Using graphing technology, graph \( f(x)=2^{x}, f(x)=2^{x}-1 \), and \( f(x)=2^{x}-2 \) and determine which statement is true. \begin{tabular}{l}\( f(x)=2^{x}-k \), for \\ \( k>0 \), will shift the graph \\ up. \\ \( \begin{array}{l}f(x)=2^{x}-k \text {, for } \\ k>0, \text { will shift the graph } \\ \text { down. }\end{array} \) \\ \( \begin{array}{l}f(x)=2^{x}-k, \text { for } \\ k>0 \text {, will shift the graph } \\ \text { right. }\end{array} \) \\ \( \begin{array}{l}f(x)=2^{x}-k \text {, for } \\ \text { left. }\end{array} \) \\ \hline\end{tabular}

When graphing the functions \( f(x) = 2^{x} \), \( f(x) = 2^{x} - 1 \), and \( f(x) = 2^{x} - 2 \), you'll find that \( f(x) = 2^{x} - k \) for \( k > 0 \) will shift the graph downward. This is because subtracting a constant \( k \) from the function lowers every point on the graph by that same constant, leading to a vertical downward shift! To clarify further, whenever you see a function in the form \( f(x) = 2^{x} - k \) where \( k \) is positive, you can expect the basic exponential curve to descend along the y-axis. So, if you plotted those functions, you would witness that each successive function \( f(x) = 2^{x} - 1 \) and \( f(x) = 2^{x} - 2 \) is simply a downward shift of the previous function. Isn't it fascinating how simple transformations can lead to such different graphs?