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tangent f(x)= 1/(x+8),\at x=9

解答

切线

f (x) = \frac{1}{x + 8}, at x = 9

解答

y = - \frac{1}{289} x + \frac{26}{289}

求解步骤

找到切点:

(9, \frac{1}{17})

计算

f (x) = \frac{1}{x + 8}

的斜率:

\frac{df ( x )}{d x} = - \frac{1}{( x + 8 ) ^{2}}

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 = - \frac{1}{289}

找到斜率 m=

- \frac{1}{289}

且穿过

(9, \frac{1}{17})

的直线:

y = - \frac{1}{289} x + \frac{26}{289}

y = - \frac{1}{289} x + \frac{26}{289}

作图

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