Calculating an operational amplifier with changing power source
Typically, over a fairly wide range, you can approximate the output voltage when railed at the positive or negative supply at Vout as Vcc-kp or GND + kn, where kp and kn are some voltage like 0.1V or 1.5V and are dependent on the output circuit of the op-amp (and loading). 



It's a voltage divider. Ignoring output drop (which you can't do with the shown op-amp) &amp; ignoring zener impedance and assuming it's biased appropriately by R3, we can write the equations by inspection. 

When Vout is low it's Vin+ = \$ \frac{R1}{R1+R2}V_Z\$ 

When Vout is high, it's Vin+ = \$(\frac{Vdd}{R1} +\frac{Vz}{R2})(R1||R2)\$. 

The first equation applies when Vout is low, so it's only of concern at the lower Vdd switching point in your design. The second applies when Vout is high, so it's only of concern at the higher Vdd switching point. 

For example, if Vdd is 10V, Vz is 5V, R1 is 100K and R2 is 10K, Vin+ will be 4.545V or 5.454V depending on your intent. 

That answers your question- to actually design this circuit for given voltages involves some algebra to re-arrange in terms of desired switching voltage, which I will leave to you.  



As an aside- designing undervoltage protection circuits is actually non-trivial and perhaps you should just buy one. Before the zener is properly biased (low Vdd) the voltages at the inputs do not follow the above equations and there may be plenty of mischief a circuit can get into below voltages that will bias a zener or normal voltage reference. A commercial circuit will have guaranteed output stated down to below some fairly low voltage (probably &lt; 1V), which is more than adequate to protect most power devices and often is enough to prevent NV memory corruption.