What is the tangent of f(x)=2x^3+5x^2,\at x=-2 ?

The tangent of f(x)=2x^3+5x^2,\at x=-2 is y=4x+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)=2x^3+5x^2,\at x=-2

tangent of

f (x) = 2 x^{3} + 5 x^{2}, at x = - 2

y = 4 x + 12

Solution steps

Find the tangent point:

(- 2, 4)

Find the slope of

f (x) = 2 x^{3} + 5 x^{2} : \frac{df ( x )}{d x} = 6 x^{2} + 10 x

Compute the slope of

f (x) = 2 x^{3} + 5 x^{2}

x = - 2 : m = 4

Find the line with slope m=

4

and passing through

(- 2, 4) : y = 4 x + 12

y = 4 x + 12

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

\frac{\partial}{\partial x} (2 x^{2} y e^{x} + 2)

x \frac{d y}{d x} + (2 x + 1) y = e^{- 2 x}

\int \frac{1}{5 x ^{2} + 20 x - 10} d x

x \to \frac{π}{3} lim (\frac{1}{sec ( x )})

What is the tangent of f(x)=2x^3+5x^2,\at x=-2 ?
The tangent of f(x)=2x^3+5x^2,\at x=-2 is y=4x+12