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

The tangent of f(x)=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 f(x)=x^2+1,(-2,5)

tangent of

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

y = - 4 x - 3

Solution steps

Find the slope of

f (x) = x^{2} + 1 : \frac{df ( x )}{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

\int \frac{x}{( 1 + x ^{3} )} d x

x \to 0 + lim (sin (x))

\int (a x^{3} - b x^{2} - c x + d) d x

\int_{- 2}^{4} ∣ x ∣ d x

\frac{d y}{d x} = x e^{y}

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