#y=1/3x^2+1/2x#
#y# is a quadratic function of the form: #ax^2+bx+c#
Where, #a=1/3#, #b=1/2# and #c=0#
Therefore we know that the graph of #y# will be a parabola.
Also, since #a>0# we know that #y# will have a singular minimum value where #x=(-b)/(2a)# which is the axis of symmetry.
#:. y_min = y(-3/4)#
#= 1/3(-3/4)^2+1/2(-3/4)#
#= 3/16-3/8 = -3/16#
We can next find the #x-#intercepts where #y=0#
#-> 1/3x^2+1/2x =0#
#x(1/3x+1/2)=0 -> x=0 or x=-3/2# which are the #x-#intercepts
Note also the #y# has a single #y-# intercept at #(0,0)#
Using these critical points we can create the graph of #y# as below.
graph{1/3x^2+1/2x [-3.08, 3.08, -1.528, 1.552]}
#y# is defined #forall x in RR#
Therefore the domain of #y# is: #(-oo,+oo)#
#y# has a minimum value of #-3/16# and no finite upper bound.
Therefore the range of #y# is: #(-3/16, +oo)#
