Inductor Voltage and Current After Switches Open and Close
You might recall this formula: -
$$\boxed{V = L\cdot\dfrac{di}{dt}}$$
At t = 0, you say that the inductor is carrying 10 amps. At that moment S1 closes and the current will ramp upwards linearly at a rate of: -
$$\dfrac{di}{dt} = \dfrac{V}{L} = 30\text{ amps per second}$$
So, after 1 second the current will be 40 amps (due to the initial current of 10 amps). The current rise is linear (as with any perfect inductor connected to an ideal voltage source).
When the switches change (at t = 1 second) to apply -1 volt to the inductor, the current will ramp down at 10 amps per second so, after a further 2 seconds, the current will linearly fall from 40 amps (at 1 second) to 20 amps at 3 seconds.

Should my graph immediately change from 40A to 20A or will there be a
linear decrease?

As with all perfect inductors and ideal voltages, the current never changes instantly but changes as a ramp.