What is the tangent of y=x^2-3,(-1,-2) ?

The tangent of y=x^2-3,(-1,-2) is y=-2x-4

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

tangent of y=x^2-3,(-1,-2)

tangent of

y = x^{2} - 3, (- 1, - 2)

y = - 2 x - 4

Solution steps

Find the slope of

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

Compute the slope of

y = x^{2} - 3

(- 1, - 2) : m = - 2

Find the line with slope m=

- 2

and passing through

(- 1, - 2) : y = - 2 x - 4

y = - 2 x - 4

f (x) = 7 x^{3} e^{- x}

\int \frac{1}{y ^{\frac{1}{2}}} d y

\int e^{2 x} sin (e^{2 x}) d x

\int 4 t^{3} + t d t

y = 10000 + 30 x e^{\frac{x}{800}}

What is the tangent of y=x^2-3,(-1,-2) ?
The tangent of y=x^2-3,(-1,-2) is y=-2x-4