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

Dec 3, 2015

In point slope form it is
$y - {e}^{4} / \left({e}^{2} + 4\right) = {\left({e}^{2} + 4\right)}^{2} / \left({e}^{6} + 8 {e}^{4}\right) \left(x - 2\right)$.

#### Explanation:

For $f \left(x\right) = {e}^{2 x} / \left({e}^{x} + 4\right)$ at $x = 2$, we get $y = {e}^{4} / \left({e}^{2} + 4\right)$.

The slope of the tangent line is given by:

$f ' \left(x\right) = \frac{\left(2 {e}^{2 x}\right) \left({e}^{x} + 4\right) - {e}^{2 x} \left({e}^{x}\right)}{{e}^{x} + 4} ^ 2 = \frac{{e}^{3 x} + 8 {e}^{2} x}{{e}^{x} + 4} ^ 2$.

At the point where $x = 2$, the slope of the tangent is

${m}_{\tan} = \frac{{e}^{6} + 8 {e}^{4}}{{e}^{2} + 4} ^ 2$.

So the slope of the normal line is

m_(norm) = (e^2+4)^2/ (e^6+8e^4).

The equation of the line (in point slope form) through $\left(2 , {e}^{4} / \left({e}^{2} + 4\right)\right)$ with slope $m = {\left({e}^{2} + 4\right)}^{2} / \left({e}^{6} + 8 {e}^{4}\right)$ is

$y - {e}^{4} / \left({e}^{2} + 4\right) = {\left({e}^{2} + 4\right)}^{2} / \left({e}^{6} + 8 {e}^{4}\right) \left(x - 2\right)$.

Rewrite using algebra as you see fit. (Or as your grader demands.)