新高考數(shù)學(xué)一輪復(fù)習(xí) 講與練第6講 函數(shù)的圖像（解析版）

第6講函數(shù)的圖像學(xué)校____________姓名____________班級____________一、知識梳理1.利用描點法作函數(shù)的圖像步驟：(1)確定函數(shù)的定義域；(2)化簡函數(shù)解析式；(3)討論函數(shù)的性質(zhì)(奇偶性、單調(diào)性、周期性、對稱性等)；(4)列表(尤其注意特殊點、零點、最大值點、最小值點、與坐標(biāo)軸的交點等)，描點，連線.2.利用圖像變換法作函數(shù)的圖像(1)平移變換(2)對稱變換y＝f(x)的圖像eq\o(→,\s\up7(關(guān)于x軸對稱))y＝－f(x)的圖像；y＝f(x)的圖像eq\o(→,\s\up7(關(guān)于y軸對稱))y＝f(－x)的圖像；y＝f(x)的圖像eq\o(→,\s\up7(關(guān)于原點對稱))y＝－f(－x)的圖像；y＝ax(a>0，且a≠1)的圖像eq\o(→,\s\up17(關(guān)于直線),\s\do15(y＝x對稱))y＝logax(a>0，且a≠1)的圖像.(3)伸縮變換y＝f(x)eq\o(→,\s\up17(縱坐標(biāo)不變),\s\do15(各點橫坐標(biāo)變?yōu)樵瓉淼腬f(1,a)（a>0）倍))y＝f(ax).y＝f(x)eq\o(→,\s\up17(橫坐標(biāo)不變),\s\do15(各點縱坐標(biāo)變?yōu)樵瓉淼腁(A>0)倍))y＝Af(x).(4)翻折變換y＝f(x)的圖像eq\o(→,\s\up17(x軸下方部分翻折到上方),\s\do15(x軸及上方部分不變))y＝|f(x)|的圖像；y＝f(x)的圖像eq\o(→,\s\up17(y軸右側(cè)部分翻折到左側(cè)),\s\do15(原y軸左側(cè)部分去掉，右側(cè)不變))y＝f(|x|)的圖像.考點和典型例題1、函數(shù)的圖像【典例1-1】（2021·全國·高三專題練習(xí)）函數(shù)SKIPIF1<0的單調(diào)遞增區(qū)間是（

）A．SKIPIF1<0 B．SKIPIF1<0和SKIPIF1<0C．SKIPIF1<0和SKIPIF1<0 D．SKIPIF1<0和SKIPIF1<0【答案】B【詳解】SKIPIF1<0如圖所示：函數(shù)的單調(diào)遞增區(qū)間是SKIPIF1<0和SKIPIF1<0．故選：B.【典例1-2】（2022·天津·漢沽一中高三階段練習(xí)）已知函數(shù)SKIPIF1<0若SKIPIF1<0(SKIPIF1<0互不相等)，則SKIPIF1<0的取值范圍是（

）A．SKIPIF1<0 B．SKIPIF1<0C．SKIPIF1<0 D．SKIPIF1<0【答案】D【詳解】作出函數(shù)SKIPIF1<0的圖象，如圖所示:設(shè)SKIPIF1<0，則SKIPIF1<0.因為SKIPIF1<0，所以SKIPIF1<0，所以SKIPIF1<0，所以SKIPIF1<0，即SKIPIF1<0.當(dāng)SKIPIF1<0時，解得SKIPIF1<0或SKIPIF1<0，所以SKIPIF1<0.設(shè)SKIPIF1<0，因為函數(shù)SKIPIF1<0在SKIPIF1<0上單調(diào)遞增，所以SKIPIF1<0，即SKIPIF1<0，所以SKIPIF1<0.故選：D.【典例1-3】（2021·全國·高三專題練習(xí)）如圖，太極圖是由黑白兩個魚形紋組成的圖案，俗稱陰陽魚，太極圖展現(xiàn)了一種相互轉(zhuǎn)化，相對統(tǒng)一的和諧美，定義：能夠?qū)AO的周長和面積同時等分成兩個部分的函數(shù)稱為圓O的一個“太極函數(shù)”，則（）A．函數(shù)SKIPIF1<0是圓O：SKIPIF1<0的一個太極函數(shù)B．函數(shù)SKIPIF1<0不是圓O：SKIPIF1<0的太極函數(shù)C．函數(shù)SKIPIF1<0不是圓O：SKIPIF1<0的太極函數(shù)D．函數(shù)SKIPIF1<0不是圓O：SKIPIF1<0的太極函數(shù)【答案】A【詳解】解：兩曲線的對稱中心均為點SKIPIF1<0，且兩曲線交于兩點，所以SKIPIF1<0能把圓SKIPIF1<0一分為二，如圖：故A正確；同理易知B，C不正確；函數(shù)SKIPIF1<0為奇函數(shù)，且SKIPIF1<0，SKIPIF1<0，如圖：所以函數(shù)SKIPIF1<0是圓O：SKIPIF1<0的一個太極函數(shù)，故D不正確，故選：A．【典例1-4】（2022·浙江紹興·模擬預(yù)測）已知函數(shù)的圖象如下圖1，則如下圖2對應(yīng)的函數(shù)有可能是（

）A．SKIPIF1<0 B．SKIPIF1<0C．SKIPIF1<0 D．SKIPIF1<0【答案】A【詳解】圖1：當(dāng)SKIPIF1<0時，SKIPIF1<0，當(dāng)SKIPIF1<0時，SKIPIF1<0當(dāng)SKIPIF1<0時SKIPIF1<0，于圖2不符合，故排除C、D．∵SKIPIF1<0恒成立，于圖2不符合，故排除B．故選：A．【典例1-5】（2022·安徽淮南·二模（文））函數(shù)SKIPIF1<0的部分圖象可能是（

）A． B． C． D．【答案】B【詳解】記SKIPIF1<0,則SKIPIF1<0,故SKIPIF1<0,SKIPIF1<0是奇函數(shù),故圖像關(guān)于原點對稱.此時可排除A,C,取SKIPIF1<0,排除D.故選:B2、圖像的平移和變換【典例2-1】（2022·四川綿陽·三模（理））已知函數(shù)SKIPIF1<0，則（

）A．SKIPIF1<0在SKIPIF1<0上單調(diào)遞增 B．SKIPIF1<0的圖象關(guān)于點SKIPIF1<0對稱C．SKIPIF1<0為奇函數(shù) D．SKIPIF1<0的圖象關(guān)于直線SKIPIF1<0對稱【答案】D【詳解】化簡得SKIPIF1<0，SKIPIF1<0的可以看作是函數(shù)SKIPIF1<0先向左平移一個單位，再向下平移一個單位得到，先畫出SKIPIF1<0的圖象，再進(jìn)行平移畫出SKIPIF1<0的圖象，明顯可見，對于原函數(shù)SKIPIF1<0，為奇函數(shù)，關(guān)于點SKIPIF1<0對稱，且在SKIPIF1<0和SKIPIF1<0上為單調(diào)減函數(shù)，所以，SKIPIF1<0經(jīng)過平移后變成的SKIPIF1<0在SKIPIF1<0上單調(diào)遞減，關(guān)于SKIPIF1<0對稱，非奇函數(shù)也非偶函數(shù)，圖象關(guān)于直線SKIPIF1<0對稱，所以，D正確；A、B、C錯誤.故選：D【典例2-2】（2022·浙江紹興·模擬預(yù)測）在同一直角坐標(biāo)系中，函數(shù)SKIPIF1<0，SKIPIF1<0，且SKIPIF1<0的圖象可能是（

）A． B．C． D．【答案】C【詳解】解：因為函數(shù)SKIPIF1<0的圖象與函數(shù)SKIPIF1<0的圖象關(guān)于SKIPIF1<0軸對稱，所以函數(shù)SKIPIF1<0的圖象恒過定點SKIPIF1<0，故選項A、B錯誤；當(dāng)SKIPIF1<0時，函數(shù)SKIPIF1<0在SKIPIF1<0上單調(diào)遞增，所以函數(shù)SKIPIF1<0在SKIPIF1<0上單調(diào)遞減，又SKIPIF1<0在SKIPIF1<0和SKIPIF1<0上單調(diào)遞減，故選項D錯誤，選項C正確.故選：C.【典例2-3】（2022·全國·高三專題練習(xí)）將曲線SKIPIF1<0上所有點的橫坐標(biāo)不變，縱坐標(biāo)縮小為原來的SKIPIF1<0，得到曲線SKIPIF1<0，則SKIPIF1<0上到直線SKIPIF1<0距離最短的點坐標(biāo)為（

）A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<0【答案】B【詳解】將SKIPIF1<0化為SKIPIF1<0，則將曲線SKIPIF1<0上所有點的橫坐標(biāo)不變，縱坐標(biāo)縮小為原來的SKIPIF1<0，得到曲線SKIPIF1<0，即SKIPIF1<0，要使曲線SKIPIF1<0上的點到直線SKIPIF1<0的距離最短，只需曲線SKIPIF1<0上在該點處的切線和直線SKIPIF1<0平行，設(shè)曲線SKIPIF1<0上該點為SKIPIF1<0，因為SKIPIF1<0，且SKIPIF1<0的斜率為SKIPIF1<0，所以SKIPIF1<0，解得SKIPIF1<0或SKIPIF1<0（舍），即該點坐標(biāo)為SKIPIF1<0.故選：B.【典例2-4】（2021·北京四中高三期中）為了得到函數(shù)SKIPIF1<0的圖像，只需把函數(shù)SKIPIF1<0的圖像（

）A．向左平移1個單位長度 B．向右平移1個單位長度C．向左平移SKIPIF1<0個單位長度 D．向右平移SKIPIF1<0個單位長度【答案】C【詳解】要得到函數(shù)SKIPIF1<0的圖象，則只需要把函數(shù)SKIPIF1<0的圖象向左平移SKIPIF1<0個單位長度，即可.故選：C．【典例2-5】（2021·甘肅·靜寧縣第一中學(xué)高三階段練習(xí)（文））已知函數(shù)SKIPIF1<0，則下列圖象錯誤的是（

）A． B．C． D．【答案】D【詳解】當(dāng)SKIPIF1<0時，SKIPIF1<0，表示一條線段，且線段經(jīng)過SKIPIF1<0和SKIPIF1<0兩點．當(dāng)SKIPIF1<0時，SKIPIF1<0，表示一段曲線．函數(shù)SKIPIF1<0的圖象如圖所示．SKIPIF1<0的圖象可由SKIPIF1<0的圖象向右平移一個單位長度得到，故A正確；SKIPIF1<0的圖象可由SKIPIF1<0的圖象關(guān)于SKIPIF1<0軸對稱后得到，故B正確；由于SKIPIF1<0的值域為SKIPIF1<0，故SKIPIF1<0，故SKIPIF1<0的圖象與SKIPIF1<0的圖象完全相同，故C正確；很明顯D中SKIPIF1<0的圖象不正確．故選：D．3、圖像的綜合應(yīng)用【典例3-1】（2022·福建寧德·模擬預(yù)測）函數(shù)SKIPIF1<0的圖象如圖所示，則f（x）的解析式可能是（

）A．SKIPIF1<0 B．SKIPIF1<0C．SKIPIF1<0 D．SKIPIF1<0【答案】B【詳解】A函數(shù)為遞減的，錯誤；C函數(shù)的值域大于等于0，錯誤；D函數(shù)為二次函數(shù)，錯誤，只有B符合.故選：B.【典例3-2】（2022·天津南開·一模）函數(shù)SKIPIF1<0的圖象可能是（

）A． B．C． D．【答案】C【詳解】由題意，函數(shù)SKIPIF1<0，因為SKIPIF1<0，即函數(shù)SKIPIF1<0的圖象過點SKIPIF1<0，可排除A、B項；又因為SKIPIF1<0，可排除D項，故選：C.【典例3-3】（2022·浙江嘉興·二模）已知函數(shù)SKIPIF1<0的圖象如圖所示，則SKIPIF1<0的解析式可能是（

）（SKIPIF1<0是自然對數(shù)的底數(shù)）A．SKIPIF1<0 B．SKIPIF1<0C．SKIPIF1<0 D．SKIPIF1<0【答案】B【詳解】解：對于A，函數(shù)SKIPIF1<0的定義域為SKIPIF1<0，由SKIPIF1<0，所以函數(shù)SKIPIF1<0為奇函數(shù)，不符合題意；對于B，函數(shù)SKIPIF1<0的定義域為SKIPIF1<0，由SKIPIF1<0，所以函數(shù)SKIPIF1<0為偶函數(shù)，符合題意；對于C，函數(shù)SKIPIF1<0，則SKIPIF1<0，得SKIPIF1<0且SKIPIF1<0，故函數(shù)SKIPIF1<0的定義域為SKIPIF1<0且SKIPIF1<0，結(jié)合函數(shù)圖像可知，不符題意；對于D，函數(shù)SKIPIF1<0的定義域為SKIPIF1<0且SKIPIF1<0，結(jié)合函數(shù)圖像可知，不符題意.故選：B.【典例3-4】（2022·安徽·安慶一中模擬預(yù)測（文））已知函數(shù)SKIPIF1<0在SKIPIF1<0上的圖象如圖所示，則函數(shù)SKIPIF1<0的解析式可能為（

）A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<0【答案】D【詳解】當(dāng)SKIPIF1<0時，SKIPIF1<0，則SKIPIF1<0，故排除AB.當(dāng)SKIPIF1<0時，則SKIPIF1<0，令SKIPIF1<0，得SKIPIF1<0或SKIPIF1<0，當(dāng)SKIPIF1<0或SKIPIF1<0時，SKIPIF1<0，當(dāng)SKIPIF1<0時，SKIPIF1<0，所以SKIPIF1<0是函數(shù)的極小值點，SKIPIF1<0是函數(shù)的極大值點，故C錯誤；當(dāng)SKIPIF1<0時，則SKIPIF1<0，令SKIPIF1<0，得SKIPIF1<0或SKIPIF1<0，當(dāng)SKIPIF1<0或SKIPIF1<0時，SKIPIF1<0，當(dāng)SKIPIF1<0時，SKIPIF1<0，所以SKIPIF1<0是函數(shù)的極大值點，SKIPIF1<0是函數(shù)的極小值點，故D正確故選：D.【典例3-5】（2022·江西上饒·二模（理））函數(shù)SKIPIF1<0的大致圖像為（

）A． B．C． D．【答案】B【詳解】當(dāng)SKIPIF1<0，SKIPIF1<0，函數(shù)為奇函數(shù)，排除C；SKIPIF1<0，排除AD；故選：B.【典例3-6】（2022·安徽師范大學(xué)附屬中學(xué)模擬預(yù)測（理））雙曲函數(shù)在實際生活中有著非常重要的應(yīng)用，比如懸鏈橋.在數(shù)學(xué)中，雙曲函數(shù)是一類與三角函數(shù)類似的函數(shù)，最基礎(chǔ)的是雙曲正弦函數(shù)SKIPIF1<0，和雙曲余弦函數(shù)SKIPIF1<0.下列結(jié)論錯誤的是（

）A．雙曲正弦函數(shù)圖象關(guān)于原點中心對稱，雙曲余弦函數(shù)圖象關(guān)于y軸對稱B．若直線SKIPIF1<0與雙曲余弦函數(shù)圖象SKIPIF1<0和雙曲正弦函數(shù)圖象SKIPIF1<0共有三個交點，則SKIPIF1<0C