# Question #a733b

May 29, 2017

$d \left(x\right) = \frac{1}{3} x - \frac{16}{3}$

#### Explanation:

I haven't really heard the term "normal" all that much, but if I remember rightly, it can interpretted as the line that cuts another line with an angle of 90 degrees.

First write the polynomial in standard form (optional):

$\left({x}^{2} - 2 x\right) \left(3 x + 2\right) = x \left(x - 2\right) \left(3 x + 2\right) = x \left(3 {x}^{2} - 4 x - 4\right) = 3 {x}^{3} - 4 {x}^{2} - 4 x$

Therefore $f \left(x\right) = 3 {x}^{3} - 4 {x}^{2} - 4 x$

Second:
f(x) in itself does not consist of a single line that can be cut at a 90 degrees angle, but a tangent on the graph in the point $P \left(1 , - 5\right)$ on the other hand. Now that can be cut at a 90 degrees angle!

So we need to find the tangent on the graph in the point "P".
To do this we find the derivative of f(x).
$\frac{d}{\mathrm{dx}} f \left(x\right) = f ' \left(x\right) = \left(3 {x}^{3}\right) ' - \left(4 {x}^{2}\right) ' - \left(4 x\right) ' = 9 {x}^{2} - 8 x - 4$

This is the general "inclination" (Don't know what it is called in english) of the tangent for all x.
Therefore the tangent will have the inclination $f ' \left(1\right) = 9 \cdot {1}^{2} - 8 \cdot 1 - 4 = - 3$ in the point $P \left(1 , - 5\right)$.

To find the tangent on the graph we use the formula:
$h \left(x\right) = f ' \left({x}_{\text{0")(x-x_"0")+f(x_"0}}\right)$ , ${x}_{\text{0}} = 1$
Therefore:
$h \left(x\right) = - 3 \left(x - 1\right) - 5 = - 3 x - 2$

Third (Had written a sort of proof for this step, but it got too long. I have included the pictures though, so you can maybe get an idea of where I was going from): Normally, we write the inclination of linear equations as $a = \frac{\Delta y}{\Delta x}$. The inclination of the tangent in this case is $- 3 = - \frac{3}{1}$ , meaning -3 y for every 1 x. Using the fourth picture, this means $a = 1$ and $b = 3$, we can see that to get a line that cuts this line with a 90 degrees angle, we just need to flip the fraction $- \frac{3}{1}$ to $\frac{1}{3}$. Instead of going out 1 down 3, we go out 3 up 1.

Generally the rule is:
If a linear equation is formed by $f \left(x\right) = a x + b$ , then the equation $g \left(x\right) = - {a}^{-} 1 x + c$ is a normal on f(x).

Fourth:
Last, we define a third function d(x) that will be a normal on h(x), and that will go through $P \left(1 , - 5\right)$.
We get:
$d \left(1\right) = - {a}^{-} 1 \cdot 1 + c = \frac{1}{3} + c = - 5 \iff c = - \frac{16}{3}$

Therefore:
$d \left(x\right) = \frac{1}{3} x - \frac{16}{3}$

Here are all the graphs:

$f \left(x\right) = 3 {x}^{3} - 4 {x}^{2} - 4 x$
graph{3x^3-4x^2-4x [-8.12, 11.88, -7.71, 2.71]}
$h \left(x\right) = - 3 x - 2$
graph{-3x-2 [-14.92, 25.08, -14.34, 6.5]}
$d \left(x\right) = \frac{1}{3} x - \frac{16}{3}$
graph{1/3x-16/3 [-7.42, 32.58, -15.1, 5.74]}