What is the tangent of 3x^2+4x+1 ?

The tangent of 3x^2+4x+1 is y=(6a_{0}+4)x-3a_{0}^2+1

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

tangent of 3x^2+4x+1

tangent of

3 x^{2} + 4 x + 1

y = (6 a_{0} + 4) x - 3 a_{0}^{2} + 1

Solution steps

Compute the tangent line to the general point

x = a_{0}

Find the tangent point:

(a_{0}, 3 a_{0}^{2} + 4 a_{0} + 1)

Find the slope of

y = 3 x^{2} + 4 x + 1 : \frac{d y}{d x} = 6 x + 4

Compute the slope of

y = 3 x^{2} + 4 x + 1

x = a_{0} : m = 6 a_{0} + 4

Find the line with slope m=

6 a_{0} + 4

and passing through

(a_{0}, 3 a_{0}^{2} + 4 a_{0} + 1) : y = (6 a_{0} + 4) x - 3 a_{0}^{2} + 1

y = (6 a_{0} + 4) x - 3 a_{0}^{2} + 1

\frac{d}{d x} (\frac{x}{2 x - 1})

\frac{d y}{d t} = - 4 y

\int 100 x d x

4 π r^{2}

2^{3 x}

What is the tangent of 3x^2+4x+1 ?
The tangent of 3x^2+4x+1 is y=(6a_{0}+4)x-3a_{0}^2+1