What is the tangent of f(x)=x^2-1,(-1,0) ?

The tangent of f(x)=x^2-1,(-1,0) is y=-2x-2

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,(-1,0)

tangent of

f (x) = x^{2} - 1, (- 1, 0)

y = - 2 x - 2

Solution steps

Find the slope of

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

Compute the slope of

f (x) = x^{2} - 1

(- 1, 0) : m = - 2

Find the line with slope m=

- 2

and passing through

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

y = - 2 x - 2

\int_{0}^{π} 8 cos^{4} (x) sin (x) d x

\int 3 x^{2} + 1 d x

\frac{d}{d x} (\frac{x}{x ^{2} + 1} - 1)

\frac{d y}{d x} = \frac{x ^{2} y - y}{y + 1}

What is the tangent of f(x)=x^2-1,(-1,0) ?
The tangent of f(x)=x^2-1,(-1,0) is y=-2x-2