Richard Francisco

Parabolas as the Locus of Vertices


Graphical Introduction To Quadratics:

Can You see a relationship between the family of 'positive' parabolas and the negative one?


The purpose of this investigation is to determine the locus of the vertices of the set of parabolas of the form y = x^2 + bx +1 where b is varied, and extend the result to all parabolas.

The vertices of the family of parabolas above form a quadratic themselves, and this parabola is given by -(x^2) + 1.

This result can be calculated using brute force algebra. Completing the square of the given equation is a well-known algebraic method, and gives:

Students should use knowledge about transformations of graphs to verify that the given vertex is correct. (Remember, a horizontal shift takes opposite sign!) In our particular problem, we are given that a and c are fixed, (and in this cas equal to one) so the vertex is:

Thinking about this pair as (x, f(x)) gives us f(x) = 1- 1*x^2, which is indeed the answer we were asked to find.


Generalizing for ax^2 + bx + c = y with fixed a, c, and b varying, gives the final solution that the resulting parabola is expressed by:

y = c - ax^2