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}}

Compute the slope of

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

x = a_{0} : 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}}

y = x, (0, 6)

\int_{0}^{4} 2 t + 4 d t

n = 0 \sum \infty \frac{1}{( n ( n + 2 ) )}

x + y = 4, x - y = 0, y + 3 x = 4

\frac{d}{d x} (\frac{4 x}{x ^{2} - 4})

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}