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受欢迎的微积分 >

tangent (x+1)(4x^2-4x-8)

解答

切线

(x + 1) (4 x^{2} - 4 x - 8)

解答

y = (12 a_{0}^{2} - 12) x - 8 a_{0}^{3} - 8

求解步骤

计算在一般点

x = a_{0}

的切线

找到切点:

(a_{0}, (a_{0} + 1) (4 a_{0}^{2} - 4 a_{0} - 8))

计算

y = (x + 1) (4 x^{2} - 4 x - 8)

的斜率:

\frac{d y}{d x} = 12 x^{2} - 12

EN : T i tl e G e n er a l Eq u a t i o n Sl o p e A t P o in t 2 Eq : m = 12 a_{0}^{2} - 12

找到斜率 m=

12 a_{0}^{2} - 12

且穿过

(a_{0}, (a_{0} + 1) (4 a_{0}^{2} - 4 a_{0} - 8))

的直线:

y = (12 a_{0}^{2} - 12) x - 8 a_{0}^{3} - 8

y = (12 a_{0}^{2} - 12) x - 8 a_{0}^{3} - 8

作图

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