# What is the equation of the line normal to f(x)=x^2+8x - 12 at x=3?

Jun 28, 2016

$x + 32 y = 675$

#### Explanation:

Given:
$\textcolor{w h i t e}{\text{XXX}} f \left(x\right) = {x}^{2} + 8 x - 12$

At $x = 3$
$\textcolor{w h i t e}{\text{XXX}} f \left(3\right) = {3}^{2} + 8 \left(3\right) - 12 = 21$

The slope of $f \left(x\right) = {x}^{2} + 8 x - 12$ for an arbitrary $x$ is given by the derivative:
$\textcolor{w h i t e}{\text{XXX}} {m}_{x} = f ' \left(x\right) = 2 x + 8$

At $x = 3$ the slope is
$\textcolor{w h i t e}{\text{XXX}} {m}_{3} = f ' \left(3\right) = 2 \left(3\right) + 8 = 32$

If $f \left(x\right)$ has a slope of $m$
then a line perpendicular to it (which the normal is) has a slope of $\left(- \frac{1}{m}\right)$

Therefore the normal to $f \left(x\right)$ at $x = 3$
passes through the point $\left(3 , 21\right)$ and a has a slope of $\left(- \frac{1}{32}\right)$

Using the slope-point form for a line
$\textcolor{w h i t e}{\text{XXX}} y - 21 = - \frac{1}{32} \left(x - 3\right)$

Converting to standard form:
$\textcolor{w h i t e}{\text{XXX}} 32 y - 672 = - x + 3$

$\textcolor{w h i t e}{\text{XXX}} x + 32 y = 675$

graph{(x^2+8x-12-y)(x+32y-675)=0 [-66, 65.6, -29, 36.85]}