I'm really struggling with this and could use all of the that i can get!

First of all, you need to know how to evaluate a function. You need to be able to do it confidently and flawlessly, over and over. All this is asking you to do is repeatedly evaluate different functions and make observations about what happens when you do.

I find it convenient to use "technology" for repeated evaluation of a function. Many graphing calculators have the ability to use a previous answer as input to a new calculation. This makes iteration very easy.

1. For f(x) = 2x, the value doubles each time you iterate the function. The result of 10 iterations of f(1) will be
.. 2, 4, 8, 16, 32, 64, 128, 256, 512, 1024

2. Same deal with -1. The result of 5 iterations will be
.. -2, -4, -8, -16, -32

3. As the number of iterations increases, the value approaches 2.

4. Same deal. The value approaches 2, from below.
See the first figure for the iterations described by questions 3 and 4.

5. Let g(x) = (-1/4)x -1. Iterations of it starting at x = ±4 are shown in the second figure. In each case, the iteration converges toward -0.8.

6. The iteration repeats.
.. g(4) = -2, g(-2) = 4, g(4) = -2 . . . . ad nauseum
.. g(2) = 0, g(0) = 2, g(2) = 0 . . . . . . ad nauseum
.. g(1) = 1, g(1) = 1, g(1) = 1 . . . . . . . . ad nauseum
See the third figure for a table.

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You can conclude that when the magnitude of the slope is less than 1, the iteration converges; when it is not, the iteration does not converge, and may diverge. (This can be handy if you want to develop an iteration function that converges to a particular solution that you want, such as the root of a polynomial.)

Period is the time it takes for one “cycle” to complete. in the parent sine function, period is 2*pi since sin(0) = 0, sin(pi/2) = 1, sin(pi) = 0, sin(3pi/2) = -1, and sin(2pi) = 0. at that point, the sequence repeats itself. i’m not sure if your graph is similar to the sine function since you didn’t provide a picture, but you can find the period by looking at a graph.