Cut / Boost of this Shelving-Filter
There are several options to tackle a circuit like this one: the brute-force algebra implying heavy KCL and KVL equations or the fast analytical techniques or FACTs. What is nice with the FACTs is that you don't need to understand in details how the circuit works to determine its transfer function. For instance, with this filter, it is a first-order circuit and its transfer function obeys the following expression:

\$H(s)=H_0\frac{1+s\tau_2}{1+s\tau_1}\$ or \$H(s)=H_{\infty}\frac{1+\frac{1}{s\tau_2}}{1+\frac{1}{s\tau_1}}\$ 

In this expression, \$H_0\$ represents the quasi-static gain obtained for \$s=0\$ (redraw the circuit with open-circuited capacitors or shorted inductors) while \$H_\infty\$ is the gain of the same circuit but observed when \$s\$ approaches infinity (caps are shorted and inductors open circuited).

The terms \$\tau_1\$ and \$\tau_2\$ respectively designate the time constants involving the considered energy-storing element (here it is \$C_1\$) when the circuit is observed with a zeroed stimulus (\$V_{in}=0\;V\$, short the source) and when the response \$V_{out}\$ is nulled (0 V despite the stimulus presence).

We can start with the dc gain determined when the cap. is removed from the circuit. The below simulation shows that the gain in this case is \$H_0=-\frac{R_2}{R_1}\$:



Now, to determine \$\tau_1\$, disconnect capacitor \$C_1\$ and determine the resistance "seen" from its connecting terminals while the stimulus \$V_{in}\$ is replaced by a short circuit. The drawing is shown below:



Equations are here:



The resistance in this case is \$48.42\;k\Omega\$ and the time constant is \$\tau_1=48.42\;k\Omega\times 10\;nF = 484.21\;µs\$

For the second time constant, it is slightly more complicated but not that much. You bring the excitation back and you determine the resistance "seen" from \$C_1\$'s connecting terminals while the output is nulled, equal to 0 V. Practically speaking, you connect a current source \$I_T\$ across \$C_1\$'s connecting terminals and you tweak it until \$V_{out} = 0\;V\$. The resistance you want is \$\frac{V_T}{I_T}\$ with \$V_T\$ the voltage across the current source in this mode. See the below drawing in action for a better explanation:



All the equations to determine these value are here:



By adjusting the value of \$k\$ you change the value of \$RV1_a\$ and \$RV1_b\$ and modify the transfer function. When \$k=0.5\$, the response is flat as the pole and the zero are coincident. Here are a few curves gathered for different \$k\$ values:



\$k=0.3\$



\$k=0.7\$



This is it. The FACTs are an excellent tool when you need to determine transfer functions swiftly and, more importantly, obtain the result in a so-called low-entropy form meaning the result is well formated with apparent poles and zeroes. In brief, "Vive Les FACTs" !