Claim:
For fixed a and b and varying c, the function y = ax^2 + bx + c will have a locus of vertices that is a line with equation x = -b/2a.
A Proof:
Use parametric equations. That is, we know that the locus of vertices must always have x = -b/2a and y = -b^2/4a + c. [Note that calculus can be used to show that x = -b/2a is the x-coordinate of the vertex of a parabola in the form y = ax^2 + bx + c; plug in the x-value to find the corresponding y-value, -b^2/4a + c.]
Now, solve both equations for c:
But we can't solve x = -b/2a for c! That implies that x = -b/2a always, no matter the value of c.
Therefore, the locus must be a vertical line at x = -b/2a.
Significance:
The locus of vertices crosses the x-axis (obviously) at x = -b/2a, which is the x-value for which the graph of y = ax^2 + bx + c will be tangent to the x-axis. Therefore, x = -b/2a provides the single real root for a specific value of c in the quadratic equation ax^2 + bx + c = 0.
Since subsitution is relatively simple in this situation, substitute x = -b/2a into ax^2 + bx + c = 0 and solve for c to find that when c = -3b^2/4a, the quadratic equation ax^2 + bx + c = 0 will have one real zero.