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

The tangent of y=((x-2))/((x-3)) is y=-1/((a_{0)-3)^2}x+(a_{0}^2-4a_{0}+6)/((a_{0)-3)^2}

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

tangent of

y = \frac{( x - 2 )}{( x - 3 )}

y = - \frac{1}{( a _{0} - 3 ) ^{2}} x + \frac{a _{0}^{2} - 4 a _{0} + 6}{( a _{0} - 3 ) ^{2}}

Solution steps

Compute the tangent line to the general point

x = a_{0}

Find the tangent point:

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

Find the slope of

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

Compute the slope of

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

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

Find the line with slope m=

- \frac{1}{( a _{0} - 3 ) ^{2}}

and passing through

(a_{0}, \frac{a _{0} - 2}{a _{0} - 3}) : y = - \frac{1}{( a _{0} - 3 ) ^{2}} x + \frac{a _{0}^{2} - 4 a _{0} + 6}{( a _{0} - 3 ) ^{2}}

y = - \frac{1}{( a _{0} - 3 ) ^{2}} x + \frac{a _{0}^{2} - 4 a _{0} + 6}{( a _{0} - 3 ) ^{2}}

6 e^{5 t} cos (2 t) - e^{- 7 t}

\int_{1}^{1000} \frac{1}{x ^{3}} d x

\frac{v ^{3} + 1}{v}

\frac{\partial}{\partial u} (ln (u cos (v)))

f (t) = r^{2} + 6 r - 3

What is the tangent of y=((x-2))/((x-3)) ?
The tangent of y=((x-2))/((x-3)) is y=-1/((a_{0)-3)^2}x+(a_{0}^2-4a_{0}+6)/((a_{0)-3)^2}