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

Jul 18, 2016

$\approx - 0.01$

#### Explanation:

We know that the slope of the tangent at $x = 2$ is $f ' \left(2\right)$. Now

$f ' \left(x\right) = 2 {e}^{2 x} - 3 {x}^{2}$

so that

$f ' \left(2\right) = 2 {e}^{4} - 12 \approx 97.2$

Thus the slope of the line normal to this is $- \frac{1}{97.2} \approx - 0.01$