1. **State the problem:** We need to find the equation of a transformed exponential function of the form $$y = A \cdot 2^x + k$$ based on the given graph. 2. **Recall the base function:** The original function is $$y = 2^x$$, which is an increasing exponential function passing through points like $(0,1)$ and $(1,2)$. 3. **Analyze the transformation:** The graph shows a curve that decreases as $x$ increases, starting near $y=2$ at $x=-5$ and dropping below $y=-5$ near $x=0$. This suggests a reflection and vertical shift. 4. **Reflection:** Since the original $2^x$ increases but the graph decreases, the function is reflected over the x-axis. This means $$A < 0$$. 5. **Vertical shift:** The graph appears shifted downward because it goes below $y=0$. The vertical shift is represented by $k$. 6. **Use a point to find $A$ and $k$:** Use the point near $(-2,1)$ from the graph. Substitute into the equation: $$1 = A \cdot 2^{-2} + k = A \cdot \frac{1}{4} + k$$ 7. **Use another point:** Use the point near $(-5,2)$: $$2 = A \cdot 2^{-5} + k = A \cdot \frac{1}{32} + k$$ 8. **Solve the system:** From the two equations: $$1 = \frac{A}{4} + k$$ $$2 = \frac{A}{32} + k$$ Subtract second from first: $$1 - 2 = \frac{A}{4} - \frac{A}{32}$$ $$-1 = A \left( \frac{1}{4} - \frac{1}{32} \right) = A \cdot \frac{7}{32}$$ So, $$A = -1 \cdot \frac{32}{7} = -\frac{32}{7}$$ 9. **Find $k$:** Substitute $A$ back into one equation: $$1 = \frac{-32/7}{4} + k = -\frac{8}{7} + k$$ So, $$k = 1 + \frac{8}{7} = \frac{15}{7}$$ 10. **Final equation:** $$y = -\frac{32}{7} \cdot 2^x + \frac{15}{7}$$ This matches the decreasing shape and vertical shift observed. **Answer:** $$y = -\frac{32}{7} \cdot 2^x + \frac{15}{7}$$