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

The tangent of f(x)=2x^2-4x,\at x=-1 is y=-8x-2

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

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

tangent of

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

y = - 8 x - 2

Solution steps

Find the tangent point:

(- 1, 6)

Find the slope of

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

Compute the slope of

f (x) = 2 x^{2} - 4 x

x = - 1 : m = - 8

Find the line with slope m=

- 8

and passing through

(- 1, 6) : y = - 8 x - 2

y = - 8 x - 2

f (x) = 4 x - x^{2}, (2, 5)

\int (2 x - 5)^{11} d x

\int_{- 1}^{2} \frac{x}{( x + 1 ) ^{2}} d x

\int sin (n t) d t

x^{^{'}} = 6 c_{1} - \frac{3 x}{5}

What is the tangent of f(x)=2x^2-4x,\at x=-1 ?
The tangent of f(x)=2x^2-4x,\at x=-1 is y=-8x-2