Michael Thomason's write-up for assignment three, investigation one.

is the general
form of the equation for a parabola (with *a* , *b* , and c
constants) along with the graphs for *a = c = *1 and *b* = {-3, -2, -1, 0, 1, 2, 3}:

`> `**plot([x^2-3*x+1,x^2-2*x+1,x^2-1*x+1,x^2+1,x^2+1*x+1,x^2+2*x+1,x^2+3*x+1],x=-5..5,y=-2..5,color=[blue,yellow,black,green,red,coral,khaki]);**

As *b*
goes from -3 to 3, the vertex of the graph
"moves" while the rest of the graph seems to maintain its curvature
relative to its vertex. Do the vertices travel along a set path? Let's find the
coordinates for the vertex. Taking the derivative of *f* ( *x*
), setting it equal to zero, and solving should give
the *x* -coordinate of the vertex, which
can then be used to find the correspondng *y*
-coordinate.

`> `**f:=x->x^2+b*x+1;
D(f)(x);
solve(D(f)(x)=0,x);
f(solve(D(f)(x)=0,x));**

So
the coordinates of the vertex for this case of , namely when *a
= c = *1, are ( , ). So then, what path does the vertex trace? ie, What is the locus of the
vertices of the parabolas? From the coordinates here we have:

Plug
into the *y* -coordinate:

Now let's graph this along with the graph from before (the locus is colored differently from the parabolas):

`> `**plot([x^2-3*x+1,x^2-2*x+1,x^2-1*x+1,x^2+1,x^2+1*x+1,x^2+2*x+1,x^2+3*x+1,-x^2+1],x=-5..5,y=-2..5,color=[blue,blue,blue,blue,blue,blue,blue,khaki]);**

does indeed trace the vertices of the graphs of *.* What traces the vertices of the graphs of
? Let's follow the
steps from before and see.

`> `**f:=x->a*x^2+b*x+c;
D(f)(x);
solve(D(f)(x)=0,x);
f(solve(D(f)(x)=0,x));**

So the coordinates of the vertex for the general form of a parabola, , are ( , ). We can follow the steps from the previous case to see the general equation for the locus of the vertices of the parabolas:

So
the locus of the set of vertices of the set of parabolas given by is the parabola . Here is a graph for , and *b* = {-15,
-10, -5, 0, 5, 10, 15} (the locus is colored differently from the parabolas):

`> `**a:=3;c:=6;y:=x->-a*x^2+c;plot([y(x),3*x^2-15*x+6,3*x^2-10*x+6,3*x^2-5*x+6,3*x^2+6,3*x^2+5*x+6,3*x^2+10*x+6,3*x^2+15*x+6],x=-5..5,y=-20..20,color=[khaki,blue,blue,blue,blue,blue,blue,blue]);**

Here
is another graph, this time with *a* =
-2, *c* = 4, and
*b* = {-6, -4, -2, 0, 2, 4, 6} (the locus
is colored differently from the parabolas):

`> `**a:=-2;c:=4;y:=x->-a*x^2+c;plot([y(x),-2*x^2-6*x+4,-2*x^2-4*x+4,-2*x^2-2*x+4,-2*x^2+4,-2*x^2+2*x+4,-2*x^2+4*x+4,-2*x^2+6*x+4],x=-3..3,y=-5..10,color=[khaki,blue,blue,blue,blue,blue,blue,blue]);**

So, I have found the locus of the set of vertices of the set of parabolas given by to be the parabola . I have also checked a couple cases and graphed them.