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tangent f(x)=9x^2+6x-6

解答

切线

f (x) = 9 x^{2} + 6 x - 6

解答

y = (18 a_{0} + 6) x - 9 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}, 9 a_{0}^{2} + 6 a_{0} - 6)

计算

f (x) = 9 x^{2} + 6 x - 6

的斜率:

\frac{df ( x )}{d x} = 18 x + 6

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 = 18 a_{0} + 6

找到斜率 m=

18 a_{0} + 6

且穿过

(a_{0}, 9 a_{0}^{2} + 6 a_{0} - 6)

的直线:

y = (18 a_{0} + 6) x - 9 a_{0}^{2} - 6

y = (18 a_{0} + 6) x - 9 a_{0}^{2} - 6

作图

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