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

The tangent of f(x)=x^2-2x-8,\at x=-2 is y=-6x-12

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

tangent of f(x)=x^2-2x-8,\at x=-2

tangent of

f (x) = x^{2} - 2 x - 8, at x = - 2

y = - 6 x - 12

Solution steps

Find the tangent point:

(- 2, 0)

Find the slope of

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

Compute the slope of

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

x = - 2 : m = - 6

Find the line with slope m=

- 6

and passing through

(- 2, 0) : y = - 6 x - 12

y = - 6 x - 12

y = \frac{( 3 x + 1 )}{2}, y = x^{2} - 2

x y \frac{d y}{d x} + y^{2} = 32 x

\int \frac{x}{4 - x ^{2}} d x

\int 2 x + 9 d x

2 t y^{^{'}} - 3 y = 0

What is the tangent of f(x)=x^2-2x-8,\at x=-2 ?
The tangent of f(x)=x^2-2x-8,\at x=-2 is y=-6x-12