Exploration of Parabolas


 Gayle Gilbert

WeÕre going to explore the equation of a parabola: y=ax2+bx+c for different values of a, b, and c.  First, letÕs look at the graph of a basic parabola y=x2, where a=1, b=0, and c=0:



Notice the graph opens up, the vertex is at x=0, and the y-intercept is at y=0.  LetÕs vary the value of a to determine how the graph changes.  LetÕs graph y=x2 (blue), y=¼x2 (green), y=½x2 (purple), y=2x2 (red), and y=4x2 (black) on the same axes.




For all these positive values of a, the graph still opens up.  Notice when 0<a<1, the graph appears to be stretched horizontally, and when a>1, the graph appears to be compressed horizontally.  What happens when a<0?  Well, we know if a=0, then y=0, which is the x-axis, so we can make assumptions about what the graph will look like when a is negative.  We can guess that the graph will reflect about the x-axis when a is negative.  LetÕs graph y=x2 (blue), y=-x2 (aqua), y=-¼x2 (green), y=-½x2 (purple), y=-2x2 (red), and y=-4x2 (black) on the same axes.




Notice that our hypothesis is correct: when -1<a<0, the graph appears to be stretch horizontally below the x-axis, and when a<-1, the graph appears to be compressed horizontally below the x-axis.  Thus, when a is negative, the graph opens down.


Now, letÕs look at different values of b while fixing a=1 and c=0 in the equation y=ax2+bx+c.  First, letÕs graph y=x2+x (red), where b=1, on the same graph with y=x2 (blue):




How does the graph change?  What hypothesis do we want to make about the effects of the value of b on the function?  Well, we notice that the graph has shifted to the left ½ and down ¼; however, the shape is not affected. LetÕs explore further by graphing y=x2+x (red), y=x2-x (black), y=x2+½x (blue), y=x2-½x (purple), y=x2+3x (aqua), and y=x2-3x (green) on the same graph. 




Notice as b increases from 0, the vertex shifts to the left and down from the origin; as b decreases from 0, the vertex shifts to the right and down from the origin.  The graphs with negative values of b are just reflections about the y=axis of the graphs with positive values of x.  When b=1, the vertex is at x= -½ and when b=-1, the vertex is at x=½; when b=½, the vertex is at x=-¼ and when b=-½, the vertex is at x=¼; when b=3, the vertex is at x=-1.5 and when b=-3, the vertex is at x=1.5.  From this, it appears that the vertex is at x=-b/2.  LetÕs explore further to see if our hypothesis is correct.


What happens when we graph y=2x2+bx for different values of b?  LetÕs graph y=2x2-x (red) and y=2x2-4x (blue):


The vertex when a=2 and b=-1 is at x=¼, and the vertex when a=2 and b=-4 is x=1.  Now, we see that our hypothesis is not completely correct, for –b/2 for the first graph would give us ½, but we have ¼.  Instead, it must be –b/2a since a=1 on our previous graphs and that works here as well. 


Finally, letÕs see what happens to the graph when we change our value of c.  LetÕs graph y=x2+x+c for different values of c (note: our a=b=1).  Here is the graph of y=x2-x (blue), x2-x+2 (red), x2-x+ ½ (green), x2-x-3 (purple):




Notice that the c is the y-intercept of each graph. 



What do you think the graph of y=-¼x2+3x-1 will look like?  Well from our discoveries, we can say that the graph will open down since a is negative.  We also know that the graph will be fairly wide since -1<a<0.  The vertex will be at x=-3/(2*(-¼))=6, so the point is at (6,8).  The y-intercept will be at y=-1.  LetÕs graph to check.




Yes!  That is correct.  I think we now have a pretty good understanding of parabolas.  For a recap, when a is positive, the graph opens up.  When a is negative, the graph opens down.  When |a|<1, then the graph stretches horizontally; when |a|>1, the graph compresses horizontally.  The value of b helps us determine our vertex.  The x-coordinate of our vertex is x=-b/2a.  Finally, the value of c gives us the y-intercept.