Curve fitter
This problem challenges you to find cubic equations which satisfy different conditions.
Problem
This problem challenges you to find cubic equations which satisfy different conditions. You may like to use Desmos to help you investigate possible cubics.
Part 1
Can you find a cubic which passes through $(0,0)$ and the points $(1, 2)$ and $(2,1)$?
Can you find more than one possible cubic?
How can you use this together with the information provided in the question?
Try using this GeoGebra page to investigate possible cubics.
Part 2 (a)
Can you find a cubic which passes through $(0,0)$ and the points $(1, 2)$ and $(2,1)$, and where the point $(1,2)$ is a turning point of the cubic?
Can you find more than one cubic satisfying all the conditions?
Part 2 (b)
Can you find a cubic which passes through $(0,0)$ and the points $(1, 2)$ and $(2,1)$, and where the point $(2,1)$ is a turning point of the cubic?
Can you find more than one cubic satisfying all the conditions?
Part 3
Can you find a cubic which passes through $(0,0)$ and where the points $(1, 2)$ and $(2,1)$ are both turning points?
Your answers to Part 2 may make you suspect that it is impossible to find a cubic which satisfies the last set of conditions. Can you prove that it is impossible?
You can use this proof sorter to show one way of proving that there are no cubics that satisfy the last set of conditions. Alternatively, this second proof sorter shows another possible way. Can you work out a third proof?
Student Solutions
Part 1
Sammy from Lancing College in the UK used the general form of a cubic equation to find some examples:
Mahdi from Mahatma Gandhi International School in India found the general form of a cubic which passes through the three points:
Mahdi also created a video showing all of the possible cubics. Click here to download Mahdi's video. Notice how similar this is to what happens on the GeoGebra page in the link in the hint of the problem.
Part 2
Siddhant from Singapore International School in India and Sammy used the same setup Sammy used in part 1 to find a cubic that works for part 2a. This is Siddhant's work:
The general form of a cubic curve is
$(0,0) (1,2) (2,1)$
To find the values of $a, b, c$ and $d,$ we just need to substitute the values of $x$ and $y$ in each of the three coordinates, and build a system of three equations. These equations are:
$(0,0)$
$0 = d$
$(1,2)$
$2 = a + b + c$
$(2,1)$
$1 = 8a + 4b + 2c$
We also know that $(1,2)$ is a stationary point. This tells us that the first derivative of the general equation is equal to $0$ at $x=1.$
first derivative: $3ax^2 + 2bx + c$
Therefore: $3a(1)^2 + 2b(1) + c = 0$
$3a + 2b + c = 0$
Thus, the three equations we have are,
1. $a + b + c = 2$
2. $8a + 4b + 2c = 1$
3. $3a + 2b + c= 0$
We can multiply equation 3 by $2,$ and then subtract that from equation 2.
This gives us,
$2a = 1$
$a = \frac12$
From this, we get the two equations:
$b + c = \frac32$
$4b + 2c = -3$
By using elimination method, the values of $b$ and $c$ are,
$b = -3\
c = 4.5$
Therefore, the cubic equation we get is:
$y = \frac12x^3 -3x^2 + 4.5x$
Mahdi used the general cubic from part 1 and also combined this with differentiation:
Equation: $y = ax^3 + \dfrac{-3-6a}{2}x^2 + \dfrac{7+4a}{2}x $
Which also gives the equation $y = \frac12 x^3 -3x^2 + \frac92 x$
Mahdi also found a curve for part 2b:
Which gives the equation $y = 1\frac14 x^3 -5\frac14 x^2 + 6x$
Part 3
Mahdi explained why this is not possible using the results from part 2:
This is Sammy's work for part 3. Sammy began with the general equation for a cubic which passes through $(0,0)$ and then chose the turning points to be when $x=1$ and when $x=2.$ However, Sammy's conclusion is not correct (see explanation below):
Between lines A and B, Sammy loses information, because $12a+2b$ was supposed to be the positive square root whereas $6a+2b$ was the negative square root. This information is lost when both sides are squared.
In line C, having lost the information about the negative square root, Sammy gets that $12a+2b=6a+2b$ and notices that this would give $a=0,$ which isn't allowed. However, using the negative square root information, this should be $12a+2b=-(6a+2b)\Rightarrow18a=-4b\Rightarrow b=\frac{-9}{2}a.$
You can substitute this back into Sammy's earlier equations to find $c$ in terms of $a$ (or vice versa) and therefore find a family of cubics which go through $(0,0)$ and have turning points at $x=1$ and $x=2.$ However, as Siddhant and Mahdi have shown, it won't include one with $y=2$ when $x=1$ and $y=1$ when $x=2.$ Can you prove it using this method?
Teachers' Resources
Why do this problem?
This problem looks at how the number of constraints can reduce the number of possible answers, by considering cubic graphs. It develops the idea that a general cubic $y=ax^3+bx^2+cx+d$ has 4 parameters (or 4 degrees of freedom) and so three pieces of information are not enough to fully determine all of the coefficients.
Possible approach
This problem featured in an NRICH Secondary webinar in January 2021.
Starting from the general form of the cubic, the information given by the points on the cubic can be used to generate equations. These can be used to find the values of some of the parameters, or to write some parameters in terms of another parameter and so reduce the number of "degrees of freedom".
As the problem progresses more pieces of information are given, so more equations can be generated which narrows down the options until the problem is "over-constrained" and it is impossible to satisfy all of the constraints at once.
Desmos can be used to help students investigate the graphs, or to check their solutions.
Students could also look at cubic graphs which satisfy different sets of conditions, such as a cubic graph that has turning points at $(1,2)$ and $(2,1)$ which does not pass through the origin.
The thinking that informed the creation of this problem is discussed in this video (after the introductory 3min 20sec):
Key questions
- What does the equation of a cubic curve look like?
- The curve passes through the origin. What does this tell us?
- How can we use the points that the curve passes through?
- If the curve has a turning point at $(1, 2)$, what information does that give us?
Possible support
Students could be introduced to this problem by first asking them to find a quadratic which passes through the three points $(0,0)$, $(1, 2)$ and $(2,1)$. They could also be asked to show that it is impossible to find a straight line that passes through all three points.
For Part 1, students should find that there is more than one curve that passes through the three given points. This GeoGebra file shows how a family of cubics can pass through $(0,0)$, $(1, 2)$ and $(2,1)$. By varying $a$ it can be shown that either $(1,2)$ or $(2,1)$ can be a turning point, but not both at the same time.
Possible extension
Students could be asked to consider what questions might have been appropriate for parts 2(c), 2(d) and 2(e), and could find cubics which satisfy these. This GeoGebra file which shows a family of cubics which pass through $(0,0)$ and has turning points at $x=1$ and $x=2$ might help students visualise how four of the five conditions can be met at any one time. Students might also like to consider how this graph has been created.