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

Oct 17, 2017

$y + {e}^{4} = \frac{x - 2}{{e}^{4} - 4}$

#### Explanation:

First, we take the derivative of the function:

$f ' \left(x\right) = \frac{d}{\mathrm{dx}} \left({x}^{2} - 4\right) - \frac{d}{\mathrm{dx}} \left({e}^{x + 2}\right)$

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

Therefore, at $x = \textcolor{red}{2}$, the slope of the tangent line is:

$f ' \left(\textcolor{red}{2}\right) = 2 \left(\textcolor{red}{2}\right) - {e}^{\textcolor{red}{2} + 2}$

$f ' \left(\textcolor{red}{2}\right) = 4 - {e}^{4}$

Since the normal line is perpendicular to the tangent line, its slope will be the negative reciprocal of the tangent line's slope. Let's call that slope $m$.

$m = \frac{- 1}{f ' \left(\textcolor{red}{2}\right)} = \frac{- 1}{4 - {e}^{4}} = \frac{1}{{e}^{4} - 4}$

We know the slope of the line; now, let's find the point where it intersects the curve. This is at $x = 2$, so:

$f \left(\textcolor{red}{2}\right) = \left({\textcolor{red}{2}}^{2} - 4\right) - {e}^{\textcolor{red}{2} + 2}$

$f \left(\textcolor{red}{2}\right) = 0 - {e}^{4}$

$f \left(\textcolor{red}{2}\right) = - {e}^{4}$

So the curve passes through the point $\left(\textcolor{red}{2} , \textcolor{b l u e}{- {e}^{4}}\right)$. Therefore, using the point-slope formula, the equation of the normal line is:

$\left(y - {y}_{1}\right) = m \left(x - {x}_{1}\right)$

$\left(y - \left(\textcolor{b l u e}{- {e}^{4}}\right)\right) = \frac{1}{{e}^{4} - 4} \left(x - \textcolor{red}{2}\right)$

$y + {e}^{4} = \frac{x - 2}{{e}^{4} - 4}$

You can manipulate this however you want to get whatever form you need.