How do you graph the function y=1/3x^2+1/2xy=13x2+12x and identify the domain and range?

1 Answer
May 31, 2018

See graph below.
Domain: (-oo,+oo)(,+) Range: (-3/16, +oo)(316,+)

Explanation:

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)