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

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

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

tangent of 4x^2-3x+2

tangent of

4 x^{2} - 3 x + 2

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

Find the slope of

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

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 = 8 a_{0} - 3

Find the line with slope m=

8 a_{0} - 3

and passing through

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

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

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