What is the tangent of f(x)=4x^2+4x,(-2,8) ?

The tangent of f(x)=4x^2+4x,(-2,8) is y=-12x-16

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

tangent of f(x)=4x^2+4x,(-2,8)

tangent of

f (x) = 4 x^{2} + 4 x, (- 2, 8)

y = - 12 x - 16

Solution steps

Find the slope of

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

Compute the slope of

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

(- 2, 8) : m = - 12

Find the line with slope m=

- 12

and passing through

(- 2, 8) : y = - 12 x - 16

y = - 12 x - 16

\int - 8 x e^{x} d x

\int sin (x) \cdot sin (2 x) \cdot sin (3 x) d x

\int \frac{x}{x - 17} d x

\int (2 - x)^{- \frac{2}{3}} d x

(1 + x^{5}) y^{^{'}} + 5 x^{4} y = 13 x^{4}, y (0) = 2

What is the tangent of f(x)=4x^2+4x,(-2,8) ?
The tangent of f(x)=4x^2+4x,(-2,8) is y=-12x-16