Which of the following expressions represents a function?
A. x=1
B. {(3,2), (3, -2), (4, 5), (4,-5)}
C. y= 4x-1
D.4x^2+y^2=16

- anonymous

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- jamiebookeater

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- anonymous

Please help, I'm stuck!

- anonymous

In order for the expression to be a function, there must exist, for each value in the domain i.e. every x-value, one and only one value in the range i.e. y-value. Which one do you think it is?

- jdoe0001

how can you tell a function from just a relation?
hint: recall your DOMAIN and RANGE sets

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## More answers

- anonymous

Ehhh I need a little more of a description on how to do it, I'm still confused

- anonymous

For every x-value there is exactly one y-value. For example {(0,1), (1,5), (0, 3), (2, 4)} is not a function because for the x-value of 0, there are two y-values, 1, and 3.

- anonymous

I get it, so it's B!

- anonymous

By the example provided above, B cannot be a function because for the x-values of 3 and 4, there is more than one associated y-value. Check the others. What do you think?

- anonymous

It couldn't be A? Because it's only a x-value right? So that'd just leave C and D

- anonymous

That's right. For a the x-value of 1 has an infinite number of y-values. Check the others.

- anonymous

I have a feeling that the answer is D, if that's right could you explain to me why please?

- anonymous

Let's look at D. Are you able to rearrange it to solve for y?

- anonymous

Maybe, 16+4x^2=y?

- anonymous

Not quite\[4x^2 + y^2 = 16\]\[y^2 = 16 - 4x^2\]\[y=\pm \sqrt{16-4x^2}\]Are you able to follow that?

- anonymous

I don't really understand where you got the subtraction sign from when you first changed the equation. Can you explain what you did in this equation please?

- anonymous

The goal is to isolate y. Therefore, in the first step, I subtracted 4x^2 from both sides of the equation. Then to solve for y, I took the square root of both sides. Make sense?

- anonymous

Yeah, so what does this tell us? That there is only a value for the y?

- anonymous

OK. We've rearranged D to solve for y. So, choose an x-value - 0 would be a good choice. If you substitute 0 in for x, how many y-values do you get?

- anonymous

0 I guess? Sorry, I'm just not getting it as quickly as others probably would

- anonymous

Take the rearranged equation D and substitute 0 in for x\[y=\pm \sqrt{16-4x^2}\]\[y=\pm \sqrt{16-4\left( 0 \right)^2}\]\[y= \pm \sqrt{16}\]What's the answer for y?

- anonymous

You can do this. What's the square root of 16?

- anonymous

Oh that's easy, 4

- anonymous

Right. So, in equation D, if x=0, then y has to be +4 or -4. There are two y-values associated with x=0. Is this expression a function?

- Mertsj

I think the square root of 16 is -4

- anonymous

Ummmm, no...no it's not...

- anonymous

You're incorrect @Mertsj . The square root of 16 is 4.

- Mertsj

Must be because (-4)(-4)=16

- Mertsj

Maybe 16 has two square roots.

- anonymous

Anyways, yes, it does represent a function

- anonymous

You're screwing up this help session. There is a mathematical distinction between answering
a) what number, when squared, equals 16
b) what is the square root of 16
The answer to the first question is =4 and -4
The answer to the second question is 5.

- Mertsj

The square root of a number is a number which multiplied by itself, gives you the original number.

- anonymous

The definition of a function is a relation in which, for every x-value there is ONE AND ONLY ONE associated y-value. We just determined that for x=0, there are two associated y-values, +4 and -4. Therefore, this expression cannot be a function. Do you understand.

- Mertsj

Here is an easier definition of a function: A function is a set of ordered pairs in which no two ordered pairs have the same first number.

- anonymous

@Mertsj , what about {(1, 4), (3, 8), (1, 4), (1, 4)} ? You had better go back to school, chum.

- anonymous

@Mertsj Oh dieses Zeug langweilt mich bis auf die Knochen .

- Mertsj

(1,4) and (1,4) are not two ordered pairs. It is one ordered pair written down twice. If you write your name twice, does that make you two people?

- Mertsj

@RavenDarkwood200 Me too.

- Mertsj

@RavenDarkwood200 Wie kann der Blinde den Blinden führt

- anonymous

@Mertsj Worüber redest du? Ich verstehe nicht. Ich bin deutscher sehen Sie?

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