Moving b in a Quadratic Equation


Rob Walsh

A quadratic equation is a second degree polynomial with at most three terms and two roots. It's graphic representation is parabolic in shape and by definition, is the locus of points that is equidistant from a given point (known as the focus) and a line (known as the directrix.). Constructions of certain parabolas have led to the common standard equation representation y = ax2 + bx + c, where x and y are the coordinates for each of the locus points and a, b, and c are parameters that describe the parabola's shape and location on the Cartesian Plane. If we fix our y value to zero, then some minor exploration shows us how parameter a and parameter c define a parabola. However, in this investigation, we are interested in how different values for b change a parabola's shape.

We explore y = x2 + bx + 1, where a and c are held constant. Let's graph this quadratic for b = -3, -2, -1, 0, 1, 2, 3.


There are a few things of interest here:

  1. The parabolas share the same y-intercept. This is not too surprising, though, since this point is where our value for x = 0, hence, the y-intercept is always our parameter c. Since we have fixed c = 1, it stands to reason that all of our parabolas will cross the y-axis at (0, 1).
  2. When b < -2 and b > 2, our parabolas cross the x-axis in two places. this tells us that at these values of b, our equations have two real roots. Similarly, when x = -2, 2, the parabolas touch the x-axis in exactly one place (they are tangent) which says that the equations have exactly one real root. Finally, for all values -2 < b < 2, there are no real roots (which implies that there are complex roots to these equations), since they do not touch the x-axis at all.
  3. The location of our parabola seems to follow some uniform path. If we focus on one particular point, say, the vertex of our parabola when b = -3, we can observe how it "moves up and to the right" for awhile and then "back down" again. In general, we acquire the coordinates of a parabola's vertex as such: x = -b/2a and y is solved for with this value of x. We produce the vertices of our parabolas here:


If we plot these coordinate pairs, the vertices start to take a very obvious shape:

If we consider the locus of the vertices of the set of parabolas graphed from y = ax2 + bx + c, what we find is that we get a parabola that opens opposite of our set:

Let's finish off by finding out the equation of this locus. We start with the general equation y = ax2 + bx + c where we know that our values for a and c are both 1. In our locus graph, since it opens opposite our set of parabolas, we take the opposite of a. We can further see why c = 1, because it is our y-intercept (when x = 0) and therefore, all terms with a variable x go to zero. Now, we have y = -x2 + bx + 1. What is left to figure out is our value for b. Recall, that the x-coordinate of a parabola's vertex is found by taking -b/2a. In this case, this value is 0 and we already have that a = -1. So, we can solve for b as follows:

x = -b/2a

0 = -b/2(-1)

0 = -b/-2

(0)(-2) = -b

0 = -b

So, we can generalize the equation of this locus to be the parabola y = -x2 + 1.

One final observation is that the vertex of our locus is the one common point of our set of parabolas. Therefore, we can further generalize the locus equation as such:

Given a uniform set of equations y = ax2 + bx + c, we determine the equation of the locus of their vertices to be y = dx2 + ex + fx, where d = -a, e is the x-coordinate of the sets common intersection point and f is the y-coordinate of the common intersection point. Let's observe one more time!



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