What is the tangent of y=x^3-3x+2,(3,20) ?

The tangent of y=x^3-3x+2,(3,20) is y=24x-52

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

tangent of y=x^3-3x+2,(3,20)

tangent of

y = x^{3} - 3 x + 2, (3, 20)

y = 24 x - 52

Solution steps

Find the slope of

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

Compute the slope of

y = x^{3} - 3 x + 2

(3, 20) : m = 24

Find the line with slope m=

24

and passing through

(3, 20) : y = 24 x - 52

y = 24 x - 52

\int \frac{x - 1}{x ^{2} - 9} d x

\int_{0}^{2} \frac{3}{x ^{2}} d x

\int 4 x^{3} y^{3} d x

\int (y) d y

\int 9 tan^{3} (x) d x

What is the tangent of y=x^3-3x+2,(3,20) ?
The tangent of y=x^3-3x+2,(3,20) is y=24x-52