# What is the equation of the normal line of f(x)=-4x-10  at x=-2 ?

##### 1 Answer
Jun 9, 2016

$y + 2 = \frac{1}{4} \left(x + 2\right)$
or
$y = \frac{x}{4} - \frac{3}{2}$

#### Explanation:

The normal line at $x = - 2$ is the line perpendicular to the tangent line at this point. The slope of the tangent line is simply the derivative evaluated at $x = - 2$.

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

The slope of the line normal to the tangent line is the negative reciprocal of this i.e. $m = \frac{1}{4}$. Now we have a slope, but we still need a point to write the line. So, evaluate the function at $x = - 2$.

$f \left(- 2\right) = - 4 \left(- 2\right) - 10 = - 2$

We now have the slope and a point, so use point slope form to acquire the equation of the normal line.

$y + 2 = \frac{1}{4} \left(x + 2\right)$
or
$y = \frac{x}{4} - \frac{3}{2}$