新高考数学一轮复习精讲精练8.2 解析式（基础版）（原卷版）

8.2解析式（精讲）（基础版）思维导图思维导图考点呈现考点呈现例题剖析例题剖析考点一待定系数法求解析式【例1】（2022·全国·高三专题练习）（多选）已知函数SKIPIF1<0是一次函数，满足SKIPIF1<0，则SKIPIF1<0的解析式可能为（

）A．SKIPIF1<0 B．SKIPIF1<0C．SKIPIF1<0 D．SKIPIF1<0【一隅三反】1．（2022·全国·高三专题练习）已知f(x)是一次函数，且满足3f(x＋1)－2f(x－1)＝2x＋17，则f(1)＝____．2．（2022·全国·高三专题练习）已知SKIPIF1<0，且SKIPIF1<0为一次函数，求SKIPIF1<0_________3（2022·全国·高三专题练习）已知SKIPIF1<0是一次函数，且满足SKIPIF1<0，求SKIPIF1<0_____．考点二换元法求解析式【例2】（2022·全国·高三专题练习）若SKIPIF1<0，则SKIPIF1<0的解析式为（

）A．SKIPIF1<0 B．SKIPIF1<0C．SKIPIF1<0 D．SKIPIF1<0【一隅三反】1．（2022·全国·高三专题练习）若函数SKIPIF1<0满足SKIPIF1<0，则SKIPIF1<0的解析式是（

）A．SKIPIF1<0 B．SKIPIF1<0C．SKIPIF1<0 D．SKIPIF1<0或SKIPIF1<0【一隅三反】1．（2022·全国·高三专题练习）已知函数SKIPIF1<0，则SKIPIF1<0的解析式为_______2．（2022·全国·高三专题练习）若函数SKIPIF1<0满足SKIPIF1<0，则SKIPIF1<0__．3．（2022·全国·高三专题练习）已知SKIPIF1<0，则SKIPIF1<0的解析式为______________．4．（2022·全国·高三专题练习）已知函数SKIPIF1<0在定义域SKIPIF1<0上单调，且SKIPIF1<0时均有SKIPIF1<0，则SKIPIF1<0的值为（

）A．3 B．1 C．0 D．SKIPIF1<0考点三解方程组求解析式【例3】（2022·全国·高三专题练习）若函数SKIPIF1<0满足SKIPIF1<0，则SKIPIF1<0（

）A．0 B．2 C．3 D．SKIPIF1<0【一隅三反】1．（2022·浙江·高三专题练习）已知函数f(x)满足SKIPIF1<0，则f(x)的解析式为（

）A．SKIPIF1<0 B．SKIPIF1<0C．SKIPIF1<0 D．SKIPIF1<02．（2022·全国·高三专题练习）已知定义域为R的函数SKIPIF1<0满足SKIPIF1<0，则SKIPIF1<0___________.3（2022·全国·高三专题练习）若函数SKIPIF1<0，SKIPIF1<0满足SKIPIF1<0，且SKIPIF1<0，则SKIPIF1<0________．4．（2022·全国·高三专题练习）已知SKIPIF1<0，则函数f(x)的解析式为___________.考点四配凑法【例4】（2022·全国·高三专题练习）已知函数f（x﹣1）＝x2+2x﹣3，则f（x）＝（）A．x2+4x B．x2+4 C．x2+4x﹣6 D．x2﹣4x﹣1【一隅三反】1．（2022·浙江·高三专题练习）已知SKIPIF1<0，则SKIPIF1<0_______．2．（2022·全国·高三专题练习）已知SKIPIF1<0，则SKIPIF1<0的值等于___.3．（2022·全国·高三专题练习）已知f(x－SKIPIF1<0)＝x2＋SKIPIF1<0，则f（x+SKIPIF1<0）＝________.8.2解析式（精练）（基础版）题组一题组一待定系数法求解析式1．（2022·全国·高三专题练习）若SKIPIF1<0是SKIPIF1<0上单调递减的一次函数，若SKIPIF1<0，则SKIPIF1<0__．2．（2022·全国·高一课时练习）已知SKIPIF1<0是一次函数，SKIPIF1<0，SKIPIF1<0，则SKIPIF1<0（

）A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<03.．（2022·江苏·）（1）已知SKIPIF1<0是一次函数，且SKIPIF1<0，求SKIPIF1<0；（2）已知SKIPIF1<0是二次函数，且满足SKIPIF1<0，求SKIPIF1<0.4．（2022·云南）（1）已知f（x）是一次函数，且满足f（x＋1）－2f（x－1）＝2x＋3，求f（x）的解析式．（2）若二次函数g（x）满足g（1）＝1，g（－1）＝5，且图象过原点，求g（x）的解析式．题组二题组二换元法求解析式1．（2022·全国·高三专题练习）已知SKIPIF1<0，且SKIPIF1<0，则SKIPIF1<0的值为_________．2．（2022·全国·高一专题练习）已知SKIPIF1<0，则有（

）A．SKIPIF1<0 B．SKIPIF1<0C．SKIPIF1<0 D．SKIPIF1<03．（2022·全国·课时练习）已知SKIPIF1<0，则SKIPIF1<0（

）．A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<04．（2023·全国·高三专题练习）已知SKIPIF1<0，则SKIPIF1<0（

）A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<05．（2022·河南·临颍县第一高级中学高二阶段练习（文））已知SKIPIF1<0，则SKIPIF1<0（

）．A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<06．（2022·山西运城·高二阶段练习）已知函数SKIPIF1<0满足SKIPIF1<0，则SKIPIF1<0（

）A．1 B．9 C．SKIPIF1<0 D．SKIPIF1<07．（2023·全国·高三专题练习）设SKIPIF1<0，SKIPIF1<0，则SKIPIF1<0（

）A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<08．（2022·江苏）设函数SKIPIF1<0，则SKIPIF1<0的表达式为（

）A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<09．（2022·甘肃·甘南藏族自治州合作第一中学）已知f（SKIPIF1<0x-1）=2x-5，且f（a）=6，则a等于（）A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<010．（2022·陕西·略阳县天津高级中学二模（理））若SKIPIF1<0，则SKIPIF1<0等于（

）A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<011（2022·全国·高三专题练习）已知函数SKIPIF1<0，求SKIPIF1<0的解析式．12．（2022·全国·课时练习）（多选）若函数SKIPIF1<0，则（

）A．SKIPIF1<0 B．SKIPIF1<0C．SKIPIF1<0 D．SKIPIF1<013（2022·黑龙江）若函数SKIPIF1<0，则SKIPIF1<0__________.题组三题组三解方程组求解析式1（2022·广东）已知函数f(x)满足f(x)＋2f(3－x)＝x2，则f(x)的解析式为（

）A．f(x)＝x2－12x＋18B．f(x)＝SKIPIF1<0－4x＋6C．f(x)＝6x＋9D．f(x)＝2x＋32．（2021·陕西安康）已知函数SKIPIF1<0满足SKIPIF1<0，则SKIPIF1<0（

）A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<03．（2022·广西）若函数SKIPIF1<0满足SKIPIF1<0，则SKIPIF1<0（

）A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<04．（2021·全国·课时练习）已知SKIPIF1<0，求SKIPIF1<0的解析式.5（2022广西）若对任意实数SKIPIF1<0，均有SKIPIF1<0，求SKIPIF1<0=6．（2022·全国·高三专题练习）已知函数f