Let us draw a graph of y=x^2-3x, by choosing a few values of x say {-6,-3,0,1.5,3,6,9} and putting these values we get corresponding values of y which are {54,18,0,-2.25,0,18,54} and then drawing the points i.e. (-6,54),(-3,18),(0,0),(1,-2.25),(3,0),(6,18),(9,54)
and joining them, we get a graph as follows
graph{x^2-3x [-18.42, 21.58, -4.32, 15.68]}
Observe that it divides the entire Cartesian plane in three parts - one on the line, which is a parabola, - two inside the parabola and - three outside the parabola.
As we have y < x^2-3x, points on parabola do not satisfy this and hence the parabola is not included.
Now select a point inside parabola say (1,1) and it is apparent that while y=1, x^2-3x=-2 and hence y > x^2-3x. Therefore points inside the parabola also do not satsfy the equality.
Selecting a point outside parabola say (-1,0), we have that y=0 and x^2-3x=1+3=4 and we have y < x^2-3x and hence area outside parrabola satisfies the given inequality.
Hence graph of y < x^2-3x is as follows:
graph{y < x^2-3x [-18.42, 21.58, -4.32, 15.68]}
Note - The line parabola has been kept dotted as points on the line are not included. Had it been y <= x^2-3x, we would have kept it complete i.e. not dotted.
