Question

Which of these sets of points define a function (points below)?

a. `{(-2,-2),(-2,-1),(-2,0),(-2,1),(-2,2)}`
b. `{(1,0),(-1,0),(2,1),(-2,1),(3,2),(-3,2)}`
c. `{(-1,-1),(-1,3),(0,2),(4,3),(2,3)}`
d. `{(-3,-3),(-3,2),(-3,5),(1,0),(1,-2),(1,3)}`

Answer 1

a. NO
b. YES
c. NO
d. NO

Explanation:

A relation is only a function when there is only one output for every input. In other words, you can't have the same `x` for different `y` values. You can however have the same `y` value for different `x` values

For a, `-2` is repeated for every `y` value. This cannot be so this relation is not a function.

For b, each `x` value has a only one unique value so this is a function.

For c, this is not a function because `-1` is used more than once

For d, This is also not a function since `-3` is used three times and `1` is used three times.

Essentially, check to see if there are any repeating `x` values. If there are, then it is not a function. If no `x` values repeat, then it is a function.

Answer 2

An alternate, more visual, way of answering the question using the vertical line test.

Explanation:

Another way to view this problem is to graph the points and use the "vertical line test".

The vertical line test is a visual way to see if, for any `x` value, there are more than 1 `y` values. If the vertical line intersects more than one point, then the equation isn't a function.

For instance, for the set of points a. {(-2, -2),(-2, -1),(-2, 0),(-2, 1),(-2, 2)}:

graph{((x+2)^2+(y+2)^2-.1)((x+2)^2+(y+1)^2-.1)((x+2)^2+(y+0)^2-.1)((x+2)^2+(y-1)^2-.1)((x+2)^2+(y-2)^2-.1)(x-0y+2)=0}

I've drawn a vertical line through more than 1 point and so the relation that created this set of points is not a function.

Contrast that with set b. {(1, 0),(-1, 0),(2, 1),(-2, 1),(3, 2),(-3, 2)}

graph{((x-1)^2+(y+0)^2-.1)((x+1)^2+(y+0)^2-.1)((x-2)^2+(y-1)^2-.1)((x+2)^2+(y-1)^2-.1)((x-3)^2+(y-2)^2-.1)((x+3)^2+(y-2)^2-.1)(x-0y+2)(x-0y+1)(x-0y+3)(x-0y-2)(x-0y-1)(x-0y-3)=0}

Each of vertical lines goes through only 1 point and so the relation that created this set of points is a function.