Answer :
Answer: it is not a function
Step-by-step explanation:
Given :
[tex]y^{2}[/tex] = 3x + 1
The first thing is to make y the subject of the formula, that means we will take the square root of both sides
y = [tex]\sqrt{3x + 1}[/tex]
since y = [tex]\sqrt{3x + 1}[/tex] , the the value of y could be positive or negative.
For example , [tex]\sqrt{4}[/tex] = ± 2 , since -2 x -2 will also give + 4
For y = [tex]\sqrt{3x + 1}[/tex] to be a function , there must be a unique value for y for each value of x , but this is not the case as y could be positive or negative for a single value of x , so it is not a function.