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tangent y=2x^3+3x^2-12x+6

解答

切线

y = 2 x^{3} + 3 x^{2} - 12 x + 6

解答

y = (6 a_{0}^{2} + 6 a_{0} - 12) x - 4 a_{0}^{3} - 3 a_{0}^{2} + 6

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} + 3 a_{0}^{2} - 12 a_{0} + 6)

计算

y = 2 x^{3} + 3 x^{2} - 12 x + 6

的斜率:

\frac{d y}{d x} = 6 x^{2} + 6 x - 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 = 6 a_{0}^{2} + 6 a_{0} - 12

找到斜率 m=

6 a_{0}^{2} + 6 a_{0} - 12

且穿过

(a_{0}, 2 a_{0}^{3} + 3 a_{0}^{2} - 12 a_{0} + 6)

的直线:

y = (6 a_{0}^{2} + 6 a_{0} - 12) x - 4 a_{0}^{3} - 3 a_{0}^{2} + 6

y = (6 a_{0}^{2} + 6 a_{0} - 12) x - 4 a_{0}^{3} - 3 a_{0}^{2} + 6

作图

流行的例子

(d^2y)/(dt^2)+3(dy)/(dt)+y=1

\frac{d ^{2} y}{d t ^{2}} + 3 \frac{d y}{d t} + y = 1

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x \to 2 lim (- 10 x + (10))

derivative f(x)=28x-0.01x^2

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