# What is the equation of the normal line of f(x)=e^-x+x^3 at x=-5?

Oct 31, 2016

$y = - \frac{x}{{e}^{5} - 75} + {e}^{5} - 125 - \frac{5}{{e}^{5} - 75}$

#### Explanation:

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

When $x = - 5 \implies f \left(- 5\right) = {e}^{5} - 125$

Differentiating wrt $x$;
$f \left(x\right) = - {e}^{-} x + 3 {x}^{2}$
so When $x = - 5 \implies f ' \left(- 5\right) = - {e}^{5} + 75 = 75 - {e}^{5}$

This is the slope of the tangent when $x = - 5$, as the normal is perpendicular to the tangent, their product is $- 1$

So, the slope of the normal is $- \frac{1}{75 - {e}^{5}} = \frac{1}{{e}^{5} - 75}$

Using $y - {y}_{1} = m \left(x - {x}_{1}\right)$, the normal equation is;

$y - \left({e}^{5} - 125\right) = \frac{1}{{e}^{5} - 75} \left(x - \left(- 5\right)\right)$
$y - {e}^{5} + 125 = \frac{1}{{e}^{5} - 75} \left(x + 5\right)$
$y - {e}^{5} + 125 = \frac{x}{{e}^{5} - 75} + \frac{5}{{e}^{5} - 75}$
$y = \frac{x}{{e}^{5} - 75} + {e}^{5} - 125 + \frac{5}{{e}^{5} - 75}$