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tangent f(x)=8x^2,\at x=10

解答

切线

f (x) = 8 x^{2}, at x = 10

解答

y = 160 x - 800

求解步骤

找到切点:

(10, 800)

计算

f (x) = 8 x^{2}

的斜率:

\frac{df ( x )}{d x} = 16 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 = 160

找到斜率 m=

160

且穿过

(10, 800)

的直线:

y = 160 x - 800

y = 160 x - 800

作图

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