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

Jul 14, 2017

The normal has equation $x = 2$.

#### Explanation:

If we plug the value of $x$ into the function, we get:

$f \left(2\right) = 2 \left(2\right) - \frac{4}{\sqrt{2 - 1}} = 4 - \frac{4}{1} = 0$

We now find the derivative.

$f ' \left(x\right) = 2 - \frac{4}{2 \sqrt{x - 1}}$

$f ' \left(x\right) = 2 - \frac{2}{\sqrt{x - 1}}$

So at $x = 2$, the tangent would have slope.

$f ' \left(2\right) = 2 - \frac{2}{\sqrt{2 - 1}} = 2 - 2 = 0$

This would be a horizontal line. Since the normal line is perpendicular to the tangent, the normal line will be vertical, in the form $x = a$. Our initial $x$ value was $x = 2$, which will be the equation of the normal.

Hopefully this helps!