It has now become a rather standard exercise, with availble
technology, to construct
graphs to consider the equation
and to overlay several graphs of
for different values of a, b, or c as the other two are held
constant. From these
graphs discussion of the patterns for the roots of the quadratic can be followed. For example, if we set y= x^2 +bx +1 for b = -3, -2, -1, 0, 1, 2, 3, and overlay the graphs, the following picture is
We can discuss the "movement" of a parabola as b is changed. The parabola always passes through the same point on the y-axis ( the point (0,1) with this equation). This is due is because for every parabola when x=0, the y-intercept is 1. For b < -2 the parabola will intersect the x-axis in two points with positive x values (i.e.the original equation will have two real roots, both positive). Consider the quadratic formula
Since a=1 and c=1, the discriminant becomes b^2-4. For 2 real roots to exist b^2-4 must be greater than zero, for one real root--b^2-4 must equal zero, and for no real roots b^2-4 must be less than zero. Having no real roots puts a negative under a radical which would only yeild imaginary solutions. Therefore, b must be greater than or equal to two for a parabola to intersect (or be tangent to) the x-axis. When the discriminant equals zero, the trinomial is a perfect square which is why only one root exists.
For b = -2, the parabola is tangent to the x-axis and so the
original equation has one real and positive root at the point
of tangency. For -2< b< 2, the parabola does not intersect
the x-axis -- the original equation has no real roots. Similarly
for b = 2 the parabola
is tangent to the x-axis (one real negative root) and for b > 2, the parabola intersects the x-axis twice to show two negative real roots for each b.
Since the equation for the axis of symmetry is
it changes along with b, which explains the horizontal shift.
Now consider the locus of the vertices of the set of parabolas graphed from y=x^2 +bx+1. The axis of symmetry is x=-b/2. Plugging this value back in for x will give the y- coordinate of the vertex. With the system of equations
and , to get a final parabola in terms of x and y, use the substitution method to eliminate b. Solving both equations for b yeilds the following:
This then becomes the quadratic, shown on the following graph in pink. Since the equation has a as negative the parabola opens downward.
Hence, this is the equation for the locus of vertices of the parabolas y=x^2+bx+1.
Graphs in the xb plane.
Consider again the equation x^2+bx+1=0.
Now graph this relation in the xb plane, that is consider the y-axis to be the b-axis. We get the following graph.
If we take any particular value of b, say b = 3, and overlay this equation on the graph we add a line parallel to the x-axis. Finding where x^2+bx+1=0 and b=3 intersect would indicate the roots of the parabola when b=3. If it intersects the curve in the xb plane the intersection points correspond to the roots of the original equation for that value of b. We have the following graph.
For each value of b we select, we get a horizontal line. It is clear on a single graph that we get two negative real roots of the original equation when b > 2, one negative real root when b = 2, no real roots for -2 < b < 2, One positive real root when b = -2, and two positive real roots when b < -2.
Consider the case when c = - 1 rather than + 1. The graph of x^2+bx-1=0 in the x-b plane then becomes
Notice, no matter what value of b is chosen, the line b=some constant will always intersect the graph twice, meaning there are two roots. This is because the discriminant b^2-4ac becomes b^2+4 and is therefore, always positive. Looking at the graphs of y=x^2 +bx-1 for various values of b as before, since the y-intercept is negative and the parabola opens upwards, it will always have two roots.
b=3, b=2, b=1, b=0, b=-1, b=-2, b=-3
Graphs in the xc plane.
In the following example the equation
is considered. If the equation is graphed in the xc plane,
it is easy to see that the
curve will be a parabola. For each value of c considered, its graph will be a line
crossing the parabola in 0, 1, or 2 points -- the intersections being at the roots of the orignal equation at that value of c. (Just like in the x-b plane b=n would yeild the roots to x^2+nx+1=0.) In the graph, the graph of c = 1 is shown. The
equation x^2 +5x+1=0 will have two negative roots -- approximately -0.2 and -4.8. These roots can be found by tracing the graph and also by the quadratic formula.
At the vertex, or maximum of the parabola, there would be only one root. This is when x^2+5x+c=0 becomes a perfect square, that is . This value of c can be determined by finding the axis of symmetry and vertex of c=-x^2+5x. The axis of symmetry is x=-b/(2a)=5/2. Substituting this back in as x yeilds c= 25 /4 = 6.25. Again due to discriminant b^2-4ac, when c is less than 6.25, the discriminant will be postive and have two roots. When c is greater than 6.25, the discriminant will be negative and therefore, no real roots will exist.
Graphs in the x-a plane
Graphs in the x-a plane are a bit more complex. Consider the equation , and its graph
If lines where a=n (some constant) are graphed like before, again this would reveal the roots at the point(s) of intersection.
When a=-1, the parabola would have two roots. The point of tangency, a.k.a. when the parabola only has one root is at a=-9 /4 = -2.25. Any values of a less than this and the parabola will have no real roots; values of a greater than this will yeild two real roots. There is an exception, however, at a=0. This is the point at which the x squared term is eliminated and the equation is no longer a quadratic, but a linear equation. Hence, on the graph in the x-a plane, the a=0 (the x-axis) will always be an assymptote.