安徽省合肥市蜀山區(qū)合肥市琥珀中學(xué)2023-2024學(xué)年九年級(jí)上學(xué)期月考數(shù)學(xué)試題-試卷下載-教習(xí)網(wǎng)

合肥市琥珀中學(xué)教育集團(tuán)2024屆九年級(jí)第一次質(zhì)量調(diào)研檢測(cè)數(shù)學(xué)試題卷注意事項(xiàng)：1．你拿到的試卷滿分為150分，考試時(shí)間為120分鐘。2．本試卷包括“試題卷”和“答題卷”兩部分?！霸囶}卷”共4頁(yè)，“答題卷”共6頁(yè)。3．請(qǐng)務(wù)必在“答題卷”上答題，在“試題卷”上答題是無(wú)效的。4．考試結(jié)束后，請(qǐng)將“試題卷”和“答題卷”一并交回。一、選擇題（本大題共10小題，每小題4分，滿分40分）每小題都給出A，B，C，D四個(gè)選項(xiàng)，其中只有一個(gè)是符合題目要求的．1．二次函數(shù)圖象的頂點(diǎn)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．將拋物線向下平移2個(gè)單位，所得拋物線的表達(dá)式為：（）A． B．C． D．3．某種蓄電池的電壓U（單位：V）為定值，使用蓄電池時(shí)，電流I（單位：A）與電阻R（單位：Ω）是反比例函數(shù)關(guān)系．當(dāng)時(shí)，，則當(dāng)時(shí)，I的值是：（）A．4 B．9 C．32 D．04．根據(jù)下列表格中二次函數(shù)（，a，b，c為常數(shù)）的自變量x與函數(shù)值y的對(duì)應(yīng)值，判斷方程的一個(gè)解x的范圍是：（）x0.010.04A． B．C． D．5．若點(diǎn)，，都在反比例函數(shù)的圖象上，則，，的大小關(guān)系是：（）A． B． C． D．6．下列函數(shù)中，y的值隨x值的增大而減小的是：（）A． B． C． D．7．小勇、小冠、小明、小天四人共同探究函數(shù)的值的情況，各自通報(bào)探究的結(jié)論，其中錯(cuò)誤的是：（）A．小勇認(rèn)為只有當(dāng)時(shí)，函數(shù)值為2B．小冠認(rèn)為找不到實(shí)數(shù)x，使函數(shù)值為0C．小明認(rèn)為拋物線開(kāi)口向上D．小天認(rèn)為拋物線與x軸有兩個(gè)交點(diǎn)8．在平面直角坐標(biāo)系中，拋物線與x軸交于，兩點(diǎn)，其中．將此拋物線向下平移，與x軸交于，兩點(diǎn)，其中，下面結(jié)論正確的是：（）A．當(dāng)時(shí)，，B．當(dāng)時(shí)，，C．當(dāng)時(shí)，，D．當(dāng)時(shí)，，9．如圖，正方形對(duì)稱(chēng)中心在原點(diǎn)O，四個(gè)頂點(diǎn)分別位于兩個(gè)反比例函數(shù)和的圖象的四個(gè)分支上，則實(shí)數(shù)k的值為：（）A． B． C． D．410．已知二次函數(shù)的圖象如圖所示，則二次函數(shù)與正比例函數(shù)的圖象大致為：（）A． B．C． D．二、填空題（本大題共4小題，每小題5分，滿分20分）11．拋物線與y軸的交點(diǎn)坐標(biāo)為________．12．一名學(xué)生推鉛球，鉛球行進(jìn)高度y（單位：m）與水平距離x（單位：m）之間的關(guān)系是，則該學(xué)生推鉛球的水平距離為________m．13．如圖，在平面直角坐標(biāo)系中，點(diǎn)O為坐標(biāo)原點(diǎn)，矩形（）交反比例函數(shù)（）的圖象于點(diǎn)D，E．點(diǎn)D的坐標(biāo)為．連接，，．若，，則k的值為________．14．定義：平面直角坐標(biāo)系中，點(diǎn)，點(diǎn)，若，，其中k為常數(shù)，且，則稱(chēng)點(diǎn)Q是點(diǎn)P的“k級(jí)變換點(diǎn)”．例如，點(diǎn)是點(diǎn)的“級(jí)變換點(diǎn)”．（1）若函數(shù)的圖象上存在點(diǎn)的“k級(jí)變換點(diǎn)”，則k的值為_________；（2）若關(guān)于x的二次函數(shù)（）的圖象上恰有兩個(gè)點(diǎn)，這兩個(gè)點(diǎn)的“1級(jí)變換點(diǎn)”都在直線上，則n的取值范圍是_________．三、（本大題共2小題，每小題8分，滿分16分）15．已知二次函數(shù)的圖象經(jīng)過(guò)點(diǎn)，求這個(gè)二次函數(shù)的表達(dá)式．16．已知拋物線．求證：無(wú)論m為何值時(shí)，拋物線與x軸總有兩個(gè)交點(diǎn)．四、（本大題共2小題，每小題8分，滿分16分）17．如圖在平面直角坐標(biāo)系中，直線分別交x軸、y軸于點(diǎn)A，B，形狀相同的拋物線：（，……）的頂點(diǎn)在直線上，其對(duì)稱(chēng)軸與x軸交點(diǎn)的橫坐標(biāo)依次是2，3，5，8，13，…，根據(jù)上述規(guī)律解決以下問(wèn)題：（1）拋物線的頂點(diǎn)坐標(biāo)是________；（2）求拋物線線中b，c的值．18．已知一個(gè)二次函數(shù)圖象上部分點(diǎn)的橫坐標(biāo)x與縱坐標(biāo)y的對(duì)應(yīng)值如表所示：x…01…y…00…（1）求這個(gè)二次函數(shù)的解析式；（2）在給定的平面直角坐標(biāo)系中畫(huà)出這個(gè)二次函數(shù)的圖象；（3）當(dāng)時(shí)，y的取值范圍為_______．五、（本大題共2小題，每小題10分，滿分20分）19．跨學(xué)科整合學(xué)習(xí)中為研究某種化學(xué)試劑的揮發(fā)情況，某研究小組在兩種不同的場(chǎng)景下做對(duì)比實(shí)驗(yàn)，收集了該試劑揮發(fā)過(guò)程中剩余質(zhì)量y（克）隨時(shí)間x（分鐘）變化的數(shù)據(jù)（），并分別繪制在平面直角坐標(biāo)系中，如圖所示．（1）從（），（），中，選擇適當(dāng)?shù)暮瘮?shù)模型分別模擬兩種場(chǎng)景下y隨x變化的函數(shù)關(guān)系，并求出相應(yīng)的函數(shù)表達(dá)式；（2）查閱文獻(xiàn)可知，該化學(xué)試劑發(fā)揮作用的最低質(zhì)量為3克，在上述實(shí)驗(yàn)中，該化學(xué)試劑在哪種場(chǎng)景下發(fā)揮作用的時(shí)間更長(zhǎng)？20．探究某種商品的售價(jià)、銷(xiāo)量、利潤(rùn)之間的變化關(guān)系，小明整理出該商品的相關(guān)數(shù)據(jù)如下表所示．時(shí)間x（天）售價(jià)（元/件）70每天銷(xiāo)量（件）已知該商品的進(jìn)價(jià)為每件10元，設(shè)銷(xiāo)售該商品的每天利潤(rùn)為y元．（1）求y與x的函數(shù)關(guān)系式；（2）銷(xiāo)售該商品第幾天時(shí)，當(dāng)天銷(xiāo)售利潤(rùn)最大，最大利潤(rùn)是多少？六、（本大題滿分12分）21．如圖，在平面直角坐標(biāo)系中，直線與y軸交于點(diǎn)，與x軸交于點(diǎn)，與反比例函數(shù)在第三象限內(nèi)的圖象交于點(diǎn)．（1）求反比例函數(shù)的表達(dá)式；（2）當(dāng)時(shí)，求x的取值范圍；（3）當(dāng)點(diǎn)P在y軸上，的面積為6時(shí)，直接寫(xiě)出點(diǎn)P的坐標(biāo)．七、（本大題滿分12分）22．鷹眼技術(shù)助力杭州亞運(yùn)，提升球迷觀賽體驗(yàn)如圖分別為足球比賽中某一時(shí)刻的鷹眼系統(tǒng)預(yù)測(cè)畫(huà)面（如圖1）和截面示意圖（如圖2），攻球員位于點(diǎn)O，守門(mén)員位于點(diǎn)A，的延長(zhǎng)線與球門(mén)線交于點(diǎn)B，且點(diǎn)A，B均在足球軌跡正下方，足球的飛行軌跡可看成拋物線．已知，，足球飛行的水平速度為，水平距離s（）與離地高度h的鷹眼數(shù)據(jù)如表：…912151821……4.24.854.84.2… 圖1 圖2（1）根據(jù)表中數(shù)據(jù)預(yù)測(cè)足球落地時(shí)，________m；（2）求h關(guān)于s的函數(shù)解析式；（3）守門(mén)員在攻球員射門(mén)瞬間就作出防守反應(yīng)，當(dāng)守門(mén)員位于足球正下方時(shí)，足球離地高度不大于守門(mén)員的最大防守高度視為防守成功一次防守中守門(mén)員面對(duì)足球后退，已知后退過(guò)程中守門(mén)員速度為，最大防守高度為2.6m．①求守門(mén)員后退到足球正下方所需時(shí)間；②這次守門(mén)員能否防守成功？試通過(guò)計(jì)算說(shuō)明．八、（本大題滿分14分）23．已知拋物線（a，c為常數(shù)，）過(guò)點(diǎn)，頂點(diǎn)為點(diǎn)P．（1）當(dāng)時(shí)，求此拋物線頂點(diǎn)P的坐標(biāo)；（2）當(dāng)時(shí)，若的面積為4，求此拋物線的解析式；（3）將拋物線向左平移2個(gè)單位，向下平移個(gè)單位（），得到新拋物線的頂點(diǎn)為A，與y軸交點(diǎn)為B，點(diǎn)M在直線上，點(diǎn)N在直線上，當(dāng)四邊形的周長(zhǎng)最小時(shí)，恰好有，求a的值．合肥市琥珀中學(xué)教育集團(tuán)2024屆九年級(jí)第一次質(zhì)量調(diào)研檢測(cè)數(shù)學(xué)參考答案一、選擇題（本大題共10小題，每小題4分，滿分40分）題號(hào)12345678910答案BBACDDDCAB二、填空題（本大題共4小題，每小題5分，滿分20分）11．； 12．10；13． 14．，且．（第一空2分；第二空3分，對(duì)一個(gè)答案1分，兩個(gè)全對(duì)3分，有錯(cuò)不得分）三、（本大題共2小題，每小題8分，滿分16分）15．解：把代入二次函數(shù)得，··························································（3分）解得，···········································································（6分）所以拋物線解析式為．即．····························································（8分）16．證明：，，．···································································（1分）·················································································（3分）．·············································································（5分）∵．∴，·············································································（7分）∴無(wú)論m為何值時(shí)，拋物線與x軸總有兩個(gè)交點(diǎn)．············································（8分）四、（本大題共2小題，每小題8分，滿分16分）17．解：（1）·····································································（3分）（2），當(dāng)時(shí)，拋物線的頂點(diǎn)坐標(biāo)是，····················································（5分）由頂點(diǎn)式得：，展開(kāi)得．········································································（7分）∴，．···········································································（8分）18．解：（1）由題意可得二次函數(shù)的頂點(diǎn)坐標(biāo)為，·········································（1分）設(shè)二次函數(shù)的解析式為：，····························································（2分）把點(diǎn)代入，得，····································································（3分）故拋物線解析式為，即·······························································（4分）（2）如圖所示：·················································································（6分）（3）．··········································································（8分）五、（本大題共2小題，每小題10分，滿分20分）19．解：（1）觀察兩種場(chǎng)景可知，場(chǎng)景A為，場(chǎng)景B為（），·····································································（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）場(chǎng)景A的函數(shù)表達(dá)式為，場(chǎng)景B的函數(shù)表達(dá)式為；（2）當(dāng)時(shí)，場(chǎng)景A中，，·······································································（8分）場(chǎng)景B中，，解得，··································································（9分），化學(xué)試劑在場(chǎng)景A下發(fā)揮作用的時(shí)間更長(zhǎng)．··············································（10分）20．解：（1）當(dāng)時(shí)，，······························································（2分）當(dāng)時(shí)，，··········································································（4分）綜上所述：y與x的函數(shù)關(guān)系式為；···············································································（5分）（2）當(dāng)時(shí)，二次函數(shù)，∵，∴當(dāng)時(shí)，，········································································（7分）當(dāng)時(shí)，中y隨x的增大而減小，∴當(dāng)時(shí)，，·······················································（9分）綜上所述，該商品第10天時(shí)，當(dāng)天銷(xiāo)售利潤(rùn)最大，最大利潤(rùn)是3200元．···························（10分）六、（本大題滿分12分）21．解：（1）將，代入得：，解得：，··········································································（1分）∴一次函數(shù)表達(dá)式為：，·····························································（2分）將代入得：，······································································（3分）∴，·············································································（4分）將代入得：，······································································（5分）∴反比例函數(shù)的表達(dá)式為：；··························································（6分）（2）設(shè)一次函數(shù)與反比例函數(shù)在第二象限交于點(diǎn)D，聯(lián)立，解得：或，···································································（7分）∴，·············································································（8分）由圖象可知：當(dāng)時(shí)，或；·····························································（10分）（3）點(diǎn)P的坐標(biāo)為：或．·····························································（12分）22．略23．略

相關(guān)試卷

相關(guān)試卷更多

相關(guān)試卷

相關(guān)試卷 更多

相關(guān)試卷更多