What is the tangent of y=x^3-3x+1,(2,3) ?

The tangent of y=x^3-3x+1,(2,3) is y=9x-15

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

tangent of y=x^3-3x+1,(2,3)

tangent of

y = x^{3} - 3 x + 1, (2, 3)

y = 9 x - 15

Solution steps

Find the slope of

y = x^{3} - 3 x + 1 : \frac{d y}{d x} = 3 x^{2} - 3

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 = 9

Find the line with slope m=

9

and passing through

(2, 3) : y = 9 x - 15

y = 9 x - 15

What is the tangent of y=x^3-3x+1,(2,3) ?
The tangent of y=x^3-3x+1,(2,3) is y=9x-15