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

The tangent of y= 1/(x-2),\at x=6 is y=-1/16 x+5/8

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tangent of y= 1/(x-2),\at x=6

tangent of

y = \frac{1}{x - 2}, at x = 6

y = - \frac{1}{16} x + \frac{5}{8}

Solution steps

Find the tangent point:

(6, \frac{1}{4})

Find the slope of

y = \frac{1}{x - 2} : \frac{d y}{d x} = - \frac{1}{( x - 2 ) ^{2}}

Compute the slope of

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

x = 6 : m = - \frac{1}{16}

Find the line with slope m=

- \frac{1}{16}

and passing through

(6, \frac{1}{4}) : y = - \frac{1}{16} x + \frac{5}{8}

y = - \frac{1}{16} x + \frac{5}{8}

\int e^{5} d x

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

x \to i lim (\frac{z + i}{z - i})

f (x) = - 2 x^{2} + 8

\int (ln (x) + 4 x) d x

What is the tangent of y= 1/(x-2),\at x=6 ?
The tangent of y= 1/(x-2),\at x=6 is y=-1/16 x+5/8