It has now become a rather standard exercise, with available technology, to construct graphs to consider the equation
ax^{2} + bx + c = 0
and to overlay several graphs of
y = ax^{2} + bx + c
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
ax^{2} +bx + c = 0
can be followed.For example, if we set
y = x^{2} + bx + 1
for b = 3 (magenta), 2 (red), 1 (blue), 0 (green), 1 (aqua), 2 (yellow), 3 (gray) 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 yaxis (the point (0, 1) with this equation). For b < 2 the parabola will intersect the xaxis in two points with positive x values (i.e. the original equation will have two real roots, both positive). For b = 2, the parabola is tangent to the xaxis and so the original equation has one real and one positive root at the point of tangency.For 2 < b < 2, the parabola does not intersect the xaxis  the original equation has no real roots. Similarly for b=2 the parabola is tangent to the xaxis twice to show two negative real roots for each b.
Here's
a picture of a GSP sketch that animates the movement
of the vertex of this family of parabolas.
If
you'd like an explanation of the construction, click
here. The locus is clearly
another parabola. To get the
equation of that parabola, start with the equations for the x and y coordinates
of the vertex. For the general
case, that'sx
= b/2a and y = f(b/2a). In
this equation, a = b = c = 1, so the equation xcoordinate is x = b/2. The
ycoordinate is given by y = (b/2)^{2} + b(b/2a)
+ 1,
or y = (b^{2}/4)  b^{2}/2 + 1, which simplifies to b^{2}/4 + 1 or ? (b/2)^{ 2 }+ 1. Therefore, the equation of the vertex is y = x^{2 }+ 1.
If we take particular values of a and overlay the equation on the graph we add lines parallel to the xaxis.If the lines intersect the graph in the xa plane the intersection points correspond to the roots of the equation for that value of a.Here are a few values of a.
The red line is a = .25, the green is a = 0 and the dark blue is a = 3.
Value
of a

Number/type
of roots

a
> .25

no
real roots

a
= .25

one
negative real root

0
< a < .25

two
negative real roots

a
< 0

one
positive, one negative real root

So, is a = .25 a lucky guess?Not quite.It's not as easy to see here as it is on a more 'normallooking' graph, but take the discriminant, set it equal to zero and solve for a:
b^{2}  4ac = 0, or 1  4a = 0 simplifies to a = 1/4.
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 xaxis. 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 selected value of b, we get a horizontal line.It
is clear on a single graph that we get:
Value
of b

Number/type
of roots

b
< 2

two
positive real roots

b
= 2

one
positive real root

2
< b < 2

no
real roots

b
= 2

one
real negative root

b
> 2

two
negative real roots

Consider the case when c = 1 rather than +1, which is the magenta graph.It is clear that there are always two distinct real roots, one positive and one negative.
The vertex of this parabola is at 5/2, 25/4, or 2.5 and 6.25. Since this is a downward opening parabola, the roots are as follows.
Value
of c

Number/type
of roots

c
> 6.25

no
real roots

c
= 6.25

one
positive real root

0
<= c < 6.25

two
real negative roots

c
< 0

one
real positive, one real negative root
