What is the tangent of y=2x^3,(-2,-16) ?

The tangent of y=2x^3,(-2,-16) is y=24x+32

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

tangent of y=2x^3,(-2,-16)

tangent of

y = 2 x^{3}, (- 2, - 16)

y = 24 x + 32

Solution steps

Find the slope of

y = 2 x^{3} : \frac{d y}{d x} = 6 x^{2}

Compute the slope of

y = 2 x^{3}

(- 2, - 16) : m = 24

Find the line with slope m=

24

and passing through

(- 2, - 16) : y = 24 x + 32

y = 24 x + 32

\int_{1}^{- 1} (12 x + 11)^{2} d x

\frac{d}{d x} (arcsin (x ln (x)))

\int_{0}^{1} \frac{5}{1 + x ^{2}} d x

(1 + \frac{y}{x ^{2}}) d x - \frac{1}{x} d y = 0

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

What is the tangent of y=2x^3,(-2,-16) ?
The tangent of y=2x^3,(-2,-16) is y=24x+32