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The locus of vertices of y = ax^2 + bx + c

for varying values of a.


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/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 a:

-b/(2x) = a

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

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

"Cross-multiplying," we have

-b4y = 2x(-b^2 + 4ac) = - 2xb^2 + 8xac

Dividing by -4b, we have

y = bx/2 - 2xac/b

But , -2xa = b. Subsitute to find that

y = bx/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/2a, -b^2/4a + c).

Find the slope of the line:

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

So, the equation for the line must be y = bx/2 + c


The zero of y = bx/2 + c, i.e., x = -2c/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 = -2c/b into ax^2 + bx + c = 0 and solve for a to find that when a = b^2/4c, the quadratic equation ax^2 + bx + c = 0 will have one real zero.