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

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

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+3x+1,(-1,-1)

tangent of

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

y = x

Solution steps

Find the slope of

f (x) = x^{2} + 3 x + 1 : \frac{df ( x )}{d x} = 2 x + 3

Compute the slope of

f (x) = x^{2} + 3 x + 1

(- 1, - 1) : m = 1

Find the line with slope m=

1

and passing through

(- 1, - 1) : y = x

y = x

\int \frac{2}{( x + 1 ) ^{3}} d x

\frac{d}{d y} (x^{2} y^{2})

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

\frac{\partial}{\partial x} (x e^{2 t} - 4 x)

x, x - 2, 0

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