Assignment 2: ďExplorations with y = ax2 + bx + cĒ

By: David Drew

EMAT 6680, J. Wilson

 

 

 

††††††††† The main question weíre asking here is what happens to the graph of y = ax2 + bx + c as a is varied where we have b = 1 and c = 2. Letís first plot the most basic graph of this function where a = 1. So we have y = x2 + x + 2.

 

 

This produces a parabola with symmetry slightly to the left of x = 0. There is a whole world of study dealing with the parabola. There are many everyday objects that have parabolic properties that most people have no idea about. Such objects include telescopes, and headlights just to name a couple. Visit here to find out much more. For now letís just explore the graphís properties. Weíll increase a so that the graph reads y = 5x2 + x + 2.

 

 

We can see that when we put a constant greater than one in for a the graph seems to get more narrow. Letís increase it a little bit more and see if the same assumption holds true. Hereís the graph for y = 10x2 + x + 2.

 

 

The assumption holds for an increasing value of a. Can we assume that the graph will stretch out for a value less than one? Letís plot the graph y = (1/4)x2 + x + 2.

 

 

Our guess was right. So we can say that as we increase the value of a then the graph gets skinny and as we decrease the value of a then the graph gets fatter. But as we can see if we graph several functions on the same graph something is similar with all the graphs.

 

 

A zoomed in view of all the graphs shows that each one passes through the point (0, 2). This is due to the value of the constant c. As long as c does not have a variable of x multiplying it then any time x = 0 the y value will simply by the constant c. In our case c = 2, so it should be no surprise that all the graphs pass through (0, 2). Even when a = 0 we can see that the graph will pass through (0, 2). Hereís the graph y = (0)x2 + x + 2

 

When a = 0 we basically have a function that says y = x. The only difference is we have added a constant c so that y = x + c. And whatever c is will be the y intercept, or in other words where the graph crosses the y axis.

 

 

But wait, a new problem has arisen. What if our constants stay the same and our value of a becomes less than 0? Will our intercept be the same and will the parabola still be shaped the way that it has been in the past?

 

Letís plug in an arbitrary negative value for a. Since we started with 1 above then letís go with a negative one value. So our equation will read y = -x2 + x + 2. Hereís the picture.

 

It seems that our graph is flipped upside down. What does this mean? It really doesnít mean anything that we havenít discovered in the previous examples. The parabola simply opens up going toward the negatives and it extends into negative infinity. But perhaps this was a fluke so letís try a larger value for our negative x. Weíll try y = -10x2 + x + 2.

 

 

Again we see that the parabola is shifting down. So our assumption was right that a negative value for x has the exact opposite effect as our positive value. Hereís a nice picture with many values of 10 both positive and negative. As an exercise can you verify which one goes with which value. Iíll give you a head start and tell you what values I used.

 

 

That sure is neat looking. It almost looks like the graphs dip up and then go back down. But donít be fooled all of these graphs only have one dip in them. If they had more than one sign change then weíd have to have a greater value for the exponent in our first value of x. Any way in the graph above I used the values -10, 10, 1, -1, -5, 5, -2, 2, and 0. Can you figure out which of these graphs belong to each value? If you have Graphing Calculator you can see for yourself. Just click on Graph. You can watch the graph move as the values of n change. Just hit the play button.

 

 

 

Write Up by David Drew

Back to my Homepage