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Next, the switching points must be determined, and they'll be the voltages at the R4-R5 junction (V3) when V2 is equal to minus 3 volts and when it's equal to 5 volts.

$$ V3 = \frac{(V1-V2)\times R5}{R4+R5} +V2 $$

$$ V3 = \frac{(V1-V2)\times R5}{R4+R5} +V2 = \frac{13V\times 10k\Omega}{20k \Omega} -3V = 3.5\text { volts} $$

Similarly, when V2 is equal to 5 volts, V3 will be equal to 7.5 volts

Since V2 and V3 vill now both be at 5 volts, U2+ will also be at +5 volts, and since U2 is a voltage follower, its output will be at 5 volts, which will be connected to R10 through the now closed S2.

Since R8 and R9 have equal resistances, as V2 rises above 5 volts, the voltage on U2 will rise at half that rate, satisfying the requirements for the slope above 5 volts.

R8 and R9 comprise a voltage divider with their junction

So, the result of all this is that V2 will be connected to R8 from the time it's between -10 volts and -3 volts, then as it goes more positive than -3 volts R8 will be disconnected from V2 until V2 gets to 5 volts, when R8 will again be connected and will stay connected for as long as V2 stay more positive than 5 volts.

Here's the drawing, and if you want to play around with the circuit, here's the LTspice .asc file so you can simulate it.

Next, the switching points must be determined, and they'll be the voltages at the R4-R5 junction (V4) when V2 is equal to minus 3 volts and when it's equal to 5 volts.

$$ V4 = \frac{(V1-V2)\times R5}{R4+R5} +V2 $$

$$ V4 = \frac{(V1-V2)\times R5}{R4+R5} +V2 = \frac{13V\times 10k\Omega}{20k \Omega} -3V = 3.5\text { volts} $$

Similarly, when V2 is equal to 5 volts, V4 will be equal to 7.5 volts

Since V2 and V3 vill now both be at 5 volts, U2+ will also be at +5 volts, and since U2 is a unity-gain voltage follower, its output will be at 5 volts, which will be connected to R10 through the now closed S2.

R8 and R9 comprise a 2:1 voltage divider because R8 and R9 have equal resistances, and with its tap connected to U2+, as V2 rises above 5 volts, the voltage on U2+ will rise at half that rate, satisfying the requirement for the slope in Vout above 5 volts.

So, the result of all this is that V2 will be connected to R8 from the time V2 is between -10 volts and -3 volts. After that, when V2 goes more positive than -3 volts, R8 will be disconnected from V2 and Vout will snap up to zero volts until V2 gets to 5 volts, when R8 will be connected To U2's output and will snap upt to +5 volts. Thereafter, V2 will stay connected to U2's output for as long as V2 stays more positive than 5 volts.

The schematic follows, and if you want to play around with the circuit, here's the LTspice .asc file so you can simulate it.

U1A and U1B are the two comparators comprising an LT1017, and their inputs aren't allowed to go more than 300 millivolts below ground, so if a 10 volt supply is used for V1, then with R4 and R5 both equal to 10k, when V2 is at -10 volts the voltage at the R4-R5 junction will be zero volts.

U1A and U1B are the two comparators comprising an LT1018, and their inputs aren't allowed to go more than 300 millivolts below ground, so if a 10 volt supply is used for V1, then with R4 and R5 both equal to 10k, when V2 is at -10 volts the voltage at the R4-R5 junction will be zero volts.