Can You Find ... Cubic Edition

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!