What is the tangent of y=(x-1)/(x-2)+3 ?

The tangent of y=(x-1)/(x-2)+3 is y=-1/((a_{0)-2)^2}x+(a_{0}^2-2a_{0}+2)/((a_{0)-2)^2}+3

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tangent of y=(x-1)/(x-2)+3

tangent of

y = \frac{x - 1}{x - 2} + 3

y = - \frac{1}{( a _{0} - 2 ) ^{2}} x + \frac{a _{0}^{2} - 2 a _{0} + 2}{( a _{0} - 2 ) ^{2}} + 3

Solution steps

Compute the tangent line to the general point

x = a_{0}

Find the tangent point:

(a_{0}, \frac{a _{0} - 1}{a _{0} - 2} + 3)

Find the slope of

y = \frac{x - 1}{x - 2} + 3 : \frac{d y}{d x} = - \frac{1}{( 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{1}{( a _{0} - 2 ) ^{2}}

Find the line with slope m=

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

and passing through

(a_{0}, \frac{a _{0} - 1}{a _{0} - 2} + 3) : y = - \frac{1}{( a _{0} - 2 ) ^{2}} x + \frac{a _{0}^{2} - 2 a _{0} + 2}{( a _{0} - 2 ) ^{2}} + 3

y = - \frac{1}{( a _{0} - 2 ) ^{2}} x + \frac{a _{0}^{2} - 2 a _{0} + 2}{( a _{0} - 2 ) ^{2}} + 3

n = 0 \sum \infty e^{\frac{- 3 n}{2}}

\frac{d y}{d x} = 2 x^{6} y^{2}

\int \frac{csc ( 7 x )}{csc ( 7 x ) - sin ( 7 x )} d x

\int_{- 1}^{1} \frac{1}{x ( x + 4 )} d x

\frac{d}{d x} (8 x - x^{2})

What is the tangent of y=(x-1)/(x-2)+3 ?
The tangent of y=(x-1)/(x-2)+3 is y=-1/((a_{0)-2)^2}x+(a_{0}^2-2a_{0}+2)/((a_{0)-2)^2}+3