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.