note: close this window to return to Amy's Write-Up #3

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

for varying values of b.


For fixed a and c and varying b, the function y = ax^2 + bx + c will have a locus of vertices that is also a parabola with equation y = - ax^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 b:

-2ax = b

+ or - sqrt(4a(c - y)) = b

and set them equal to each other: -2ax = + or - sqrt(4a(c - y))

Squaring both sides, we have

4a^2x^2 = 4a(c - y)

ax^2 = c - y

y = - ax^2 + c.


The zeros of y = - ax^2 + c will be real only when - a and c are opposite in sign, i.e., when a and c are the same sign. When a and c ARE either both positive or both negative, the zeros of y = - ax^2 + c are x = sqrt(c/a) and x = -sqrt(c/a). These zeros are the values of x that are single real roots for some specific values of b in the quadratic equation ax^2 + bx + c = 0. That is, these zeros indicate the x-values for the values of b that make the graph of y = ax^2 + bx + c tangent to the x-axis.