What is the tangent of f(x)=sqrt(x-3),(12,3) ?

The tangent of f(x)=sqrt(x-3),(12,3) is y= 1/6 x+1

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

tangent of f(x)=sqrt(x-3),(12,3)

tangent of

f (x) = x - 3, (12, 3)

y = \frac{1}{6} x + 1

Solution steps

Find the slope of

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

EN : T i tl e G e n er a l Eq u a t i o n Sl o p e A t P o in t 2 Eq : m = \frac{1}{6}

Find the line with slope m=

\frac{1}{6}

and passing through

(12, 3) : y = \frac{1}{6} x + 1

y = \frac{1}{6} x + 1

What is the tangent of f(x)=sqrt(x-3),(12,3) ?
The tangent of f(x)=sqrt(x-3),(12,3) is y= 1/6 x+1