Text description - screencast for exercise 3

[The narrator reads out the onscreen text.]

NARRATOR: Exercise 3. Not all cubic polynomial graphs are obtained by dilating and translating the standard cubic y equals x cubed. Equivalently, not all cubic polynomials are of the form f of x equals a times (x minus h) all cubed plus k. This exercise explains why, purely algebraically.

Part A. Explain how the graph of y equals a times (x minus h) cubed plus k is related to the standard cubic graph y equals 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 equals x cubed. H is to do with translating the graph to the left or right. If h is positive, then (x minus h) all cubed is a translation to the right by h.

[When h is bigger than zero]

NARRATOR: And (x plus h) all cubed is a translation to the left by h.

[When h is bigger than zero]

NARRATOR: K is to do with translations up and down, and if k is a positive number then x cubed plus k is a translation up by k and x cubed minus k would be a translation down by k.

Part B. Show that if a cubic polynomial f of x equals ax cubed plus bx squared plus cx plus d can be rewritten in the form f of x equals a times (x minus h) all cubed plus k, then b squared minus 3ac must equal zero. We first note that if a cubic equation can be rewritten in the form a times (x minus h) all cubed plus 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 plus bx squared plus cx plus d, which gives us 3ax squared plus 2bx plus 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 minus 4 times 3a times c, which simplifies to 4b squared minus 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 minus 3ac equals 0. So b squared minus 3ac equals 0 will ensure that f of x equals ax cubed plus bx squared plus cx plus d has only one stationary point and hence that will allow us to write it in the form a times x minus h all cubed plus k.

Part C. Find an example of a cubic polynomial f of x equals ax cubed plus bx squared plus cx plus d where b squared minus 3ac is not equal to zero.
From Part B we know that when b squared minus 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 minus 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 plus 6x squared plus 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 plus 6x squared plus 2x plus 5 would be an example of a polynomial where b squared minus 3ac does not equal zero. That is, it's a polynomial that has either two stationary points or no stationary points.