A general parabolic form

Doug Griffin

Now we wish to take a closer look at the graps of the equations produced by varying b in the general form

Lets begin by taking a look at graphs when b={-3,-2,-1,0,1,2,3}.

These graphs appear to produce parabolas which overlap each other and form a nice pattern. For instance if we look at the graphs of those equations produced by b and -b they appear to reflect around the x axis, that is to be symmetric with respect to it.

If we focus in on the vertices of these parabolas we notice that plotting the vertices only that indeed these vertex points seem to describe a parabolic equation. This parabola has vertex point at (0,2) and opens down. From previous discussions about parabolas we conjecture that the parabola that passes through the vertex points of the parabolas above is described by the equation . Graphing this equation confirms this conjecture.

Observing the relationship between the parabola that passes throught the vertex points and the family of parabolas that produce those vertex points we are ready to establish another conjecture. For the graphs of parabolas formed by where c is any constant value and b varies, the equation which graphs as a parabola that passes through the vertices of these parabolas is . Lets try another example to test our conjecture.

This result seems to confirm our conjecture. We will look later at trying to prove that these parabolas always intersect at the vertex points.

What happens to our hypothesis as a varies from 1? Lets look at an example of this.

So to take our conjecture one step further, describes a set of parabolas as a and c are some fixed values and b varies. The equation of the parabola which will pass through the vertex points of all such parabolas is . how may we begin to prove such a conjecture?

Lets begin by looking at an concrete example of what we are trying to show. Using our last example above lets find the points of intersection between and

. This equates to solving the following equation;

=

x(4x+1)=0

x=0 0r x=-1/4. Therefore, substituting for x and solving for y, the points of intersection are (0,-2) and (-1/4,-17/8). Is (-1/4,-17/8) the vertex of the parabola? Lets find it by algebraic methods.

The vertex of the parabola is indeed (-1/4,-17/8). the general case and the proof are much similar to the specific case. Lets first find the vertex point for the parabola

So the vertex point is defined (-b/2a,c-bsquared/4a) Will this always be the point of intersection?

So x=0 or (2ax+b)=0. If 2ax+b=0, then 2ax=-b. Therefore x=-b/2a. Substituting we get

for y.

Return