How do you find the domain and range of #y=3x^2+2x+1#?

2 Answers
May 1, 2017

Answer:

Since this is a quadratic equation, the domain is ] #-oo,oo# [

Explanation:

For the range, one possible method is to express the equation in vertex form, which gives the coordinates of the minimum point. The y-coordinate of the minimum point would give you the smallest value of the range, and the largest value y can take should be #oo# as a quadratic equation does not have asymptotes.

May 1, 2017

Answer:

Domain: #(-oo, +oo)# Range: #[2/3, +oo)#

Explanation:

#y = 3x^2 +2x +1#

#y# is a quadratic function defined #forall x in RR#

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

Since the coefficient of #x^2# is positive #y# will have a minimum value where #y' =0#

#y' = 6x+2# [Power rule]

#6x+2=0 -> x=-1/3#

#y_"max" = y_"(-1/3)"= 3*1/9 -2*1/3+1#

#= 1/3-2/3+1 = 2/3#

Since #y# has no upper bound, the range of #y# is #[2/3, +oo)#

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

graph{3x^2 +2x +1 [-5.7, 5.4, -0.846, 4.704]}