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WOMAN: In this interactive
we consider cubic graphs

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

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In this interactive we're able to
change the values of 'b' and 'c'

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using this graph
in the left-hand side

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by selecting pairs of values
of 'b' and 'c'

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relative to the equation
b squared – 3c = 0.

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So first I'd like to think about

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where that equation
b squared – 3c = 0 comes from.

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So if we look at graphs of the form
y = x cubed + bx squared + cx

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and we consider
their stationary points,

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we know we do that by solving
the derivative equal to zero.

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So the derivative is 3x squared + 2bx
+ c and that equal to zero.

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We now have a quadratic equation,

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and if we're looking at the number
of stationary points

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that we might get
from cubic graphs of this form,

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we'd be looking at the discriminant
of this quadratic equation,

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and the discriminant
works out to be 4b squared – 12c.

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So we know that equations of this
form would have two stationary points

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if 4b squared – 12c
is bigger than zero,

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and that is if b squared – 3c
is bigger than zero,

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we'd have one stationary point
if b squared – 3c is equal to zero,

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and no stationary points
if b squared – 3c is less than zero.

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So in fact, what we're seeing
in this interactive

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is a way to manipulate the number
of stationary points that are cubic.

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So we should find if we select points

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on the line b squared – 3c,

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we should be getting cubic graphs

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with just one stationary point.

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If we select points that
are within this shaded region,

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that is, where b squared – 3c
is less than zero,

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we see that we're getting
no stationary points,

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which agrees with the algebra
that we saw earlier.

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And lastly, if we select points

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in the region where b squared – 3c
is bigger than zero,

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we see that we do indeed
get cubic graphs

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with two stationary points.