Can you find ... cubic edition
Can you find cubic functions which satisfy each condition?
Age
16 to 18
Challenge level
Problem
Can you find a cubic curve that...
(a) ... passes through the $x$-axis at $x=1$ and $x=-1$?
(b) ... passes through the origin and touches the $x$-axis at $x=-3$?
(c) ... touches the $x$-axis at $x=2$ and crosses the $y$-axis at $12$?
(d) ... crosses the $y$-axis at $-6$ and has three integer roots?
(e) ... crosses the $y$-axis at $y=5$ and touches the $x$-axis at $x=1$?
Are any of the curves described above unique?
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Visit the site at undergroundmathematics.org to find more resources, which also offer suggestions, solutions and teacher notes to help with their use in the classroom.
Underground Mathematics is hosted by Cambridge Mathematics. The project was originally funded by a grant from the UK Department for Education to provide free web-based resources that support the teaching and learning of post-16 mathematics.
Visit the site at undergroundmathematics.org to find more resources, which also offer suggestions, solutions and teacher notes to help with their use in the classroom.
Student Solutions
Here is Louis from EMS' solution to part $a)$ of the question:
We can express all cubics in the form $y = (x-a)(x-b)(x-c)$, where the x-intercepts will be $a$ , $b$ and $c$, and the y-intercept will be $-abc$.
So one possible solution for this question is:
$y=(x)(x-1)(x+1)$
$= x(x^2-1)$
$= x^3-x$.
This is not unique. The initial bracket that contains
just an '$x$' can be multipled by any non-zero real number to give an answer
that fits the description, but is a vertically stretched/compressed version
of the curve I gave.
Here is Vatsal from Wilson's School's solution to parts b) and c) of the question:
$(b)$ The question informs us that the cubic must pass through $(0,0)$ and touches the $x$ axis at $x= -3$. The fact that the curve touches the $x$ axis at $-3$ rather than cuts through it informs us that we can consider this point to be either a maximum or minimum point on the curve, and this point can be considered to 'count' for two roots. Hence a possible solution is: $y = x(x+3)^2$, but this is not unique as it can be multiplied by a constant.
$(c)$ The question tells us that the function must touch the $x$ axis at $x=2$ and cross the $y$ axis at $12$. Since the curve touches the $x$ axis at $x=2$, we know the curve is of the form: $y = (x+a)(x-2)^2$ To find $a$, we can use the $y$ intercept ($12$) to form the equation $a \times (-2)^2 =12$. Therefore, $4a=12$, giving us $a=3$. So one answer is: $y = (x+3)( x-2)^2$.
This is not unique as $y=3(x+1)(x-2)^2$ is also a solution.
Here is Aneesh from West Island School Hong Kong's solution to parts d) and e) of the question. Aneesh thought about cubic equations of the form $(x-a)(x-b)(x-c)d$, with roots at $a$, $b$ and $c$ and $y$-intercept $-abcd$.
$(d)$ $(x+a)(x+b)(x+c)d= -6$ at $x=0$, so $abcd=-6$. Rearranging, $abc=-6/d$ . As $a$,$b$,$c$ are all integers, we also have $abc$ is an integer and so $-6/d$ is an integer.
If $d$ is an integer: the only positive factors of $6$ are $1,6,2,3$. So $d$ may only be $\pm 1,\pm 6, \pm-2, \pm 3$.
For each possible integer $d$, here is a set of integers $(a, b, c)$ where $abcd = -6$ is satisfied:
$d=1$: $(1,2,-3)$
$d = -1$: $(1,2,3)$
$d = 2$: $(1,-1,3)$
$d = -2$: $(1,1,3)$
$d = 3$: $(-1,1,2)$
$d = -3$: $(1,1,2)$
$d = 6$: $(-1,1,1)$
$d = -6$: $(1,1,1)$
These are not the only solutions for each $d$. And if d is in the closed interval $[0,1]$ (values such as $\frac{1}{2}$, $\frac{1}{3}$ etc.. then there are even more possibilities. So cubics that satisfy these conditions are NOT unique.
$(e)$ This is the same as condition $(c)$, with different values of $x$ and $y$ intercepts. Solving similarly, we have $(x-1)^2(x+a)b = 5$ at $x=0$. This yields $ab=5$, where $a$ and $b$ are real numbers. So the solutions are $a,b$ are not zero, $b$ cannot be equal to $-5$ ( as then $a=-1$ and we have a saddle point at $x=1$). So for any $a,b$ that satisfy the above conditions, the conditions of the question are fulfilled.
Thank you to everybody who submitted a solution to this problem!
We can express all cubics in the form $y = (x-a)(x-b)(x-c)$, where the x-intercepts will be $a$ , $b$ and $c$, and the y-intercept will be $-abc$.
So one possible solution for this question is:
$y=(x)(x-1)(x+1)$
$= x(x^2-1)$
$= x^3-x$.
This is not unique. The initial bracket that contains
just an '$x$' can be multipled by any non-zero real number to give an answer
that fits the description, but is a vertically stretched/compressed version
of the curve I gave.
Here is Vatsal from Wilson's School's solution to parts b) and c) of the question:
$(b)$ The question informs us that the cubic must pass through $(0,0)$ and touches the $x$ axis at $x= -3$. The fact that the curve touches the $x$ axis at $-3$ rather than cuts through it informs us that we can consider this point to be either a maximum or minimum point on the curve, and this point can be considered to 'count' for two roots. Hence a possible solution is: $y = x(x+3)^2$, but this is not unique as it can be multiplied by a constant.
$(c)$ The question tells us that the function must touch the $x$ axis at $x=2$ and cross the $y$ axis at $12$. Since the curve touches the $x$ axis at $x=2$, we know the curve is of the form: $y = (x+a)(x-2)^2$ To find $a$, we can use the $y$ intercept ($12$) to form the equation $a \times (-2)^2 =12$. Therefore, $4a=12$, giving us $a=3$. So one answer is: $y = (x+3)( x-2)^2$.
This is not unique as $y=3(x+1)(x-2)^2$ is also a solution.
Here is Aneesh from West Island School Hong Kong's solution to parts d) and e) of the question. Aneesh thought about cubic equations of the form $(x-a)(x-b)(x-c)d$, with roots at $a$, $b$ and $c$ and $y$-intercept $-abcd$.
$(d)$ $(x+a)(x+b)(x+c)d= -6$ at $x=0$, so $abcd=-6$. Rearranging, $abc=-6/d$ . As $a$,$b$,$c$ are all integers, we also have $abc$ is an integer and so $-6/d$ is an integer.
If $d$ is an integer: the only positive factors of $6$ are $1,6,2,3$. So $d$ may only be $\pm 1,\pm 6, \pm-2, \pm 3$.
For each possible integer $d$, here is a set of integers $(a, b, c)$ where $abcd = -6$ is satisfied:
$d=1$: $(1,2,-3)$
$d = -1$: $(1,2,3)$
$d = 2$: $(1,-1,3)$
$d = -2$: $(1,1,3)$
$d = 3$: $(-1,1,2)$
$d = -3$: $(1,1,2)$
$d = 6$: $(-1,1,1)$
$d = -6$: $(1,1,1)$
These are not the only solutions for each $d$. And if d is in the closed interval $[0,1]$ (values such as $\frac{1}{2}$, $\frac{1}{3}$ etc.. then there are even more possibilities. So cubics that satisfy these conditions are NOT unique.
$(e)$ This is the same as condition $(c)$, with different values of $x$ and $y$ intercepts. Solving similarly, we have $(x-1)^2(x+a)b = 5$ at $x=0$. This yields $ab=5$, where $a$ and $b$ are real numbers. So the solutions are $a,b$ are not zero, $b$ cannot be equal to $-5$ ( as then $a=-1$ and we have a saddle point at $x=1$). So for any $a,b$ that satisfy the above conditions, the conditions of the question are fulfilled.
Thank you to everybody who submitted a solution to this problem!