What is the tangent of y= 6/(x-1),(3,3) ?

The tangent of y= 6/(x-1),(3,3) is y=-3/2 x+15/2

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

tangent of y= 6/(x-1),(3,3)

tangent of

y = \frac{6}{x - 1}, (3, 3)

y = - \frac{3}{2} x + \frac{15}{2}

Solution steps

Find the slope of

y = \frac{6}{x - 1} : \frac{d y}{d x} = - \frac{6}{( x - 1 ) ^{2}}

Compute the slope of

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

(3, 3) : m = - \frac{3}{2}

Find the line with slope m=

- \frac{3}{2}

and passing through

(3, 3) : y = - \frac{3}{2} x + \frac{15}{2}

y = - \frac{3}{2} x + \frac{15}{2}

\int \frac{1}{y ^{2} + 9} d y

(14 x - x^{2})^{3}

f (x) = \frac{1}{x}, at x = 9

\frac{4}{e ^{2 x}}

x \to 0 lim ((\frac{5}{6})^{\frac{1}{x}})

What is the tangent of y= 6/(x-1),(3,3) ?
The tangent of y= 6/(x-1),(3,3) is y=-3/2 x+15/2