Design a circuit that would turn off output between two input voltage levels
Take a look at Zener diodes to provide the +5 and -3 V references.  You can arrange them to subtract from the input voltage.  For example, a 5 V zener can be used to create (Vin - 5).

After subtracting off the 5 V and 3 V offsets, you can use resistor dividers to get any slope &lt; 1.

Added:

It's been long enough and the OP has apparently checked out, so here is the answer:



Here D3 and D4 are ideal diodes.  If you actually want to build this, then use something like 1N4148 for D3 and D4, and make the zener voltages about 600 mV less to compensate for the diode drops.

Added in response to comments by EM Fields

The OP asked for switching at "exactly" -3 and +5 V.  That's no spec at all.  If he truly meant exactly, then no circuit suffices.  Since everything by necessity will violate that spec, we have no way to know how far is OK.  You don't know if a few 100 mV is good enough or not.  Again, "exactly" is a meaningless spec, so can only be ignored.

To your second point, I think you are interpreting the graph incorrectly.  The graph is quite clear in that Vout should be 0 for Vin from -3 to +5 volts.  Specifically, here is the function described by the graph and that the circuit above realizes:

Vout = Vin + 3  &nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;(Vin &lt; 3)
Vout = 0  &nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;(-3 &lt;= Vin &lt;= 5)
Vout = (Vin - 5)/2 &nbsp;&nbsp;&nbsp;(Vin > 5)

You say the OP's drawing is wrong, but I don't see how you can know that since it's the only spec we've got, and it is quite clear by itself.