What is the tangent of y=8x^2+8,(2,40) ?

The tangent of y=8x^2+8,(2,40) is y=32x-24

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

tangent of y=8x^2+8,(2,40)

tangent of

y = 8 x^{2} + 8, (2, 40)

y = 32 x - 24

Solution steps

Find the slope of

y = 8 x^{2} + 8 : \frac{d y}{d x} = 16 x

Compute the slope of

y = 8 x^{2} + 8

(2, 40) : m = 32

Find the line with slope m=

32

and passing through

(2, 40) : y = 32 x - 24

y = 32 x - 24

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

\frac{d}{d x} (5 x + x^{2})

x \to 5 + lim (5 x - 5)

\frac{\partial}{\partial t} (t^{2} e^{- x})

\frac{1}{s ^{2} + s + 1} \cdot \frac{1}{s}

What is the tangent of y=8x^2+8,(2,40) ?
The tangent of y=8x^2+8,(2,40) is y=32x-24