**Locus of a Parabolic Function**

By Thuy Nguyen

In
this investigation we want to consider the locus of the vertices of the set of
parabolas graphed from the function f(x) = x^{2} + bx +1.

We
want to show that the locus of the vertices of the set of parabolas graphed
from f(x) = x^{2} + bx +1 is the parabola g(x) = -x^{2} + 1.

LetŐs first examine the graphs of the function for b = -3, -2, -1, 0, 1, 2, 3:

It will be useful for us to find the minimum for each of these parabolas, so letŐs do so by first taking the derivative of f(x) and set it equal to zero:

fŐ(x) = 2x + b = 0

x = -b/2

Now substitute this back into f(x):

f(-b/2) = (-b/2)^{2}
+ b(-b/2) + 1 = -b^{2}/4 + 1

So the vertex of each
parabola is at (-b/2,1 – b^{2}/4).

Now letŐs consider the
function g(x) = -x^{2} + 1:

We notice that the parabola formed by this function seems to intersect each of the previous parabolas at its vertex. LetŐs see if this is true:

f(x) = g(x) when

x^{2} + bx + 1 = -x^{2}
+ 1

2x = -b

x = -b/2

Substituting into g(x):
g(-b/2) = -b^{2}/4 + 1. Thus g(x) intersects each parabola at its
vertex.

__Generalization__: The
locus of the vertices of the set of parabolas graphed from y = x^{2} +
bx + c is the parabola y = -x^{2} + c.

HereŐs an example when c = 3: