# What is the equation of the line that is normal to f(x)=(x+4)^2-e^x at  x=-3 ?

Jun 10, 2018

$y = \frac{{e}^{3} x}{1 - 2 {e}^{3}} + \frac{3 {e}^{3}}{1 - 2 {e}^{3}} + \frac{{e}^{3} - 1}{e} ^ 3$

or " "y = (e^3x)/(1-2e^3) + (e^6 + 3e^3 - 1)/(e^3(1-2e^3)

#### Explanation:

Given: $f \left(x\right) = {\left(x + 4\right)}^{2} - {e}^{x} \text{ at } x = - 3$

First find the slope of the curve at $x = - 3$.

The first derivative is called the slope function. We need to find the first derivative:

$f ' \left(x\right) = 2 {\left(x + 4\right)}^{1} \left(1\right) - {e}^{x}$

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

slope = $m = f ' \left(- 3\right) = 2 \left(- 3\right) + 8 - {e}^{-} 3 = 2 - \frac{1}{e} ^ 3 = \frac{2 {e}^{3} - 1}{e} ^ 3$

A line that is normal is a perpendicular line. m_("perpendicular") = -1/m: " "-e^3/(2e^3 - 1)

$f \left(- 3\right) = {\left(- 3 + 4\right)}^{2} - {e}^{-} 3 = 1 - \frac{1}{e} ^ 3 = \frac{{e}^{3} - 1}{e} ^ 3$

To find the Normal line, use the point $\left(- 3 , \frac{{e}^{3} - 1}{e} ^ 3\right)$:

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

$y - \frac{{e}^{3} - 1}{e} ^ 3 = - {e}^{3} / \left(2 {e}^{3} - 1\right) \left(x + 3\right)$

$y - \frac{{e}^{3} - 1}{e} ^ 3 = - {e}^{3} / \left(2 {e}^{3} - 1\right) x - \frac{3 {e}^{3}}{2 {e}^{3} - 1}$

$y = - {e}^{3} / \left(2 {e}^{3} - 1\right) x - \frac{3 {e}^{3}}{2 {e}^{3} - 1} + \frac{{e}^{3} - 1}{e} ^ 3$

$y = \frac{{e}^{3} x}{1 - 2 {e}^{3}} + \frac{3 {e}^{3}}{1 - 2 {e}^{3}} + \frac{{e}^{3} - 1}{e} ^ 3$

Find a common denominator for the last two terms:

y = (e^3x)/(1-2e^3) + (e^6 + 3e^3 - 1)/(e^3(1-2e^3)