河北省保定市順平縣2024-2025學(xué)年九年級(jí)上學(xué)期期中數(shù)學(xué)試題

2024—2025學(xué)年第一學(xué)期期中調(diào)研考試九年級(jí)數(shù)學(xué)人教版（本卷為閉卷考試，試卷頁(yè)數(shù)：8頁(yè)，考試時(shí)間：120分鐘，卷面分：5分，總分：125分）一、選擇題（本大題共12個(gè)小題，每小題3分，共36分.在每小題給出的四個(gè)選項(xiàng)中，只有一項(xiàng)是符合題目要求的）1.把方程化成一般式，則的值是（）A. B.7 C. D.12.下列圖形中，既是中心對(duì)稱(chēng)又是軸對(duì)稱(chēng)圖形的是（）A. B. C. D.3.將拋物線向左平移1個(gè)單位，再向上平移5個(gè)單位，得到拋物線的函數(shù)表達(dá)式為（）A. B.C. D.4.在平面直角坐標(biāo)系中，點(diǎn)G的坐標(biāo)是，連接，將線段繞原，點(diǎn)O旋轉(zhuǎn)180°，得到對(duì)應(yīng)線段，則點(diǎn)G的坐標(biāo)為（）A. B. C. D.5.用配方法解方程時(shí)，配方后所得的方程是（）A. B. C. D.6.正比例函數(shù)和二次函數(shù)的圖象一致的是（）A. B. C. D.7.已知關(guān)于x的一元二次方程的一個(gè)根是0，則m的值為（）A. B. C.1 D.以上三個(gè)都可以8.在平面直角坐標(biāo)系中，將拋物線沿y軸向下平移2個(gè)單位.則平移后得到的拋物線的頂點(diǎn)一定在（）A.第一象限 B.第二象限 C.第三象限 D.第四象限9.某廠家2024年1—5月份銷(xiāo)售的電車(chē)數(shù)量如圖所示.設(shè)從2月份到4月份，該廠家電車(chē)銷(xiāo)售的平均月增長(zhǎng)率為x，根據(jù)題意可得方程（）A. B.C. D.10.對(duì)于一元二次方程，下列四個(gè)結(jié)論中錯(cuò)誤的是（）A.若，當(dāng)時(shí)，沒(méi)有實(shí)數(shù)根B.若4是方程的一個(gè)根，那么是方程的另一個(gè)根C.若方程的兩根符號(hào)相同，那么方程的兩根符號(hào)也相同D.若沒(méi)有實(shí)數(shù)根，則二次函數(shù)與直線沒(méi)有交點(diǎn)11.如圖，O是等邊內(nèi)一點(diǎn)，，，，將線段以點(diǎn)A為旋轉(zhuǎn)中心逆時(shí)針旋轉(zhuǎn)60°，得到線段.嘉嘉：由圖知，琪琪：.以下說(shuō)法，正確的是（）A.嘉嘉正確，琪琪錯(cuò)誤 B.嘉嘉錯(cuò)誤，琪琪正確C.兩人都正確 D.兩人都錯(cuò)誤12.已知二次函數(shù)，其中k，m為常數(shù)，則下列說(shuō)法正確的是（）A.若，，則二次函數(shù)y的最小值小于0B.若，，則二次函數(shù)y的最大值小于0C.若，，則二次函數(shù)y的最大值大于0D.若，，則二次函數(shù)的最小值小于0二、填空題（本大題共4個(gè)小題，每小題3分，共12分，把答案寫(xiě)在題中橫線上）13.若2是方程的一個(gè)根，則______.14.生物興趣小組的同學(xué)，將自己收集的標(biāo)本向本組其他成員各贈(zèng)送一件，全組共贈(zèng)送了90件，則全組共有______名同學(xué).15.小明推鉛球，鉛球行進(jìn)高度與水平距離之間的關(guān)系式為，當(dāng)鉛球行進(jìn)的高度為時(shí)，鉛球行進(jìn)的水平距離______m.16.如圖，在平面直角坐標(biāo)系中，等腰直角三角形的頂點(diǎn)A，B分別在x軸和y軸上，將繞點(diǎn)O逆時(shí)針旋轉(zhuǎn)30°后得到，如果點(diǎn)A的坐標(biāo)為，那么點(diǎn)的坐標(biāo)為_(kāi)_____.三、解答題（本大題共8個(gè)小題，共72分.解答應(yīng)寫(xiě)出文字說(shuō)明、證明過(guò)程或演算步驟）17.（本小題滿分7分）已知某二次函數(shù)的圖象的頂點(diǎn)為，且過(guò)點(diǎn).（1）求此二次函數(shù)的解析式（2）判斷點(diǎn)是否在這個(gè)二次函數(shù)的圖象上，并說(shuō)明理由.18.（本小題滿分8分）按要求解一元二次方程，（1）（配方法） （2）（因式分解法）19.（本小題滿分8分）已知關(guān)于x的一元二次方程.（1）求證：無(wú)論k為何實(shí)數(shù)，方程總有兩個(gè)實(shí)數(shù)根；（2）若方程的兩個(gè)實(shí)數(shù)根，，滿足，求k的值.20.（本小題滿分8分）開(kāi)展鹽堿地綜合利用對(duì)保障國(guó)家糧食安全、端牢中國(guó)飯碗具有重要戰(zhàn)略意義.我市某農(nóng)業(yè)科技小組對(duì)A，B兩個(gè)小麥品種進(jìn)行抗鹽堿實(shí)驗(yàn)種植對(duì)比研究.去年A、B兩個(gè)品種各種植了20畝，收獲后發(fā)現(xiàn)，B品種的平均畝產(chǎn)量比A品種高100千克，且A、B兩個(gè)品種總產(chǎn)量為14000千克.（1）求A、B兩個(gè)品種去年平均畝產(chǎn)量分別是多少千克？（2）今年，科技小組優(yōu)化了小麥的種植方法，預(yù)計(jì)A、B兩個(gè)品種平均畝產(chǎn)量將在去年的基礎(chǔ)上分別增加30千克和千克.由于B品種深受市場(chǎng)歡迎，今年決定對(duì)兩個(gè)品種的種植面積進(jìn)行調(diào)整，B品種種植面積比去年增加m畝，且保持總種植面積與去年相同，使總產(chǎn)量達(dá)到15200千克，求m的值.21.（本小題滿分9分）如圖，在平面直角坐標(biāo)系中，的三個(gè)頂點(diǎn)分別是，，.（1）作關(guān)于軸對(duì)稱(chēng)的圖形，記為，畫(huà)出，并寫(xiě)出點(diǎn)，的坐標(biāo)；（2）請(qǐng)畫(huà)出關(guān)于原點(diǎn)O成中心對(duì)稱(chēng)的；（3）在x軸上找一點(diǎn)P，使得點(diǎn)P到點(diǎn)A，C的距離最小，并求出P點(diǎn)坐標(biāo).22.（本小題滿分9分）已知拋物線（a，b，c是常數(shù)，）中x與y的部分對(duì)應(yīng)值如下表.x…012…y…0…（1）根據(jù)以上信息，可知______0（選填“<”“>”或“=”）；（2）求拋物線的解析式并在圖中畫(huà)出的圖象； 備用圖（3）地物線的解析式為.設(shè)直線與拋物線、都有兩個(gè)交點(diǎn)，交點(diǎn)從左到右依次為，，，，請(qǐng)根據(jù)圖象直接寫(xiě)出線段的值.23.（本小題滿分11分）如圖1所示，將線段繞點(diǎn)A逆時(shí)針旋轉(zhuǎn)得到線段，在線段上找一點(diǎn)D，將線段繞點(diǎn)A逆時(shí)針旋轉(zhuǎn)得到線段，連接，，.圖1 圖2 圖3（1）求證：；（2）如圖2所示，：將繞點(diǎn)A逆時(shí)針旋轉(zhuǎn)一定的角度，（1）中的結(jié)論是否依然成立？若成立，請(qǐng)加以證明；若不成立，請(qǐng)說(shuō)明理由.（3）點(diǎn)D為的中點(diǎn)，，，在繞點(diǎn)A逆時(shí)針旋轉(zhuǎn)過(guò)程中，若點(diǎn)B，D，E恰好第一次在一條直線上，如圖3，直接寫(xiě)出線段的長(zhǎng).24.（本小題滿分12分）拋物線與x軸交于點(diǎn)A，C（點(diǎn)A在點(diǎn)C的右側(cè)），與y軸交于點(diǎn)B.一次函數(shù)經(jīng)過(guò)點(diǎn)A，B.圖1 圖2（1）求k，b的值；（2）如圖1，過(guò)點(diǎn)C的直線交線段于點(diǎn)M，若，直接寫(xiě)出點(diǎn)M的坐標(biāo)；（3）如圖2，點(diǎn)D是第一象限內(nèi)拋物線上的一個(gè)動(dòng)點(diǎn)，過(guò)點(diǎn)D作軸交于點(diǎn)E，，垂足為F.當(dāng)時(shí)，求點(diǎn)F的坐標(biāo).

2024—2025學(xué)年第一學(xué)期期中調(diào)研考試九年級(jí)數(shù)學(xué)（人教版）參考答案一、選擇題（本大題共12個(gè)小題，每小題3分，共36分）1—5BDBCB 6—10DCABA 11.C12.D二、填空題（本大題共4個(gè)小題，每小題3分，共12分）13.4 14.10 15.2或6 16.三、解答題（本大題共8個(gè)小題，共72分）17.解：（1）∵此二次函數(shù)圖象的頂點(diǎn)為∴設(shè)拋二次函數(shù)的解析式為：∵它的圖象過(guò)點(diǎn)∴，解得∴此二次函數(shù)的關(guān)系式為······························································4分（2）點(diǎn)不在這個(gè)二次函數(shù)的圖象上.························································5分理由：當(dāng)時(shí)，.∴點(diǎn)不在這個(gè)二次函數(shù)的圖象上······························································7分18.解：（1）（配方法）∴∴，····································································································4分（2）（因式分解法）∴或∴，····································································································8分19.解：（1）∵無(wú)論k為何實(shí)數(shù)，∴∴無(wú)論k為何實(shí)數(shù)，方程總有兩個(gè)實(shí)數(shù)根···································································5分（2）由根與系數(shù)的關(guān)系得出解得········································································································8分20.解：（1）設(shè)A品種去年平均畝產(chǎn)量為x千克，B品種去年平均畝產(chǎn)量為千克解得：答：A、B兩個(gè)品種去年平均畝產(chǎn)量分別為300千克和400千克·········································3分（2）解得：，（舍去）∴m的值為5.··········································································································8分21.解：（1）即為所求，，·········································4分（2）即為所求···························································································6分（3）點(diǎn)P即為所求································································································7分設(shè)直線所在直線解析式為：將和代入得解得∴當(dāng)時(shí)，∴P點(diǎn)坐標(biāo)為····························································································9分22.解：（1）<······································································································2分（2）由表格知拋物線的頂點(diǎn)坐標(biāo)為，所以設(shè)拋物線的解析式為：∵它的圖象過(guò)點(diǎn)∴，解得.∴拋物線解析式為，圖象如下圖·························································7分（3）·······································································································9分23.證明：（1）由題意可知：，，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）當(dāng)時(shí)，，解得，∴，，·······································································2分∵經(jīng)過(guò)點(diǎn)A，B，將，代入，得解得，·································································································4分（2）········································································································6分（3）過(guò)點(diǎn)F作軸于點(diǎn)G，過(guò)點(diǎn)E作于點(diǎn)H由（1）可知，∴一次函數(shù)解析式為：因?yàn)辄c(diǎn)E在直線上∴設(shè)E點(diǎn)坐標(biāo)∴D點(diǎn)坐標(biāo)為∴，解得，∴E點(diǎn)坐標(biāo)為或····································