What would this curve be called?

Since it is a resistor-capacitor circuit, this means that the differential equation describing your system must be of the following type:

$$R\frac{dQ}{dt}+\frac{Q}{C}=U \; ,$$

in which $R$ is the resistance, $C$ the capacity, $Q$ the charge stored in the capacitor at some time, $U$ the voltage applied on the system by some external source like a DC battery.

Solving this system gives

$$Q(t)=CU+(Q_0-CU)e^{-\frac{t}{RC}} \; ,$$

in which $Q_0$ is the charge stored at time $0$. If you need the voltage stored, just divide by $C$ to get:

$$V(t)=U+(V_0-U)e^{-\frac{t}{RC}} \; ,$$

with $V_0=Q_0/C$. The figure you show is giving the current flowing through the system, but that is related to the voltage by $V=IR$. Also the $V$ in the picture is what I called $U$.

As Oltarus already mentioned, this is an exponential curve, and the phenomenon it describes is often termed "exponential relaxation".