What is the tangent of (5x+2)/(6x-4) ?

The tangent of (5x+2)/(6x-4) is y=-8/((3a_{0)-2)^2}x+(15a_{0}^2+12a_{0}-4)/(2(3a_{0)-2)^2}

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tangent of (5x+2)/(6x-4)

tangent of

\frac{5 x + 2}{6 x - 4}

y = - \frac{8}{( 3 a _{0} - 2 ) ^{2}} x + \frac{15 a _{0}^{2} + 12 a _{0} - 4}{2 ( 3 a _{0} - 2 ) ^{2}}

Solution steps

Compute the tangent line to the general point

x = a_{0}

Find the tangent point:

(a_{0}, \frac{5 a _{0} + 2}{6 a _{0} - 4})

Find the slope of

y = \frac{5 x + 2}{6 x - 4} : \frac{d y}{d x} = - \frac{8}{( 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{8}{( 3 a _{0} - 2 ) ^{2}}

Find the line with slope m=

- \frac{8}{( 3 a _{0} - 2 ) ^{2}}

and passing through

(a_{0}, \frac{5 a _{0} + 2}{6 a _{0} - 4}) : y = - \frac{8}{( 3 a _{0} - 2 ) ^{2}} x + \frac{15 a _{0}^{2} + 12 a _{0} - 4}{2 ( 3 a _{0} - 2 ) ^{2}}

y = - \frac{8}{( 3 a _{0} - 2 ) ^{2}} x + \frac{15 a _{0}^{2} + 12 a _{0} - 4}{2 ( 3 a _{0} - 2 ) ^{2}}

What is the tangent of (5x+2)/(6x-4) ?
The tangent of (5x+2)/(6x-4) is y=-8/((3a_{0)-2)^2}x+(15a_{0}^2+12a_{0}-4)/(2(3a_{0)-2)^2}