安徽省合肥市蜀山区合肥市琥珀中学2023-2024学年九年级上学期月考数学试题

这是一份安徽省合肥市蜀山区合肥市琥珀中学2023-2024学年九年级上学期月考数学试题，共9页。
合肥市琥珀中学教育集团2024届九年级第一次质量调研检测数学试题卷注意事项：1．你拿到的试卷满分为150分，考试时间为120分钟。2．本试卷包括“试题卷”和“答题卷”两部分。“试题卷”共4页，“答题卷”共6页。3．请务必在“答题卷”上答题，在“试题卷”上答题是无效的。4．考试结束后，请将“试题卷”和“答题卷”一并交回。一、选择题（本大题共10小题，每小题4分，满分40分）每小题都给出A，B，C，D四个选项，其中只有一个是符合题目要求的．1．二次函数图象的顶点所在的象限是：（    ）A．第一象限 B．第二象限 C．第三象限 D．第四象限2．将抛物线向下平移2个单位，所得抛物线的表达式为：（    ）A．  B．C．  D．3．某种蓄电池的电压U（单位：V）为定值，使用蓄电池时，电流I（单位：A）与电阻R（单位：Ω）是反比例函数关系．当时，，则当时，I的值是：（    ）A．4 B．9 C．32 D．04．根据下列表格中二次函数（，a，b，c为常数）的自变量x与函数值y的对应值，判断方程的一个解x的范围是：（    ）x0.010.04A． B．C． D．5．若点，，都在反比例函数的图象上，则，，的大小关系是：（    ）A． B． C． D．6．下列函数中，y的值随x值的增大而减小的是：（    ）A． B． C． D．7．小勇、小冠、小明、小天四人共同探究函数的值的情况，各自通报探究的结论，其中错误的是：（    ）A．小勇认为只有当时，函数值为2B．小冠认为找不到实数x，使函数值为0C．小明认为抛物线开口向上D．小天认为抛物线与x轴有两个交点8．在平面直角坐标系中，抛物线与x轴交于，两点，其中．将此抛物线向下平移，与x轴交于，两点，其中，下面结论正确的是：（    ）A．当时，，B．当时，，C．当时，，D．当时，，9．如图，正方形对称中心在原点O，四个顶点分别位于两个反比例函数和的图象的四个分支上，则实数k的值为：（    ）A． B． C． D．410．已知二次函数的图象如图所示，则二次函数与正比例函数的图象大致为：（    ）A．   B．C．   D．二、填空题（本大题共4小题，每小题5分，满分20分）11．抛物线与y轴的交点坐标为________．12．一名学生推铅球，铅球行进高度y（单位：m）与水平距离x（单位：m）之间的关系是，则该学生推铅球的水平距离为________m．13．如图，在平面直角坐标系中，点O为坐标原点，矩形（）交反比例函数（）的图象于点D，E．点D的坐标为．连接，，．若，，则k的值为________．14．定义：平面直角坐标系中，点，点，若，，其中k为常数，且，则称点Q是点P的“k级变换点”．例如，点是点的“级变换点”．（1）若函数的图象上存在点的“k级变换点”，则k的值为_________；（2）若关于x的二次函数（）的图象上恰有两个点，这两个点的“1级变换点”都在直线上，则n的取值范围是_________．三、（本大题共2小题，每小题8分，满分16分）15．已知二次函数的图象经过点，求这个二次函数的表达式．16．已知抛物线．求证：无论m为何值时，抛物线与x轴总有两个交点．四、（本大题共2小题，每小题8分，满分16分）17．如图在平面直角坐标系中，直线分别交x轴、y轴于点A，B，形状相同的抛物线：（，……）的顶点在直线上，其对称轴与x轴交点的横坐标依次是2，3，5，8，13，…，根据上述规律解决以下问题：（1）抛物线的顶点坐标是________；（2）求抛物线线中b，c的值．18．已知一个二次函数图象上部分点的横坐标x与纵坐标y的对应值如表所示：x…01…y…00…（1）求这个二次函数的解析式；（2）在给定的平面直角坐标系中画出这个二次函数的图象；（3）当时，y的取值范围为_______．五、（本大题共2小题，每小题10分，满分20分）19．跨学科整合学习中为研究某种化学试剂的挥发情况，某研究小组在两种不同的场景下做对比实验，收集了该试剂挥发过程中剩余质量y（克）随时间x（分钟）变化的数据（），并分别绘制在平面直角坐标系中，如图所示．（1）从（），（），中，选择适当的函数模型分别模拟两种场景下y随x变化的函数关系，并求出相应的函数表达式；（2）查阅文献可知，该化学试剂发挥作用的最低质量为3克，在上述实验中，该化学试剂在哪种场景下发挥作用的时间更长？20．探究某种商品的售价、销量、利润之间的变化关系，小明整理出该商品的相关数据如下表所示．时间x（天）售价（元/件）70每天销量（件）已知该商品的进价为每件10元，设销售该商品的每天利润为y元．（1）求y与x的函数关系式；（2）销售该商品第几天时，当天销售利润最大，最大利润是多少？六、（本大题满分12分）21．如图，在平面直角坐标系中，直线与y轴交于点，与x轴交于点，与反比例函数在第三象限内的图象交于点．（1）求反比例函数的表达式；（2）当时，求x的取值范围；（3）当点P在y轴上，的面积为6时，直接写出点P的坐标．七、（本大题满分12分）22．鹰眼技术助力杭州亚运，提升球迷观赛体验如图分别为足球比赛中某一时刻的鹰眼系统预测画面（如图1）和截面示意图（如图2），攻球员位于点O，守门员位于点A，的延长线与球门线交于点B，且点A，B均在足球轨迹正下方，足球的飞行轨迹可看成抛物线．已知，，足球飞行的水平速度为，水平距离s（）与离地高度h的鹰眼数据如表：…912151821……4.24.854.84.2…      图1                            图2（1）根据表中数据预测足球落地时，________m；（2）求h关于s的函数解析式；（3）守门员在攻球员射门瞬间就作出防守反应，当守门员位于足球正下方时，足球离地高度不大于守门员的最大防守高度视为防守成功一次防守中守门员面对足球后退，已知后退过程中守门员速度为，最大防守高度为2.6m．①求守门员后退到足球正下方所需时间；②这次守门员能否防守成功？试通过计算说明．八、（本大题满分14分）23．已知抛物线（a，c为常数，）过点，顶点为点P．（1）当时，求此抛物线顶点P的坐标；（2）当时，若的面积为4，求此抛物线的解析式；（3）将抛物线向左平移2个单位，向下平移个单位（），得到新抛物线的顶点为A，与y轴交点为B，点M在直线上，点N在直线上，当四边形的周长最小时，恰好有，求a的值． 合肥市琥珀中学教育集团2024届九年级第一次质量调研检测数学参考答案一、选择题（本大题共10小题，每小题4分，满分40分）题号12345678910答案BBACDDDCAB二、填空题（本大题共4小题，每小题5分，满分20分）11．； 12．10；13． 14．，且．（第一空2分；第二空3分，对一个答案1分，两个全对3分，有错不得分）三、（本大题共2小题，每小题8分，满分16分）15．解：把代入二次函数得，··························································（3分）解得，···········································································（6分）所以抛物线解析式为．即．····························································（8分）16．证明：，，．···································································（1分）·················································································（3分）．·············································································（5分）∵．∴，·············································································（7分）∴无论m为何值时，抛物线与x轴总有两个交点．············································（8分）四、（本大题共2小题，每小题8分，满分16分）17．解：（1）·····································································（3分）（2），当时，抛物线的顶点坐标是，····················································（5分）由顶点式得：，展开得．········································································（7分）∴，．···········································································（8分）18．解：（1）由题意可得二次函数的顶点坐标为，·········································（1分）设二次函数的解析式为：，····························································（2分）把点代入，得，····································································（3分）故抛物线解析式为，即·······························································（4分）（2）如图所示：·················································································（6分）（3）．··········································································（8分）五、（本大题共2小题，每小题10分，满分20分）19．解：（1）观察两种场景可知，场景A为，场景B为（），·····································································（1分）把，代入得：，···············································································（2分）解得，···········································································（3分）∴···············································································（4分）把，代入得：······································································（5分），解得，··········································································（6分）∴；·············································································（7分）场景A的函数表达式为，场景B的函数表达式为；（2）当时，场景A中，，·······································································（8分）场景B中，，解得，··································································（9分），化学试剂在场景A下发挥作用的时间更长．··············································（10分）20．解：（1）当时，，······························································（2分）当时，，··········································································（4分）综上所述：y与x的函数关系式为；···············································································（5分）（2）当时，二次函数，∵，∴当时，，········································································（7分）当时，中y随x的增大而减小，∴当时，，·······················································（9分）综上所述，该商品第10天时，当天销售利润最大，最大利润是3200元．···························（10分）六、（本大题满分12分）21．解：（1）将，代入得：，解得：，··········································································（1分）∴一次函数表达式为：，·····························································（2分）将代入得：，······································································（3分）∴，·············································································（4分）将代入得：，······································································（5分）∴反比例函数的表达式为：；··························································（6分）（2）设一次函数与反比例函数在第二象限交于点D，联立，解得：或，···································································（7分）∴，·············································································（8分）由图象可知：当时，或；·····························································（10分）（3）点P的坐标为：或．·····························································（12分）22．略23．略