How do you find the domain of #y=sqrt(5x+2)#?

1 Answer
Sep 19, 2016

Answer:

#x>=-2/5# or #[-2/5,oo)#

Explanation:

#y=sqrt(5x+2)#, find the domain

In order for y to be defined, the expression under the square root must not be negative. In other words, the expression under the square root must be greater than or equal to zero.

#5x+2>=color(white)a0#
#color(white)(aa)-2color(white)(aa)-2#

#5x>=-2#

#(5x)/5>=-2/5#

#x>=-2/5# is the domain written as an inequality.

#[-2/5,oo)# is the domain written in interval notation