Rom

Moves are \((1,0),~(0,1),~(-1,0),~(0,-1)\)

We have a set of 4 moves and the sum of those must equal (0,0)

There are 36 of these sets of 4 moves that sum to (0,0)

There are a total of \(4^4 = 256\) moves.

P[returning to 0] = \(\dfrac{36}{256} = \dfrac{9}{64}\)

CoolStuffYT is correct.

according to this description

\(y = x^2 + 1\)

represents exponential growth

exponential growth occurs when the ratio of two values at different times is exponential in the difference between those times with a positive constant.

i..e

\(\dfrac{f(t_1)}{f(t_0)} \sim e^{\lambda(t_1-t_0)},~\lambda > 0\)

In your example (which is indeed exponential growth)

\(\dfrac{2^{x_2}}{2^{x_1}} = 2^{x_2-x_1} = e^{\ln(2)(x_2-x_1)}\\ \ln(2) > 0\)

well.. it's got the points (0,-5) and (1,-1) in it and it looks to have a horizontal asymptote of -7 as x goes to negative infinity.

I'd go with 

\(-7+2e^{\ln(3) x}\)

\(g(x) = f(x+1) = (x+1)^7 - 3(x+1)^3 + 2\)

Now this is some polynomial

\(g(x) = \sum \limits_{k=0}^7~c_k x^k \\ \text{and we can find the sum of the coefficients by simply finding }g(1)\)

\(g(1) = 2^7 - 3(2^3)+2 = 128 - 24+2 = 106\)

quadratic

\(f(x)=(x+3)(x-5)\)

\(y = 270000(1+2.9\%)^x = 2.7\times10^5(1.029)^x\)

\(\tan(A) = \dfrac{\sin(A)}{\cos(A)} = \\ \dfrac{2\cos(A)}{\cos(A)} = 2\)

\(V=\dfrac 4 3 \pi r^3\\ \dfrac{dV}{dr} = 4 \pi r^2\\ \text{ok a bit of magic here}\\ dV = 4\pi r^2 dr\\ \dfrac{dV}{dt} = 4\pi r^2 \dfrac{dr}{dt} \\ 2 = 4 \pi (3)^2 \dfrac{dr}{dt}\\ \dfrac{dr}{dt} = \dfrac{1}{18\pi}~in/hr\)

a function maps every element of it's domain to a single value in it's range.

the only completion of the first table is so that f is not a function is

Any other completion of this table will ensure f is a function.

For the second part.

is not a function because -1 is mapped to two different values