How do you find the domain and range for #y=4x^2 - 2x#?

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Answer 1 · 2018-05-19T09:26:14

Domain: #(-oo, +oo)#, Range: #[-1/4, +oo)#

#y = 4x^2-2x#

#y# is defined #forall x in RR#

#:.# the domain of #y# is #(-oo,+oo)#

#y# is a quadratic function of the form: #ax^2+bx+c#
Where: #a=4, b=-2, c=0#

Since #a>0#, #y# will have a minimum value where #x=-b/(2a)#

I.e. where #x= 2/(2xx4) = 1/4#

#:. y_min = y(1/4) = 4*(1/4)^2 - 2*(1/4)#

#= 1/4 - 1/2 =-1/4#

Since #y# has no finite upper bound the range of #y# is #[-1/4,+oo)#

We can infer these results from the graph of #y# below.

graph{4x^2-2x [-2.35, 3.125, -0.713, 2.024]}