# What is the slope of the line normal to the tangent line of #f(x) = 2x-4/sqrt(x-1) # at # x= 2 #?

##### 1 Answer

Jul 14, 2017

The normal has equation

#### Explanation:

If we plug the value of

#f(2) = 2(2) - 4/sqrt(2 - 1) = 4 - 4/1 = 0#

We now find the derivative.

#f'(x) = 2 - 4/(2sqrt(x - 1))#

#f'(x) = 2 - 2/(sqrt(x- 1))#

So at

#f'(2) = 2 - 2/sqrt(2 - 1) = 2 - 2 = 0#

This would be a horizontal line. Since the normal line is perpendicular to the tangent, the normal line will be vertical, in the form

Hopefully this helps!