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

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

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

tangent of

\frac{x - 1}{2 - x}

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

Solution steps

Compute the tangent line to the general point

x = a_{0}

Find the tangent point:

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

Find the slope of

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

Find the line with slope m=

\frac{1}{( 2 - a _{0} ) ^{2}}

and passing through

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

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

\int_{25}^{x} 1 + 9 t d t

\frac{d}{d x} (6 x^{2} - 6 x)

\int_{1}^{3} 15 r^{3} ln (r) d r

y^{^{''}} + 2 y^{^{'}} + y = sin (x) + 3 cos (2 x)

\int (9 x)^{5} (9 x)^{3} + 1 d x

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