Exploring Quadratic Equations Keeping a and c Constant

By Colleen Garrett

In this investigation I will explore the graph y=ax^{2}+bx+c
for values of b equaling -3,-2,

-1,0,1,2,3 while keeping a and c constant.

When =3 x has 2 real roots because the graph intersects the x-axis at 2 points.

When =2 x has 1 real root because the graph intersects the x-axis at 1 point.

When =1 x has no real roots because the graph does not intersect the x-axis at any point.

Now letŐs consider the locus of the vertices of the set of
parabolas graphed from y=x^{2}+bx+1. We know that from the quadratic formula we get that the
x-coordinate of the vertex is equal to –b/2a. Plugging this value you in for x in our equation y=x^{2}+bx+1
we get that y-coordinate of the vertex equals (–b^{2}/4)+1. We can put these values of x and y into
a parametric equation to generate all vertices for all values of b in the
equation y=x^{2}

+bx+1.

The black parabola created by the parametric equation is
y=-x^{2}+1.

LetŐs look at a few more examples. y=3x^{2}+bx+1 for values of b
equaling -3,-2,-1,0,1,2,3.

The locus of the vertices graphed from y=3x^{2}+bx+1**
**is y=-3x^{2}+1

What about y=-2x^{2}+bx+1 for values of b equaling
-3,-2,-1,0,1,2,3.

The locus of the vertices graphed from y=-2x^{2}+bx+1
is y=2x^{2}+1

Let b=0 in each of our examples. We get the following equations:

y=x^{2}+1

y=3x^{2}+1

y=-2x^{2}+1

Now take a look at the locus of vertices from all three of these equations

y=-x^{2}+1

y=-3x^{2}+1

y=2x^{2}+1

You can see that the value for c does not change and that the value of a changes to –a.

This helps us to conclude that the locus of vertices
generated by a function when b=0 is y=-ax^{2}+c