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WOMAN: This interactive allows
us to explore cubic graphs

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of the form y = x cubed
+ bx squared + cx.

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At the moment both 'b' and 'c'
are set to zero

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so we're looking
at the graph of y = x cubed.

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Let's first think about 'b'

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and we note that at the moment,
with 'b' equal to zero,

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we have a cubic graph with one
stationary point and one x-intercept.

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And we see that
as we make 'b' positive

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we get a slight
translation to the left.

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But more importantly, we're
looking at now two stationary points

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and also two x-intercepts.

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And thinking about
when we make 'b' negative,

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we see a slight
translation to the right.

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But more importantly,
we're again looking

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at a second stationary point
being introduced

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and also a second x-intercept.

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So setting that back to zero.

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Now if we look at 'c',
as we make 'c' positive ...

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'C' is more to do with changing
the slope of the graph,

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so we see that we've gone
from having one stationary point

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to now having no stationary points
and still just that one x-intercept.

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And as we make 'c' negative,

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we see that we go
from having one stationary point

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to two stationary points

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and from one x-intercept
to now three x-intercepts.

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But there are many different
combinations of things

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and what we see
is that both 'b' and 'c'

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have a slightly intertwined
sort of effect.

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So for example, we saw earlier
that making 'c' positive

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gave us no stationary points.

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But that combined with,
in this case, 'b' equal to five

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sees that we, in fact,
get two stationary points.

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So they don't have a unique
effect on the graph.

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Their effect is intertwined.

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So let's think about
the algebra behind this

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to see what's actually going on.

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So we'll first think about the
x-intercepts of graphs of this form

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and we know to find x-intercepts

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we're solving x cubed + bx squared
+ cx equal to zero,

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and factorising that shows us that,
in fact, graphs of this form

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will always have
one x-intercept at x = 0.

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And then our other x-intercepts

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are going to come from the
quadratic factor x squared + bx + c

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and how many solutions
x squared + bx + c equal to zero has.

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So we know that we get at least
one x-intercept for this graph,

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but we would get three x-intercepts

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if we were to get two solutions
from that quadratic equation,

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so that is if the discriminant

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of that quadratic equation
is positive,

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and that will occur
when b squared is bigger than 4c.

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And we could get two x-intercepts

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if we get just one solution
from the quadratic factor,

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and that's going to occur

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when the discriminant of that
quadratic factor is equal to zero,

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that is, when b squared
is equal to 4c,

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and we'll get just the one
x-intercept at x = 0

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if the quadratic factor
gives us no solutions,

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and that will occur
when b squared is less than 4c.

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Thinking now about
the stationary points.

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So we need to look at the derivative
of x cubed + bx squared + cx,

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which is 3x squared + 2bx + c,

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and we know that the number
of stationary points is ...

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Stationary points can be obtained

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by solving the derivative
equal to zero,

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and in this case we're looking
again at a quadratic equation

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so we can look at the number
of stationary points

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in relation to the discriminant
of that quadratic,

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and the discriminant in this case
is 2b squared – 12c.

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So we know that we'll get
two stationary points

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if that discriminant is positive,

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that is, when b squared
is bigger than 3c,

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and we'll get just
one stationary point

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if that discriminant
is equal to zero when b squared = 3c,

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and we'll have no stationary points

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if the discriminant
of the quadratic is negative,

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and that is when b squared
is less than 3c.

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So given these different restrictions
that we've established

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that are to do with the number
of x-intercepts

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and the number of stationary points,

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I thought we'd attempt to put that
together to get a bit of an idea

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of, well, when do we get
each of the different cases?

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And for ease,
in this particular case,

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we're going to just consider

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when both 'b' and 'c'
are positive real numbers.

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So we know that
we get three x-intercepts

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when b squared is greater than 4c

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and two stationary points
when b squared is greater than 3c,

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so with positive 'b' and 'c'

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the overlap of those two sets would
be when b squared is greater than 4c,

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and an example of when that occurs

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is when 'b' equals five
and when 'c' equals five.

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So let's consider the interactive
and see what happens.

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So if 'b' equals five
and 'c' is also equal to five,

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we do indeed see that we have
three x-intercepts

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and two stationary points.

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Sticking with
the two stationary point theme,

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that is, when b squared
is greater than 3c,

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if we were to have two stationary
points and only two x-intercepts,

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that's going to occur
when b squared is equal to 4c,

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and an example of when
that would be the case

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is when 'b' equals four
and when 'c' also equals four.

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So testing that out
with our interactive -

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when 'b' is four
and when 'c' is four -

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we indeed see that we have
two stationary points

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and also two x-intercepts.

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Sticking with the two
stationary points,

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we may only get one x-intercept

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and that occurs when b squared
is both greater than 3c

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and less than 4c,

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and an example of when that would
occur is if 'b' equals five

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and 'c' equals seven.

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So going back to our interactive ...

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If we make 'b' equal to five
and 'c' equal to seven,

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we do see that we have
two stationary points

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but only that one x-intercept.

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Now if we were to just
have one stationary point ...

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Simply thinking about
the geometry of a cubic graph

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with only one stationary point,

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it wouldn't be possible
to get three x-intercepts,

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and we're seeing that confirmed
by the algebra as well

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in that b squared cannot both equal
3c and be bigger than 4c,

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and similarly, we cannot have
one stationary point

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and two x-intercepts.

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But it's possible to have one
stationary point with one x-intercept

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and that occurs when b squared = 3c,

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and an example of that
would be when 'b' equals three

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and also when 'c' equals three.

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So if we make 'b' equal three
and 'c' equal three,

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we see that we do indeed have
just the one stationary point

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and the one x-intercept.

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And the last case that's possible -
if we have no stationary points.

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So again, thinking about the geometry
of a cubic with no stationary points,

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it's not possible for us
to have three x-intercepts

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nor two x-intercepts,

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and, again, the algebra that
we're seeing here confirms that.

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But if we had no stationary points
and only the one x-intercept,

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that's going to occur
when b squared is less than 3c,

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and an example of when
that might happen

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is if 'b' were to equal three
and 'c' were to equal seven.

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So let's test that out.

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'B' equals three
and 'c' equals seven.

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And we indeed see
that we have no stationary points

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and just that one x-intercept.

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So as you can see, there are a number
of different things that could occur

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with graphs of the form

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y = x cubed + bx squared + cx

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and we've attempted to look at the
different cases that might happen.