What is the tangent of y=x^2+1,(-2,5) ?

The tangent of y=x^2+1,(-2,5) is y=-4x-3

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+1,(-2,5)

tangent of

y = x^{2} + 1, (- 2, 5)

y = - 4 x - 3

Solution steps

Find the slope of

y = x^{2} + 1 : \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 = - 4

Find the line with slope m=

- 4

and passing through

(- 2, 5) : y = - 4 x - 3

y = - 4 x - 3

x, at 25

x^{4} - 2 x^{2} + 4

\frac{\partial}{\partial x} (\frac{5 x - 9 y}{5 x + 9 y})

\int x^{2} ln (3 x) d x

\int_{0}^{v} v d v

What is the tangent of y=x^2+1,(-2,5) ?
The tangent of y=x^2+1,(-2,5) is y=-4x-3