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tangent f(x)=(2x+3)/(3x-2)

解答

切线

f (x) = \frac{2 x + 3}{3 x - 2}

解答

y = - \frac{13}{( 3 a _{0} - 2 ) ^{2}} x + \frac{6 a _{0}^{2} + 18 a _{0} - 6}{( 3 a _{0} - 2 ) ^{2}}

求解步骤

计算在一般点

x = a_{0}

的切线

找到切点:

(a_{0}, \frac{2 a _{0} + 3}{3 a _{0} - 2})

计算

f (x) = \frac{2 x + 3}{3 x - 2}

的斜率:

\frac{df ( x )}{d x} = - \frac{13}{( 3 x - 2 ) ^{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{13}{( 3 a _{0} - 2 ) ^{2}}

找到斜率 m=

- \frac{13}{( 3 a _{0} - 2 ) ^{2}}

且穿过

(a_{0}, \frac{2 a _{0} + 3}{3 a _{0} - 2})

的直线:

y = - \frac{13}{( 3 a _{0} - 2 ) ^{2}} x + \frac{6 a _{0}^{2} + 18 a _{0} - 6}{( 3 a _{0} - 2 ) ^{2}}

y = - \frac{13}{( 3 a _{0} - 2 ) ^{2}} x + \frac{6 a _{0}^{2} + 18 a _{0} - 6}{( 3 a _{0} - 2 ) ^{2}}

作图

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