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

The tangent of y=x^3-2x+2,(2,6) is y=10x-14

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

tangent of y=x^3-2x+2,(2,6)

tangent of

y = x^{3} - 2 x + 2, (2, 6)

y = 10 x - 14

Solution steps

Find the slope of

y = x^{3} - 2 x + 2 : \frac{d y}{d x} = 3 x^{2} - 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 = 10

Find the line with slope m=

10

and passing through

(2, 6) : y = 10 x - 14

y = 10 x - 14

What is the tangent of y=x^3-2x+2,(2,6) ?
The tangent of y=x^3-2x+2,(2,6) is y=10x-14