What is the domain and range of #y=x^2 - 4x + 1#?

1 Answer
Jun 23, 2018

Answer:

Range: #y>=-3#
Domain: #x in RR#

Explanation:

Complete the square (putting the function in vertex form)

#y=(x-2)^2-4+1#
#y=(x-2)^2-3#

Hence the minimum of the function is #y=-3#, so we can say that the range is

#y>=-3#

As for the domain, any value of #x# can be passed to the function so we say that the domain is

#x in RR#