What is the tangent of y=x^2+3,(-3,12) ?

The tangent of y=x^2+3,(-3,12) is y=-6x-6

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

tangent of y=x^2+3,(-3,12)

tangent of

y = x^{2} + 3, (- 3, 12)

y = - 6 x - 6

Solution steps

Find the slope of

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

EN : T i tl e G e n er a l Eq u a t i o n Sl o p e A t P o in t 2 Eq : m = - 6

Find the line with slope m=

- 6

and passing through

(- 3, 12) : y = - 6 x - 6

y = - 6 x - 6

What is the tangent of y=x^2+3,(-3,12) ?
The tangent of y=x^2+3,(-3,12) is y=-6x-6