At time #t = 0s#, the particle is traveling at some initial velocity, #v# in the positive x-direction. We do not know whether there was an impulse that gave the particle its initial velocity at #t=0s# because the graph starts at 0s but does not include 0s. Please remember that #v# in the positive x-direction is represented by the positive slope of the graph.
At time #t = 2s#, the particle receives an impulse of magnitude, #m# in the #-x# direction, that causes the velocity of the particle to go from #v# in the positive x direction to #0#. We know that the velocity is 0 because the slope of the graph is 0.
From #t = 2s# to #t=4s# the particle remains at rest.
At #t=4s# the particle receives an impulse of magnitude, #m# in the #-x# direction, that causes the speed of the particle to go from #0# to #v# in the negative x-direction; we know this because the slope of the graph is the same magnitude but negative.
We do not know what happens at time #t= 6s# or beyond because the graph does not include time #t >= 6s#
This is alternative (1) where the particle receives two impulses of magnitude #m# in the #-x# direction.
