To determine which set would not be a function if we add the ordered pair [tex]\((4,2)\)[/tex] to it, we need to understand that a function must have unique [tex]\(x\)[/tex]-values for each [tex]\(y\)[/tex]-value. That means no [tex]\(x\)[/tex]-value should repeat within each set of ordered pairs.
Let's check each set:
Set 1: [tex]\(\{(3,2),(2,6),(1,7),(-1,-2),(-3,-5)\}\)[/tex]
- Current [tex]\(x\)[/tex]-values: [tex]\(3, 2, 1, -1, -3\)[/tex]
- Adding [tex]\( (4,2) \)[/tex] introduces [tex]\(x = 4\)[/tex], which is not already in the set.
- It remains a function.
Set 2: [tex]\(\{(1,5),(2,-5),(-5,6),(-2,1),(0,8)\}\)[/tex]
- Current [tex]\(x\)[/tex]-values: [tex]\(1, 2, -5, -2, 0\)[/tex]
- Adding [tex]\( (4,2) \)[/tex] introduces [tex]\(x = 4\)[/tex], which is not already in the set.
- It remains a function.
Set 3: [tex]\(\{(3,-1),(4,0),(-2,4),(-3,6),(5,-2)\}\)[/tex]
- Current [tex]\(x\)[/tex]-values: [tex]\(3, 4, -2, -3, 5\)[/tex]
- Adding [tex]\( (4,2) \)[/tex] introduces [tex]\(x = 4\)[/tex], which already exists in the set.
- It would no longer be a function because of the repeated [tex]\(x\)[/tex]-value.
Set 4: [tex]\(\{(3,6),(-2,-4),(5,-8),(-3,1),(2,-4)\}\)[/tex]
- Current [tex]\(x\)[/tex]-values: [tex]\(3, -2, 5, -3, 2\)[/tex]
- Adding [tex]\( (4,2) \)[/tex] introduces [tex]\(x = 4\)[/tex], which is not already in the set.
- It remains a function.
Thus, the set that would not be a function if the ordered pair [tex]\((4,2)\)[/tex] is added is:
Set 3
The correct answer is:
C Set 3
