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

The tangent of f(x)=(x-1)^2+1,\at x=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-1)^2+1,\at x=0

tangent of

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

y = - 2 x + 2

Solution steps

Find the tangent point:

(0, 2)

Find the slope of

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

Compute the slope of

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

x = 0 : m = - 2

Find the line with slope m=

- 2

and passing through

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

y = - 2 x + 2

(1 + x^{2})^{\frac{1}{4}}

\frac{10 x + 8 x ^{2}}{( 5 + 8 x ) ^{2}}

\int (6 x^{2}) (2 x^{3} + 1)^{2} d x

2 sec (θ)

\frac{d y}{d x} = - 5 y

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