Charlie Conway 7/30/2009

The general form of a quadratic equation is y=ax^2+bx+c. Let's explore the equation with a=1 and c=1 and a variable b to get y=x^2+bx+1.

What happens as we change the value of b?

As b moves from -3 to 3, the parabola moves as you can see above. The y-intercept is always 1. The vertex of the parabola crosses the x-axis at two points, x=-1 and x=1. Let's discuss the root of x=-1.

For what value of b between -3 and 0 does the equation y=x^2+bx+1 have exactly one root (x=-1)? By factoring, we see that this is true only when b=-2.

What value of b creates the equation with one root of x=1? Again, by factoring, we find that b=2 gives a single root, that being x=1.

So as b ranges from negative infinity and approaches -2, the equation has 2 real roots. At b=-2, there is one real solution. Between b=-2 and b=2, there are no real roots. When b=2, again, there is one real root. And as b>2 approaches infinity, there are two real roots.

Follow the vertex of the parabola as it moves. The locus appears to make a parabolic shape. Let us investigate by manipulating the original quadratic.

By completing the square, we get

which can be reduced to

For simplification purposes, we can substitute in p and q where

After substituting, we get

This happens to be a type of general quadratic, one that makes it very easy to find the vertex. The coordinates of the vertex are simply (p,q).

So our x and y coordinates are

If we solve for b in terms of p, we get

Substituting this in for b in the second equation gives

which simplifies to

and since p and q correspond to x and y, respectively. The equation of the quadratic that is the locus of the vertices with respect to the value of b is