The relation \(R=\{(x,y):y^2=x+1,\ x\in X,\ y\in Y\}\), where \(X=\{0,3,8\}\) and \(Y=\{-3,-2,-1,1,2,3\}\), is considered from \(X\) to \(Y\). What is the correct conclusion?

Correct answer: It is not a function because \(0\) has two images

For \(x=0\), the equation gives \(y^2=1\), so both \(y=1\) and \(y=-1\) are obtained. Thus, the relation contains both \((0,1)\) and \((0,-1)\). A function must assign exactly one image to each element of its domain, so this relation is not a function. Likewise, \(x=3\) gives \(y=\pm2\), and \(x=8\) gives \(y=\pm3\). Exam tip: to test a relation for being a function, check whether any one domain element is paired with two different elements of the codomain.