On the picture above, you'll see a circuit. The exercise is to make a transfer function and draw a bode plot. I've tried to do that, but because there's no solution, I don't know if I'm right.
Specifications: C1 and C2 are infinity, there's no value given for the two resistors (R), R1=180 ohm, R2=180k ohm, C3=22 microF
My attempt
This is an non-inverting opamp-circuit, so the attenuation is: $$\frac{V_{out}}{V_{in}}=1+\frac{R_1+Z_{C_3}}{R_1+R_2+Z_{C_3}}$$
Converting this to a transfer function: $$H(s)=\frac{R_1*C_3s+1}{C_3s(R_1+R_2)+1}$$ The zero and pole point is: $$z_1=-\frac{1}{R_1*C_3}\quad and \quad p_1=-\frac{1}{(R_1+R_2)*C_3}$$ The magnetude and break-frequencies are: $$|H(j \omega)|=\frac{\sqrt{(R_1 C_3 \omega)^2+1}}{\sqrt{(C_3 \omega (R_1+R_2))^2+1}}\\ f_{break1}=\frac{1}{2 \pi R_1 C_3}=40Hz\\ f_{break2}=\frac{1}{2 \pi (R_1+R_2) C_3}=0,04Hz$$
Can someone check if this is correct, or improve me if I'm wrong, please?
EDIT by the correction from the answers
This is an non-inverting opamp-circuit, so the attenuation is: $$\frac{V_{out}}{V_{in}}=\frac{R_1+R_2+Z_{C_3}}{R_1+Z_{C_3}}$$
Converting this to a transfer function: $$H(s)=\frac{C_3s(R_1+R_2)+1}{R_1*C_3s+1}$$ The pole and zero point is: $$p_1=-\frac{1}{R_1*C_3}\quad and \quad z_1=-\frac{1}{(R_1+R_2)*C_3}$$ The magnetude and break-frequencies are: $$|H(j \omega)|=\frac{\sqrt{(C_3 \omega (R_1+R_2))^2+1}}{\sqrt{(R_1 C_3 \omega)^2+1}}\\ f_{break1}=\frac{1}{2 \pi R_1 C_3}=40Hz\\ f_{break2}=\frac{1}{2 \pi (R_1+R_2) C_3}=0,04Hz$$