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

Jan 30, 2017

$\frac{16}{5} {e}^{- 4} x - y + \frac{1}{4} {e}^{4} = 0$ that gives $0.0586 - y + 13.65$, nearly. See the normal-inclusive Socratic graph..

#### Explanation:

$f \left(x - 4\right) = {e}^{4} {e}^{x}$. Differentiating,

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

At $P \left(0 , {e}^{4} / 4\right) ,$

$i ' = - \frac{5}{16} {e}^{4}$

The slope of the normal is

$- \frac{1}{f '} = \frac{16}{5} {e}^{- 4}$. So, the equation to the normal at $P \left(0 , \frac{1}{4} {e}^{4}\right)$ is

$y - \frac{1}{4} {e}^{4} = \frac{16}{5} {e}^{- 4} x$.
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]}