(a) The values of \(x\) are equally spaced, namely by \(\Delta x = 1\text{.}\) We have to check if all ratios between consecutive values of \(y\) are the same:
\begin{equation*}
\dfrac{2}{0.5} = 4, \quad \dfrac{8}{2} = 4, \quad \dfrac{32}{8} = 4, \quad \dfrac{128}{32} = 4, \quad \dfrac{512}{128} = 4.
\end{equation*}
All ratios are the same. Since \(\Delta x = 1\text{,}\) the ratios give us the growth factor \(b = 4\text{.}\) With each increase in \(x\) by \(1\text{,}\) the current value of the function is multiplied by the constant factor of \(4\text{.}\) The initial value is the \(y\) value at \(0\text{,}\) so \(A = 0.5\text{.}\) The table corresponds to the exponential function \(y = 0.5 \cdot 4^x\text{.}\)
(b) The ratios between consecutive outputs are
\begin{equation*}
\dfrac{4.5}{0.5} = 9, \quad \dfrac{8.5}{4.5} = 1.888, \qquad \dots
\end{equation*}
which are clearly not equal, so the function is not exponential. Observe that the values of the function start at \(y=0.5\) when \(x = 0\) and then increase by \(4\) for every unit increase in \(x\text{.}\) This is precisely the definition of a linear function, so table (b) corresponds to \(y = 0.5 + 4x\text{.}\)
(c) The ratios between consecutive outputs are
\begin{equation*}
\dfrac{1}{2} = 0.5, \quad \dfrac{0.5}{1} = 0.5, \quad \dfrac{0.25}{0.5} = 0.5, \quad \dfrac{0.125}{0.25} = 0.5
\end{equation*}
which are equal, and the inputs are evenly spaced by \(\Delta x = 1\text{,}\) so the function is exponential. The growth factor is \(b = 0.5\) and so \(y = A \cdot 0.5^x\text{.}\) The initial value \(A\) is unknown. Because the value of the function at \(0\) is known, set up an equation for \(A\) using, for example, the point \((1, 2)\) from the table. This gives the equation \(A \cdot 0.5^1 = 2\text{,}\) and so \(A = 4\text{.}\) The function represented by the table (c) is \(y = 4 \cdot 0.5^x\text{.}\)
