# What is the slope of the line normal to the tangent line of f(x) = e^(x-2)+x-2  at  x= 2 ?

Apr 1, 2018

$- \frac{1}{2}$

#### Explanation:

In this case, the slope of the tangent line will be given by $f ' \left(2\right) .$

The normal line is the line perpendicular to the tangent line; therefore, its slope will be the negative reciprocal of the slope to the tangent line, or,

${m}_{n} = - \frac{1}{f ' \left(2\right)}$

Let's first differentiate the function:

$f ' \left(x\right) = {e}^{x - 2} + 1$.

Determine the slope of the tangent:

$f ' \left(2\right) = {e}^{2 - 2} + 1 = {e}^{0} + 1 = 2$

The slope of the normal line is then

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