Like Assignment 2, this write up will focus on the parabola. We'll use the form where the coefficient on the squared term is 1. This time, we're going to focus on roots for x when y = 0.

0 = x^2 + Bx + C

One of the most memorable things from Algebra is the quadratic formula, which gives the two roots for x. Here, we defined A = 1.

x = \frac{-B \pm \sqrt{ B^2 - 4 A C}}{2 A} = \frac{-B}{2} \pm \frac{\sqrt{ B^2 - 4 C}} {2}

The value under the square root, B^2 - 4 C , is called the discriminant. If the discriminant is a positive real number, the square root will be real and there will be two real roots. If the discriminant is zero, the \pm part will be zero and there will be only one real root, repeated twice, as (x - r)^2 . Otherwise, the values of x will be complex valued. The complex plane is more complicated, so we're interested in real roots of x.

Let's simplify and restrict ourselves to cases where B and C are real values. Then, if C is zero or less than zero, the discriminant will be non-negative and all roots will always be real. On the other hand, if C is positive, there are some values of B where there will be no real roots. We can compute these things algebraically, but it would be nice to see them in graphics.

Let's start with the case C = +1, making the equation x^2 + Bx = -1 . We'll draw the graph with x as the horizontal axis, and values of B that satisfy the equation on the vertical y axis. Solutions to the equation appear in blue. The graph forms a hyperbola.

For some values of B, if |B| < 2, there are no real valued roots for x. For instance, if B = 0, we need to solve x^2 = -1 , which we cannot do without complex numbers.

At B = 2 and B = -2, there is only one real root. Algebraically, for B = 2, x^2 + 2x = -1 has a solution at x = -1: ~~ (-1)^2 + 2 (-1) = -1 . For B = -2 the solution is x = +1.

When |B| > 2, we can see two roots. For example, if B = 4, x^2 + 4x = -1 has roots at roughly x = -0.27 and x = -3.73.

I added orange lines at B = 2 and B = 4 to the picture. We can see the roots, where the horizontal line intersects the blue curve.

A logical second case flips the sign of C, making C = -1. From earlier, we know there are always two solutions for any value of B. On the graph, any horizontal line crosses the blue solution hyperbola in two points. The orange lines again appear at B = 2 and B = 4.

Taking the same cases as before, when B = 0, the solutions to x^2 = +1 are at x = +1 and x = -1.

For B = 2, x^2 + 2x = +1 has solutions around x = +0.41 and x = -2.41.

And for B = 4, x^2 + 4x = +1 has solutions around x = +0.24 and x = -4.24.

One solution is trending towards zero as B gets further from zero in absolute value, while the other is heading away.

Before we explore more general B and C, let's look at a case where B is fixed but C can vary. Let's assign B = 1. We'll plot values of x on the horizontal axis and solutions for x^2 + x = C on the vertical axis. We get a parabola. The minimum value for C occurs when x = - \frac{1}{2} ; C = - \frac{1}{4} . There are no crossings at C = -2, since x^2 + x = -2 is complex valued. At C = 4, the orange line crosses the blue solution curve at x = +1.56 and x = -2.56.

If you want to explore more in the XB or XC planes, I made Java applets with GeoGebra. If you're reading this and have Java installed, you can run them.

The XB plane web page is asmt3XB.html.

The XC plane web page is asmt3XC.html.