Assignment 3:

Exploring  Parabolas

By Nikhat Parveen, UGA.

Exploring the graph: y = ax2 + bx + c


In this assignment we will be exploring the graphs of parabola for different values of a, b and c.We will further investigate the roots of the parabola y = ax2 + bx + c and discuss by overlaying several other graphs.

We will try do this in the following steps.

Part I: Varying a

Let a vary while b and c remain fixed.

Example 1: b = 0 c = 0



For positive values of 'a' the parabola opens upward and with increase in the value of 'a', the graph of parabola gets narrower. For negative values of 'a' the parabola opens downward which means for -ve values of a the parabola changes its direction.


Example 2: Let b = 1 and c = 0,while varying a



So what do we see here?? All the graphs of the parabola still passes through the origin but there bases have moved. All the parabolas have at least one root as zero.


Example 3: Let a vary and b = -1 and c = 0

Based on the previous graph we would expect the graphs facing upward to move to the right and and the graphs facing downward to the left. why?? The positive value of b shifted the graph to the left for the upward parabolas with respect to the graph in example 1, so the negative value should do the opposite.


And yes we are in fact true!! The graph did move as expected.


Example4: Let a vary and b = 1 and c = 1



What's happening here?? All the graphs shifted vertically upward by 1 unit and passes through (0,1). Also we can notice that the graphs with the positive values of a do not intersect the x-axis --- the original equation has no real solutions when b = 1 and c =1 and a > 0.


Part II: Varying b

Let b vary while a and c remain fixed.



The parabola in the green represents the equation y = x2 + bx +1, where b =1

Let's discuss the "movement" of a parabola as b is changed.


Part III: Varying c

Let's vary 'c' keeping a and b constant

Example 4: Let a = 1 and b = 1



So the graphs with c values less than or equal to zero has real roots and two real solutions as their graphs intersects the x -axis and for  c > 0 the graphs does not intersect the x-axis and has no real solutions.


Now consider the locus of the vertices of the set of parabolas graphed from  y = x2 + bx +1

If we consider only the vertices of the parabola from example 1 of Part II

they are (-1.5, -1.25), (-1,0), (-1/2, 3/4), (0,1), (1/2,3/4), (1,0), (1/2, -1.25) and if we trace these vertices the locus these points will be a parabola y = -x2 +1 that passes through all the vertices.





Proof: Considering the equation y = x2 + bx +1

We know that for any standard equation of a parabola f(x) = ax2 + bx +c, the vertex of the graph is given by (-b/2a, f(-b/2a)) that also represents the x-coordinate and y-coordinate of the graph.

Therefore the x-coordinate of the equation  y = x2 + bx +1 is -b/2. Substituting it in the given equation and solving for y:

y = (-b/2)2 + b(-b/2) +1

    = b2/4 - b2/2 +1 = (b2-2b2)/4 + 1

    = -b2/4 +1

    = (-b/2)2 +1

y = -x2 +1; where x = -b/2

 Therefore the locus of vertices of all parabolas of the form y = x2 + bx +1 is the parabola y = -x2 +1.




Further Explorations of the graph y = x2 + bx +1.


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