WOMAN: This interactive allows us to explore cubic graphs
of the form y = x cubed + bx squared + cx.
At the moment both 'b' and 'c' are set to zero
so we're looking at the graph of y = x cubed.
Let's first think about 'b'
and we note that at the moment, with 'b' equal to zero,
we have a cubic graph with one stationary point and one x-intercept.
And we see that as we make 'b' positive
we get a slight translation to the left.
But more importantly, we're looking at now two stationary points
and also two x-intercepts.
And thinking about when we make 'b' negative,
we see a slight translation to the right.
But more importantly, we're again looking
at a second stationary point being introduced
and also a second x-intercept.
So setting that back to zero.
Now if we look at 'c', as we make 'c' positive ...
'C' is more to do with changing the slope of the graph,
so we see that we've gone from having one stationary point
to now having no stationary points and still just that one x-intercept.
And as we make 'c' negative,
we see that we go from having one stationary point
to two stationary points
and from one x-intercept to now three x-intercepts.
But there are many different combinations of things
and what we see is that both 'b' and 'c'
have a slightly intertwined sort of effect.
So for example, we saw earlier that making 'c' positive
gave us no stationary points.
But that combined with, in this case, 'b' equal to five
sees that we, in fact, get two stationary points.
So they don't have a unique effect on the graph.
Their effect is intertwined.
So let's think about the algebra behind this
to see what's actually going on.
So we'll first think about the x-intercepts of graphs of this form
and we know to find x-intercepts
we're solving x cubed + bx squared + cx equal to zero,
and factorising that shows us that, in fact, graphs of this form
will always have one x-intercept at x = 0.
And then our other x-intercepts
are going to come from the quadratic factor x squared + bx + c
and how many solutions x squared + bx + c equal to zero has.
So we know that we get at least one x-intercept for this graph,
but we would get three x-intercepts
if we were to get two solutions from that quadratic equation,
so that is if the discriminant
of that quadratic equation is positive,
and that will occur when b squared is bigger than 4c.
And we could get two x-intercepts
if we get just one solution from the quadratic factor,
and that's going to occur
when the discriminant of that quadratic factor is equal to zero,
that is, when b squared is equal to 4c,
and we'll get just the one x-intercept at x = 0
if the quadratic factor gives us no solutions,
and that will occur when b squared is less than 4c.
Thinking now about the stationary points.
So we need to look at the derivative of x cubed + bx squared + cx,
which is 3x squared + 2bx + c,
and we know that the number of stationary points is ...
Stationary points can be obtained
by solving the derivative equal to zero,
and in this case we're looking again at a quadratic equation
so we can look at the number of stationary points
in relation to the discriminant of that quadratic,
and the discriminant in this case is 2b squared – 12c.
So we know that we'll get two stationary points
if that discriminant is positive,
that is, when b squared is bigger than 3c,
and we'll get just one stationary point
if that discriminant is equal to zero when b squared = 3c,
and we'll have no stationary points
if the discriminant of the quadratic is negative,
and that is when b squared is less than 3c.
So given these different restrictions that we've established
that are to do with the number of x-intercepts
and the number of stationary points,
I thought we'd attempt to put that together to get a bit of an idea
of, well, when do we get each of the different cases?
And for ease, in this particular case,
we're going to just consider
when both 'b' and 'c' are positive real numbers.
So we know that we get three x-intercepts
when b squared is greater than 4c
and two stationary points when b squared is greater than 3c,
so with positive 'b' and 'c'
the overlap of those two sets would be when b squared is greater than 4c,
and an example of when that occurs
is when 'b' equals five and when 'c' equals five.
So let's consider the interactive and see what happens.
So if 'b' equals five and 'c' is also equal to five,
we do indeed see that we have three x-intercepts
and two stationary points.
Sticking with the two stationary point theme,
that is, when b squared is greater than 3c,
if we were to have two stationary points and only two x-intercepts,
that's going to occur when b squared is equal to 4c,
and an example of when that would be the case
is when 'b' equals four and when 'c' also equals four.
So testing that out with our interactive -
when 'b' is four and when 'c' is four -
we indeed see that we have two stationary points
and also two x-intercepts.
Sticking with the two stationary points,
we may only get one x-intercept
and that occurs when b squared is both greater than 3c
and less than 4c,
and an example of when that would occur is if 'b' equals five
and 'c' equals seven.
So going back to our interactive ...
If we make 'b' equal to five and 'c' equal to seven,
we do see that we have two stationary points
but only that one x-intercept.
Now if we were to just have one stationary point ...
Simply thinking about the geometry of a cubic graph
with only one stationary point,
it wouldn't be possible to get three x-intercepts,
and we're seeing that confirmed by the algebra as well
in that b squared cannot both equal 3c and be bigger than 4c,
and similarly, we cannot have one stationary point
and two x-intercepts.
But it's possible to have one stationary point with one x-intercept
and that occurs when b squared = 3c,
and an example of that would be when 'b' equals three
and also when 'c' equals three.
So if we make 'b' equal three and 'c' equal three,
we see that we do indeed have just the one stationary point
and the one x-intercept.
And the last case that's possible - if we have no stationary points.
So again, thinking about the geometry of a cubic with no stationary points,
it's not possible for us to have three x-intercepts
nor two x-intercepts,
and, again, the algebra that we're seeing here confirms that.
But if we had no stationary points and only the one x-intercept,
that's going to occur when b squared is less than 3c,
and an example of when that might happen
is if 'b' were to equal three and 'c' were to equal seven.
So let's test that out.
'B' equals three and 'c' equals seven.
And we indeed see that we have no stationary points
and just that one x-intercept.
So as you can see, there are a number of different things that could occur
with graphs of the form
y = x cubed + bx squared + cx
and we've attempted to look at the different cases that might happen.