What is the tangent of f(x)= 2/(x-2),(4,1) ?

The tangent of f(x)= 2/(x-2),(4,1) is y=-1/2 x+3

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tangent of f(x)= 2/(x-2),(4,1)

tangent of

f (x) = \frac{2}{x - 2}, (4, 1)

y = - \frac{1}{2} x + 3

Solution steps

Find the slope of

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

Compute the slope of

f (x) = \frac{2}{x - 2}

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

Find the line with slope m=

- \frac{1}{2}

and passing through

(4, 1) : y = - \frac{1}{2} x + 3

y = - \frac{1}{2} x + 3

f (x) = \frac{4}{x}

\int (x^{2} + 2 x) \cdot cos (x) d x

\int x^{2} e^{- x^{2}} d x

\int (4 x^{3} - 3 x^{2} + 2 x - 5) d x

What is the tangent of f(x)= 2/(x-2),(4,1) ?
The tangent of f(x)= 2/(x-2),(4,1) is y=-1/2 x+3