9.1切線方程（精講）（基礎(chǔ)版）思維導(dǎo)圖思維導(dǎo)圖考點(diǎn)呈現(xiàn)考點(diǎn)呈現(xiàn)例題剖析例題剖析考點(diǎn)一導(dǎo)數(shù)幾何意義【例1-1】（2022·日照模擬）曲線SKIPIF1<0在SKIPIF1<0處的切線的傾斜角為SKIPIF1<0，則SKIPIF1<0的值為（）A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<0【答案】B【解析】根據(jù)已知條件，SKIPIF1<0，因?yàn)榍€SKIPIF1<0在SKIPIF1<0處的切線的傾斜角為SKIPIF1<0，所以SKIPIF1<0，所以SKIPIF1<0.因?yàn)镾KIPIF1<0，SKIPIF1<0，則解得SKIPIF1<0，SKIPIF1<0，故SKIPIF1<0.故答案為：B.【例1-2】（2022·棗莊模擬）曲線SKIPIF1<0在點(diǎn)SKIPIF1<0處的切線與直線SKIPIF1<0垂直，則SKIPIF1<0的值為（）A．-1 B．0 C．1 D．2【答案】C【解析】設(shè)SKIPIF1<0，則SKIPIF1<0，直線SKIPIF1<0的斜率為SKIPIF1<0，由題意可得SKIPIF1<0，解得SKIPIF1<0.故答案為：C.【例1-3】（2022高三下·安徽期中）已知SKIPIF1<0，則曲線SKIPIF1<0在點(diǎn)SKIPIF1<0處的切線的斜率為（）A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<0【答案】D【解析】對SKIPIF1<0，求導(dǎo)可得，SKIPIF1<0，得到SKIPIF1<0，所以，SKIPIF1<0，所以，SKIPIF1<0，SKIPIF1<0故答案為：D【一隅三反】1．（2023高三上·江漢開學(xué)考）若函數(shù)SKIPIF1<0在點(diǎn)（1，f（1））處的切線的斜率為1，則SKIPIF1<0的最小值為（）A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<0【答案】A【解析】由已知SKIPIF1<0，所以SKIPIF1<0，SKIPIF1<0，當(dāng)且僅當(dāng)SKIPIF1<0時(shí)等號成立．故答案為：A．2．（2022·宜春模擬）已知函數(shù)SKIPIF1<0是定義在R上的奇函數(shù)，且SKIPIF1<0，則函數(shù)SKIPIF1<0的圖象在點(diǎn)SKIPIF1<0處的切線的斜率為（）A．-21 B．-27 C．-24 D．-25【答案】A【解析】SKIPIF1<0是奇函數(shù)，SKIPIF1<0恒成立，所以SKIPIF1<0，SKIPIF1<0，SKIPIF1<0，所以SKIPIF1<0，SKIPIF1<0，即SKIPIF1<0，SKIPIF1<0．故答案為：A．3．（2022·成都模擬）若曲線SKIPIF1<0在點(diǎn)（1，2）處的切線與直線SKIPIF1<0平行，則實(shí)數(shù)a的值為（）A．－4 B．－3 C．4 D．3【答案】B【解析】SKIPIF1<0，所以SKIPIF1<0。故答案為：B考點(diǎn)二在型求切線【例2-1】（2022·貴州模擬）曲線SKIPIF1<0在點(diǎn)SKIPIF1<0處的切線方程為（）A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<0【答案】B【解析】因?yàn)镾KIPIF1<0，所以曲線SKIPIF1<0在點(diǎn)SKIPIF1<0處的切線的斜率為SKIPIF1<0，當(dāng)x=1時(shí)，y=0，切點(diǎn)坐標(biāo)為（1，0）.故所求切線方程為SKIPIF1<0.故答案為：B【例2-2】（2022海南）曲線SKIPIF1<0：SKIPIF1<0在點(diǎn)SKIPIF1<0處的切線方程為___________【答案】SKIPIF1<0【解析】因?yàn)镾KIPIF1<0，SKIPIF1<0SKIPIF1<0，SKIPIF1<0，又SKIPIF1<0，SKIPIF1<0所求的切線方程為SKIPIF1<0，即SKIPIF1<0，故答案為：SKIPIF1<0.

【例2-3】（2022福州模擬）已知函數(shù)SKIPIF1<0為偶函數(shù)，當(dāng)x＜0時(shí)，SKIPIF1<0，則曲線SKIPIF1<0在x＝1處的切線方程為（）A．x-y＝0 B．x-y-2＝0 C．x+y-2＝0 D．3x-y-2＝0【答案】A【解析】當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0，SKIPIF1<0，又函數(shù)SKIPIF1<0為偶函數(shù)，所以SKIPIF1<0，SKIPIF1<0，所以SKIPIF1<0，SKIPIF1<0，故切線方程為SKIPIF1<0，即SKIPIF1<0.故答案為：A．【一隅三反】1．（2022高三上·杭州期末）函數(shù)SKIPIF1<0在點(diǎn)SKIPIF1<0處的切線方程是.【答案】y=1【解析】因?yàn)镾KIPIF1<0，所以切線斜率SKIPIF1<0，所以切線方程為y=1.故答案為：y=12．（2022·廣東模擬）已知SKIPIF1<0為奇函數(shù)，當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0，則曲線在點(diǎn)SKIPIF1<0處的切線方程為.【答案】y=2x-e【解析】由題意SKIPIF1<0時(shí)SKIPIF1<0，是奇函數(shù)，SKIPIF1<0時(shí)SKIPIF1<0，SKIPIF1<0，SKIPIF1<0，由點(diǎn)斜式直線方程得SKIPIF1<0，整理得y=2x-e；故答案為：y=2x-e.3．（2021·海南模擬）已知偶函數(shù)SKIPIF1<0滿足SKIPIF1<0，且在SKIPIF1<0處的導(dǎo)數(shù)SKIPIF1<0，則曲線SKIPIF1<0在SKIPIF1<0處的切線方程為（）A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<0【答案】A【解析】由條件知SKIPIF1<0，所以SKIPIF1<0，從而SKIPIF1<0，即函數(shù)SKIPIF1<0的周期為4.在SKIPIF1<0中，令SKIPIF1<0得SKIPIF1<0，所以SKIPIF1<0，又SKIPIF1<0，所以曲線SKIPIF1<0在SKIPIF1<0處的切線方程為SKIPIF1<0，即SKIPIF1<0.故答案為：A.考點(diǎn)三過型求切線【例3-1】.(2022·豫北)已知f(x)＝x2，則過點(diǎn)P(－1，0)，曲線y＝f(x)的切線方程為【答案】y＝0或4x＋y＋4＝0【解析】易知點(diǎn)P(－1，0)不在f(x)＝x2上，設(shè)切點(diǎn)坐標(biāo)為(x0，xeq\o\al(2,0))，由f(x)＝x2可得f′(x)＝2x，∴切線的斜率k＝f′(x0)＝2x0.∵切線過點(diǎn)P(－1，0)，∴k＝eq\f(xeq\o\al(2,0),x0＋1)＝2x0，解得x0＝0或x0＝－2，∴k＝0或－4，故所求切線方程為y＝0或4x＋y＋4＝0.【例3-2】(2022·江西)過點(diǎn)P(1,1)且與曲線y＝x3相切的直線的條數(shù)為【答案】2【解析】當(dāng)點(diǎn)P為切點(diǎn)時(shí)，∵y′＝3x2，∴y′|x＝1＝3，則曲線y＝x3在點(diǎn)P處的切線方程為y－1＝3(x－1)，即3x－y－2＝0.當(dāng)點(diǎn)P不是切點(diǎn)時(shí)，設(shè)直線與曲線切于點(diǎn)(x0，y0)(x0≠1)，則k＝eq\f(y0－1,x0－1)＝eq\f(x\o\al(3,0)－1,x0－1)＝xeq\o\al(2,0)＋x0＋1.∵y′＝3x2，∴y′|x＝x0＝3xeq\o\al(2,0)，∴2xeq\o\al(2,0)－x0－1＝0，∴x0＝1(舍)或x0＝－eq\f(1,2)，∴過點(diǎn)P(1,1)與曲線y＝x3相切的直線方程為3x－4y＋1＝0.綜上，過點(diǎn)P的切線有2條【一隅三反】1．（2021·永州模擬）曲線SKIPIF1<0在SKIPIF1<0處的切線SKIPIF1<0過原點(diǎn)，則SKIPIF1<0的方程是（）A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<0【答案】A【解析】曲線SKIPIF1<0SKIPIF1<0切點(diǎn)為SKIPIF1<0，所以切線SKIPIF1<0的斜率SKIPIF1<0，又直線SKIPIF1<0過原點(diǎn)，所以SKIPIF1<0，得SKIPIF1<0，SKIPIF1<0．所以SKIPIF1<0，故切線SKIPIF1<0的方程為SKIPIF1<0即SKIPIF1<0．故答案為：A．2．（2021·浙江高三專題練習(xí)）過曲線SKIPIF1<0上一點(diǎn)SKIPIF1<0的切線的斜率為SKIPIF1<0，則點(diǎn)SKIPIF1<0的坐標(biāo)為______．【答案】SKIPIF1<0或SKIPIF1<0.【解析】由SKIPIF1<0可得SKIPIF1<0，設(shè)切點(diǎn)SKIPIF1<0，由題意可得SKIPIF1<0，解得：SKIPIF1<0，即SKIPIF1<0，當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0，此時(shí)點(diǎn)SKIPIF1<0的坐標(biāo)為SKIPIF1<0當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0，此時(shí)點(diǎn)SKIPIF1<0的坐標(biāo)為SKIPIF1<0，故答案為：SKIPIF1<0或SKIPIF1<0.3．（2022·廣東佛山市）已知函數(shù)SKIPIF1<0的圖象是經(jīng)過原點(diǎn)的曲線（非直線），且在原點(diǎn)處的切線方程為SKIPIF1<0，請寫出一個(gè)符合條件函數(shù)SKIPIF1<0的解析式____________．【答案】SKIPIF1<0（或SKIPIF1<0，SKIPIF1<0，SKIPIF1<0等）.【解析】由題意可知：SKIPIF1<0，取SKIPIF1<0，此時(shí)SKIPIF1<0，故符合，故答案為：SKIPIF1<0（或SKIPIF1<0，SKIPIF1<0，SKIPIF1<0等）.考點(diǎn)四根據(jù)切線求參數(shù)【例4-1】（2022·岳陽模擬）已知a，SKIPIF1<0為正實(shí)數(shù)，直線SKIPIF1<0與曲線SKIPIF1<0相切，則SKIPIF1<0的最小值是（）A．6 B．SKIPIF1<0 C．8 D．SKIPIF1<0【答案】C【解析】設(shè)切點(diǎn)為（m，n），y＝ln（x+b）的導(dǎo)數(shù)為SKIPIF1<0，由題意可得SKIPIF1<0=1，又n＝m﹣2a，n＝ln（m+b），解得n＝0，m＝2a，即有2a+b＝1，因?yàn)閍、b為正實(shí)數(shù)，所以SKIPIF1<0，當(dāng)且僅當(dāng)SKIPIF1<0時(shí)取等號，故SKIPIF1<0的最小值為8。故答案為：C．【例4-2】（2022·柳州模擬）已知直線SKIPIF1<0是曲線SKIPIF1<0的一條切線，則b＝．【答案】2【解析】函數(shù)SKIPIF1<0的定義域?yàn)椋?，+∞），則SKIPIF1<0，令SKIPIF1<0，則x=1，

所以切點(diǎn)為（1，3），代入y=x+b，得1+b=3，所以b=2.故答案為：23．（2022·新高考Ⅰ卷）若曲線SKIPIF1<0有兩條過坐標(biāo)原點(diǎn)的切線，則a的取值范圍是.【答案】a＞0或a＜-4【解析】易得曲線不過原點(diǎn)，設(shè)切點(diǎn)為(x0，(x0+a)ex0)，則切線斜率為f(x0)=(x0+a+1)ex0，

可得切線方程為y-(x0+a)ex0=(x0+a+1)ex0(x-x0)，又切線過原點(diǎn)，

可得-(x0+a)ex0=-x0(x0+a+1)ex0，化簡得SKIPIF1<0(※)，

又切線有兩條，即方程※有兩不等實(shí)根，由判別式△=a2+4a>0，得a<-4或a>0.

故答案為：a<-4或a>0.

【一隅三反】1．（2022·重慶模擬）已知SKIPIF1<0為非零實(shí)數(shù)，直線SKIPIF1<0與曲線SKIPIF1<0相切，則SKIPIF1<0.【答案】e【解析】設(shè)切點(diǎn)坐標(biāo)為SKIPIF1<0，對函數(shù)SKIPIF1<0求導(dǎo)得SKIPIF1<0，所以切線方程為SKIPIF1<0，所以SKIPIF1<0，可得SKIPIF1<0，SKIPIF1<0.故答案為：e.2．（2022·晉中模擬）若直線y=2x+a是函數(shù)SKIPIF1<0的圖象在某點(diǎn)處的切線，則實(shí)數(shù)SKIPIF1<0.【答案】-1【解析】【解答】設(shè)切點(diǎn)坐標(biāo)為SKIPIF1<0，則SKIPIF1<0所以SKIPIF1<0.故答案為：-1.

3．（2021高三上·德州期中）函數(shù)SKIPIF1<0在SKIPIF1<0處的切線與直線SKIPIF1<0平行，則實(shí)數(shù)SKIPIF1<0的值為.【答案】1【解析】因?yàn)镾KIPIF1<0，所以SKIPIF1<0，所以SKIPIF1<0，所以函數(shù)SKIPIF1<0在SKIPIF1<0處的切線的斜率為SKIPIF1<0，又因?yàn)楹瘮?shù)SKIPIF1<0在SKIPIF1<0處的切線與直線SKIPIF1<0平行，所以SKIPIF1<0，解得SKIPIF1<0故答案為：1

4．（2021高三上·安慶月考）已知函數(shù)SKIPIF1<0的圖象在點(diǎn)SKIPIF1<0處的切線的斜率為SKIPIF1<0，則SKIPIF1<0的值為.【答案】-1【解析】由題意，函數(shù)SKIPIF1<0，可得SKIPIF1<0，所以SKIPIF1<0，因?yàn)楹瘮?shù)SKIPIF1<0的圖象在點(diǎn)SKIPIF1<0處的切線的斜率為SKIPIF1<0，可得SKIPIF1<0，即SKIPIF1<0，所以SKIPIF1<0，解得SKIPIF1<0.故答案為：-1.5．（2020高三上·內(nèi)蒙古期中）若函數(shù)SKIPIF1<0（SKIPIF1<0為常數(shù)）存在兩條均過原點(diǎn)的切線，則實(shí)數(shù)a的取值范圍是.【答案】SKIPIF1<0【解析】由題意得SKIPIF1<0的定義域?yàn)镾KIPIF1<0，且SKIPIF1<0，設(shè)切點(diǎn)坐標(biāo)為SKIPIF1<0，則過原點(diǎn)的切線斜率SKIPIF1<0，整理得SKIPIF1<0存在兩條過原點(diǎn)的切線，SKIPIF1<0SKIPIF1<0存在兩個(gè)不同的解.設(shè)SKIPIF1<0，則問題等價(jià)于SKIPIF1<0于SKIPIF1<0存在兩個(gè)不同的交點(diǎn)，又SKIPIF1<0當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0，SKIPIF1<0單調(diào)遞增，當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0，SKIPIF1<0單調(diào)遞減，SKIPIF1<0.又當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0；當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0，若SKIPIF1<0于SKIPIF1<0存在兩個(gè)不同的交點(diǎn)，則SKIPIF1<0.解得SKIPIF1<0.故答案為：SKIPIF1<09.1切線方程（精練）（基礎(chǔ)版）題組一題組一導(dǎo)數(shù)的幾何意義1．（2021高三上·煙臺期中）曲線SKIPIF1<0在SKIPIF1<0處的切線的傾斜角為SKIPIF1<0，則SKIPIF1<0（）A．-1 B．SKIPIF1<0 C．SKIPIF1<0 D．2【答案】B【解析】SKIPIF1<0，當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0，所以SKIPIF1<0，由萬能公式得：SKIPIF1<0，所以SKIPIF1<0。故答案為：B2．（2022青海）已知曲線SKIPIF1<0在點(diǎn)SKIPIF1<0處的切線與直線SKIPIF1<0垂直，則SKIPIF1<0的值為（）A．1 B．-1 C．SKIPIF1<0 D．-SKIPIF1<0【答案】A【解析】SKIPIF1<0．當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0，因?yàn)榍芯€與直線SKIPIF1<0垂直，直線斜率為SKIPIF1<0，所以切線斜率為2，即SKIPIF1<0，得：SKIPIF1<0故答案為：A3（2021·廣西模擬）函數(shù)SKIPIF1<0的圖象在點(diǎn)SKIPIF1<0處的切線斜率為（）A．-8 B．-7 C．-6 D．-5【答案】A【解析】因?yàn)镾KIPIF1<0，所以所求切線的斜率為SKIPIF1<0.故答案為：A4．（2021·東陽模擬）已知點(diǎn)P在曲線SKIPIF1<0上，SKIPIF1<0為曲線在點(diǎn)P處的切線的傾斜角，則SKIPIF1<0的取值范圍是（）A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<0【答案】D【解析】因?yàn)镾KIPIF1<0，由于SKIPIF1<0，所以SKIPIF1<0，根據(jù)導(dǎo)數(shù)的幾何意義可知：SKIPIF1<0,所以SKIPIF1<0，故答案為：D.

5．（2022·白山模擬）函數(shù)SKIPIF1<0的圖象在點(diǎn)SKIPIF1<0處的切線的斜率為（）．A．-8 B．-7 C．6 D．-5【答案】A【解析】由SKIPIF1<0，得SKIPIF1<0，則SKIPIF1<0，所以函數(shù)SKIPIF1<0的圖象在點(diǎn)SKIPIF1<0處的切線的斜率為-8。故答案為：A6．（2022·鄭州模擬）函數(shù)SKIPIF1<0的圖象在SKIPIF1<0處切線的傾斜角為．【答案】SKIPIF1<0【解析】由SKIPIF1<0求導(dǎo)得：SKIPIF1<0，則SKIPIF1<0，所以函數(shù)SKIPIF1<0的圖象在SKIPIF1<0處切線斜率為-1，傾斜角為SKIPIF1<0。故答案為：SKIPIF1<0。7．（2021·蚌埠模擬）已知曲線SKIPIF1<0在SKIPIF1<0處切線的斜率為SKIPIF1<0，則SKIPIF1<0．【答案】0【解析】對函數(shù)SKIPIF1<0求導(dǎo)得SKIPIF1<0，由已知條件可得SKIPIF1<0，解得SKIPIF1<0.故答案為：0.8．（2021·聊城模擬）曲線SKIPIF1<0在SKIPIF1<0處的切線的傾斜角為SKIPIF1<0，則SKIPIF1<0．【答案】SKIPIF1<0【解析】由題得SKIPIF1<0，所以SKIPIF1<0，所以SKIPIF1<0，所以SKIPIF1<0.故答案為：SKIPIF1<0題組二題組二在型求切線1．（2021·全國甲卷）曲線SKIPIF1<0在點(diǎn)（-1，-3）處的切線方程為?！敬鸢浮?x-y+2=0【解析】由題意得SKIPIF1<0，所以在點(diǎn)（-1，-3）處的切線斜率k=5，故切線方程為y+3=5(x+1)，即5x-y+2=0故答案為：5x-y+2=0

2．（2022德州）函數(shù)SKIPIF1<0在點(diǎn)SKIPIF1<0處的切線方程為．【答案】SKIPIF1<0【解析】∵SKIPIF1<0，∴SKIPIF1<0，又SKIPIF1<0，∴切線方程為SKIPIF1<0，即SKIPIF1<0．故答案為：SKIPIF1<0．3．（2022·西安模擬）已知傾斜角為SKIPIF1<0的直線SKIPIF1<0與曲線SKIPIF1<0相切，則直線SKIPIF1<0的方程是.【答案】x-y-2+ln2=0【解析】因?yàn)橹本€SKIPIF1<0的傾斜角為SKIPIF1<0，所以直線SKIPIF1<0的斜率為1，將曲線SKIPIF1<0求導(dǎo)，得SKIPIF1<0，SKIPIF1<0，令SKIPIF1<0，可得SKIPIF1<0，所以切點(diǎn)坐標(biāo)為SKIPIF1<0，所以直線SKIPIF1<0：SKIPIF1<0，即x-y-2+ln2=0，故答案為：x-y-2+ln2=0.4．（2022·河南省?？h第一中學(xué)）曲線SKIPIF1<0在SKIPIF1<0處的切線方程為【答案】4x+y+8＝0【解析】因?yàn)镾KIPIF1<0，所以SKIPIF1<0，所以SKIPIF1<0．又當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0，故切點(diǎn)坐標(biāo)為SKIPIF1<0，所以切線方程為SKIPIF1<0．5．（2022·河南）已知SKIPIF1<0，則曲線SKIPIF1<0在點(diǎn)SKIPIF1<0處的切線方程為【答案】SKIPIF1<0【解析】∵SKIPIF1<0，∴SKIPIF1<0，SKIPIF1<0，∴SKIPIF1<0，∴y=f(x)在SKIPIF1<0處的切線方程為：SKIPIF1<0，即SKIPIF1<0．6．（2022·安徽·蚌埠二中）已知定義域?yàn)镾KIPIF1<0的函數(shù)SKIPIF1<0存在導(dǎo)函數(shù)SKIPIF1<0，且滿足SKIPIF1<0，則曲線SKIPIF1<0在點(diǎn)SKIPIF1<0處的切線方程是【答案】SKIPIF1<0【解析】SKIPIF1<0的定義域?yàn)镾KIPIF1<0，由SKIPIF1<0可知，SKIPIF1<0是偶函數(shù)，由SKIPIF1<0可知，SKIPIF1<0周期為4，因?yàn)镾KIPIF1<0，故SKIPIF1<0關(guān)于SKIPIF1<0軸對稱，又因?yàn)镾KIPIF1<0，所以SKIPIF1<0也是SKIPIF1<0的對稱軸，因?yàn)镾KIPIF1<0在SKIPIF1<0上存在導(dǎo)函數(shù)SKIPIF1<0，所以SKIPIF1<0是SKIPIF1<0的極值點(diǎn)，即SKIPIF1<0，曲線SKIPIF1<0在點(diǎn)SKIPIF1<0處的切線斜率為0，故切線方程可能為SKIPIF1<0.題組三題組三過型求切線1．（2022·廣東茂名）已知直線l為函數(shù)SKIPIF1<0的切線，且經(jīng)過原點(diǎn)，則直線l的方程為__________.【答案】SKIPIF1<0【解析】設(shè)切點(diǎn)坐標(biāo)為SKIPIF1<0，所以直線l的斜率為SKIPIF1<0，所以直線l的方程為SKIPIF1<0又直線l過點(diǎn)SKIPIF1<0，所以SKIPIF1<0，整理得SKIPIF1<0，解得SKIPIF1<0，所以SKIPIF1<0，直線l的斜率SKIPIF1<0，所以直線l的方程為SKIPIF1<0，故答案為：SKIPIF1<0.2．（2022·四川成都）已知函數(shù)f（x）=x3－3x，則過點(diǎn)（1，－2）的切線方程為__________．【答案】SKIPIF1<0和SKIPIF1<0【解析】由函數(shù)SKIPIF1<0，則SKIPIF1<0，當(dāng)點(diǎn)SKIPIF1<0為切點(diǎn)時(shí)，則SKIPIF1<0，即切線的斜率SKIPIF1<0，所以切線的方程為SKIPIF1<0，當(dāng)點(diǎn)SKIPIF1<0不是切點(diǎn)時(shí)，設(shè)切點(diǎn)SKIPIF1<0，則SKIPIF1<0，即SKIPIF1<0，解得SKIPIF1<0或SKIPIF1<0（舍去），所以SKIPIF1<0所以切線的方程為SKIPIF1<0，即SKIPIF1<0．故答案為：SKIPIF1<0和SKIPIF1<0．3．（2022·四川成都）過點(diǎn)SKIPIF1<0的直線l與曲線SKIPIF1<0相切，則直線l的斜率為___________.【答案】3或SKIPIF1<0【解析】因?yàn)镾KIPIF1<0，所以SKIPIF1<0，SKIPIF1<0，當(dāng)SKIPIF1<0為切點(diǎn)時(shí)，SKIPIF1<0，當(dāng)SKIPIF1<0不為切點(diǎn)時(shí)，設(shè)切點(diǎn)為SKIPIF1<0，SKIPIF1<0，所以SKIPIF1<0，所以切線方程為：SKIPIF1<0，過點(diǎn)SKIPIF1<0，所以SKIPIF1<0即SKIPIF1<0，即SKIPIF1<0，解得SKIPIF1<0或SKIPIF1<0（舍），所以切點(diǎn)為SKIPIF1<0，所以SKIPIF1<0，綜上所述：直線l的斜率為3或SKIPIF1<0，故答案為：3或SKIPIF1<04．（2022·廣東·南海中學(xué)）函數(shù)SKIPIF1<0過原點(diǎn)的切線方程是_______.【答案】SKIPIF1<0.【解析】設(shè)切點(diǎn)為SKIPIF1<0，SKIPIF1<0，則SKIPIF1<0，故切點(diǎn)為SKIPIF1<0的切線方程為SKIPIF1<0，又因此切線過原點(diǎn)，所以SKIPIF1<0，解得SKIPIF1<0，所以函數(shù)SKIPIF1<0過原點(diǎn)的切線方程是SKIPIF1<0，即SKIPIF1<0.故答案為：SKIPIF1<0.題組四題組四根據(jù)切線方程求參1．（2021高三上·普寧月考）若曲線SKIPIF1<0的一條切線SKIPIF1<0與直線SKIPIF1<0垂直，則直線SKIPIF1<0的方程為．【答案】SKIPIF1<0【解析】由SKIPIF1<0得SKIPIF1<0，則由題意得切線l的斜率為SKIPIF1<0

設(shè)切點(diǎn)為(x0，y0)，又y'=ex，則k=ex0=e則x0=1，y0=e則切點(diǎn)為(1，e)則切線l為y-e=e(x-1)，即y=ex故答案為：y=ex

2．（2022·天河模擬）已知函數(shù)SKIPIF1<0，且SKIPIF1<0，則SKIPIF1<0，曲線SKIPIF1<0在SKIPIF1<0處的切線方程為.【答案】0；SKIPIF1<0【解析】由SKIPIF1<0，則SKIPIF1<0，因?yàn)镾KIPIF1<0，即SKIPIF1<0，解得SKIPIF1<0，所以SKIPIF1<0，SKIPIF1<0，所以SKIPIF1<0，SKIPIF1<0，所以曲線SKIPIF1<0在SKIPIF1<0處的切線方程為：SKIPIF1<0。故答案為：0；SKIPIF1<0。3．（2022·泰安模擬）已知直線SKIPIF1<0是曲線SKIPIF1<0的一條切線，則SKIPIF1<0.【答案】4【解析】設(shè)SKIPIF1<0，切點(diǎn)為SKIPIF1<0，因?yàn)镾KIPIF1<0，所以SKIPIF1<0，解得SKIPIF1<0，所以SKIPIF1<0，故切點(diǎn)為SKIPIF1<0，又切點(diǎn)在切線SKIPIF1<0上，故SKIPIF1<0.故答案為：44．（2022·大連模擬）已知函數(shù)f（x）=axlnx﹣bx（a，b∈R）在點(diǎn)（e，f（e））處的切線方程為y=3x﹣e，則a+b=.【答案】0【解析】∵在點(diǎn)SKIPIF1<0處的切線方程為SKIPIF1<0，SKIPIF1<0，代入SKIPIF1<0得SKIPIF1<0①.又SKIPIF1<0②.聯(lián)立①②解得：SKIPIF1<0.SKIPIF1<0.故答案為：0.5（2022·東莞模擬）已知SKIPIF1<0在SKIPIF1<0的切線方程為SKIPIF1<0，則SKIPIF1<0．【答案】2【解析】【解答】由題意得SKIPIF1<0SKIPIF1<0，∴SKIPIF1<0，∴SKIPIF1<0．故答案為：26（2022·遼寧）已知曲線SKIPIF1<0在點(diǎn)SKIPIF1<0處的切線與直線SKIPIF1<0垂直，則實(shí)數(shù)a的值為______．【答案】SKIPIF1<0【解析】由題得SKIPIF1<0，所以SKIPIF1<0，所以曲線SKIPIF1<0在點(diǎn)SKIPIF1<0處的切線斜率為3，又曲線在點(diǎn)SKIPIF1<0處的切線與直線SKIPIF1<0垂直，所以SKIPIF1<0，解得SKIPIF1<0．故答案為：SKIPIF1<0.7．（2022·湖南）已知P是曲線SKIPIF1<0上的一動點(diǎn)，曲線C在P點(diǎn)處的切線的傾斜角為SKIPIF1<0，若SKIPIF1<0，則實(shí)數(shù)a的取值范圍是【答案】SKIPIF1<0【解析】因?yàn)镾KIPIF1<0，所以SKIPIF1<0，因?yàn)榍€在M處的切線的傾斜角SKIPIF1<0，所以SKIPIF1<0對于任意的SKIPIF1<0恒成立，即SKIPIF1<0對任意SKIPIF1<0恒成立，即SKIPIF1