Frequency Characteristics of Low pass filter
What you need is a transfer function linking the response (what you observe across \$C_1\$) and the stimulus (what you inject across \$R_2\$). There are plenty of possibilities to determine this transfer function and I recommend using the fast analytical circuits techniques or FACTs. 

The principle is simple: reduce the stimulus voltage to 0 V or replace the input source by a short circuit. Then temporarily remove the capacitor from its connecting terminals and "look" through these connections to determine what is the resistance \$R\$. See the below exercise for your problem:



If you look through the connection, by inspection, you can see that the resistance is \$R_1\$. The time constant associated with this circuit is therefore \$\tau_1=R_1C_1\$. The transfer function linking \$V_{out}\$ to \$V_{in}\$ in the Laplace domain is thus expressed as \$H(s)=\frac{1}{1+\tau_1s}=\frac{1}{1+\frac{s}{\omega_p}}\$ in which \$\omega_p=\frac{1}{R_1C_1}\$.

You could also apply an impedance divider formula and say \$H(s)=\frac{\frac{1}{sC_1}}{\frac{1}{sC_1}+R_1}\$ but it takes longer time to develop and rearrange. Furthermore, you can make mistakes when developing the expression.

Once you have this formula, you need to determine its magnitude and phase in order to get the graph you shown. The magnitude will be expressed in dB while the \$x\$-axis will be log-compressed: what is horizontally plotted won't be \$f\$ but \$Log(f)\$. The magnitude of a complex number \$z=x+jy\$ is \$|z|=\sqrt{x^2+y^2}\$. If you replace \$s\$ by \$s=j\omega\$ in the expression \$H(s)\$, then you have \$H(j\omega)=\frac{1}{1+j\frac{\omega}{\omega_p}}\$ and the magnitude you want to plot in the vertical axis is simply \$|H(\omega)|=\frac{1}{\sqrt{1+(\frac{\omega}{\omega_p})^2}}\$. To plot this magnitude function, simply calculate \$y(\omega)=20Log{|H(\omega)|}\$ and you will have a vertical axis in dB. When \$\omega\$ is zero, the magnitude is 1 (no attenuation) or 0 dB. When \$\omega=\omega_p\$ the magnitude is 0.707 or -3 dB. This the so-called cutoff frequency. And as \$\omega\$ increases beyond this point, the magnitude keeps going down with a slope of -20 dB per decade (also called a -1 slope).

The phase is obtained by remembering that the argument of \$z\$ is \$arg(z)=\arctan(\frac{y}{x})\$. With a quotient as we have with \$H\$, the argument is that of the numerator minus that of the denominator. Therefore, \$arg(H(\omega))=arg(1)-\arctan(\frac{\omega}{\omega_p})=-\arctan(\frac{\omega}{\omega_p})\$. The below Mathcad sheet shows how this looks like with typical components values.