Which of the following relations is a function?
A. {(5, 2), (3, 2), (4, 2), (5, 1)}
B. {(1, 2), (1, 3), (2, 1), (3, 4)}
C. {(1, 3), (2, 6), (3, 8), (1, 7)}
D. {(1, 4), (8, 9), (10, 11), (12, 12)}

Question

Which of the following relations is a function?
	A.	{(5, 2), (3, 2), (4, 2), (5, 1)}
	B.	{(1, 2), (1, 3), (2, 1), (3, 4)}
	C.	{(1, 3), (2, 6), (3, 8), (1, 7)}
	D.	{(1, 4), (8, 9), (10, 11), (12, 12)}

e.mccormick · Answer

Do you know the definition of a function when it comes to x,y pairs?

anonymous · Answer

is a set of ordered pairs where each y is paired with exactly one x?

anonymous · Answer

wait no a x is paired with one y right

e.mccormick · Answer

The second.

For any given x there is one and only one y.  So if you have an x of say 42, and it maps to 12 in one pair and 14 in another pair, it is NOT a function.  On the other hand, if you have an x of 42 and an x of 24 and they both map to a y of 12, that is just fine!

anonymous · Answer

ohhh

anonymous · Answer

Would the answer be D?

e.mccormick · Answer

D looks good.  =)

anonymous · Answer

Wow thanks!

technodynamic · Answer

Th reason this is, it function/relation can never have more than one of the same in its sets.

technodynamic · Answer

so it's D.

e.mccormick · Answer

Once you get the basic concept, these are not too hard.  Just remember it is unique x values.  Now, when you get to one-to-one things, it needs to be both unique x and unique y.

anonymous · Answer

OK thanks i have a other question i have no idea how to do.

anonymous · Answer

Is the equation y = x3 - 6 even, odd, or neither?
	A.	even
	B.	odd
	C.	neither

anonymous · Answer

This i don't know how to do.

technodynamic · Answer

is that a cubed x? or 3x?

anonymous · Answer

yes cubed

technodynamic · Answer

if you're doing quadratic eqn then it would be -x3+y+6=0
or, if you're solving just that equation get -y on both sides.

e.mccormick · Answer

No, he is asking what type of function it is, even, odd, or neither.
$y = x^3 - 6$

Do you know what the even and odd tests are?

anonymous · Answer

Not really i don't know

e.mccormick · Answer

$f(-x)$ and then see what would happen if you follow the rules of algebra.  See, if you have $f(x)=x^2$ then $f(-x)= (-x)^2$  Well, the square gets rid of the - sign.  Therefore $f(x)=f(-x)$ and the function is even.

e.mccormick · Answer

If $-f(x)=f(-x)$ it is odd.
If $f(x)=f(-x)$ it is even.

All other cases it is neither.

e.mccormick · Answer

Some examples: http://www.purplemath.com/modules/fcnnot3.htm

anonymous · Answer

This is kinda hard but i think i got it.

e.mccormick · Answer

If the x to -x is confusing, use a different letter.  See if you put a and -a in and if they match or are opposites.  Oh, and $-f(x)$ means multiply through with -1, so any function with a constant, like +2 on the end, can NEVER be odd.

anonymous · Answer

Would the equation be something like -f(x)=x3-6?

e.mccormick · Answer

Well, for the odd test, it is this:
Test -f(x)
$f(x)=x^2+2$
$(-1)f(x)=(-1)(x^2+2)$
$-f(x)=-x^2-2$
and test f(-x)
$f(-x)=(-x)^2+2$
$f(-x)=x^2+2$
and see if they match
$-f(x)=f(-x)\implies x^2+2=-x^2-2$
Because the last is not tue, it is not odd.

e.mccormick · Answer

Usually people test even first because f(-x) is all it needs, then you reuse the result in odd... but that is when you have to show all your work.

anonymous · Answer

did you like simplify the x3 and 6? and then was x2+2?

e.mccormick · Answer

No.  I just gave an example.  Not doing your work for you.  =)

anonymous · Answer

ohh ok XD

e.mccormick · Answer

np.  With $x^n$ the $-x$ part is pretty easy.  If n is an odd number, the - stays.  If the n is an even number, the - goes by-by.

anonymous · Answer

=(-1)(x3-6)
=-x3+6

anonymous · Answer

so it be odd?

anonymous · Answer

@InYourHead

e.mccormick · Answer

Like I said, if it has a constant at the end, it can not be odd because the sign of the constant changes.

anonymous · Answer

Alright

anonymous · Answer

Can you show me a example of one being neither ?

anonymous · Answer

i get odd now. :)

e.mccormick · Answer

Well, any with a variable to an odd power is not even, right? And any with a constant term is not odd, right?  So all the ones with odd powers and a constant term are neither even nor odd.

anonymous · Answer

soo would this be neither ? :o

e.mccormick · Answer

Yah. Here is a graph of three examples of each type. https://www.desmos.com/calculator/taotr3tsip You can turn on and off any graph at any time by clicking the colored dot next to the equation. Notice how the evens are all mirroed right to left, the odds are mirroed from Q1 to Q3 and Q2 to Q4 and the nones are not mirrors at all.

anonymous · Answer

one questino in order to be negative i need a even power on a variable

anonymous · Answer

thats a cool web site

e.mccormick · Answer

negative?  You mean odd?

anonymous · Answer

yes odd*

e.mccormick · Answer

Even must have only even powers, may or may not have constants.
Odd must only have odd powers, must not have costants.

anonymous · Answer

ok i get everything now thanks soo much

e.mccormick · Answer

Yah, looking at a graph can really help at times.  It gives you something to focus on with visual memory, which is the biggest part of sensory memory.  =)