What is the tangent of y= 8/(x-2),\at x=6 ?

The tangent of y= 8/(x-2),\at x=6 is y=-1/2 x+5

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tangent of y= 8/(x-2),\at x=6

tangent of

y = \frac{8}{x - 2}, at x = 6

y = - \frac{1}{2} x + 5

Solution steps

Find the tangent point:

(6, 2)

Find the slope of

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

Compute the slope of

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

x = 6 : m = - \frac{1}{2}

Find the line with slope m=

- \frac{1}{2}

and passing through

(6, 2) : y = - \frac{1}{2} x + 5

y = - \frac{1}{2} x + 5

\frac{d}{d x} (26 e^{x^{2}} x)

\frac{\partial}{\partial y} (5 x sin (y - z))

\frac{d}{d x} ((7 x^{2} - 12 x) e^{x})

y^{^{'}} - e^{5 x + y} = 0

e^{y} + \frac{( x e ^{y} + 1 ) d y}{d x} = 0

What is the tangent of y= 8/(x-2),\at x=6 ?
The tangent of y= 8/(x-2),\at x=6 is y=-1/2 x+5