What is the equation of the normal line of f(x)=(x-1)(x-3)(x+2)  at x=3 ?

Dec 18, 2015

$x + 10 y = 3$

Explanation:

Given:
$\textcolor{w h i t e}{\text{XXX}} f \left(x\right) = \left(x - 1\right) \left(x - 3\right) \left(x + 2\right)$

$f \left(3\right) = 0$ (since $\left(x - 3\right) = 0$ when $x = 3$)

So the point of intersection is going to be at $\left(x , y\right) = \left(3 , 0\right)$

Expanding the given equation we get
$\textcolor{w h i t e}{\text{XXX}} f \left(x\right) = {x}^{3} - 2 {x}^{2} - 5 x + 6$

So
$\textcolor{w h i t e}{\text{XXX}} f ' \left(x\right) = 3 {x}^{2} - 4 x - 5$

And at $\left(3 , 0\right)$ the tangent slope will be
$\textcolor{w h i t e}{\text{XXX}} f ' \left(3\right) = 3 \left({3}^{2}\right) - 4 \left(3\right) - 5 = 10$

Therefore the slope of the normal will be $\left(- \frac{1}{10}\right)$

Using the slope-point form for the normal we have
$\textcolor{w h i t e}{\text{XXX}} \left(y - 0\right) = \left(- \frac{1}{10}\right) \left(x - 3\right)$

which can be simplified as
$\textcolor{w h i t e}{\text{XXX}} x + 10 y = 3$

(Sorry about the image quality. I've been having trouble with the graph method and this is an attempt to overlay 2 separate graphs snipped from the screen).