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

解答

切线

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

解答

y = - 4 x + 16

求解步骤

找到切点:

(3, 4)

计算

f (x) = [(x - 2)^{2} - x]^{2}

的斜率:

\frac{df ( x )}{d x} = 2 (x^{2} - 5 x + 4) (2 x - 5)

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 = - 4

找到斜率 m=

- 4

且穿过

(3, 4)

的直线:

y = - 4 x + 16

y = - 4 x + 16

作图

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