What is the tangent of x^2+4x-8 ?

The tangent of x^2+4x-8 is y=(2a_{0}+4)x-a_{0}^2-8

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

tangent of x^2+4x-8

tangent of

x^{2} + 4 x - 8

y = (2 a_{0} + 4) x - a_{0}^{2} - 8

Solution steps

Compute the tangent line to the general point

x = a_{0}

Find the tangent point:

(a_{0}, a_{0}^{2} + 4 a_{0} - 8)

Find the slope of

y = x^{2} + 4 x - 8 : \frac{d y}{d x} = 2 x + 4

Compute the slope of

y = x^{2} + 4 x - 8

x = a_{0} : m = 2 a_{0} + 4

Find the line with slope m=

2 a_{0} + 4

and passing through

(a_{0}, a_{0}^{2} + 4 a_{0} - 8) : y = (2 a_{0} + 4) x - a_{0}^{2} - 8

y = (2 a_{0} + 4) x - a_{0}^{2} - 8

x \to 0 + lim (x^{5} ln (x))

x \to 1 - lim (6 x - 2)

\frac{\partial}{\partial x} (3 x^{2} y^{2} + \frac{x}{y})

y = sin (3 x + 2 y)

\int (5 - 3 x + x^{2}) d x

What is the tangent of x^2+4x-8 ?
The tangent of x^2+4x-8 is y=(2a_{0}+4)x-a_{0}^2-8