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

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

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

tangent of

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

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

Solution steps

Find the tangent point:

(10, 1)

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}

x = 10 : 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 22 sec^{4} (θ) tan^{2} (θ) d θ

(4.1) (1.1)

x \to \frac{π}{3} lim (2 sin^{2} (x))

\frac{d}{d x} (3 tan (π - x))

\frac{d}{d x} (\frac{x ^{3} e ^{x}}{x})

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