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

f(x) = a*x^2+b*x+cis 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]);

[Maple Plot]

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));

f := proc (x) options operator, arrow; x^2+b*x+1 en...

2*x+b

-1/2*b

-1/4*b^2+1

So the coordinates of the vertex for this case of f(x) = a*x^2+b*x+c, namely when a = c = 1, are ( -b/2, -b^2/4+1). 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:

x = (-b)/2

2*x = -b

-2*x = b

Plug into the y -coordinate:

y = -(-2*x)^2/4+1

y = -4*x^2/4+1

y = -x^2+1

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]);

[Maple Plot]

y = -x^2+1does indeed trace the vertices of the graphs of f(x) = x^2+b*x+1. What traces the vertices of the graphs of f(x) = a*x^2+b*x+c? 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));

f := proc (x) options operator, arrow; a*x^2+b*x+c ...

6*x+b

-1/6*b

-1/12*b^2+6

So the coordinates of the vertex for the general form of a parabola, f(x) = a*x^2+b*x+c, are ( -b/(2*a), -b^2/(4*a)+c). We can follow the steps from the previous case to see the general equation for the locus of the vertices of the parabolas:

x = -b/(2*a)

b = -2*a*x

y = -b^2/(4*a)+c

y = -(-2*a*x)^2/(4*a)+c

y = -4*a^2*x^2/(4*a)+c

y = -a*x^2+c

So the locus of the set of vertices of the set of parabolas given by f(x) = a*x^2+b*x+cis the parabola y = -a*x^2+c. Here is a graph for a = 3, c = 6, 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]);

a := 3

c := 6

y := proc (x) options operator, arrow; -a*x^2+c end...

[Maple Plot]

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]);

a := -2

c := 4

y := proc (x) options operator, arrow; -a*x^2+c end...

[Maple Plot]

So, I have found the locus of the set of vertices of the set of parabolas given by f(x) = a*x^2+b*x+cto be the parabola y = -a*x^2+c. I have also checked a couple cases and graphed them.

Return to my main page.