WOMAN: Exercise 3.
Not all cubic polynomial graphs are obtained
by dilating and translating the standard cubic y = x cubed.
Equivalently, not all cubic polynomials are of the form
f(x) = a(x – h) all cubed + k.
This exercise explains why, purely algebraically.
Part A - explain how the graph
of y = a(x – h) cubed + k
is related to the standard cubic graph y = x cubed.
There is an interactive available on the website,
which allows you to explore the transformations
caused by 'a', 'h' and 'k'.
For now, I will simply tell you the results.
The magnitude or size of 'a'
causes a dilation by a factor of 'a' from the x-axis.
The sign of 'a' - that is, if 'a' is negative -
we see a reflection in the x-axis
of the graph of y = x cubed.
'h' is to do with translating the graph to the left or right.
If 'h' is positive,
then (x – h) all cubed is a translation to the right by 'h'
and (x + h) all cubed is a translation to the left by 'h'.
'K' is to do with translations up and down,
and if 'k' is a positive number
then x cubed + k is a translation up by 'k'
and x cubed – k would be a translation down by 'k'.
Part B - show that if a cubic polynomial
f(x) = ax cubed + bx squared + cx + d
can be rewritten in the form
f(x) = a(x – h) all cubed + k,
then b squared – 3ac must equal zero.
We first note that if a cubic equation can be rewritten
in the form a(x – h) all cubed + k,
then it is because it has only one stationary point.
In order to find stationary points,
we look at when the derivative of a function is equal to zero,
and in this case, for us to have just one stationary point
we need to consider when the derivative equal to zero
has only one solution.
And so first we calculate the derivative
of ax cubed + bx squared + cx + d,
which gives us 3ax squared + 2bx + c,
and we want that to equal zero.
in order for us to calculate stationary points.
In order for there to be only one stationary point
we want this equation to have just one solution,
and given that it's a quadratic equation
we can think about the discriminant, and for one solution
we need the discriminant of this quadratic equation to equal zero.
So we first calculate the discriminant,
which is equal to the coefficient of 'x' squared minus four
times the coefficient of 'x' squared and times the constant term.
So in this case that gives us (2b) all squared – 4(3a)(c),
which simplifies to 4b squared – 12ac.
And so we need for that to equal zero
in order for there to be only one solution,
and when we simplify that equation
we get the equation that b squared – 3ac = 0.
So b squared – 3ac = 0 will ensure
that f(x) = ax cubed + bx squared + cx + d
has only one stationary point
and hence that will allow us to write it in the form
a(x – h) all cubed + k.
Part C - find an example of a cubic polynomial
f(x) = ax cubed + bx squared + cx + d
where b squared – 3ac is not equal to zero.
From Part B we know that when b squared – 3ac is equal to zero
we get a cubic polynomial with only one stationary point.
So in this example, we're looking to find a cubic polynomial
which doesn't have only one stationary point.
That is, it could have two stationary points
or it could have no stationary points.
There are infinitely many solutions to this question,
but I'm going to start by letting 'b' equal six,
which will mean that 3ac cannot be equal to 36,
so I'm going to choose 'a' equal to three and 'c' equal to two.
So this means that b squared – 3ac
will equal six squared minus three times three times two,
which is 36 minus 18
and so will be equal to 18, which is indeed non-zero.
So if 'b' is six, 'a' is three and 'c' is two,
then our cubic polynomial has equation
3x cubed
+ 6x squared + 2x ...
And then we know that 'd' is only to do with vertical translations,
so it's not affecting the number of stationary points that the cubic has.
So 'd' could be anything and still give us a cubic polynomial
that fits the required restrictions.
So I'm going to choose 'd' equal to 5.
So the cubic polynomial 3x cubed + 6x squared + 2x + 5
would be an example of a polynomial
where b squared – 3ac does not equal zero.
That is, it's a polynomial that has either two stationary points
or no stationary points.