What is the tangent of ((2x+1))/(x-4) ?

The tangent of ((2x+1))/(x-4) is y=-9/((a_{0)-4)^2}x+(2a_{0}^2+2a_{0}-4)/((a_{0)-4)^2}

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tangent of ((2x+1))/(x-4)

tangent of

\frac{( 2 x + 1 )}{x - 4}

y = - \frac{9}{( a _{0} - 4 ) ^{2}} x + \frac{2 a _{0}^{2} + 2 a _{0} - 4}{( a _{0} - 4 ) ^{2}}

Solution steps

Compute the tangent line to the general point

x = a_{0}

Find the tangent point:

(a_{0}, \frac{2 a _{0} + 1}{a _{0} - 4})

Find the slope of

y = \frac{2 x + 1}{x - 4} : \frac{d y}{d x} = - \frac{9}{( x - 4 ) ^{2}}

Compute the slope of

y = \frac{2 x + 1}{x - 4}

x = a_{0} : m = - \frac{9}{( a _{0} - 4 ) ^{2}}

Find the line with slope m=

- \frac{9}{( a _{0} - 4 ) ^{2}}

and passing through

(a_{0}, \frac{2 a _{0} + 1}{a _{0} - 4}) : y = - \frac{9}{( a _{0} - 4 ) ^{2}} x + \frac{2 a _{0}^{2} + 2 a _{0} - 4}{( a _{0} - 4 ) ^{2}}

y = - \frac{9}{( a _{0} - 4 ) ^{2}} x + \frac{2 a _{0}^{2} + 2 a _{0} - 4}{( a _{0} - 4 ) ^{2}}

\frac{d}{d x} (cos^{3} (1 - 5 x^{2}))

y = arcsin (5 x)

\frac{6 x}{1 - cot ( x )}

\int \frac{( x + 1 ) ^{3}}{( x ) ^{10}} d x

\frac{x}{99}

What is the tangent of ((2x+1))/(x-4) ?
The tangent of ((2x+1))/(x-4) is y=-9/((a_{0)-4)^2}x+(2a_{0}^2+2a_{0}-4)/((a_{0)-4)^2}