What is the tangent of y=(-8x)/(x^2+1),(1,-4) ?

The tangent of y=(-8x)/(x^2+1),(1,-4) is y=-4

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

tangent of y=(-8x)/(x^2+1),(1,-4)

tangent of

y = \frac{- 8 x}{x ^{2} + 1}, (1, - 4)

y = - 4

Solution steps

Find the slope of

y = \frac{- 8 x}{x ^{2} + 1} : \frac{d y}{d x} = - \frac{8 ( - x ^{2} + 1 )}{( x ^{2} + 1 ) ^{2}}

EN : T i tl e G e n er a l Eq u a t i o n Sl o p e A t P o in t 2 Eq : m = 0

Find the line with slope m=

0

and passing through

(1, - 4) : y = - 4

y = - 4

What is the tangent of y=(-8x)/(x^2+1),(1,-4) ?
The tangent of y=(-8x)/(x^2+1),(1,-4) is y=-4