Question

曲线y=3x^2-4x+2上以点M(1,1)为切点的切线方程是

Answer 1

有两种方法:
[方法一]:
设过点M的切线方程为:y - 1 = k(x - 1)
整理:y = k(x - 1) + 1 ~ (1)
联立切线方程与曲线方程,
代入得, k(x - 1) + 1 = 3x² - 4x + 2
整理:3x² - (4 + k)x + 1 + k = 0
∵它们只有一个交点(切点)
∴△ = 0
即[- (4 + k)]² - 4 * 3 * (1 + k) = 0
整理:k² - 4k + 4 = 0
(k - 2)² = 0
解得:k = 2
代入(1)式, 得切线方程为y = 2x - 1

[方法二]:
求曲线方程的导数,
y' = 6x - 4 (此为曲线在某一点的切线的斜率)
∴当x = 1时, 曲线在x = 1处的切线的斜率为y' = 6 * 1 - 4 = 2
利用点斜式, 得y - 1 = 2 * (x - 1)
也可得切线方程为y = 2x - 1