What is the tangent of ((x+1))/((5-x)) ?

The tangent of ((x+1))/((5-x)) is y= 6/((5-a_{0))^2}x+(-a_{0}^2-2a_{0}+5)/((-a_{0)+5)^2}

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tangent of ((x+1))/((5-x))

tangent of

\frac{( x + 1 )}{( 5 - x )}

y = \frac{6}{( 5 - a _{0} ) ^{2}} x + \frac{- a _{0}^{2} - 2 a _{0} + 5}{( - a _{0} + 5 ) ^{2}}

Solution steps

Compute the tangent line to the general point

x = a_{0}

Find the tangent point:

(a_{0}, \frac{a _{0} + 1}{5 - a _{0}})

Find the slope of

y = \frac{x + 1}{5 - x} : \frac{d y}{d x} = \frac{6}{( 5 - x ) ^{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{6}{( 5 - a _{0} ) ^{2}}

Find the line with slope m=

\frac{6}{( 5 - a _{0} ) ^{2}}

and passing through

(a_{0}, \frac{a _{0} + 1}{5 - a _{0}}) : y = \frac{6}{( 5 - a _{0} ) ^{2}} x + \frac{- a _{0}^{2} - 2 a _{0} + 5}{( - a _{0} + 5 ) ^{2}}

y = \frac{6}{( 5 - a _{0} ) ^{2}} x + \frac{- a _{0}^{2} - 2 a _{0} + 5}{( - a _{0} + 5 ) ^{2}}

What is the tangent of ((x+1))/((5-x)) ?
The tangent of ((x+1))/((5-x)) is y= 6/((5-a_{0))^2}x+(-a_{0}^2-2a_{0}+5)/((-a_{0)+5)^2}