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

Jul 20, 2016

Slope of the normal to $f \left(x\right)$ at $x = 0$ is $\frac{1}{2}$

#### Explanation:

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

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

To find the slope of $f \left(x\right)$ at $x = 0$:
$f ' \left(0\right) = - 1 + 0 - 1 = - 2$

Hence slope of the tangent $\left(m\right)$ to $f \left(x\right)$ at $x = 0$ is $- 2$

Since the slope of the normal at the same point = $- \frac{1}{m}$
Therefore the slope of the normal to $f \left(x\right)$ at $x = 0 = - \frac{1}{- 2} = \frac{1}{2}$