Assignment 3

Explorations of the X-B Plane

Drew Wilson

Consider the equation

If we consider what this equation should look like we can use the quadradic formula to determine the solutions to this equation.  We know that the solutions to this equation are :

,

If we look closely at the discriminant, we will notice there are possible values of b that will yield imaginary solutions.  Therefore, we know that when b=2 or b=-2, there will be one real solution at -b/2 because the discriminant will be zero.  When -2<b<2 there are imaginary solutions, and when b>2 and b<-2 there are two real solutions. By looking at the discriminant we know the graph will not exist in the interval -2<b<2. We also know that when b>2, there will be two real solutions and both solutions will be negative because of the quadratic formula and because of the properties of addition (when you subtract from a negative, you change it to addition and change the sign of the number being subtracted from the negative number). Similarly, when b<-2 we know that there will be two real solutions and both solutions will be positive.  Therefore, we know that there will be a vertical asymptote at x=0 and the part of the graph shaped like a "U" will be in quadrant 2 with the horizontal tangent line at point b=2, and the part shaped like a "n" will be in quadrant 4 with the horizontal tangent line at point b=-2. The graph of this equation is provided at the right.

Now if we change the value of the constant in this equation to 1, 3, 5, 7 and graph these equations at the same time, we will notice that the same occurs for each of these equations and the interval of wh ich there are imaginary solutions is: (the different graphs are shown in the color red). Let's also consider graphs when the value of the constant is negative, say -1, -3, -5, -7 (the different graphs are shown in the color blue).  Then we will notice that there will always be two real solutions because the discriminant will never be zero or less than zero.  If we consider the equation when c=0, then we will get the equation x=-b and x=0 which are the equations of the asymptotes of the hyperbolas (the equation x=-b is shown in the graph by the color green, x=0 is the b-axis). If we consider the roots of these equations, we can find the roots of each equation by graphing horizontal lines b=k, where k is some constant.  If the graphcrosses the hyperbola zero times then there are no real roots.  If the graph crosses the hyperbola once then there is one real root and the discriminant is zero. If the graph crosses the hyperbola two times then there are two real roots. This holds true for both sets of hyperbolas, the ones in red and the ones in blue.  The set of hyperbolas in blue will always have two real roots, therefore, the graph b=k will always cross the blue hyperbola two times, which is why these hyperbolas are horizontal because their graph encompasses all values of b. The graph at the right is an animation of the equation b=k as k changes. Let's consider, for example provided below, that b=3 and we have the equations: and . First notice that the equation b=3 intersects both equations two times, so we should have two real roots for both equations.  Now, let's use the quadratic formula to determine the roots of both equations.  By the quadratic formula we know that the roots of the first equation are and , and the roots of the second equation are and . Therefore, you may notice that the when the equation b=k intersects one of the graphs, the point of intersection will correspond to the roots of the original equation for that value of b. You may also consider the possible solutions and how they may be represented in the graph.  We know that the equation b=k will cross each graph either 0, 1, or 2 times.  If b=k crosses the graph 0 times, then there are zero real roots and two imaginary roots.  If b=k crosses the graph one time, then there is one real root and the discriminant is zero.  If b=k crosses the graph two times, then there are two real roots and zero imaginary roots. The graph of the example above is provided below.

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