已知曲线y=13x3+2x-23.(1)求曲线在点P(2,6)处的切线方程;(2)求曲线过点P(2,6)的切线方程.-数学

题目简介

已知曲线y=13x3+2x-23.(1)求曲线在点P(2,6)处的切线方程;(2)求曲线过点P(2,6)的切线方程.-数学

题目详情

已知曲线 y=
1
3
x3+2x-
2
3

(1)求曲线在点P(2,6)处的切线方程;
(2)求曲线过点P(2,6)的切线方程.
题型:解答题难度:中档来源:不详

答案

(1)由题意得,y′=x2+2,
∴在点P(2,6)处的切线的斜率k=y′|x=2=6,
∴在点P(2,6)处的切线方程为:y-6=6(x-2)
即 6x-y-6=0,
(2)设曲线y=class="stub"1
3
x3+2x-class="stub"2
3
与过点P(2,6)的切线相切于点A(x0,class="stub"1
3
x30
+2x0-class="stub"2
3
)

则切线的斜率k=y|x=x0=
x20
+2,
∴切线方程为y-(class="stub"1
3
x30
+2x0-class="stub"2
3
)=(
x20
+2)(x-x0)

y=(
x20
+2)x-class="stub"2
3
x30
-class="stub"2
3
  ①,
∵点P(2,6)在切线上,∴6=2(
x20
+2)-class="stub"2
3
x30
-class="stub"2
3

x30
-3
x20
+4=0
,∴
x30
+
x20
-4
x20
+4=0

x20
(x0+1)-4(x0+1)(x0-1)=0
,化简得(x0+1)(x0-2)2=0
解得x0=-1或x0=2,代入①得,y=3x或y=6x-6,
故所求的切线方程为3x-y=0,6x-y-6=0.

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