What is your reasoning? You should have a reason for saying that. Suppose for I, $y = Ab^x$. Then look at the values when $x = -1$ and when $x = 1$. You have $Ab^{-1} = 20$ and $Ab = 5$. From the first equation, multiply both sides by $b$ and you find $A=20b$. Plug that in to the second equation: $(20b)b = 5$. Solving for $b$, you find $b = \pm \dfrac{1}{2}$. Obviously, that is less than 1, so it does not satisfy $b>1$.

For II, at $x=0, y = 2$. At $x=1, y \approx 4$, and at $x=2, y\approx 8$. So, let's do the same thing. Assume $y=Ab^x$. Plugging in when $x=0,y=2$ and when $x=1,y=4$, you get: $Ab^0 = 2$, $Ab = 4$. This means $A=2$ and $2b=4$, so $b=2>1$. This does, indeed, meet the requirements of the problem.

For III, once more, you assume $y = Ab^x$. Again, we try plugging in when $x=\pm 1$. We find two equations: $-5 = Ab^{-1}$ and $-20 = Ab$. So, from the first equation, we find $A = -5b$. Plugging that into the second, we find $-20 = (-5b)b = -5b^2$. Solving for $b$, we find $b = \pm 2$. If $b=-2$, then it is less than one, so assume $b=2$. Then, $A = -5(2) = -10$ and the equation becomes $y = -10(2)^x$. This is the second possibly correct function. Let's try the remaining points. At $x=-2$, you have $y = -10(2)^{-2} = -2.5$, which matches the chart. At $x=2$, you have $y = -10(2)^2 = -40$, and again, this matches the chart, so you might, indeed, have a function of that form.