Question #a733b

1 Answer

#d(x) = 1/3x-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):

#(x^2-2x)(3x+2)=x(x-2)(3x+2)=x(3x^2-4x-4)=3x^3-4x^2-4x#

Therefore #f(x)=3x^3-4x^2-4x#

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(1,-5)# 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).
#d/dxf(x)=f'(x)=(3x^3)'-(4x^2)'-(4x)'=9x^2-8x-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'(1)=9*1^2-8*1-4=-3# in the point #P(1,-5)#.

To find the tangent on the graph we use the formula:
#h(x)=f'(x_"0")(x-x_"0")+f(x_"0")# , #x_"0"=1#
Therefore:
#h(x)=-3(x-1)-5=-3x-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):
enter image source here

Normally, we write the inclination of linear equations as #a=(Deltay)/(Deltax)#. The inclination of the tangent in this case is #-3=-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 #-3/1# to #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(x)=ax+b# , then the equation #g(x)=-a^-1x+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(1,-5)#.
We get:
#d(1)=-a^-1*1+c=1/3+c=-5 iff c = -16/3#

Therefore:
#d(x)=1/3x-16/3#

Here are all the graphs:

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