第=page11頁(yè)，共=sectionpages22頁(yè)瀘縣四中學(xué)區(qū)2025年秋初中九年級(jí)數(shù)學(xué)學(xué)科半期定時(shí)訓(xùn)練題參考答案一．選擇題題號(hào)12345678910答案AACBBBCADD題號(hào)1112答案CA二．填空題13．414．15．202716．①②③三．解答題17．解：∵，∴，∴，·························································2分∴，·······························································4分解得：．······························································5分18．解：········································································2分············································································4分或··········································································5分，．·············································································6分19．解：∵拋物線(xiàn)經(jīng)過(guò)點(diǎn)，∴，············································································3分解得；·················································································5分拋物線(xiàn)的關(guān)系式為，······························································6分20．（1）解：設(shè)二次函數(shù)表達(dá)式為∵圖像經(jīng)過(guò)（-2，5）∴5=··················································································1分∴·····················································································2分············································································3分（2）解：令y=0，即=0····························································4分解得：x=3或x=-1··········································································6分故此拋物線(xiàn)與x軸的交點(diǎn)坐標(biāo)為（3，0），（-1，0）··············································7分21．（1）解：如圖，即為所求··································································3分（2）如圖，即為所求···································································5分（3）由（1）圖可知，，由（2）圖可知，······································7分22．（1）證明：∵關(guān)于的一元二次方程中，，····················································3分∴無(wú)論為何值，方程總有兩個(gè)不相等的實(shí)數(shù)根；··············································4分（2）解：由題意知，，················································5分∵，∴，即，··························································7分解得，∴的值是1．···············································································8分23．（1）解：將等腰直角三角形繞點(diǎn)A旋轉(zhuǎn)得到，∴．··································································1分∵是等腰直角三角形，···································································2分∴，····································································3分∴，·∴；·································································4分（2）解：∵是等腰直角三角形，∴，··································································5分根據(jù)勾股定理，得，·····················································6分由（1）得，，········································7分根據(jù)勾股定理，得．·····················································8分24．（1）解：設(shè)每件襯衫的售價(jià)上漲x元，由題意得：且（即），·········································3分解得：（舍棄），·······························································5分∴，答：每件襯衫的售價(jià)90元···································································6分（2）解：每件襯衫的售價(jià)上漲x元，月利潤(rùn)是y元，則，······························8分∴，開(kāi)口向下，，∴當(dāng)時(shí)，y有最大值，最大值為；·················································10分此時(shí)每件襯衫的售價(jià)為（元），·····················································11分答：每件襯衫的售價(jià)定為100元時(shí)，每個(gè)月可獲得最大利潤(rùn)，最大月利潤(rùn)7200元．··················12分25．（1）解：設(shè)拋物線(xiàn)的解析式為，則，解得，·····························································3分∴拋物線(xiàn)的解析式為；························································4分（2）解：∵拋物線(xiàn)的解析式為，∴拋物線(xiàn)的對(duì)稱(chēng)軸為直線(xiàn)，···············································5分作點(diǎn)C關(guān)于拋物線(xiàn)對(duì)稱(chēng)軸的對(duì)稱(chēng)點(diǎn)，連接，交拋物線(xiàn)對(duì)稱(chēng)軸于點(diǎn)Q，則，，此時(shí)取最小值，··············································6分設(shè)直線(xiàn)的解析式為，則，解得，∴直線(xiàn)的解析式為，·····························································7分將代入，則，∴；················································································8分（3）解：∵P點(diǎn)的橫坐標(biāo)為m，∴，過(guò)點(diǎn)P作軸的垂線(xiàn)，交于點(diǎn)，設(shè)直線(xiàn)的解析式為，則，解得，∴直線(xiàn)的解析式為，··············