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

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

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tangent of y= 2/(x-2)

tangent of

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

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

Solution steps

Compute the tangent line to the general point

x = a_{0}

Find the tangent point:

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

Find the slope of

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

Compute the slope of

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

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

Find the line with slope m=

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

and passing through

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

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

\int \frac{1 - 2 x}{e ^{- 2 x}} d x

\int (\frac{1}{2} + \frac{1}{x}) d x

\frac{d y}{d x} = \frac{( x - 4 )}{y + 3}

(1, - 7), (2, - 1)

\int e^{\frac{x}{3}} d x

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