Gloria L. Jones


Assignment 3


The expression b²-4ac in the quadratic formula is called the discriminant.  From this number we can determine the nature of the solutions of a quadratic equation.  We can also tell how many real number solutions the equation has.  Letıs look at the quadratic equation ax²+bx+c=0, with a‚0, and all coefficients real numbers.  It is obvious that each variable a, b and c play a role in the determination of the roots of the quadratic equation.  With available technology it is easy to overlay several graphs of y=ax²+bx+c for different values of a, b, or c as the other two as the other two values are held constant.  Discussion of the patterns for the roots of y=ax²+bx+c can be followed by examining the x-intercepts of each graph as understood to be where y=0.


For example, if we graph



for b=-3,-2,-1,0,1,2,3 and overlay the graphs, the following sketch is obtained:



One notices from the graphs that for b<-2, the equation has two positive real roots (since it crosses the x-axis at two positive values), for b=-2 there is one positive real root (at the point of tangency), for -2<b<2 no real roots exist since the graph does not cross the x-axis, for b=2 one real negative root and b>2 two negative real roots.


Now, letıs consider the locus of the vertices of the set of parabolas graphed:



We can see that the locus of the vertices is also a parabola and since (0,1) is the vertex and (1,0) is a point on this parabola, then we obtain y=a(x-0)²+1 and substituting (1,0) for (x,y), we get 0=a+1 thus a=(-1) and the equation of the parabola is y=-x²+1.  In general, since (0,c) will always be the vertex of this parabola and its direction is opposite that of the original graphs, then


is the locus of points of the vertices.


Graphs in the xb plane

If we are concerned with examining the roots of an equation, we know that y would be held constant at zero.  Thus the previously discussed equation y=ax²+bx+1 can be reduced to 0=x²+bx+1 which is an equation in two variables, x and b.  Solving for b, we obtain


and graphing in the xb plane allows for an alternative analysis of the roots based on infinitely many values for b.  We will no longer need separate graphs to determine the role that different values of b play in the determination of the roots:


 Horizontal lines represent a particular value for b and points of intersection with the graph represent the roots to the equation y=x²+bx+1.  It is clear that the range is not defined for -2<b>2 thus indicating no real roots, for b>2 the graphs would cross at two points with negative x-values (thus two negative real roots), for b=2 the graphs would cross at a point of tangency with a negative x-value (thus one negative real root), for

b= -2 a point of tangency with a positive x-value and for b< -2 two points of intersection with positive x-values.


Next, consider the same equation

      y=x²+bx+1 and


with c= -1 rather than c=1, we set y=0 and solve for b to get b=(-x²+1)/x:



One notices from the resulting hyperbola that no matter what value of b is chosen there will always be two real roots (one positive and one negative).  Our discovery is confirmed by a discriminant calculation of b²+4 which can never equal zero or a negative number.


Graphs in the xc plane

Similarly, one can take a fixed a and b value and graph in the xc plane to analyze the role that c plays in the nature of the roots.  First consider the graphs of y=x²+5x+c for c=1,-2, -1, -2, 4, 6, and 8:

It is clear that c plays a role in the vertical translation of the graph and that as c increases the graph shifts upward and will eventually reach a point where there are no real roots.


Next, consider x²+5x+c=0.  Solving for c, we obtain c=-x²-5x which yields a downward parabola with vertex at (-2.5, 6.25):



Since the vertex has a value of c=6.25, we realize that for c=6.25, the equation will have one real root.  For c>6.25, there are no points of intersection and thus no real roots and for c<6.25 there are two real roots (more specifically when 0<c<6.25 there are two negative real roots and for c<0 we get one positive and one negative real root).


Graphs in the xa plane

Consider the graphs of y=ax²+2x+1 for a= -2, -1, 0, 1, 2:



It is clear to see that the sign of (a) influences the direction of a parabola and that when a=0 the quadratic function reduces to a line and thus only has one real root.  As (a) increases it appears the parabola will continue to rise and eventually ascend beyond the

x-axis thus giving no real roots.  Looking at the graph it appears to occur at c=1.


We can investigate more closely the effect that (a) has on the roots by fixing b and c and solving for (a).  Consider ax²+2x+1=0.  Solving for (a), we obtain




which yields the following hyperbola:



This confirms our speculation.  For c>1 there are no real roots, for c=1, one real negative root exists, for 0<c<1 two negative real roots exist, for c=0 one negative real root and c<0, one negative and one positive real root exist.