# What is the slope of the line normal to the tangent line of f(x) = e^(x-2)+x-2  at  x= 0 ?

Apr 13, 2018

${e}^{2} / \left(2 - {e}^{2}\right)$

#### Explanation:

Differentiating,

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

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

Slope of the tangent at $x = 0$ is $\frac{{e}^{2} - 2}{e} ^ 2$

Slope of the normal at $x = 0$ is ${e}^{2} / \left(2 - {e}^{2}\right)$