What is the tangent of y= 1/3 x^3-3x^2+5x ?

The tangent of y= 1/3 x^3-3x^2+5x is y=(a_{0}^2-6a_{0}+5)x-2/3 a_{0}^3+3a_{0}^2

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

tangent of y= 1/3 x^3-3x^2+5x

tangent of

y = \frac{1}{3} x^{3} - 3 x^{2} + 5 x

y = (a_{0}^{2} - 6 a_{0} + 5) x - \frac{2}{3} a_{0}^{3} + 3 a_{0}^{2}

Solution steps

Compute the tangent line to the general point

x = a_{0}

Find the tangent point:

(a_{0}, \frac{1}{3} a_{0}^{3} - 3 a_{0}^{2} + 5 a_{0})

Find the slope of

y = \frac{1}{3} x^{3} - 3 x^{2} + 5 x : \frac{d y}{d x} = x^{2} - 6 x + 5

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 = a_{0}^{2} - 6 a_{0} + 5

Find the line with slope m=

a_{0}^{2} - 6 a_{0} + 5

and passing through

(a_{0}, \frac{1}{3} a_{0}^{3} - 3 a_{0}^{2} + 5 a_{0}) : y = (a_{0}^{2} - 6 a_{0} + 5) x - \frac{2}{3} a_{0}^{3} + 3 a_{0}^{2}

y = (a_{0}^{2} - 6 a_{0} + 5) x - \frac{2}{3} a_{0}^{3} + 3 a_{0}^{2}

What is the tangent of y= 1/3 x^3-3x^2+5x ?
The tangent of y= 1/3 x^3-3x^2+5x is y=(a_{0}^2-6a_{0}+5)x-2/3 a_{0}^3+3a_{0}^2