What is the tangent of 1/(x-8) ?

The tangent of 1/(x-8) is y=-1/((a_{0)-8)^2}x+(2a_{0}-8)/((a_{0)-8)^2}

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tangent of 1/(x-8)

tangent of

\frac{1}{x - 8}

y = - \frac{1}{( a _{0} - 8 ) ^{2}} x + \frac{2 a _{0} - 8}{( a _{0} - 8 ) ^{2}}

Solution steps

Compute the tangent line to the general point

x = a_{0}

Find the tangent point:

(a_{0}, \frac{1}{a _{0} - 8})

Find the slope of

y = \frac{1}{x - 8} : \frac{d y}{d x} = - \frac{1}{( x - 8 ) ^{2}}

Compute the slope of

y = \frac{1}{x - 8}

x = a_{0} : m = - \frac{1}{( a _{0} - 8 ) ^{2}}

Find the line with slope m=

- \frac{1}{( a _{0} - 8 ) ^{2}}

and passing through

(a_{0}, \frac{1}{a _{0} - 8}) : y = - \frac{1}{( a _{0} - 8 ) ^{2}} x + \frac{2 a _{0} - 8}{( a _{0} - 8 ) ^{2}}

y = - \frac{1}{( a _{0} - 8 ) ^{2}} x + \frac{2 a _{0} - 8}{( a _{0} - 8 ) ^{2}}

r = cos (2 θ)

\frac{\partial}{\partial y} (y^{4} sin (3 x))

\int \frac{1}{x ^{2} + 2 x + 6} d x

\int (e^{2 t} + \frac{1}{( t + 1 ) ^{2}}) d t

\frac{s - 3}{s ^{2} + 10 s + 9}

What is the tangent of 1/(x-8) ?
The tangent of 1/(x-8) is y=-1/((a_{0)-8)^2}x+(2a_{0}-8)/((a_{0)-8)^2}