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受欢迎的微积分 >

tangent f(x)=2x^3+42x^2+270x+4

解答

切线

f (x) = 2 x^{3} + 42 x^{2} + 270 x + 4

解答

y = (6 a_{0}^{2} + 84 a_{0} + 270) x - 4 a_{0}^{3} - 42 a_{0}^{2} + 4

求解步骤

计算在一般点

x = a_{0}

的切线

找到切点:

(a_{0}, 2 a_{0}^{3} + 42 a_{0}^{2} + 270 a_{0} + 4)

计算

f (x) = 2 x^{3} + 42 x^{2} + 270 x + 4

的斜率:

\frac{df ( x )}{d x} = 6 x^{2} + 84 x + 270

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 = 6 a_{0}^{2} + 84 a_{0} + 270

找到斜率 m=

6 a_{0}^{2} + 84 a_{0} + 270

且穿过

(a_{0}, 2 a_{0}^{3} + 42 a_{0}^{2} + 270 a_{0} + 4)

的直线:

y = (6 a_{0}^{2} + 84 a_{0} + 270) x - 4 a_{0}^{3} - 42 a_{0}^{2} + 4

y = (6 a_{0}^{2} + 84 a_{0} + 270) x - 4 a_{0}^{3} - 42 a_{0}^{2} + 4

作图

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