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

The tangent of f(x)=(x-5)^2,\at x=1 is y=-8x+24

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-5)^2,\at x=1

tangent of

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

y = - 8 x + 24

Solution steps

Find the tangent point:

(1, 16)

Find the slope of

f (x) = (x - 5)^{2} : \frac{df ( x )}{d x} = 2 (x - 5)

Compute the slope of

f (x) = (x - 5)^{2}

x = 1 : m = - 8

Find the line with slope m=

- 8

and passing through

(1, 16) : y = - 8 x + 24

y = - 8 x + 24

\int_{0}^{l n (4)} 3 (4 - x e^{x}) d x

(1 + x^{2}) y^{^{'}} = x y + x^{2} y^{2}

f (x) = \frac{1}{7 x}, at x = 9

y^{^{''}} - 2 x y^{^{'}} = 0

\int_{- 2}^{62} 2 π (y) (2) d y

What is the tangent of f(x)=(x-5)^2,\at x=1 ?
The tangent of f(x)=(x-5)^2,\at x=1 is y=-8x+24