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tangent f(x)=-3x^2-2x^2-1,\at x=-1

解答

切线

f (x) = - 3 x^{2} - 2 x^{2} - 1, at x = - 1

解答

y = 10 x + 4

求解步骤

找到切点:

(- 1, - 6)

计算

f (x) = - 3 x^{2} - 2 x^{2} - 1

的斜率:

\frac{df ( x )}{d x} = - 10 x

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 = 10

找到斜率 m=

10

且穿过

(- 1, - 6)

的直线:

y = 10 x + 4

y = 10 x + 4

作图

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