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tangent 1.12x^3-6.3x^2-11.2x+1.2

解答

切线

1.12 x^{3} - 6.3 x^{2} - 11.2 x + 1.2

解答

y = (3.36 a_{0}^{2} - 12.6 a_{0} - 11.2) x - 2.24 a_{0}^{3} + 6.3 a_{0}^{2} + 1.2

求解步骤

计算在一般点

x = a_{0}

的切线

找到切点:

(a_{0}, 1.12 a_{0}^{3} - 6.3 a_{0}^{2} - 11.2 a_{0} + 1.2)

计算

y = 1.12 x^{3} - 6.3 x^{2} - 11.2 x + 1.2

的斜率:

\frac{d y}{d x} = 3.36 x^{2} - 12.6 x - 11.2

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 = 3.36 a_{0}^{2} - 12.6 a_{0} - 11.2

找到斜率 m=

3.36 a_{0}^{2} - 12.6 a_{0} - 11.2

且穿过

(a_{0}, 1.12 a_{0}^{3} - 6.3 a_{0}^{2} - 11.2 a_{0} + 1.2)

的直线:

y = (3.36 a_{0}^{2} - 12.6 a_{0} - 11.2) x - 2.24 a_{0}^{3} + 6.3 a_{0}^{2} + 1.2

y = (3.36 a_{0}^{2} - 12.6 a_{0} - 11.2) x - 2.24 a_{0}^{3} + 6.3 a_{0}^{2} + 1.2

作图

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