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

1 Answer
Apr 27, 2017

Answer:

Domain: #(-\infty,\infty)#
Range: #[2,\infty)#

Explanation:

To get the domain, we have to find any instances of division by zero and negatives under radicals. We don't have any division or radicals so the domain is all values or #(-\infty,\infty)#.

For the range, we need to find the lowest value of the function (in this case). Since #x# is being squared it will never be negative. The smallest non-negative number is zero. If we plug that into the function we get
#y(0)=4(0)^2+2=4(0)+2=0+2=2#

So the lowest value is 2. Since our domain goes to infinity and the function is never negative, then all the values must go to infinity. So our range is #[2,\infty)#.