Assignment 3


By: Robin Kirkham, J. Matt Tumlin, & Cara Haskins


Examine Parabolas



We start with the quadratic equation, y = ax2 + bx + c.


Let us use Graphing Calculator 3.2 to examine the effects of using different values for a, b, and c.


Our first looks at the parabola

when a=1, c=1, and varying the b.




We discuss the "movement" of a parabola as b is changed.  The parabola always passes through the same point on the y-axis (the point (0,1) with this equation).  For b<-2, the parabola intersects the x-axis in two points with positive x values (ie. the original equation will have two real roots, both positive). For b=-2, the parabola intersects the x-axis in one point with a positive x value.  For b>2, the parabola intersects the x-axis in two points with negative x values.  For b=2, the parabola intersects the x-axis in one point with a negative x value.  


Let's first start by looking at different values of a. We set b=1 and c=1 for y = ax2 + bx + c.


Now, we explore what happens when a = -3, -2, -1, 0, 1, 2, 3.




From the graph above, we see that the equation when a=0 is tangent to all of the parabolas with a as a different values. We can also see that the line of tangency always cross the y-axis at c with a slope of b since the equation of the line will be y = bx + c.


When b=1 and c=1, our original equation has two roots if a is negative. If a=0 our original equation will have one root. For each negative a value there are 2 roots, one positive root and one negative root. Notice if we changed the value of c then the a values that have roots would change. Now, it appears that our original equation will not have roots for positive a, but take a look at the next group of equations.




Here, we see that the our equation becomes tangent to the x-axis at (-2, 0) giving us one negative root for a=0.25 and two negative roots for 0<a<0.25. 


Consider the equation

0= ax2 +x + 1, or a = –x – 1



Now graph this relation in the xa plane. We get the following graph.



If we take any particular value of a, say a = -2, and overlay this equation on the graph we add a line parallel to the x-axis. If it intersects the curve in the xa plane, the intersection points correspond to the roots of the original equation for that value of a. We have the following graph.



For each value of a we select, we get a horizontal line. It is clear on a single graph that we get a negative real root and a positive real root of the original equation when a<0, one negative real root when 0 < a < 0.25, one negative real root when a=0.25, and no real roots for a>0.25. This is exactly what we discovered in our previous exploration of y= ax2 +x + 1.


Next, let's explore b again. We set a=1 and c=1.  So, if we set

b = -3, -2, -1, 0, 1, 2, 3, and overlay the graphs, the following picture is obtained.




Now, consider the locus of the vertices of the set of parabolas graphed from y = ax2 + bx + c.


The vertices are as follows:


(1.5, -1.25) for b=-3


(1, 0) for b=-2


(0.5, 0.75) for b=-1


(0, 1) for b=0


(-0.5, 0.75) for b=1


(-1, 0) for b=2


(-1.5, -1.25) for b=3




As you can see the locus of the vertices appears to be parabolic.



To find the equation of the parabola we first go back to the original form of a parabola y = ax2 + bx + c.


Now, we see that the parabola is concave down by looking at the vertices above. Thus, our b= -1 and we get y = -1x2 + bx + c.


We also see that the roots of the parabola are 1 and -1 from the points (0,1) and (-1, 0). We use these roots to form the following equation in factored form, y= (x + 1)(x – 1).


Now, we see that setting each factor equal to 0 give us the roots 1 and -1. When simplifying we get

y = -1x2 + 1.



Therefore, the locus of the vertices when a=1, c=1, is the parabola y= -1x2 + 1.



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