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Mathematics 18 Online
OpenStudy (anonymous):

Determine whether each of the following equations defines y as a function of x. (Do not graph.) x + 3 = y^2

OpenStudy (anonymous):

how do I go about figuring problems like these out? I have a few examples

OpenStudy (agent0smith):

If you see a y^2, it's not a function of x... since once you take a square root, you get a plus or minus. Post more examples.

OpenStudy (anonymous):

every time I see a ^ 2? not a ^3? and does it have to be y or can it be x?

OpenStudy (anonymous):

3x^4 − 2xy = 5x

OpenStudy (agent0smith):

y^3 is fine. y^4 is not. y^5 is fine. y^6 is not. Notice a pattern?

OpenStudy (agent0smith):

x can be anything.

OpenStudy (anonymous):

putting it simply...a function should have only a single value of y for every x

OpenStudy (agent0smith):

This is fine: 3x^4 − 2xy = 5x since you only have a y, it can be rearranged to get a single y= ...

OpenStudy (anonymous):

ohh okay. 3x^4 − 2xy = 5x is no?

OpenStudy (anonymous):

ohh only 1 y.

OpenStudy (anonymous):

you can try plugging in a number and see if u are getting a single value for y or not. Note that taking square root of a number will lead 2 values...positive and negative

OpenStudy (anonymous):

what about 39. |y| |− 2 = x

OpenStudy (agent0smith):

y^3 is fine. y^4 is not. y^5 is fine. y^6 is not. y^7 is fine. y^8 is not. y^9 is fine. Notice a pattern? An even power on the y means when you take a root, you get a plus or minus, meaning two values of y for every x - not a function. Odd powers on y are fine.

OpenStudy (agent0smith):

|y| − 2 = x ?

OpenStudy (agent0smith):

|y| − 2 = x becomes |y| = x + 2 once you drop the absolute value signs, just like a square root, you get a plus or minus, like y = x+2 OR y = -(x+2)

OpenStudy (agent0smith):

So an absolute value on a y = not a function. Absolute value of x is a function.

OpenStudy (anonymous):

since absolute values have two answers will they always be no?

OpenStudy (agent0smith):

Absolute values on y, yes.

OpenStudy (anonymous):

not necessarily....only for y

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