What is the tangent of y=4x^2-x^3,(1,3) ?

The tangent of y=4x^2-x^3,(1,3) is y=5x-2

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

tangent of y=4x^2-x^3,(1,3)

tangent of

y = 4 x^{2} - x^{3}, (1, 3)

y = 5 x - 2

Solution steps

Find the slope of

y = 4 x^{2} - x^{3} : \frac{d y}{d x} = 8 x - 3 x^{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 = 5

Find the line with slope m=

5

and passing through

(1, 3) : y = 5 x - 2

y = 5 x - 2

(0.1) (1.1)

x \to a lim (0)

f (x) = x x + 3

\int \frac{5 x}{x} d x

\int 5 sec^{5} (x) d x

What is the tangent of y=4x^2-x^3,(1,3) ?
The tangent of y=4x^2-x^3,(1,3) is y=5x-2