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

The tangent of f(x)=x^2+3,\at x=-1 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+3,\at x=-1

tangent of

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

y = - 2 x + 2

Solution steps

Find the tangent point:

(- 1, 4)

Find the slope of

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

Compute the slope of

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

x = - 1 : m = - 2

Find the line with slope m=

- 2

and passing through

(- 1, 4) : y = - 2 x + 2

y = - 2 x + 2

f (x) = (3 x^{6} + 2 x^{3})^{4}

\frac{d y}{d x} = y (3 y - 6)

\frac{d}{d x} (\frac{1}{3 x ^{4} + 1})

\int sec^{2} (2 x) - 1 d x

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