1. **State the problem:** We need to find the equation of an exponential function of the form $$y = a(2)^x + d$$ given that the graph is a vertical reflection and shift of $$y = 2^x$$, passes through points (0, -1) and (1, -4), and has a horizontal asymptote at $$y = 2$$. 2. **Identify the horizontal shift:** The horizontal asymptote $$y = d = 2$$ tells us the vertical shift is $$d = 2$$. 3. **Use the point (0, -1) to find $$a$$:** Substitute $$x=0$$ and $$y=-1$$ into the equation: $$-1 = a(2)^0 + 2$$ $$-1 = a + 2$$ Solve for $$a$$: $$a = -1 - 2 = -3$$. 4. **Verify with the point (1, -4):** Substitute $$x=1$$ and $$y=-4$$: $$-4 = -3(2)^1 + 2$$ $$-4 = -3 \times 2 + 2 = -6 + 2 = -4$$ This confirms our values. 5. **Write the final equation:** $$y = -3(2)^x + 2$$ This equation represents an exponential decay (since $$a$$ is negative) reflected vertically and shifted up by 2 units.