What is the tangent of y= 2/(x-8) ?

The tangent of y= 2/(x-8) is y=-2/((a_{0)-8)^2}x+(4a_{0}-16)/((a_{0)-8)^2}

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tangent of y= 2/(x-8)

tangent of

y = \frac{2}{x - 8}

y = - \frac{2}{( a _{0} - 8 ) ^{2}} x + \frac{4 a _{0} - 16}{( a _{0} - 8 ) ^{2}}

Solution steps

Compute the tangent line to the general point

x = a_{0}

Find the tangent point:

(a_{0}, \frac{2}{a _{0} - 8})

Find the slope of

y = \frac{2}{x - 8} : \frac{d y}{d x} = - \frac{2}{( x - 8 ) ^{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{2}{( a _{0} - 8 ) ^{2}}

Find the line with slope m=

- \frac{2}{( a _{0} - 8 ) ^{2}}

and passing through

(a_{0}, \frac{2}{a _{0} - 8}) : y = - \frac{2}{( a _{0} - 8 ) ^{2}} x + \frac{4 a _{0} - 16}{( a _{0} - 8 ) ^{2}}

y = - \frac{2}{( a _{0} - 8 ) ^{2}} x + \frac{4 a _{0} - 16}{( a _{0} - 8 ) ^{2}}

x \to 1 lim (\frac{1}{1 - x ^{2}})

\int 4 x^{2} + 4 x + 1 d x

\int \frac{1}{( y - 3 ) ^{2}} d y

\frac{\partial}{\partial z} (4 e^{x} cos (yz))

\frac{\partial}{\partial x} (x (x + y))

What is the tangent of y= 2/(x-8) ?
The tangent of y= 2/(x-8) is y=-2/((a_{0)-8)^2}x+(4a_{0}-16)/((a_{0)-8)^2}