by

Merve Nur Kursav

Let examine graphs for the parabola y= ax2+bx+c for different values of a, b, and c. (a, b, c can be any rational numbers).

Figure I

Please see the videos of parabola for   a, b, and c.

Fix the values for b and c, vary a. Make at least 5 graphs on the same axes as you vary a.

Figure II

Please see the video while a is varying, another video while b is varying, and the video while c is varying.

Is there a common point to all graphs? What is it?

As it is seen in the Figure II, there is a common point where all of the graphs are passing through it. The equation in general terms is y= ax2+x+2 and a is varying. When x= 0, then y= a.0+0+2= 2, so the point representing this is (0, 2) as it is shown in the Figure III.

Figure III

What is the significance of the graph where a = 0?

When a= 0, the equation of y= ax2+bx+c will be y= 0.x2+bx+c. If b varies and a=0, please see the first and second videos of the corresponding graph while b is varying. Furthermore, please see the first and second videos of the corresponding graph while c is varying.

When b=0, the equation of y= ax2+bx+c will be y= ax2+0.x+c. If a varies and b=0, please see the video of the graph.

Fix the values for a and b, vary c. Make at least 5 graphs on the same axes as you vary c.

While c is varying in the interval [-4, 4],  the figure below can be obtined. Please look at the Figure IV.

Figure IV

Additionally, you can examine the video while c is varying.

In order to prove this, let us we have any different values of c such that c and c' where c is an element of the integers. Then the equations will be y= x2+x+c and y'= x2+x+c'. y-y'= y= (x2+x+c)- (x2+x+c'). Then y-y'= y= x2+x+c-x2-x-c'. Thus, y-y'= c-c'. Also, c-c' is an element of integeres since c & c' are integers. The difference between y and y' is only the change in hte c while c is varying as an integer number, so the only change on the graph will be the place of the grapg of the equation.

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