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

## a. $\left\{\begin{matrix}- 2 & - 2 \\ - 2 & - 1 \\ - 2 & 0 \\ - 2 & 1 \\ - 2 & 2\end{matrix}\right\}$ b. $\left\{\begin{matrix}1 & 0 \\ - 1 & 0 \\ 2 & 1 \\ - 2 & 1 \\ 3 & 2 \\ - 3 & 2\end{matrix}\right\}$ c. $\left\{\begin{matrix}- 1 & - 1 \\ - 1 & 3 \\ 0 & 2 \\ 4 & 3 \\ 2 & 3\end{matrix}\right\}$ d. $\left\{\begin{matrix}- 3 & - 3 \\ - 3 & 2 \\ - 3 & 5 \\ 1 & 0 \\ 1 & - 2 \\ 1 & 3\end{matrix}\right\}$

Apr 20, 2017

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.

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.