y=1/3x^2+1/2xy=13x2+12x
yy is a quadratic function of the form: ax^2+bx+cax2+bx+c
Where, a=1/3a=13, b=1/2b=12 and c=0c=0
Therefore we know that the graph of yy will be a parabola.
Also, since a>0a>0 we know that yy will have a singular minimum value where x=(-b)/(2a)x=−b2a 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)