# What is the equation of the normal line of #f(x)=e^(4+x)/(4-x)# at #x=0#?

##### 1 Answer

Jan 30, 2017

#### Explanation:

f'(4-x)-f-e^4e^x=0.

At

The slope of the normal is

graph{(y(4-x)-e^(4+x))(y-13.65)(x^2+(y-13.65-.09x)^2-.1)=0 [-32.76, 32.71, 0, 32]}