What is the tangent of y= 1/(x-8),(-1,-1/9) ?

The tangent of y= 1/(x-8),(-1,-1/9) is y=-1/81 x-10/81

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tangent of y= 1/(x-8),(-1,-1/9)

tangent of

y = \frac{1}{x - 8}, (- 1, - \frac{1}{9})

y = - \frac{1}{81} x - \frac{10}{81}

Solution steps

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}

(- 1, - \frac{1}{9}) : m = - \frac{1}{81}

Find the line with slope m=

- \frac{1}{81}

and passing through

(- 1, - \frac{1}{9}) : y = - \frac{1}{81} x - \frac{10}{81}

y = - \frac{1}{81} x - \frac{10}{81}

\frac{d}{d x} (x^{2} - 2 x - 3)

x \to 0 lim (\frac{sin ( m x )}{x})

\int x^{3} + 2 x^{2} - 3 x d x

y^{^{'}} = 3 + t - y

\frac{d}{d x} (10 e^{- x})

What is the tangent of y= 1/(x-8),(-1,-1/9) ?
The tangent of y= 1/(x-8),(-1,-1/9) is y=-1/81 x-10/81