If I have a graph displaying the equation of elastic potential energy, why does the graph increase quadratically? I was wondering why the graph is quadratic. If one could also link me to an external source to explain this, I would be just as grateful.
 A: Elastic potential energy is $$U=\frac 12 k x^2$$ Quadratic with extension, yes. 
This is because the elastic force (the spring force) is not constant:
$$F=kx$$
It increases with extension. If the force pulled equally hard at all extensions,  then you would expect double as much energy to be stored for double the displacement, since the force will pull equally much but double as far. A linear relationship. This is how gravity works. Gravity is constant at all (not too heigh) heights, and the gravitational potential energy $U=mgh$ is also linear with distance. But the elastic force is not constant, and so the elastic potential energy is not linear. 
In general, you calculate potential energy due to a force as an integral:
$$U=-\int F\;dx$$
If $x$ is included in the expression for $F$, then the result will contain $x^2$, quadratic.
This expression fundamentally comes from the work formula, which you might know, because potential energy is just the work that s conservative force will do:
$$W=F\Delta x\to\int F \;dx$$
A: This is an addition to @Steeven's answer.
Your graph plots the distance a marker is catapulted vs. the pull-back distance on the rubber bands. We can find a theoretical prediction for how far your projectiles should go.
Assuming the markers land at the same height at which they are launched, they will travel a distance of
$$d = \frac{v^2\sin2\theta}{g},$$
where $v$ is the launch speed, $\theta$ is the angle of launch (zero begin horizontal), and $g$ is the acceleration due to gravity. All of the marker's kinetic energy comes from the rubber bands, so we can write
$$K = U$$
$$\frac{1}{2}mv^2 = \frac{1}{2}kx^2$$
$$v^2 = \frac{kx^2}{m}$$
where $K$ is the kinetic energy, $U$ is the potential energy, $m$ is the mass of the marker, $k$ is the spring constant of the rubber bands, and $x$ is the pullback distance. Substituting this $v^2$ into the range equation above results in
$$d = \frac{kx^2\sin2\theta}{mg}.$$
So, we expect the range of the marker to be a quadratic function of the pullback distance.
