What is the tangent of f(x)=x^3-6x,(2,-4) ?

The tangent of f(x)=x^3-6x,(2,-4) is y=6x-16

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

tangent of f(x)=x^3-6x,(2,-4)

tangent of

f (x) = x^{3} - 6 x, (2, - 4)

y = 6 x - 16

Solution steps

Find the slope of

f (x) = x^{3} - 6 x : \frac{df ( x )}{d x} = 3 x^{2} - 6

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 = 6

Find the line with slope m=

6

and passing through

(2, - 4) : y = 6 x - 16

y = 6 x - 16

\int_{- 1}^{3} (2 x^{2} + 1) d x

\frac{\partial}{\partial y} (ln (x^{3} + y^{4}))

\frac{\partial}{\partial t} (r sin (t))

f (x) = 2 x^{2} + 5, at x = 1

\frac{d}{d x} ((1 + 3 x^{4})^{5})

What is the tangent of f(x)=x^3-6x,(2,-4) ?
The tangent of f(x)=x^3-6x,(2,-4) is y=6x-16