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

The tangent of f(x)= 1/(x-2),(-2,-1/4) is y=-1/16 x-3/8

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

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

tangent of

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

y = - \frac{1}{16} x - \frac{3}{8}

Solution steps

Find the slope of

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

Compute the slope of

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

(- 2, - \frac{1}{4}) : m = - \frac{1}{16}

Find the line with slope m=

- \frac{1}{16}

and passing through

(- 2, - \frac{1}{4}) : y = - \frac{1}{16} x - \frac{3}{8}

y = - \frac{1}{16} x - \frac{3}{8}

\frac{d}{d x} (10 sec (x))

\frac{d}{d x} (\frac{4}{x} + \frac{2}{x} + \frac{x}{2})

\int \frac{( x - 1 ) ( x + 1 )}{x} d x

x \to π lim (15)

x \to 3 lim (5 x^{3} - 2 x^{2} + x - 7)

What is the tangent of f(x)= 1/(x-2),(-2,-1/4) ?
The tangent of f(x)= 1/(x-2),(-2,-1/4) is y=-1/16 x-3/8