Write Up #3 :: Parabolic Transformations
and a Locus of Points
Clay Kitchings
Consider the
parabolas formed by f(x) = x^{2 }+
bx + 1, where b = 33.
After some
observations of what happens as the bvalues change, one may see a pattern
forming with the vertices of each of the parabolas in the image above. The image below illustrates the
potential pattern with the dashed line (curve):
Conjecture: The
dashed line is itself a parabola that passes through the vertices of each of
the original parabolas. How could
we show that this is indeed the case?
Consider the
vertices of each of the original f(x) functions. We could find these using a variety of methods. I used the
derivative of f(x) to find the minimum xvalue, and from there I found the
corresponding yvalue. The table below
contains each of the functions and their corresponding vertices:
f(x): 
xvaluevertex 
yvaluevertex 
Vertex: 
x^{2}3x+1 
1.50 
1.25 
(1.5, 7.75) 
x^{2}2x+1 
1.00 
0.00 
(1, 0) 
x^{2}x+1 
0.50 
0.75 
(0.5, 1.75) 
x^{2}+1 
0.00 
1.00 
(0, 1) 
x^{2}+1x+1 
0.50 
0.75 
(0.5, 0.75) 
x^{2}+2x+1 
1.00 
0.00 
(1, 0) 
x^{2}+3x+1 
1.50 
1.25 
(1.5, 1.25) 
One test for
whether or not a function is quadratic is to determine finite differences. If the secondorder differences are
constant, then the function is quadratic.
f(x): 
yvaluevertex 
1st differences 
2nd differences 
x^23x+1 
1.25 


x^22x+1 
0.00 
1.25 

x^2x+1 
0.75 
0.75 
0.50 
x^2+1 
1.00 
0.25 
0.50 
x^2+1x+1 
0.75 
0.25 
0.50 
x^2+2x+1 
0.00 
0.75 
0.50 
x^2+3x+1 
1.25 
1.25 
0.50 
An evaluation of
the differences in yvalues shows a pattern of constant differences at the
second level. This implies that
the dashed curve above is quadratic, and therefore a parabola. Now, how do we determine a specific
equation for this parabola?
Any three points
determine a parabola. (Why?) We
also assume these three points are noncollinear as that is a trivial case (a
degenerate parabola or a line). Any three of the vertices above are
noncollinear, so we may proceed to find this equation.
I will choose the
three vertices with somewhat nice numbers. I will choose the vertices (1, 0) (0, 1) and (1, 0).
We can generate
equations using each of the points mentioned in the last sentence from the
equation of a parabola in standard form: f(x) = ax^{2} + bx + c. We will substitute values of x and y
into the function to generate three distinct equations:
Vertex (1, 0): f(1) = 0 = a + b + c a
+ b + c = 0
Vertex (0,
1): f(0) = 1 = c c
= 1
Vertex (1, 0):
f(1) = 0 = a – b + c a
– b + c = 0
Notice that we now
have three equations with three unknowns. Therefore we can solve a system of
equations by any appropriate preferred method. I will use a 3x4 augmented matrix and row reduce it to find
the values of a, b, and c. The reduced
rowechelon form of the augmented matrix is:
Therefore, a = 1,
b = 0, and c = 1. Furthermore, if we substitute these values in for f(x) = ax^{2}
+ bx + c, we obtain the equation:
f(x) = x^{2}
+ 1