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

Mar 7, 2017

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

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

$f ' \left(- 2\right) = 8 - \frac{2}{e} ^ 2 \leftarrow$ This is the gradient of the tangent.

$\text{m"_"n} = - \frac{1}{{m}_{\tan}} = - \frac{1}{8 - \frac{2}{e} ^ 2} = - \frac{{e}^{2}}{8 {e}^{2} - 2}$

$f \left(- 2\right) = - 8 - \frac{2}{e} ^ 2 = - \frac{8 {e}^{2} - 2}{e} ^ 2$

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

$y + \frac{8 {e}^{2} - 2}{e} ^ 2 = - \frac{{e}^{2}}{8 {e}^{2} - 2} \left(x + 2\right)$

$y + \frac{8 {e}^{2} - 2}{e} ^ 2 = - \frac{x {e}^{2}}{8 {e}^{2} - 2} - \frac{2 {e}^{2}}{8 {e}^{2} - 2}$

$y = \frac{x {e}^{2}}{1 - 8 {e}^{2}} - \frac{33 {e}^{4} - 16 {e}^{2} + 2}{{e}^{2} \left(2e-1\right) \left(2e+1\right)}$