Quadratic and Cubic Equations

by Brook Buckelew



If we set

for b = -3,-2,-1,0,1,2,3, and overlay the graphs, the following picture is obtained.


Now we can find the locus of each parabla.

(1.5, -1.25)

(1, 0)

(.5, .75)

(0, 1)

(-.5, .75)

(-1, 0)

(-1.5, -1.25)


Now let's plot the locus point only of each parabola.


Notice that the points form what looks like a concave down parabola with the locus at (0,1). We can see from the picture that the roots of the new parabola are x=-1 and x=1. Let's try to find the equation of the new parabola.

First set y equal to the roots.


y=x^2 -1

But because this is a concave down parabola the x^2 must be negative, so let's see what happens when we graph y=-x^2+1.



Here is what we get when everything is graphed together.





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