Freshman Algebra 1 >_< ?

So far all the questions on the HW have been extremely easy, but i dont know what it is about these 3, if im getting caught up in the wording or what?

Anyone mind doing them AND explaining how to it for test reference? I figure it should be fairly easy for you smart folks

Determine whether each relation is a function. If the relation is a function, state the domain and range

27- ( trying to make a box for these, sorry for my failness at it ! )

lX l Yl
l1 l -3l
l6 l -2l
l9 l -1l
l1 l 3l ​

28-

lX l Yl
l0 l 2l
l3 l 1l
l3 l -1l
l5 l 3l​

29-

lX l Yl
l-4l-4l
l-1l-4l
l0l-4l
l3l-4l​

Help is much appreciated!

Also, mind explaining the whole " is this relation a function " thing >_<


EDIT: Some over ones i cant figure out


Use the functions f(x)=2x and g(x) = x to the second power + 1

43
f(3) + g(4)

45
f(5) - 2g(1)

46
fg=(g(3))

 

The first two are not functions because from the table, say the first one. When x=1 y= -3 and 3. When you have this, it is not considered a function. The third one is and the function is y=-4, because whatever x is, y is -4 regardless.

For the second batch, for your reference ^2 means power of 2.

f(3) means plug in 3 for f(x). So, f(x) = 2x
f(3) = 2(3) = 6
You can use this to solve it and the second one. For the third solve the right side and using algebra you have something like
2x*(x^2 + 1) = whatever
It turns into
2x^3 + 2x = whatever

Now that I think about it do you mean f(g)? because this seems way too far ahead.

 

I remember this stuff its really easy koot has the idea on it, just for your info DONT TAKE TRIG. it will ruin your life , take pre calculus instead

 

When you say this, it could mean g(x) = (x^2) +1 or g(x) = (x^(2+1)).

Which one is it that you actually need?

And yea, Koot is correct with his explanations of the functions. To make determining whether something is a function or not easy, you can use the Vertical Line Test. Basically, you run along the x-axis and make an imaginary line for every x value. If at any moment you hit 2 points with your imaginary line, the graph is not a function.
http://en.wikipedia.org/wiki/Vertical_line_test

It doesn't work as well with tables of values, but it's good to know.

/EDIT:
3,333 post. Fuck yea.

 

A function is only allowed to have an X value point to a single Y value. In tabular form, this means that you can not have the same X value twice, with each pointing to a different Y. In graphical form, you must be able to draw a vertical line anywhere on the graph and only hit one point (the X point coordinate and the single Y it points to).

Domain is the possible X values. For these, I'm assuming its all the X values in the table.

Range is the possible Y values.

f(3) + g(4): Find the answer to f(3) and add it to g(4).

f(5) - 2g(1) : Fine the answer to f(5) and subtract it from the answer of g(1) multiplied by 2.

I don't really know what the third one is. >_>.

 

This is somewhat off-topic, but I've found that making a table in Word, screenshotting it, and posting the pic is the easiest thing to do when I need to make a table.

Anyway OT:

1 and 2 are not functions because they have two of the same x input. Functions can't have two points that are on top of eachother (i.e., two of the same x value).

3 is a function. The domain is all possible X values, so it is X is greater than or equal to -4 and less than1. The range is all possible y values, so it is -4.




EDIT: The third one in the last part is f(g(3)), which means "f of the number which is g(3)'s solution..." basically, g(3) is 3^2+1, or 10. Then do f(10), which is 20.

 

if your looking at a graph, you can tell if its a function by the vertical line test. In other words, if you see two y points at the same x value and one of the y values isn't an: O. O representing a hole in the graph and meaning that part of the function doesn't exist.(usually for piecewise functions.)

 

Ive gotten them all but 43,45, and 46.

Thanks for trying to help though

 

I just took these, if you're still unsure add me, [email protected]

So I can explain easier.

 

If f(x)=2x and g(x) = x^(2) + 1


43
f(3) + g(4)

2(3) + 4^(2) + 1 = 6 + 16 +1 = 6 + 17 = 23

45
f(5) - 2g(1)

2(5) - 2(1^(2) + 1) = 10 - 2(1 + 1) = 10 - 2(2) = 10 - 4 = 6

46
fg=(g(3))
__________

I don't understand your notation use in the third one, sorry.

 

got it, thanks everyone.

And apparently Determinate, in #3 your suppose to do G3 first, and FG= is just a fancy way of saying Y

 

Yes, F(g) or F(x) is the same as saying 'Y=...'