State the inputs and outputs of the following relation. Is it a function? { ( − 3 , 5 ) , ( 0 , 2 ) , ( 3 , 5 ) \left(-3,5\right),\left(0,2\right),\left(3,5\right) ( − 3 , 5 ) , ( 0 , 2 ) , ( 3 , 5 ) }

Okay. So let's give this problem a try. So in this problem, we are told to see the inputs and outputs of the following relation, which is the relation we have right here. We're also asked to determine whether or not this is a function. Now what I'm first going to do is draw the inputs and outputs for this, and recall that the inputs correspond to the x values, whereas the outputs correspond to the y values. Now the inputs that I see are going to be each of these x coordinates or basically the first coordinate for each of these. So doing this in ascending order, we have negative 3, then we have 0, and then we have positive 3. And these are going to be all of the inputs right here. Now for the outputs, I can see that we have positive 5, I see that we have 2, and I see that we have 5 again, but I don't need to write 5 twice because I already wrote it once in this little bubble. So these are all of the inputs and outputs. Now to determine whether or not this is a function, let's see how these all relate. Well, I can see that negative 3 relates to 5, so this is going to rate relate to that. I can see that 0 relates to 2, and then I can see that 3 relates to 5. And in order for something to be a function, you cannot have an input related to more than one output. I can see this negative 3 only goes to 5, so that's good. I can see the 0 only goes to 2, and then I can see this 3 only goes to 5. Well, by the looks of this, each input has only one output. So we could say that this is an example of a function. Now you might have looked at this situation that we have where the 5 is related to both negative 33, but we can have an output related to multiple inputs. We just can't have an input related to multiple outputs. So this is perfectly okay. So we would say that this right here is the diagram for our inputs and outputs, and this relation, in fact, is a function.