Given that an orbit is (at least in a simple two-body setup) an ellipse, a fairly regular path for which the angle (of the vector from the planet to the moon relative to some suitable starting point) increases continually over time, then it is not hard to see that it must rotate a full angle (360 degrees, $2\pi$ radians), for every orbit to keep the same face as best as such a path will permit, since it must return to that face upon return to its starting point and moreover must not complete any more cycles in the interim.
As a free body, the rotation of the Moon itself is more or less just usual rotation (i.e. ignoring other perturbing forces at this level of detail), which is created by its possession of rotational angular momentum. With no external torques, the angular momentum of an object does not change. Angular momentum is related to rotation speed by
$$L = I\omega$$
where $\omega$ is the angular velocity (rad/s). (The rotational period is $T = \frac{2\pi}{\omega}$, derived by considering that a full angle is $2\pi$ rad.). With no change in angular momentum $L$, the rotational speed does not change either. It therefore executes this motion, with such uniform rotational speed, while traversing the orbit ellipse, independently of the fact of that elliptical traversal. (The whole reason it rotates at the same rate as it orbits is that it was effectively "torqued down" a long time ago, thus now that torque has ceased.)
Thus for your simple game, you can model the Moon as a circular object which rotates once per sidereal (true orbital) month (2.36 Ms, or about 27 days, which you will probably not be running as literally that period of course :) ), that is, you just define its angle by
$$\theta_\mathrm{orientation}(t) := 2\pi \frac{t}{T}$$
in radians, where $T$ is the period (which you can take as whatever value you need for your simulation) and $t$ the current simulation time. You can use whatever favorite rotation method you want to use to rotate the graphic by this angle (e.g. matrices, quaternions, etc.).
For the actual movement of the Moon along the orbit, however, this is a bit more complicated. In general, the value of orbital angle ("true anomaly", i.e. the angle I just mentioned formed by the Moon's position vector, where the zero point is taken as the periapse) over time is not a simple, even elementary function: in theory, you can write it using a sort of "generalized Lambert W function", but it is best computed numerically.
One final point I want to point out is that not all "other such objects" have this synchronous rotation characteristic. It is the case for the Moon, but moons of other planets may have other rotation patterns, even entirely decoupled from their orbit. The trick is whether or not there are strong enough gravitational interactions and of the right type - here due to the tides - to create some kind of orbital position-dependent torque that will then bring the body into both a rotation and orbit pattern that are related in some fashion. If you look at a planet like Jupiter, you will find the four famous "Galilean" moons have a complex "resonance" pattern caused by the fact that they essentially apply bursts of force to each other during their orbits as they get close. The outer moons of Jupiter, which are very small, effectively large asteroids, and thus neither exert any strong forces on each other nor have the spatial extent to have significant tides raised on them by Jupiter, have effectively decoupled rotational and orbital periods, as is the case with Earth and the Sun (e.g. the moon Himalia orbits at about 11 500 Mm from Jupiter, is about 150 km in size, orbits with a period of 21.6 Ms, about 2/3 of a year, but rotates far faster, very roughly a thousand times, with a period of only 28 ks.).
ADD: It seems that another desired piece of information here is related to finding just how exactly the Moon itself slows down to reach synchronous rotation. This is a bit more involved to explain.
Basically, what happens is that, thanks to Newton's law of universal gravitation, i.e.
$$F_G = G\frac{m_1m_2}{r^2}$$
the gravitational force from a planet, like the Earth, is not uniform with increasing distance, but rather falls off. (This is fortunate, if it weren't, then the whole Universe would have never been able to expand out of the Big Bang! For us and more practically, it also is what allows us to "escape" a gravity well with finite energy - if gravity kept tugging as strong as ever no matter how far away you were, you would eventually stop and fall back down even if projected out very fast, and you would have to continually be burning engines to climb higher and higher.)
What that also means, however, is that if we are given something like the Moon, because it is not a point mass but rather has some spatial extent - in particular, about 3474 km from front to back, it will experience more force ($r$ is smaller) on the side facing the Earth than on the side facing away. This results in effectively a force gradient across the Moon, which has the effect of stretching it like a piece of clay. As a result, it becomes slightly oblong, not a perfect sphere. (The same thing happens to the Earth as a result of the Moon, and this is why we have the tides, and moreover, why this is called a "tidal" force.) From the Moon's point of view, the apparent forces change (similar to how centrifugal force works) and it sees this as though it has a force being applied to its front and back which are pulling in opposite directions, stretching it out, while another force is applied around the "waist" between these two points, trying to squeeze it in. [If you've read anything ahead about black holes, you may have come across the idea of 'spaghettification'. This is a far more extreme example of this same thing, and conversely, this can be considered the effect in an extremely incipient and immature form.]
As the Moon rotates around the Earth in its orbit, since it's rotating at the same speed (ignoring also the slight, but ultimately netting zero due to equal reversal each period, irregularity due to the slight ellipticity of the Moon's orbit) as it orbits (i.e. the periods are equal), this bulge points toward the Earth at all times.
However, were the Moon in rotation, this would not quite be the case: in particular, if I were to apply a small impulse to the Moon to give it a "kick" that spins it slightly ahead, the "bulge" won't simply ripple through it right away and remain aligned because the Moon's material takes time to deform under the realignment of forces. In particular, this is as a result of both inertial (the natural property of matter to resist changing its state of motion in response to a force as in Newton's first and second laws which btw are not the same thing), and also mechanical (i.e. the natural stiffness of the material) effects. Thus, what will happen is the bulge will end up at an awkward angle, with part facing slightly behind the Moon in its orbit and part facing ahead.
However, both parts will be being pulled toward the center of the Earth, or by the Moon's view they will be being tugged apart. Effectively, the force situation looks like this:
If you were to think of these forces as being provided by two people engaged in a tug-of-war with respect to the bar in the middle, you would easily expect it would straighten out, right? And indeed, that's what's going on here. The two forces constitute a couple, and the result is to produce a net torque. Torque (also called "moment"), after all, is defined by
$$\mathbf{\tau} = \mathbf{r} \times \mathbf{F}$$
where $\mathbf{r}$ is the vector to the point of force application from the center (here along the bar), and $\mathbf{F}$ is the applied force. Thanks to the right-hand rule for the cross product, two such equal and contrary forces will combine their torques and thus produce a net torque that will tend to rotate the object.
Moreover, since the Moon is always orbiting, it actually can't stop being angled so long as it rotates faster, and as a result, it continues to experience torque. The torque thus acts to retard the rotation until it finally no longer makes an angle with its bulge toward the gravitator (Earth). If it had initially no rotation (a scenario that would allow us to actually see the back side of the Moon), the same effect would do the opposite and torque it up, speeding it up until it reached synchronous rotation "from below".
It is a bit involved to compute the effects of the torque in this situation, however the relevant equation is
$$\frac{d \omega}{dt} = \frac{45}{8} k G \frac{M_E^2 A^3}{M_M r^6} \sin(2\alpha)$$
where $M_E$ is the mass of the Earth, $M_M$ the mass of the Moon, $r$ the orbital distance, $\alpha$ the angle the bulge forms with the line between Earth and Moon, and $k$ is something known as the "tidal Love number" which takes account of the composition of the object. $A$ is the Moon's cross-sectional area.
Now the angle $\alpha$ itself depends on the rotation speed $\omega$ - in particular, we could imagine $\alpha = C (\omega_\mathrm{orbit} - \omega)$ where $C$ is some inertial constant (the harder the Moon would rotate, the further "back" it can keep its "bulge"), and thus technically the equation has an $\omega$ in the sine, i.e.
$$\frac{d \omega}{dt} = \frac{45}{8} k G \frac{M_E^2 A^3}{M_M r^6} \sin(C [\omega_\mathrm{orbit} - \omega])$$
and thus is nonlinear. If however $\alpha$ is suitably small (under about 500 mrad), we can approximate the sine as $\alpha$ itself and so if the Moon isn't rotating too fast (the inaccuracy should not matter for a game), we get
$$\frac{d \omega}{dt} = \frac{45}{8} k G \frac{M_E^2 A^3}{M_M r^6} \cdot C (\omega_\mathrm{orbit} - \omega)$$
Which has the generic form
$$\frac{d\omega}{dt} = K \omega + J$$
and we recognize this as an exponential equation for $\omega$. $K$ will be negative if the rotation is slowing, so over the long term, $\omega$ will decay exponentially - providing justification for @CR Drost's answer.
Note, however, that this also depends very crucially on the orbital radius $r$, and in fact that is not constant. In particular, the loss of angular momentum ($L = I\omega$ so $\frac{dL}{dt} = I \frac{d\omega}{dt}$) to rotation ends up as a gain in angular momentum by the rest of the system - both the Earth and the Moon's orbit. Thus technically speaking we actually have a coupled system of differential equations, and it will not be possible to solve exactly, but it can be numerically integrated.
That said, I'd suspect that CR Drost's suggestion to just use the exponential will suffice for simplistic modeling in game physics. It will though be rather poor if you are considering the Moon all the way back to the point of formation. If you want to get nitpicky, one possible way to improve on it is to fake (still approximate - see a trend here? This is physics! This is why that there's the long-standing joke about the "spherical cow", you try and toss away what you can and when you can't you still try to toss away as much as you can and still get away with it! If the cow can't be perfectly spherical, it better be an ellipsoid. If it can't be that, make it triaxial. Etc.!) it with an assumed circular orbit, i.e. take for the Moon that
$$G \frac{M_E M_M}{r^2} = M_M r \omega_\mathrm{orbit}^2$$
and since for the orbital angular momentum we have that $L_\mathrm{orbit} = (M_M r^2) \omega_\mathrm{orbit}$ that
$$G \frac{M_E M_M}{r^2} = \frac{L_\mathrm{orbit}^2}{M_M r}$$
you find how $L_\mathrm{orbit}$ affects $r$, the orbital distance. If we then also use another assumption that all lost angular momentum goes into the orbit, then $\frac{dL_\mathrm{orbit}}{dt} = -\frac{dL}{dt}$ where $L$ is the Moon's rotational angular momentum from before, and also that the rotation speed of the Moon $\omega$ is related to $L$ by $L = I\omega$, we can combine these to get a differential equation for $\omega$ taking into account the orbital outspiral. This will not be exponential in its early stages but contain a rapidly dropping-off power term, I believe (at least - our favorite word - approximately!). You can numerically integrate this (e.g. Runge-Kutta) for your simulation. Though personally, I'd say just stick to the exponential :)
That said, I should also point out what was mentioned in the comments that this also affects the Earth as well, and in fact the Earth is slowing down. More accurately, it is slowing right now (mind the $\frac{1}{r^6}$) at a rate of about 0.54 ns/s, equiv. 0.54 ms/Gs (milliseconds per Gigasecond). As a result, while we usually think of a day as 86 400 s - or equivalently, most commonly as 24 hours - and use this interval to define it in clocks, it is actually slightly longer: it is now 86 400.002 s: or about two milliseconds longer. This is because the current definition of a second is a fixed time unit based on the atomic clock, and it was established to refer to 1/86 400 of the day of the year 1900. There have now been 108 years, or 3.4 gigaseconds, since then, and so the day has grown by 0.54 x 3.4 = 1.8 ms, which I rounded to 2 ms (it's actually a bit more irregular than that due to other, more complex, factors.). Over your lifetime - typically 2.5 Gs (79 years) for the rich world, 2.2 Gs for the global average - you can expect it will lose another 1.3-1.4 ms or so. A whole second will then be had by around 2000 Gs from now (over 60,000 years), and the day will be 86 401 s long. The Earth's rotation will never get a chance to slow to the ~2.4 Ms length of the Moon's orbit: by this formula (we will actually need the full exponential to be more accurate, and that actually means, since $r$ is increasing, it will be longer, so this is a lower bound) it would take - using that 0.54 ms/Gs = 0.54 Ms/Es (know your SI prefixes well - three up from milli is mega, three up from giga is exa!) 4.4 Es for that to happen. But one exasecond is over twice the elapsed lifetime of the Universe, and the Sun will die long before then. In its death throe, the Earth and Moon will be dragged into the greatly expanded stellar atmosphere and be vaporized.
Contrary to what H.G. Wells imagined in his old novel "The Time Machine" (where his time traveler traveled far into the future and reported the slowing down of the Earth's rotation), there would not be time for this to occur in reality. (At his time, much less was understood about astrophysics and especially stellar dynamics than now.)
(Keep in mind this is the mean solar day, not the sidereal day: the difference is the latter is the true rotation period, the former is the time between two noons or midnights. The two are not equal due to the effects of the motion of the Earth in its orbit around the Sun.)
