陕西省渭南市韩城市新城区第四初级中学 2022-2023学年九年级上学期第一阶段性数学作业(含答案)

这是一份陕西省渭南市韩城市新城区第四初级中学 2022-2023学年九年级上学期第一阶段性数学作业(含答案)，共9页。试卷主要包含了选择题，填空题，解答题等内容，欢迎下载使用。
C（人教版）2022~2023学年度第一学期第一次阶段性作业九年级数学（建议完成时间：120分钟   满分：120分）一、选择题（共8小题，每小题3分，计24分．每小题只有一个选项是符合题意的）1．方程的解是（    ）A． B． C． D．2．已知二次函数，当时，则函数y的值为（    ）A．6 B． C．9 D．3．解一元二次方程，配方得到，则a的值为（    ）A．2 B． C．1 D．4．二次函数的图象可能是下列图象中的（    ）A． B． C． D．5．在平面直角坐标系中，已知抛物线与关于y轴对称，则的值为（    ）A．6 B． C． D．6．已知关于x的一元二次方程的两实数根分别为，则的值为（    ）A． B．1 C．5 D．7．如图，某校在操场东边开发出一块边长分别为18米、11米的矩形菜园，作为劳动教育系列课程的实验基地之一．为了便于管理，现要在中间开辟一纵两横三条等宽的小道，要使种植面积为96平方米．设小道的宽为x米，可列方程为（    ）A． B．C． D．8．已知二次函数（a为常数，），则该函数图象的顶点位于（    ）A．第一象限 B．第二象限 C．第三象限 D．第四象限二、填空题（共5小题，每小题3分，计15分）9．请写出一个关于二次函数图象或性质的结论________．（写出一条即可）10．已知是关于x的方程的一个根，则a的值为________．11．已知二次函数的图象与x轴有两个交点，则a的取值范围是________．12．点，在二次函数的图象上，若，，则与的大小关系是________（填“>”“<”或“=”）．13．若等腰三角形三边的长分别是a，b，3，且a，b是关于x的一元二次方程的两个根，则满足上述条件的m的值为________．三、解答题（共13小题，计81分．解答应写出过程）14．（5分）已知关于x的方程的一个解是，求代数式的值．15．（5分）已知二次函数，当函数值时，求此时x的值．16．（5分）已知，是抛物线上不同的两点，且，求m的值．17．（5分）已知关于x的一元二次方有两个不相等的实数根a，b．若，求k的值．18．（5分）已知抛物线．（1）当时，请判断点是否在该抛物线上；（2）抛物线经过点，求m的值．19．（5分）已知二次函数的图象如图所示；（1）写出对称轴是________；（2）直接写出抛物线与x轴的交点坐标；（横坐标均是整数）（3）利用图象直接写出当x为何值时，函数值y大于0？20．（5分）已知二次函数的图象为抛物线C．将抛物线C先向左平移1个单位长度，再向上平移2个单位长度，得到抛物线，求出抛物线的解析式以及抛物线与y轴交点纵坐标．21．（6分）解方程：（1）；      （2）．22．（7分）已知抛物线经过，和三点．（1）若该抛物线的顶点恰为点B，求此时n的值，并判断抛物线的开口方向；（2）当时，确定这个抛物线的解析式，并判断抛物线的开口方向．23．（7分）关于x的一元二次方程．（1）求证：方程总有两个实数根；（2）若方程有一个根为1，求m的值和另一个根．24．（8分）某街心花园的喷水池中心O有一喷水管，从A点向外喷水，喷出的水柱为抛物线．如图，以水平方向为x轴，点O为原点建立平面直角坐标系，点A在y轴上，x轴上的点D为水柱的落水点，水柱所在抛物线的函数解析式为．（1）求O、D之间的距离；（2）若在上离O点10m的E处竖立标杆，，且标杆的顶端F刚好碰到水柱，求标杆的高．25．（8分）随着我国数字化阅读方式的接触率和人群持续增多，数字阅读凭借独有的便利性成为了更快获得优质内容的重要途径，目前，数字阅读已经成为当下更环保、更年轻的阅读方式，2019年某市数字阅读市场规模为400万元，2021年为576万元．（1）求2019年到2021年该市数字阅读市场规模的年平均增长率；（2）若年平均增长率不变，预计2022年该市数字阅读市场规模是否可以达到700万元？26．（10分）如图，抛物线与x轴交于点A和点，与y轴交于点，连接，，点E是对称轴上的一个动点．（1）求抛物线的解析式；（2）请在抛物线的对称轴上找一点P，使的周长最小，并求此时点P的坐标；（3）当时，求点E的坐标．2022~2023学年度第一学期第一次阶段性作业九年级数学参考答案及评分标准一、选择题（共8小题，每小题3分，计24分．每小题只有一个选项是符合题意的）1．B   2．D   3．A   4．C   6．B   7．D   8．A二、填空题（共5小题，每小题3分，计15分）9．对称轴为（答案不唯一）   10．3    11．    12．>    13．4或3三、解答题（共13小题，计81分．解答应写出过程）14．解：∵是方程的解，∴，·················································································（2分）∴，∴原式．··············································································（5分）15．解：当时，则，·····································································（2分）解得．故x的值为1或5．········································································（5分）16．解：∵，是抛物线上不同的两点，把代入得，，··········································································（2分）把代入得，，∵，∴，∴或（舍），∵．·················································································（5分）（其他方法正确不扣分，可参照给分）17．解：∵关于x的一元二次方程有两个实数根a，b．∴，，，··············································································（2分）∴．∵，∴，∴，·················································································（4分）∴．·················································································（5分）18．解：（1）当时，抛物线为，····························································（2分）将代入得，∴点不在该抛物线上．···································································（3分）（2）∵抛物线经过点，∴，∴．·················································································（5分）19．解：（1）．········································································（2分）（2）二次函数的图象与x轴有两个交点，交点坐标为和；·········································（4分）（3）由图象可得：当时，函数值y大于0．·····················································（5分）20．解：∵，··········································································（1分）∵将抛物线C先向左平移1个单位长度，再向上平移2个单位长度，得到抛物线，∴，·················································································（3分）令代入得，，∴抛物线与y轴交点的纵坐标为3．···························································（5分）21．解：（1）移项得，因式分解得，··········································································（2分），∴．·················································································（3分）（2），，···················································································（4分）∴，∴．·················································································（6分）22．解：（1）∵抛物线的顶点为，由抛物线的对称性可知，即．······························································（2分）∴设抛物线的解析式为：，∵抛物线经过原点，∴将代入，得：，解得：，∴抛物线开口向下．·····································································（4分）（2）∵，∴∵抛物线经过，∴，把，代入，得：解得··············································································（6分）∴这个抛物线的解析式为：，∵，∴抛物线开口向上．··································································（7分）23．（1）证明：∵．····································································（2分），∴方程总有两个实数根．··································································（4分）（2）解：把代入方程得：，解得：，把代入得：，··········································································（5分）解得：，所以另一根为．·········································································（7分）24．解：（1）时，，解得：（舍去），，·····································································（3分）∴点D的坐标为，∴O、D之间的距离为11m．································································（4分）（2）由标杆的顶端F刚好碰到水柱且，可知点F在抛物线上，且横坐标为10，当时，，··············································································（6分）∴点，∴．∴标杆的高为．·········································································（8分）25．解：（1）设2019年到2021年该市数字阅读市场规模的年平均增长率为x．根据题意可得，·········································································（3分）解得（舍），所以．∴2019年到2021年该市数字阅读市场规模的年平均增长率为20%．···································（5分）（2）由题意得：万元．···································································（7分）∵，∴预计2022年该市数字阅读市场规模不可以达到700万元．········································（8分）26．解：（1）∵抛物线经过点，，解得∴该抛物线的解析式为．··································································（3分）（2）∵，∴抛物线对称轴为直线．点A关于对称轴的对称点是点B，交对称轴直线于点P，点P就是使的周长最小的点．设直线的解析式为，把点，代入，得：解得：∴直线的解析式为．·····································································（5分）当时，，∴．·················································································（6分）（3）∵点A与关于直线对称，∴，∴，∴，·················································································（7分）设，则，∴，∵，∴，·················································································（9分）解得：或，∴点E的坐标为或．······································································（10分）