Curvy cubics

Thank you to everyone who has submitted solutions to this problem.

Jayden has answered very well the question of what makes a cubic curvy or uncurvy.

A cubic polynomial can be expressed in the form

$Ax^3 + Bx^2 + Cx+D$

The gradient function ($\frac{dy}{dx}$) of the general cubic expression is:

$3Ax^2+2Bx+C$

A stationary point is defined as a point on the curve where the gradient is 0.

This is what gives a cubic equation 'its curves' since when the gradient is 0, the part of the curve prior to the stationary point could be decreasing (i.e. have negative gradient), and as we get closer to the stationary point, the gradient converges to zero before becoming positive once we go past the stationary point (and vice versa).

Since the gradient function of a cubic is a quadratic, there are a maximum of two stationary points and a minimum of zero (since $ax^2+bx+c=0$ has a maximum of two reals solutions and a minimum of zero real solutions).

For a cubic to be 'curvy', there should be two stationary points, so the gradient function (i.e. the quadratic) must have two reals solutions when set to equal zero. Therefore the discriminant ($b^2-4ac$) must be positive so $4B^2-12AC>0$

To generate an 'uncurvy' cubic there should be no stationary points, so the gradient function must have no solutions when set to equal zero. Therefore the discriminant ($b^2-4ac$) must be negative so $4B^2-12AC<0$

Minhaj from St Ivo school has used similar ideas to work out a possible equation of the given cubic curve.

A cubic function is one of in the form $f(x) = ax^3+bx^2+cx+d$

Since the y-intercept for this cubic graph is when $y=1$, we know the equation will be of the form $y=ax^3+bx^2+cx+1$

Differentiating this function gives $\frac{dy}{dx} = 3ax^2+2bx+c$. Looking at the graph above, it can be seen that there are no stationary points, so for this graph $\frac{dy}{dx} \ne 0$.

I figured out that $c$ can't be 0, as then you would end up with $ \frac{dy}{dx} = 3ax^2+2bx$ and if you set the this equal to zero, $3ax^2+2bx$ is factorisable and hence you would always have two stationary points.

I realised that if the value of $b$ was zero then $\frac{dy}{dx}$ would be $3a x^2+c$. This could be a possible option for this graph, if the value of $c$ is positive and $a$ is positive (from looking at the graph, I can tell the graph is a positive cubic so $a$ is positive). The reason for this is if the value of $c$ is positive, there would be no real values for $x$ as you can not take the sqare root of ($\frac{-c}{3a}$) because it is a negative number, so in other words, there would be no stationary points.

From simple substitution, I shall demonstrate how I found out that $b=0$.

Looking at the graph, I can see that when $x=1$, $y=4$, so

$a+b+c = 3$

Also I can see that when $x=-1$, $y=-2$. So

$-a+b-c=-3$

Adding these equations together we see that $b$ has to equal zero for this to work.

I made an attempt to find the value of $c$: when $x=0$, then $\frac{dy}{dx} = c$.

So I then proceeded to draw an approximate tangent line at the coordinate $(0,1)$, and saw the line looks like it would have gradient of 2, suggesting $c=2$

If $c=2$ then $a$ must be 1 as $a+c=3$.

So my answer for the equation for the cubic graph is

$$y = x^3+2x+1 $$

Well done to Minhaj for getting this correct answer. There are other answers that would be acceptable for this cubic equation, as the picture does not give enough information to determine the cubic exactly, although it can be estimated as Minhaj has done. Any cubic of the form $y=ax^3+cx+1$ with $a$ positive would be fine as long as $a+c=3$, and $c$ is positive (so the cubic is 'not curvy').