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tangent f(x)=(x+2)/(x^2-9),\at x=1

解答

切线

f (x) = \frac{x + 2}{x ^{2} - 9}, at x = 1

解答

y = - \frac{7}{32} x - \frac{5}{32}

求解步骤

找到切点:

(1, - \frac{3}{8})

计算

f (x) = \frac{x + 2}{x ^{2} - 9}

的斜率:

\frac{df ( x )}{d x} = \frac{- x ^{2} - 4 x - 9}{( x ^{2} - 9 ) ^{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{7}{32}

找到斜率 m=

- \frac{7}{32}

且穿过

(1, - \frac{3}{8})

的直线:

y = - \frac{7}{32} x - \frac{5}{32}

y = - \frac{7}{32} x - \frac{5}{32}

作图

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