河北省保定市顺平县2024-2025学年九年级上学期期中数学试题

2024—2025学年第一学期期中调研考试九年级数学人教版（本卷为闭卷考试，试卷页数：8页，考试时间：120分钟，卷面分：5分，总分：125分）一、选择题（本大题共12个小题，每小题3分，共36分.在每小题给出的四个选项中，只有一项是符合题目要求的）1.把方程化成一般式，则的值是（）A. B.7 C. D.12.下列图形中，既是中心对称又是轴对称图形的是（）A. B. C. D.3.将抛物线向左平移1个单位，再向上平移5个单位，得到抛物线的函数表达式为（）A. B.C. D.4.在平面直角坐标系中，点G的坐标是，连接，将线段绕原，点O旋转180°，得到对应线段，则点G的坐标为（）A. B. C. D.5.用配方法解方程时，配方后所得的方程是（）A. B. C. D.6.正比例函数和二次函数的图象一致的是（）A. B. C. D.7.已知关于x的一元二次方程的一个根是0，则m的值为（）A. B. C.1 D.以上三个都可以8.在平面直角坐标系中，将抛物线沿y轴向下平移2个单位.则平移后得到的抛物线的顶点一定在（）A.第一象限 B.第二象限 C.第三象限 D.第四象限9.某厂家2024年1—5月份销售的电车数量如图所示.设从2月份到4月份，该厂家电车销售的平均月增长率为x，根据题意可得方程（）A. B.C. D.10.对于一元二次方程，下列四个结论中错误的是（）A.若，当时，没有实数根B.若4是方程的一个根，那么是方程的另一个根C.若方程的两根符号相同，那么方程的两根符号也相同D.若没有实数根，则二次函数与直线没有交点11.如图，O是等边内一点，，，，将线段以点A为旋转中心逆时针旋转60°，得到线段.嘉嘉：由图知，琪琪：.以下说法，正确的是（）A.嘉嘉正确，琪琪错误 B.嘉嘉错误，琪琪正确C.两人都正确 D.两人都错误12.已知二次函数，其中k，m为常数，则下列说法正确的是（）A.若，，则二次函数y的最小值小于0B.若，，则二次函数y的最大值小于0C.若，，则二次函数y的最大值大于0D.若，，则二次函数的最小值小于0二、填空题（本大题共4个小题，每小题3分，共12分，把答案写在题中横线上）13.若2是方程的一个根，则______.14.生物兴趣小组的同学，将自己收集的标本向本组其他成员各赠送一件，全组共赠送了90件，则全组共有______名同学.15.小明推铅球，铅球行进高度与水平距离之间的关系式为，当铅球行进的高度为时，铅球行进的水平距离______m.16.如图，在平面直角坐标系中，等腰直角三角形的顶点A，B分别在x轴和y轴上，将绕点O逆时针旋转30°后得到，如果点A的坐标为，那么点的坐标为______.三、解答题（本大题共8个小题，共72分.解答应写出文字说明、证明过程或演算步骤）17.（本小题满分7分）已知某二次函数的图象的顶点为，且过点.（1）求此二次函数的解析式（2）判断点是否在这个二次函数的图象上，并说明理由.18.（本小题满分8分）按要求解一元二次方程，（1）（配方法） （2）（因式分解法）19.（本小题满分8分）已知关于x的一元二次方程.（1）求证：无论k为何实数，方程总有两个实数根；（2）若方程的两个实数根，，满足，求k的值.20.（本小题满分8分）开展盐碱地综合利用对保障国家粮食安全、端牢中国饭碗具有重要战略意义.我市某农业科技小组对A，B两个小麦品种进行抗盐碱实验种植对比研究.去年A、B两个品种各种植了20亩，收获后发现，B品种的平均亩产量比A品种高100千克，且A、B两个品种总产量为14000千克.（1）求A、B两个品种去年平均亩产量分别是多少千克？（2）今年，科技小组优化了小麦的种植方法，预计A、B两个品种平均亩产量将在去年的基础上分别增加30千克和千克.由于B品种深受市场欢迎，今年决定对两个品种的种植面积进行调整，B品种种植面积比去年增加m亩，且保持总种植面积与去年相同，使总产量达到15200千克，求m的值.21.（本小题满分9分）如图，在平面直角坐标系中，的三个顶点分别是，，.（1）作关于轴对称的图形，记为，画出，并写出点，的坐标；（2）请画出关于原点O成中心对称的；（3）在x轴上找一点P，使得点P到点A，C的距离最小，并求出P点坐标.22.（本小题满分9分）已知抛物线（a，b，c是常数，）中x与y的部分对应值如下表.x…012…y…0…（1）根据以上信息，可知______0（选填“<”“>”或“=”）；（2）求抛物线的解析式并在图中画出的图象； 备用图（3）地物线的解析式为.设直线与抛物线、都有两个交点，交点从左到右依次为，，，，请根据图象直接写出线段的值.23.（本小题满分11分）如图1所示，将线段绕点A逆时针旋转得到线段，在线段上找一点D，将线段绕点A逆时针旋转得到线段，连接，，.图1 图2 图3（1）求证：；（2）如图2所示，：将绕点A逆时针旋转一定的角度，（1）中的结论是否依然成立？若成立，请加以证明；若不成立，请说明理由.（3）点D为的中点，，，在绕点A逆时针旋转过程中，若点B，D，E恰好第一次在一条直线上，如图3，直接写出线段的长.24.（本小题满分12分）抛物线与x轴交于点A，C（点A在点C的右侧），与y轴交于点B.一次函数经过点A，B.图1 图2（1）求k，b的值；（2）如图1，过点C的直线交线段于点M，若，直接写出点M的坐标；（3）如图2，点D是第一象限内抛物线上的一个动点，过点D作轴交于点E，，垂足为F.当时，求点F的坐标.

2024—2025学年第一学期期中调研考试九年级数学（人教版）参考答案一、选择题（本大题共12个小题，每小题3分，共36分）1—5BDBCB 6—10DCABA 11.C12.D二、填空题（本大题共4个小题，每小题3分，共12分）13.4 14.10 15.2或6 16.三、解答题（本大题共8个小题，共72分）17.解：（1）∵此二次函数图象的顶点为∴设抛二次函数的解析式为：∵它的图象过点∴，解得∴此二次函数的关系式为······························································4分（2）点不在这个二次函数的图象上.························································5分理由：当时，.∴点不在这个二次函数的图象上······························································7分18.解：（1）（配方法）∴∴，····································································································4分（2）（因式分解法）∴或∴，····································································································8分19.解：（1）∵无论k为何实数，∴∴无论k为何实数，方程总有两个实数根···································································5分（2）由根与系数的关系得出解得········································································································8分20.解：（1）设A品种去年平均亩产量为x千克，B品种去年平均亩产量为千克解得：答：A、B两个品种去年平均亩产量分别为300千克和400千克·········································3分（2）解得：，（舍去）∴m的值为5.··········································································································8分21.解：（1）即为所求，，·········································4分（2）即为所求···························································································6分（3）点P即为所求································································································7分设直线所在直线解析式为：将和代入得解得∴当时，∴P点坐标为····························································································9分22.解：（1）<······································································································2分（2）由表格知抛物线的顶点坐标为，所以设抛物线的解析式为：∵它的图象过点∴，解得.∴抛物线解析式为，图象如下图·························································7分（3）·······································································································9分23.证明：（1）由题意可知：，，在和中∴············································································································3分（2）仍然成立.由（1）可知，，∴即在和中∴∴············································································································9分（3）······································································································11分24.解：（1）当时，，解得，∴，，·······································································2分∵经过点A，B，将，代入，得解得，·································································································4分（2）········································································································6分（3）过点F作轴于点G，过点E作于点H由（1）可知，∴一次函数解析式为：因为点E在直线上∴设E点坐标∴D点坐标为∴，解得，∴E点坐标为或····································