Assignment
3
By: Cara Haskins, Matt Tumlin, & Robin Kirkham
Examine
Parabolas
We will start
with the quadratic equation, y = ax2 + bx + c.
Let us use Graphing Calculator 3.2 to examine the effects of using different values for a, b, and c.
Our first step
will be to look at the parabola
when a=1, c=1,
and varying the b.
We can discuss
the "movement" of a parabola as b is changed. The parabola
always passes through the same point on the y-axis (the point (0,1) with this
equation). For b<-2, the parabola will intersect the x-axis in two
points with positive x values (ie. the original equation will have two real
roots, both positive). For b=-2, the parabola will intersect the x-axis in one
point with a positive x value. For
b>2, the parabola will intersect the x-axis in two points with negative x
values. For b=2, the parabola will
intersect the x-axis in one point with a negative x value.
Let's first start by looking at different values of a. We will set b=1 and c=1 for y = ax2 + bx + c.
Now, we will explore
what happens when a = -3, -2, -1, 0, 1, 2, 3.
From the graph
above, we can see that the equation when a=0 is tangent to all of the parabolas
with a as a different values. We can also see that the line of tangency will
always cross the y-axis at c with a slope of b since the equation of the line
will be y = bx + c.
When b=1 and
c=1, our original equation will have two roots if a is negative. If a=0 our
original equation will have one root. For each negative a value there are 2
roots, one positive root and one negative root. Notice if we changed the value
of c then the a values that have roots would change. Now, it appears that our
original equation will not have roots for positive a, but take a look at the
next group of equations.
Here, we see
that the our equation becomes tangent to the x-axis at (-2, 0) giving us one
negative root for a=0.25 and two negative roots for 0<a<0.25.
Consider the
equation
0= ax2
+x + 1, or a = Ðx Ð 1
x2
Now graph this
relation in the xa plane. We get the following graph.
If we take any particular value of a, say a = -2, and overlay this equation on the graph we add a line parallel to the x-axis. If it intersects the curve in the xa plane, the intersection points correspond to the roots of the original equation for that value of a. We have the following graph.
For each value
of a we select, we get a horizontal line. It is clear on a single graph that we
get a negative real root and a positive real root of the original equation when
a<0, one negative real root when 0 < a < 0.25, one negative real root
when a=0.25, and no real roots for a>0.25. This is exactly what we
discovered in our previous exploration of y= ax2 +x + 1.
Next, let's
explore b again. We will set a=1 and c=1.
So, if we set
b = -3, -2,
-1, 0, 1, 2, 3, and overlay the graphs, the following picture is obtained.
Now, consider
the locus of the vertices of the set of parabolas graphed from y = ax2
+ bx + c.
The vertices
are as follows:
(1.5, -1.25)
for b=-3
(1, 0) for
b=-2
(0.5, 0.75)
for b=-1
(0, 1) for b=0
(-0.5, 0.75)
for b=1
(-1, 0) for
b=2
(-1.5, -1.25)
for b=3
As you can see
the locus of the vertices appears to be parabolic.
To find the equation
of the parabola we first go back to the original form of a parabola y = ax2 + bx +
c.
Now, we can
see that the parabola is concave down by looking at the vertices above. Thus,
our a will be -1 and we get y = -1x2 + bx + c.
We also see that
the roots of the parabola are 1 and -1 from the points (0,1) and (-1, 0). We
can use these roots to form the following equation in factored form, y= (x +
1)(x Ð 1).
Now, we see
that setting each factor equal to 0 will give us the roots 1 and -1. When
simplifying we get
y = -1x2
+ 1.
Therefore, the locus of
the vertices when a=1, c=1, is the parabola y= -1x2 + 1.