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

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

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

tangent of

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

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

Solution steps

Find the slope of

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

Compute the slope of

f (x) = \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 csc^{2} (2 x) d x

\int \frac{8}{x ^{2}} d x

e^{2 x - 3}, 1

x \to 2 lim (5 x^{3} - 6 x - 1)

\int (x^{- \frac{2}{7}}) d x

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