则整数a 的值是 _.10.若直线axy和直线bxy的交点坐标为(8,m),则ba_.二、选择题（每题3 分，共 24 分）11.下列函数中，与yx 表示同一个函数的是（）yx2x yx2 y(x)2 y3x312.下列关系式中，不是函数关系的是（）.y=x(x0).y=x(x0).y=x(x0)13.若 m0,n0,则一次函数y=mx+n 的图象不经过（）A.第一象限B.第二象限C.第三象限D.第四象限14.已知函数y3x+1，当自变量增加m 时，相应的函数值增加（）3m+1 3m m 3m1 15.汽车由地驶往相距120km 的 B 地，它的平均速度是30km/h，则汽车距地路程s(km)与行驶时间t(h)的函数关系式及自变量t 的取值范围是（）S=12030t(0 t4)S=12030t(t0)S=30t(0t40)S=30t(t4)16.已知函数221xy，当11x时，y 的取值范围是（）|精.|品.|可.|编.|辑.|学.|习.|资.|料.*|*|*|*|欢.|迎.|下.|载.第 1 页，共 5 页精品好资料欢迎下载A.2325yB.2523yC.2523yD.2523y17.小明的父亲饭后散步，从家中走20 分钟到一个离家900 米的报亭看10 分钟的报纸后，用 15 分钟返回家中，下列图形中表示小明父亲离家的时间与距离之间的关系是（）A.B.C.D.18.当00 b,a时，函数y=ax+b 与abxy在同一坐标系中的图象大致是（）A.B.C.D.三、解答题（第19 题 6 分，其余每题10 分，共 56 分）19.地壳的厚度约为8 到 40km，在地表以下不太深的地方，温度可按y3.5xt 计算，其中x 是深度，t 是地球表面温度，y 是所达深度的温度（）在这个变化过程中，自变量和因变量分别是什么？（）如果地表温度为2，计算当x 为 5km 时地壳的温度20.已知3y与x成正比例,且2x时,7y.(1)求y与x的函数关系式;(2)当21x时,求y的值;(3)将所得函数图象平移,使它过点(2,-1).求平移后直线的解析式.|精.|品.|可.|编.|辑.|学.|习.|资.|料.*|*|*|*|欢.|迎.|下.|载.第 2 页，共 5 页文档编码:CY2X2D3Z4A9 HK2T4O6B2F9 ZL4X1T2U2T2文档编码:CY2X2D3Z4A9 HK2T4O6B2F9 ZL4X1T2U2T2文档编码:CY2X2D3Z4A9 HK2T4O6B2F9 ZL4X1T2U2T2文档编码:CY2X2D3Z4A9 HK2T4O6B2F9 ZL4X1T2U2T2文档编码:CY2X2D3Z4A9 HK2T4O6B2F9 ZL4X1T2U2T2文档编码:CY2X2D3Z4A9 HK2T4O6B2F9 ZL4X1T2U2T2文档编码:CY2X2D3Z4A9 HK2T4O6B2F9 ZL4X1T2U2T2文档编码:CY2X2D3Z4A9 HK2T4O6B2F9 ZL4X1T2U2T2文档编码:CY2X2D3Z4A9 HK2T4O6B2F9 ZL4X1T2U2T2文档编码:CY2X2D3Z4A9 HK2T4O6B2F9 ZL4X1T2U2T2文档编码:CY2X2D3Z4A9 HK2T4O6B2F9 ZL4X1T2U2T2文档编码:CY2X2D3Z4A9 HK2T4O6B2F9 ZL4X1T2U2T2文档编码:CY2X2D3Z4A9 HK2T4O6B2F9 ZL4X1T2U2T2文档编码:CY2X2D3Z4A9 HK2T4O6B2F9 ZL4X1T2U2T2文档编码:CY2X2D3Z4A9 HK2T4O6B2F9 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19.（1）自变量是地表以下的深度x，因变量是所达深度的温度y；（2）19.5 20.（1）y=2x+3；（2）2；（3）y=2x-5 21.y=0.3x+6 22.(1)60 米；(2)300 米，小强；(3)8 分钟23.(1)y=4-x(0 x 2)(2)当 y=4-x=1.5 时，x=2.5 不在 0 x2，因此不存在点P使四边形 APCD的面积为 1.5 24.由题意得.32,1245kyxkyx解得.72,732kykx|精.|品.|可.|编.|辑.|学.|习.|资.|料.*|*|*|*|欢.|迎.|下.|载.第 4 页，共 5 页文档编码:CY2X2D3Z4A9 HK2T4O6B2F9 ZL4X1T2U2T2文档编码:CY2X2D3Z4A9 HK2T4O6B2F9 ZL4X1T2U2T2文档编码:CY2X2D3Z4A9 HK2T4O6B2F9 ZL4X1T2U2T2文档编码:CY2X2D3Z4A9 HK2T4O6B2F9 ZL4X1T2U2T2文档编码:CY2X2D3Z4A9 HK2T4O6B2F9 ZL4X1T2U2T2文档编码:CY2X2D3Z4A9 HK2T4O6B2F9 ZL4X1T2U2T2文档编码:CY2X2D3Z4A9 HK2T4O6B2F9 ZL4X1T2U2T2文档编码:CY2X2D3Z4A9 HK2T4O6B2F9 ZL4X1T2U2T2文档编码:CY2X2D3Z4A9 HK2T4O6B2F9 ZL4X1T2U2T2文档编码:CY2X2D3Z4A9 HK2T4O6B2F9 ZL4X1T2U2T2文档编码:CY2X2D3Z4A9 HK2T4O6B2F9 ZL4X1T2U2T2文档编码:CY2X2D3Z4A9 HK2T4O6B2F9 ZL4X1T2U2T2文档编码:CY2X2D3Z4A9 HK2T4O6B2F9 ZL4X1T2U2T2文档编码:CY2X2D3Z4A9 HK2T4O6B2F9 ZL4X1T2U2T2文档编码:CY2X2D3Z4A9 HK2T4O6B2F9 ZL4X1T2U2T2文档编码:CY2X2D3Z4A9 HK2T4O6B2F9 ZL4X1T2U2T2文档编码:CY2X2D3Z4A9 HK2T4O6B2F9 ZL4X1T2U2T2文档编码:CY2X2D3Z4A9 HK2T4O6B2F9 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ZL4X1T2U2T2文档编码:CY2X2D3Z4A9 HK2T4O6B2F9 ZL4X1T2U2T2文档编码:CY2X2D3Z4A9 HK2T4O6B2F9 ZL4X1T2U2T2文档编码:CY2X2D3Z4A9 HK2T4O6B2F9 ZL4X1T2U2T2文档编码:CY2X2D3Z4A9 HK2T4O6B2F9 ZL4X1T2U2T2文档编码:CY2X2D3Z4A9 HK2T4O6B2F9 ZL4X1T2U2T2精品好资料欢迎下载因为两直线交点在第四象限，所以x0，y0，即.072,0732kk解得.2,23kk故223k时，两直线交点在第四象限25.提示:先求出直线的解析式为y=x+1，再求出它与两坐标轴的交点，进而求得三角形的面积为 0.5|精.|品.|可.|编.|辑.|学.|习.|资.|料.*|*|*|*|欢.|迎.|下.|载.第 5 页，共 5 页文档编码:CY2X2D3Z4A9 HK2T4O6B2F9 ZL4X1T2U2T2文档编码:CY2X2D3Z4A9 HK2T4O6B2F9 ZL4X1T2U2T2文档编码:CY2X2D3Z4A9 HK2T4O6B2F9 ZL4X1T2U2T2文档编码:CY2X2D3Z4A9 HK2T4O6B2F9 ZL4X1T2U2T2文档编码:CY2X2D3Z4A9 HK2T4O6B2F9 ZL4X1T2U2T2文档编码:CY2X2D3Z4A9 HK2T4O6B2F9 ZL4X1T2U2T2文档编码:CY2X2D3Z4A9 HK2T4O6B2F9 ZL4X1T2U2T2文档编码:CY2X2D3Z4A9 HK2T4O6B2F9 ZL4X1T2U2T2文档编码:CY2X2D3Z4A9 HK2T4O6B2F9 ZL4X1T2U2T2文档编码:CY2X2D3Z4A9 HK2T4O6B2F9 ZL4X1T2U2T2文档编码:CY2X2D3Z4A9 HK2T4O6B2F9 ZL4X1T2U2T2文档编码:CY2X2D3Z4A9 HK2T4O6B2F9 ZL4X1T2U2T2文档编码:CY2X2D3Z4A9 HK2T4O6B2F9 ZL4X1T2U2T2文档编码:CY2X2D3Z4A9 HK2T4O6B2F9 ZL4X1T2U2T2文档编码:CY2X2D3Z4A9 HK2T4O6B2F9 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