# What is the equation of the normal line of f(x)= x^8 + 6 e^x  at x=0?

Dec 25, 2015

$y - 6 = - \frac{1}{6} \left(x - 0\right)$ in the slope point form. $y = - \frac{1}{6} x + 6$ in slope intercept form. The detailed explanation is given below.

#### Explanation:

We need to find the equation of the normal line.

To find the equation of the line we need two things.
1. Slope $m$.
2. Point $\left({x}_{1} , {y}_{1}\right)$
Then the equation can be formed by using the slope point form

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

We are to find the equation of normal at $x = 0$ let us use that to find $f \left(0\right)$ this would give our points $\left({x}_{1} , {y}_{1}\right)$
$f \left(x\right) = {x}^{8} + 6 {e}^{x}$
$f \left(0\right) = {0}^{8} + 6 {e}^{0}$
$f \left(0\right) = 0 + 6$
$f \left(0\right) = 6$

Our $\left({x}_{1} , {y}_{1}\right) = \left(0 , 6\right)$

Now to find the slope of required equation. For this, we need to find the slope of the tangent. The slope of the tangent is found by finding $f ' \left(x\right)$

$f ' \left(x\right) = 8 {x}^{7} + 6 {e}^{x}$
Slope of tangent at $x = 0$
$f ' \left(0\right) = 8 {\left(0\right)}^{7} + 6 {e}^{0}$
$f ' \left(0\right) = 6$

The slope of the normal is the negative reciprocal of slope of the tangent.

Slope of normal $m = - \frac{1}{6}$

Equation of the normal line at $x = 0$ is

$y - 6 = - \frac{1}{6} \left(x - 0\right)$ in the slope point form.

$y = - \frac{1}{6} x + 6$ in slope intercept form.