This document discusses finding a quintic (5th degree) function to approximate motion data. It explains that while the calculator can do quadratic, cubic, and quartic regression, it cannot do quintic regression. To find the quintic function, six points are needed to create six equations with the six unknown coefficients. The key points chosen are points of extrema (maximums and minimums) and inflection, which are used to set up a system of six equations that can be solved for the coefficients.

2. In part one of Limits and the Motion of Objects you
plotted data that you were given and used limits to
approximate the velocity of the object at a specific point
by finding the slope of the graph.
In part II of the project you will use your calculator to
find a function that approximates the data. You will then
use that function and the analytic capabilities of your
calculator to find the velocity of the object at a specific
point by finding the slope of the function

3. So far, if you connected the plotted points, you should
have a graph that looks something like this,

4. But the plotted points you were given were
incomplete. Adding additional points gives you the
graph,

5. Clearly this is a fifth degree or quintic function. This is
unfortunate because although your calculator has
quadratic, cubic and quartic regression capabilities, it
does not have quintic regression capabilities.

6. A quintic function is of the form,
𝑓 𝑥 = 𝑎5 𝑥5
+ 𝑎4 𝑥4
+ 𝑎3 𝑥3
+ 𝑎2 𝑥2
+ 𝑎1 𝑥 + 𝑎0
To find this particular function we need to find,
𝑎5, 𝑎4, 𝑎3, 𝑎2, 𝑎1, and 𝑎0
Six unknowns will require six equations and six equation
will require six points on the line. But which six points
should be chosen?

7. The point you want to chose would be the critical points.
In calculus we will rigorously define critical points, but for
now we will define them as the points of extrema and the
points of inflection.
For the graph drawn we have extrema at (-1.4, 161),
(0.9, 0.0), (2.4, 34.1), and (4.0, -32.7).
For the graph drawn we have points of inflections at
(0.0, 45.5), (1.6, 15.7), and (3.2, 4.1).
This give us 7 points, one more then we need so solve
the system.

8. I will chose to disregard the maximum at 𝑥 = 2.4 mainly
because the two adjacent points of inflection guarantee
the maximum will exist in the solution to the system of
equations. So the system becomes,
𝑎5 −1.4 5 + 𝑎4 −1.4 4 + 𝑎3 −1.4 3 + 𝑎2 −1.4 2 + 𝑎1 −1.4 + 𝑎0 = 161
𝑎5 0.0 5 + 𝑎4 0.0 4 + 𝑎3 0.0 3 + 𝑎2 0.0 2 + 𝑎1 0.0 + 𝑎0 = 45.5
𝑎5 0.9 5 + 𝑎4 0.9 4 + 𝑎3 0.9 3 + 𝑎2 0.9 2 + 𝑎1 0.9 + 𝑎0 = 0.0
𝑎5 1.6 5 + 𝑎4 1.6 4 + 𝑎3 1.6 3 + 𝑎2 1.6 2 + 𝑎1 1.6 + 𝑎0 = 15.7
𝑎5 3.2 5 + 𝑎4 3.2 4 + 𝑎3 3.2 3 + 𝑎2 3.2 2 + 𝑎1 3.2 + 𝑎0 = 4.1
𝑎5 4.0 5 + 𝑎4 4.0 4 + 𝑎3 4.0 3 + 𝑎2 4.0 2 + 𝑎1 4.0 + 𝑎0 = −32.7