WOMAN: In this interactive we consider cubic graphs
of the form y = x cubed + bx squared + cx.
In this interactive we're able to change the values of 'b' and 'c'
using this graph in the left-hand side
by selecting pairs of values of 'b' and 'c'
relative to the equation b squared – 3c = 0.
So first I'd like to think about
where that equation b squared – 3c = 0 comes from.
So if we look at graphs of the form y = x cubed + bx squared + cx
and we consider their stationary points,
we know we do that by solving the derivative equal to zero.
So the derivative is 3x squared + 2bx + c and that equal to zero.
We now have a quadratic equation,
and if we're looking at the number of stationary points
that we might get from cubic graphs of this form,
we'd be looking at the discriminant of this quadratic equation,
and the discriminant works out to be 4b squared – 12c.
So we know that equations of this form would have two stationary points
if 4b squared – 12c is bigger than zero,
and that is if b squared – 3c is bigger than zero,
we'd have one stationary point if b squared – 3c is equal to zero,
and no stationary points if b squared – 3c is less than zero.
So in fact, what we're seeing in this interactive
is a way to manipulate the number of stationary points that are cubic.
So we should find if we select points
on the line b squared – 3c,
we should be getting cubic graphs
with just one stationary point.
If we select points that are within this shaded region,
that is, where b squared – 3c is less than zero,
we see that we're getting no stationary points,
which agrees with the algebra that we saw earlier.
And lastly, if we select points
in the region where b squared – 3c is bigger than zero,
we see that we do indeed get cubic graphs
with two stationary points.