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tangent f(x)=2x^2+8x,\at x=4

解答

切线

f (x) = 2 x^{2} + 8 x, at x = 4

解答

y = 24 x - 32

求解步骤

找到切点:

(4, 64)

计算

f (x) = 2 x^{2} + 8 x

的斜率:

\frac{df ( x )}{d x} = 4 x + 8

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 = 24

找到斜率 m=

24

且穿过

(4, 64)

的直线:

y = 24 x - 32

y = 24 x - 32

作图

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