The task is to use transformations on the graph of $y = f(x) = 2^x$ to obtain the graph of $g(x) = 2^{-x}$. Here's how it works:

Step-by-Step Transformation:

1. Reflection across the $y$-axis:
   The transformation from $2^x$ to $2^{-x}$ involves reflecting the graph of $2^x$ across the $y$-axis.

   Mathematically, replacing $x$ with $-x$ in the function reflects the graph over the $y$-axis. Thus, $g(x) = 2^{-x}$.

Explanation of the Transformation:

• The graph of $y = 2^x$ grows exponentially as $x$ increases, starting very close to 0 for large negative $x$, passing through $(0, 1)$, and rising sharply for large positive $x$.
• When you reflect $y = 2^x$ across the $y$-axis to get $y = 2^{-x}$, the behavior reverses:
  • The graph approaches 0 as $x$ increases (for large positive $x$).
  • It rises sharply for large negative $x$.

Would you like further clarification or a graph to illustrate this?

Related Questions:

1. How does replacing $x$ with $-x$ transform a general function $f(x)$?
2. What other transformations (e.g., shifts or stretches) can be applied to exponential functions like $2^x$?
3. What is the domain and range of $g(x) = 2^{-x}$?
4. How do the asymptotes of $f(x) = 2^x$ and $g(x) = 2^{-x}$ compare?
5. Can this transformation process be applied to logarithmic functions as well?

Tip:

For any function $y = f(x)$, replacing $x$ with $-x$ reflects the graph across the $y$-axis, while replacing $y$ with $-y$ reflects it across the $x$-axis.