What is the tangent of 4x^3+8x^2+3x+2 ?

The tangent of 4x^3+8x^2+3x+2 is y=(12a_{0}^2+16a_{0}+3)x-8a_{0}^3-8a_{0}^2+2

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

tangent of 4x^3+8x^2+3x+2

tangent of

4 x^{3} + 8 x^{2} + 3 x + 2

y = (12 a_{0}^{2} + 16 a_{0} + 3) x - 8 a_{0}^{3} - 8 a_{0}^{2} + 2

Solution steps

Compute the tangent line to the general point

x = a_{0}

Find the tangent point:

(a_{0}, 4 a_{0}^{3} + 8 a_{0}^{2} + 3 a_{0} + 2)

Find the slope of

y = 4 x^{3} + 8 x^{2} + 3 x + 2 : \frac{d y}{d x} = 12 x^{2} + 16 x + 3

Compute the slope of

y = 4 x^{3} + 8 x^{2} + 3 x + 2

x = a_{0} : m = 12 a_{0}^{2} + 16 a_{0} + 3

Find the line with slope m=

12 a_{0}^{2} + 16 a_{0} + 3

and passing through

(a_{0}, 4 a_{0}^{3} + 8 a_{0}^{2} + 3 a_{0} + 2) : y = (12 a_{0}^{2} + 16 a_{0} + 3) x - 8 a_{0}^{3} - 8 a_{0}^{2} + 2

y = (12 a_{0}^{2} + 16 a_{0} + 3) x - 8 a_{0}^{3} - 8 a_{0}^{2} + 2

e^{x^{3}}

\int \frac{1}{( 1 + x )} d x

y^{^{''}} = y^{^{'}} + 2 y + 9 e^{2 x}

\int 11 sin^{2} (x co) s^{2} x d x

\int 2 x^{3} e^{- x^{2}} d x

What is the tangent of 4x^3+8x^2+3x+2 ?
The tangent of 4x^3+8x^2+3x+2 is y=(12a_{0}^2+16a_{0}+3)x-8a_{0}^3-8a_{0}^2+2