# What is the equation of the line that is normal to f(x)=(x+4)^2e^x at  x=-3 ?

Feb 8, 2017

$\frac{1}{3} {e}^{3} x + y - {e}^{- 3} + {e}^{3} = 0$. See normal-inclusive Socratic graph

#### Explanation:

graph{(e^x(x+4)^2-y)(1/3e^3x+y-e^(-3)+e^3)=0 [-4, -2, -0.5, 0.5]}

y > 0.

As $x \to = \infty , y \to 0$, revealing that $y = 0 \leftarrow$ is the asymptote.

At $x = - 3 , y = {e}^{- 3}$.

So, the foot of the normal is $P \left(- 3 , {e}^{- 3}\right)$

$y ' = {e}^{x} \left({\left(x + 4\right)}^{2} + 2 \left(x + 4\right)\right) = 3 {e}^{- 3}$, at $x = - 3$.

The slope of the normal is $- \frac{1}{y '} = - \frac{1}{3} {e}^{3}$

And so, the equation to the normal at P is

$y - {e}^{- 3} = - \frac{1}{3} {e}^{3} \left(x + 3\right)$, giving

$\frac{1}{3} {e}^{3} x + y - {e}^{- 3} + {e}^{3} = 0$