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WOMAN: Exercise 3.

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Not all cubic polynomial graphs
are obtained

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by dilating and translating
the standard cubic y = x cubed.

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Equivalently, not all cubic
polynomials are of the form

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f(x) = a(x – h) all cubed + k.

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This exercise explains why,
purely algebraically.

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Part A - explain how the graph

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of y = a(x – h) cubed + k

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is related to the standard
cubic graph y = x cubed.

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There is an interactive
available on the website,

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which allows you to explore
the transformations

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caused by 'a', 'h' and 'k'.

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For now, I will simply
tell you the results.

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The magnitude or size of 'a'

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causes a dilation by a factor
of 'a' from the x-axis.

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The sign of 'a' -
that is, if 'a' is negative -

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we see a reflection in the x-axis

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of the graph of y = x cubed.

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'h' is to do with translating
the graph to the left or right.

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If 'h' is positive,

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then (x – h) all cubed is a
translation to the right by 'h'

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and (x + h) all cubed is a
translation to the left by 'h'.

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'K' is to do with
translations up and down,

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and if 'k' is a positive number

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then x cubed + k
is a translation up by 'k'

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and x cubed – k would be
a translation down by 'k'.

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Part B - show that
if a cubic polynomial

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f(x) = ax cubed + bx squared + cx + d

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can be rewritten in the form

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f(x) = a(x – h) all cubed + k,

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then b squared – 3ac must equal zero.

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We first note that if a cubic
equation can be rewritten

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in the form a(x – h) all cubed + k,

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then it is because it has
only one stationary point.

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In order to find stationary points,

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we look at when the derivative
of a function is equal to zero,

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and in this case, for us
to have just one stationary point

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we need to consider
when the derivative equal to zero

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has only one solution.

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And so first we calculate
the derivative

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of ax cubed + bx squared + cx + d,

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which gives us 3ax squared + 2bx + c,

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and we want that to equal zero.

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in order for us to calculate
stationary points.

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In order for there to be only
one stationary point

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we want this equation
to have just one solution,

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and given that
it's a quadratic equation

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we can think about the discriminant,
and for one solution

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we need the discriminant of this
quadratic equation to equal zero.

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So we first calculate
the discriminant,

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which is equal to the coefficient
of 'x' squared minus four

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times the coefficient of 'x' squared
and times the constant term.

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So in this case that gives us
(2b) all squared – 4(3a)(c),

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which simplifies
to 4b squared – 12ac.

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And so we need for that to equal zero

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in order for there
to be only one solution,

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and when we simplify that equation

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we get the equation that
b squared – 3ac = 0.

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So b squared – 3ac = 0 will ensure

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that f(x) = ax cubed + bx squared
+ cx + d

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has only one stationary point

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and hence that will allow us
to write it in the form

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a(x – h) all cubed + k.

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Part C - find an example
of a cubic polynomial

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f(x) = ax cubed + bx squared + cx + d

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where b squared – 3ac
is not equal to zero.

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From Part B we know that
when b squared – 3ac is equal to zero

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we get a cubic polynomial
with only one stationary point.

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So in this example, we're looking
to find a cubic polynomial

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which doesn't have only one
stationary point.

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That is, it could have
two stationary points

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or it could have
no stationary points.

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There are infinitely many solutions
to this question,

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but I'm going to start
by letting 'b' equal six,

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which will mean that
3ac cannot be equal to 36,

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so I'm going to choose 'a' equal
to three and 'c' equal to two.

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So this means that b squared – 3ac

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will equal six squared
minus three times three times two,

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which is 36 minus 18

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and so will be equal to 18,
which is indeed non-zero.

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So if 'b' is six, 'a' is three
and 'c' is two,

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then our cubic polynomial
has equation

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3x cubed

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+ 6x squared + 2x ...

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And then we know that 'd' is only
to do with vertical translations,

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so it's not affecting the number of
stationary points that the cubic has.

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So 'd' could be anything
and still give us a cubic polynomial

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that fits the required restrictions.

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So I'm going to choose
'd' equal to 5.

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So the cubic polynomial
3x cubed + 6x squared + 2x + 5

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would be an example of a polynomial

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where b squared – 3ac
does not equal zero.

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That is, it's a polynomial
that has either two stationary points

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or no stationary points.