What is the tangent of y=x^3-6x^2+7x+3 ?

The tangent of y=x^3-6x^2+7x+3 is y=(3a_{0}^2-12a_{0}+7)x-2a_{0}^3+6a_{0}^2+3

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

tangent of y=x^3-6x^2+7x+3

tangent of

y = x^{3} - 6 x^{2} + 7 x + 3

y = (3 a_{0}^{2} - 12 a_{0} + 7) x - 2 a_{0}^{3} + 6 a_{0}^{2} + 3

Solution steps

Compute the tangent line to the general point

x = a_{0}

Find the tangent point:

(a_{0}, a_{0}^{3} - 6 a_{0}^{2} + 7 a_{0} + 3)

Find the slope of

y = x^{3} - 6 x^{2} + 7 x + 3 : \frac{d y}{d x} = 3 x^{2} - 12 x + 7

Compute the slope of

y = x^{3} - 6 x^{2} + 7 x + 3

x = a_{0} : m = 3 a_{0}^{2} - 12 a_{0} + 7

Find the line with slope m=

3 a_{0}^{2} - 12 a_{0} + 7

and passing through

(a_{0}, a_{0}^{3} - 6 a_{0}^{2} + 7 a_{0} + 3) : y = (3 a_{0}^{2} - 12 a_{0} + 7) x - 2 a_{0}^{3} + 6 a_{0}^{2} + 3

y = (3 a_{0}^{2} - 12 a_{0} + 7) x - 2 a_{0}^{3} + 6 a_{0}^{2} + 3

\int \frac{( e ^{x} - e ^{- x} )}{e ^{x} + e ^{- x}} d x

\frac{x d y}{d x} = 5 x e^{x} - y + 9 x^{2}

\int 5 x^{2} e^{6 x} d x

\int sec (t) d t

\frac{d y}{d x}, 4 x y = 3

What is the tangent of y=x^3-6x^2+7x+3 ?
The tangent of y=x^3-6x^2+7x+3 is y=(3a_{0}^2-12a_{0}+7)x-2a_{0}^3+6a_{0}^2+3