What is the tangent of f(x)=x^4-4x^3+2x^2-6,\at x=1 ?

The tangent of f(x)=x^4-4x^3+2x^2-6,\at x=1 is y=-4x-3

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

tangent of f(x)=x^4-4x^3+2x^2-6,\at x=1

tangent of

f (x) = x^{4} - 4 x^{3} + 2 x^{2} - 6, at x = 1

y = - 4 x - 3

Solution steps

Find the tangent point:

(1, - 7)

Find the slope of

f (x) = x^{4} - 4 x^{3} + 2 x^{2} - 6 : \frac{df ( x )}{d x} = 4 x^{3} - 12 x^{2} + 4 x

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

Find the line with slope m=

- 4

and passing through

(1, - 7) : y = - 4 x - 3

y = - 4 x - 3

What is the tangent of f(x)=x^4-4x^3+2x^2-6,\at x=1 ?
The tangent of f(x)=x^4-4x^3+2x^2-6,\at x=1 is y=-4x-3