What is the tangent of f(x)=2x^2+8x-2 ?

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

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

tangent of f(x)=2x^2+8x-2

tangent of

f (x) = 2 x^{2} + 8 x - 2

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

Solution steps

Compute the tangent line to the general point

x = a_{0}

Find the tangent point:

(a_{0}, 2 a_{0}^{2} + 8 a_{0} - 2)

Find the slope of

f (x) = 2 x^{2} + 8 x - 2 : \frac{df ( x )}{d x} = 4 x + 8

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 = 4 a_{0} + 8

Find the line with slope m=

4 a_{0} + 8

and passing through

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

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

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