What is the range of the function #y=4x^2+2#?

Question

2468 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2018-05-08T06:50:52

See explanation.

Graph of this function is a parabola with vertex at #(0,2)#. The function's values go to #+oo# if #x# goes to either #-oo# or #+oo#, so the range is:

#r=(2,+oo)#

The graph is:

graph{4x^2+2 [-10, 10, -5, 5]}

Answer 2 · 2018-05-08T06:55:25

Range: #[+2,+oo)#

#y = 4x^2+2#

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

#y# will have a parabolic graph with axis of symmetry where #x=-b/(2a)#

#:. x=0#

Since #a>0# #y# will have a minimum value at #x=0#

#:. y_min = +2#

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