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

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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

Mathematics
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Please help, I'm stuck!
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?
how can you tell a function from just a relation? hint: recall your DOMAIN and RANGE sets

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Ehhh I need a little more of a description on how to do it, I'm still confused
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.
I get it, so it's B!
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?
It couldn't be A? Because it's only a x-value right? So that'd just leave C and D
That's right. For a the x-value of 1 has an infinite number of y-values. Check the others.
I have a feeling that the answer is D, if that's right could you explain to me why please?
Let's look at D. Are you able to rearrange it to solve for y?
Maybe, 16+4x^2=y?
Not quite\[4x^2 + y^2 = 16\]\[y^2 = 16 - 4x^2\]\[y=\pm \sqrt{16-4x^2}\]Are you able to follow that?
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?
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?
Yeah, so what does this tell us? That there is only a value for the y?
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?
0 I guess? Sorry, I'm just not getting it as quickly as others probably would
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?
You can do this. What's the square root of 16?
Oh that's easy, 4
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?
I think the square root of 16 is -4
Ummmm, no...no it's not...
You're incorrect @Mertsj . The square root of 16 is 4.
Must be because (-4)(-4)=16
Maybe 16 has two square roots.
Anyways, yes, it does represent a function
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.
The square root of a number is a number which multiplied by itself, gives you the original number.
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.
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.
@Mertsj , what about {(1, 4), (3, 8), (1, 4), (1, 4)} ? You had better go back to school, chum.
@Mertsj Oh dieses Zeug langweilt mich bis auf die Knochen .
(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?
@RavenDarkwood200 Wie kann der Blinde den Blinden führt
@Mertsj Worüber redest du? Ich verstehe nicht. Ich bin deutscher sehen Sie?

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