*Claim:*

For fixed **b** and **c** and varying **a**,
the function **y =** **ax^2 +
bx + c** will have a locus of vertices that is a line
with equation **y =** **bx/2 + c**.

*A Proof:*

Use parametric equations. That is, we
know that the locus of vertices must always have x = -**b**/2**a** and y = -**b**^2/4**a** + **c**. [Note that
calculus can be used to show that x = -**b**/2**a** is the x-coordinate of the vertex of a parabola
in the form y = **a**x^2
+ **b**x + **c**; plug in the x-value to find the corresponding
y-value, -**b**^2/4**a** + **c**.]

Now, solve both equations for **a**:

-**b**/(2x) = **a**

(-**b**^2 + 4**ac**)/4y = **a**

and set them equal to each other: -**b**/(2x) = (-**b**^2 + 4**ac**)/4y

"Cross-multiplying," we have

-**b**4y = 2x(-**b**^2 + 4**ac**) = - 2x**b**^2 + 8x**ac**

Dividing by -4**b**, we have

y = **b**x/2 - 2x**ac**/**b**

But , -2x**a** = **b**. Subsitute to find that

y = **b**x/2 + **c**

*Here's another way to see it, if
you assume that the locus is a line:*

Note that the line must pass through
(0, **c**)
on the y-axis and that it also must pass through every other vertex
along the way, which have coordinates of the form (-**b**/2**a**, -**b**^2/4**a** + **c**).

Find the slope of the line:

slope = (-**b**^2/4**a** + **c** - **c**)/(-**b**/2**a** - 0) = (-**b**^2/4**a**)/(-**b**/2**a**) = **b**/2

So, the equation for the line must be
y = **b**x/2
+ **c**

*Significance:*

The zero of **y =** **bx/2 +** **c**, i.e., x = -2**c**/**b**, will be the x-value for which the graph of
**y =** **ax^2 + bx + c** will be tangent to the x-axis. Therefore, this
zero is the value of x that is the single real root for a specific
value of **a**
in the quadratic equation **ax^2
+ bx + c** **= 0**.

Since subsitution is relatively simple
in this situation, substitute x = -2**c**/**b** into **ax^2
+ bx + c** **= 0** and solve for **a** to find that when **a** = **b**^2/4**c**, the quadratic equation **ax^2
+ bx + c** **= 0** will have one real zero.