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

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

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

tangent of

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

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

Solution steps

Find the slope of

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

Compute the slope of

y = \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

\frac{d}{d t} (t^{2} + t)

f (x) = \frac{3 cos ( x )}{3 x ^{2} + 3 x - 1}

\frac{d}{d x} (ln (c x))

\frac{d}{d x} (a r (x) c, (ln (x)))

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