The purpose of this assignment is to produce several graphs of y
= ax^2 on the same axis. In order to do this, we
will use several values of a.
First, we choose a = 1. Since we will be graphing a family
of graphs where all the graphs are related to the first, we will call y
= x^2 the parent graph. The parent graph appears as follows:
Notice the graph is a parabola which opens up and has its vertex at the
origin (0,0).
When we change a so that a = 2, we see the following graph:
Notice that the graph is still a parabola which opens up and has its vertex
at the origin. The difference between the two graphs is the 'fatness'. The
second graph is expanded upward as if we took hold of each side of the parabola
and pulled upward. This is called a stretch.
Following the same pattern, we can predict that if a = 3,
then the graph of y = 3x^2 would be a parabola that opens
up with its vertex at the origin, and more of a stretch is produced. Let's
show the graph.
As can be seen, our predictions have proven true. (Note that the size
of the scale blocks are different. Even so, one can determine the x- coordinate
of both graphs at y = 1 to observe the effect of the stretch. The x- coordinate
of the first stretch surpasses 1 while the x- coordinate of the second stretch
approaches but does not exceed 1.)
To confirm our conclusions, we will assign a = 4.
As can be seen, parabola opens up, has its vertex at the origin, and
stretches so that the x-coordinate at y = 1 approaches but does not exceed
1/2.
The following graph allows easy comparison of the above parent graph and
its stretches.
It is easy to see the original parent graph (red) being stretched upward,
seemingly making it 'skinnier'.
Next, we will assign a = 1/2.
The graph still opens upward and has its vertex at the origin. Note the
'fatness' of this graph. It is expanded outward. Where y = 1, the x- coordinate
is 3/2.
We will now let a = 1/3.
Notice, again, that the parabola opens up and has the vertex at the origin.
It is also expanded outward further. The x- coordinate at y = 1 is more
than 3/2.
We can predict that if a = 1/4, then the parabola will open
up with its vertex at (0,0), and expands even further outward.
As can be seen, our predictions have been verified. The 'expanding outward'
is called a shrink. Overlaying the parent graph (red) and its shrinks, we
can see the expansions.
So far, we have chosen values of a > 0. What would
happen if a < 0 ? Lets try
The vertex of the parabola is still at the origin, but it opens downward.
For a = -2, we will predict that the parabola will stretch
but open downward from the origin. For a = - 1/2, we will
predict that the parabola will shrink but open downward from the origin.
As can be seen, we were correct in our predictions.
From this point, I would ask my students to further explore the parent graph
of y = ax^2 with
various other points of a < 0, a = 0,
and a > 0. When finished, I would have them formalize
their conjectures. (Conjectures should include coordinates of vertex, shape
of graph, stretches and shrinks, and relation of a to all
of these.)