# What is the equation of the normal line of f(x)=-4x-1 at x=2?

Jan 10, 2016

$y = \frac{x}{4} - \frac{19}{2}$

#### Explanation:

If the excercise does not clearly request the derivation use, you could also think:

$y = - 4 x - 1$

is a line, then the tangent line to it is the line itself

$y = m x + q$

with

$m = - 4$

and

$q = - 1$

${m}_{n} = - \frac{1}{m} = - \frac{1}{- 4} = \frac{1}{4}$

The normal line is

$\left(y - {y}_{0}\right) = {m}_{n} \left(x - {x}_{0}\right)$

with
${x}_{0} = 2$
${y}_{0} = f \left({x}_{0}\right) = y \left({x}_{0}\right) = - 4 \cdot 2 - 1 = - 8 - 1 = - 9$

then:

the normal line becomes:

$\left(y - \left(- 9\right)\right) = \left(\frac{1}{4}\right) \left(x - 2\right)$

$y + 9 = \frac{1}{4} x - \frac{1}{2}$
$y = \frac{x}{4} - \frac{1}{2} - 9 = \frac{x}{4} - \left(\frac{1 + 18}{2}\right) = \frac{x}{4} - \frac{19}{2}$