Wave Project

ROOTS OF PARABOLAS

By Ann Webbink

Initialization

Needs["Graphics`Colors`"]

Getting Started

When studying the parabola, many students become confused about what is happening when the roots are no longer real roots.  This program will attempt to show that there are always two roots to the quadratic equation, whether the roots be real, one root of multiplicity two, or two imaginary roots.

The parabola with two real roots

First let's look at a parabola with two real roots.

Clear[f,x];
f[x_]:=x^2 -2x -8
Plot[f[x],{x,-5,5},AxesLabel->{"x","y"},
	PlotRange->{-9,9},
		PlotStyle->{Red}];

As we can see, the graph of the parabola crosses the x-axis in two places, thus giving us two real roots for this quadratic equation.  Also notice that the sign of the function changes at these points since the graph crosses the x-axis at the points.

The parabola with one root of multiplicity two

Now let's examine what happens when a quadratic has only one root.  When this happens what would you expect the graph of the parabola to look like?  How many times should the graph cross the x-axis?

Clear[f,x];
f[x_]:=x^2-2x+1
Plot[f[x],{x,-5,5},AxesLabel->{"x","y"},
	PlotRange->{-5,5},AspectRatio->1,PlotStyle->{Red}];

Notice that this graph only touches the x-axis, and that the function does not change signs at this point.  If we had factored this quadratic equation to solve for the roots we would have found the following equation:  x \(TraditionalForm\`\^2\)-2x +1 = 0.  In factored form we would have
 (x - 1)(x - 1) = 0, giving us two identical roots at x = 1.  This is called a root of multiplicity two since it occurs twice.

A quadratic equation with two imaginary roots

Now, what would you expect the graph of the quadratic that does not have any real roots to look like? Notice that in the two previous examples, we have changed only the value of the constant in the quadratic equation and have changed the nature of the roots of the equation.  Let's try this one more time and examine the graph of the resulting parabola.

Clear[f,x];
f[x_]:=x^2 +2x +3;
Plot[f[x],{x,-5,5},AxesLabel->{"x","y"},PlotRange->{-5,5},
	AspectRatio->1,PlotStyle->{Red}];

Did you notice that the graph no longer even touches the x-axis?  Does that mean that there are suddenly no roots for this quadratic equation?  Would that make any sense?  We can find roots if we use the quadratic formula, but these roots would be imaginary. First, let's look at the roots when we let c vary from -2 to 4 in increments of one.

Clear[f,x,c,comrt,sol,g,parabgraph ,i];
f[x_,c_]:= x^2 -2x +c;
sol = Table[N[Solve[f[x,c]==0,x]],{c,-2,4}]

Notice that the first four roots are real, with the fourth root being a root of multiplicity two, or a double root.   The last three roots are imaginary.  To see what is happening, let's examine the graph of these successive quadratic equations.  The graph to the right is the graph of the parabola on the real plane.  The graph to the left is the graph of only the roots  on the imaginary plane.

comrt=Table[{Re[sol[[j,i,1,2]]],Im[sol[[j,i,1,2]]]},
      {j,1,Length[sol]},
      {i,1,2}];
      
g[i_]:=ListPlot[comrt[[i]],
   PlotRange->{{-5.0,5.0},{-5.0,5.0}},
   AxesLabel->{"a","b"},
   PlotStyle->{Blue,PointSize[0.05]},
   AspectRatio->1,DisplayFunction -> Identity];
   
parabgraph := Plot[f[x,c],{x,-5,5},AxesLabel->{"x","y"},
PlotRange->{-5,5},AspectRatio->1,PlotStyle->{Red},
DisplayFunction->Identity];
Table[Show[GraphicsArray[{g[c+3],parabgraph},
	DisplayFunction->$DisplayFunction]],{c,-2,4}];

Did you notice that when the graph of the parabola "lifted" off the x=axis, the roots on the imaginary plane left the real axis and switched to the imaginary axis?  This shows that we will always have two roots to a quadratic equation, whether they be two real roots, one root of multiplicity two, or two imaginary roots.