# What is the equation of the normal line of #f(x)=xe^-x-x# at #x=2#?

##### 1 Answer

Feb 4, 2017

#### Explanation:

At x = 2, f =-2(1-e^(-2))=-1.72933, nearly.

The foot of the normal is P(2, -1.729), nearly

The slope of the normal at P is -1/f'= 0.7124, nearly.

So, the equation to the normal at P is

#y+1.729=0.7124(x-2), giving

graph{(x(e^(-x)-1)-y)((x-2)^2+(y+1.729)^2-.01)(0.7124x-y-3.1538)=0 [-10, 10, -5, 5]}