The problem asks for the equation of the function $f(x) = 2^{x-3} + 1$ after a rotation of $-90^\circ$ about the origin $(0, 0)$.

Steps to solve:

1. Understand the rotation effect:
   A rotation of $-90^\circ$ about the origin switches the coordinates $(x, y)$ to $(-y, x)$.

2. Initial equation:
   The given function is: $f(x) = 2^{x-3} + 1$ which is in the form $y = 2^{x-3} + 1$.

3. Apply rotation: When rotating the graph of a function by $-90^\circ$, the point $(x, y)$ transforms into $(y, -x)$. This changes the relationship between $x$ and $y$.

   Substitute $x$ for $y$ and $-y$ for $x$: $x = 2^{-y-3} + 1$

4. Solve for $y$: To rewrite this equation in terms of $y$, first isolate the exponential term: $x - 1 = 2^{-y-3}$ Take the logarithm (base 2) of both sides: $\log_2(x - 1) = -y - 3$ Solving for $y$: $y = -(\log_2(x - 1) + 3)$ Simplify: $y = -\log_2(x - 1) - 3$

Thus, the equation of the rotated function is: $y = -\log_2(x - 1) - 3$

Would you like further details or clarifications?

Related Questions:

1. What is the effect of a $90^\circ$ rotation instead of $-90^\circ$?
2. How does the function $f(x) = 2^{x-3} + 1$ behave graphically before rotation?
3. What is the transformation rule for other degrees of rotation (e.g., $180^\circ$)?
4. How can we verify the result of a rotated function using geometry?
5. What would be the function equation after a reflection across the x-axis?

Tip:

When transforming functions by rotation, remember that the coordinate changes (like $(x, y) \to (-y, x)$ for $-90^\circ$) help guide the process of rewriting the function equation.