Question

Find the domain and range of the relation.
Determine whether the relation is a function.

{(1/6,8), (7,1/6), (-7,1/6), (1/6,-2)}

Answer 1

okay what is most confusing to you , and what have you tried

Answer 2

I don't know how to do it. It's all new to me

Answer 3

Since you have only points and no continuous function, then your domain is just the x-values, right?
And the range only the y-values.

Answer 4

It's not a function, it's just a set of points. A function must have one or more variables--a set of points has none.

Answer 5

How do I find the range?

Answer 6

the relation is not a function

Answer 7

right but what is the range of the relation?

Answer 8

@jdoe0001

Answer 9

Or how do you know if it's a function or not. I'm still unsure of how to figure that out.

Answer 10

The y-values are the range of the function.
The x-values are the domain.

In order for a relation to be a function it must include one or more variables are pass a vertical line test--a set of points does not meet this criteria.

Answer 11

and* rather

Answer 12

How do I find out if it passes the vertical line test?

Answer 13

Find the unique values of your x-coordinates, that's your domain
Find the unique values of your y-coordinates, that's your range
If for every x there is a unique y-value, then it is a function

Answer 14

I still don't get it but thanks

Answer 15

Let's take the domain first:

{(1/6,8), (7,1/6), (-7,1/6), (1/6,-2)}

What are all the x-coordinates for these sets of points?

Answer 16

I understand the domain and range now but I don't understand how to tell if it's a function

Answer 17

Ok, that's just as easy.

let's look at each x-coordinate and make sure that there isn't another point with a DIFFERENT y-value. This ensures that the given relation is a function.

Take 1/6. What are the possible values for y given this x-value?

Answer 18

8,1/6,1/6,-2

Answer 19

There are two points with 1/6 as the x-value:

{(1/6,8),(1/6,-2)}

What are the possible y-values here?

Answer 20

8,-2

Answer 21

That's right. So for x=1/6 y is either 8 or -2. This can NOT happen for functions. We expect to get a unique output for every x- that's the definition of a function.

Make sense?

Answer 22

So what would make it a function? I don't know why I'm so confused lol

Answer 23

I function would be something like the set above, but WITHOUT the point (1/6,-2), because then we would have a set of points with unique y-values for every possible x-value.

Answer 24

hmm ok

Answer 25

For example, this is a function:
{(1/6,8), (7,1/6), (-7,1/6)}

Every x value has a unique y-value.

So your original relation is NOT a function.

Answer 26

Ok thank you

Answer 27

You're Welcome.

Functions can be confusing in their definition, but if you try to just think about what its saying, then it makes it a little easier.

BTW, the equation of a circle x^2 + y^2 = 1 is NOT a function, because there are many points along x that give 2 y values:



This is where the horizontal test would immediately clue you that there are two values of y for a single value of x.

Hope this helps. Good luck!

Answer 28

*Vertical line test (not horizontal)