There are more current efficient techniques to achieve the same function, especially if the 9V supply is a battery: one way is to cofigure inverting opamp input with 1MOhm to 9v supply and 1MOhm to gnd. This offsets the input to 4.5v, which gives you good headroom with 0v-9v rails. This option would require input and output coupling caps. Another method that won't decrease your input impedance is to offset the non-inverting input to mid-supply.
If you want to design opamp circuits, there are three idealized rules you can use to make your life easier. These rules work in the majority of general purpose design cases.
1) Opamp input impedance is infinite. That is opamp draws no input current.
2) Opamp output impedance is zero.
3) Opamp open loop gain is infinite. That is there is no difference in + and - inputs during normal operation.
So, how do these help you? In the design case you provided, since the + input is connected to ground, opamp will force - input to 0V via the feedback resistor. The + input is called virtual ground here.
Since this is true, your inverting opamp circuit's input resistance will be the 470K resistor value.
Another example using these rules with your circuit is: input=+2V, therefore with + input forced to 0V, input current = 2V/470K= 4.26uA, but using these rules, 100% of input current flows through feedback resistor. Now to calculate output voltage. This current flows in feedback resistor, and given sign, -4.26uA*470K= -2V. Gain = -1.
Sorry for the long winded response, but IMO, these idealized rules will help you design and understand opamps like Bob Pease
![Smile :)](./images/smilies/icon_smile.gif)
cheers,
rob