It has now become a rather standard exercise, with available 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
can be followed. For example, if we set
for b = -3, -2, -1, 0, 1, 2, 3, and overlay the graphs, the following
picture is obtained.
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).
When 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), as for b > 2, the parabola intersects the x-axis twice to show two negative real roots for each b.
When 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. Similarly for b = 2 the parabola is tangent to the x-axis (one real negative root).
For -1 < b < 2, the parabola does not intersect the x-axis -- the original equation has no real roots.
Now consider the locus of the vertices of the set of parabolas graphed from
Show that the locus is the parabola
By taking the derivative of y= ax^2 + bx +c (which provides the slope of the tangent line at any given point), we can use the locus of vertices of the from the set of parabolas graphed above in the form y= x^2 +bx +1. From this we will be able to attain the equation y = -x^2 +1.
The derivative of y= ax^2 + bx +c is y` = 2ax +b. From this we can solve for x and get x= -b/2a which will allow us to obtain the points in order to graph the equation of y = -x^2 +1.
By letting a=1, b= -3,-2,-1,0,1,2,3, x= -b/2a, and obtaining y by substitution into y= x^2 +bx +1.
By plotting (x,y) we obtain the graph y= -x^2+1. The graph is shown in the following picture.
To verify that the graph above is indeed y= -x^2 +1, the following three point (1,0), (0,1), and (-1,0) algebraically.
y = ax^2 +bx +c:
0 = a + b +c
1 = c
0 = -a - b + c.
Therefore, a +b =-1 and a-b = -1. Using the addition method, we obtain
that -2 = 2a. Thus, a=-1 and b = 0. By substitution we obtained y = -x^2
Consider again the equation
Now graph this relation in the xb plane. 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. 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.
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 original equation at that value
of c. In the graph, the graph of c = 1 is shown. The equation
will have two negative roots -- approximately -0.2 and -4.8.
There is one value of c where the equation will have only 1 real root -- at c = 6.25.
For c > 6.25 the equation will have no real roots because the parallel line to the x-axis will be above our graph, and therefore will not cross our graph.
When c < 6.25 the equation will have two roots, both negative for 0 < c < 6.25, one negative and one 0 when c = 0 and one negative and one positive when c < 0.