How do you graph the function #y=-1/3x^2-4x+1# and identify the domain and range?

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Answer 1 · 2017-06-11T13:40:31

Domain: #(-oo, +oo)#
Range: #(-oo, 13]#

#y = -1/3x^2-4x+1#

#y# is defined #forall x in RR#
Hence the domain of #y# is #(-oo, +oo)#

Consider the standard form of the quadratic: #ax^2+bx+c#
It is known that the critical point will be where #x=(-b)/(2a)#

In this case, since the coefficient of #x^2 <0#, the critical value will be a maximum.

#:.# for #y_"max"# #x= 4/(-(2/3))#

#= -6#

#y_"max" = y(-6) = -12+24+1 = 13#

Hence the range of #y# is (-oo, 13]

This can be seen from the graph of #y# below.

graph{(-x^2)/3-4x+1 [-41.1, 41.1, -20.6, 20.5]}