What is the tangent of y= 2/(x-8),(10,1) ?

The tangent of y= 2/(x-8),(10,1) is y=-1/2 x+6

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tangent of y= 2/(x-8),(10,1)

tangent of

y = \frac{2}{x - 8}, (10, 1)

y = - \frac{1}{2} x + 6

Solution steps

Find the slope of

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

Compute the slope of

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

(10, 1) : m = - \frac{1}{2}

Find the line with slope m=

- \frac{1}{2}

and passing through

(10, 1) : y = - \frac{1}{2} x + 6

y = - \frac{1}{2} x + 6

\int \frac{2 x}{5 - x ^{4}} d x

y = 2 x - x^{2}, (2, 0)

\int_{- 1}^{1} x^{- 3} d x

\frac{\partial}{\partial x} (\frac{e ^{x} - e ^{y}}{e ^{x y}})

\frac{\partial}{\partial u} (2 cos (u))

What is the tangent of y= 2/(x-8),(10,1) ?
The tangent of y= 2/(x-8),(10,1) is y=-1/2 x+6