# What is the equation of the line normal to f(x)=x^3 - 7x^2-x  at x=4?

May 9, 2016

$x - 17 y = 880$

#### Explanation:

The slope of an equation is given by its derivative.

If $f \left(x\right) = {x}^{3} - 7 {x}^{2} - x$
then $f ' \left(x\right) = 3 {x}^{2} - 14 x - 1$
and
at $x = 4$
the slope is $m = f ' \left(4\right) = 3 \left({4}^{2}\right) - 14 \left(4\right) - 1 = - 17$

If a line has a slope of $m$ then any line perpendicular to it (i.e. its "normal") has a slope of $- \frac{1}{m}$

At $x = 4$
$\textcolor{w h i t e}{\text{XXX}} f \left(4\right) = {4}^{3} - 7 \left({4}^{2}\right) - 4 = - 52$

So the required normal line
$\textcolor{w h i t e}{\text{XXX}}$has a slope of $\frac{1}{17}$
and
$\textcolor{w h i t e}{\text{XXX}}$passes through the point $\left(4 , - 52\right)$

Using the slope-point form:
$\textcolor{w h i t e}{\text{XXX}} y - \left(- 52\right) = \frac{1}{17} \left(x - 4\right)$

$\textcolor{w h i t e}{\text{XXX}} \rightarrow 17 y + 884 = x - 4$

or, in standard form:
$\textcolor{w h i t e}{\text{XXX}} x - 17 y = 880$