1. **State the problem:** We are given the function $$y = - \left(\frac{1}{2}\right)^{x+3} - 2$$ and a table of values for $x = -6, -5, -4, -3, -2, -1$. We want to understand how to compute $y$ for these values and describe the behavior of the function. 2. **Formula and rules:** The function is an exponential function with base $\frac{1}{2}$ raised to the power $x+3$, then reflected across the x-axis (due to the negative sign in front), and shifted down by 2 units. 3. **Calculate intermediate values:** - Compute $x+3$ for each $x$. - Compute $\left(\frac{1}{2}\right)^{x+3}$. - Multiply by $-1$ to get $-\left(\frac{1}{2}\right)^{x+3}$. - Subtract 2 to get $y$. 4. **Example calculations:** - For $x = -6$: $x+3 = -3$, so $$\left(\frac{1}{2}\right)^{-3} = 2^3 = 8$$ - Then, $$y = -8 - 2 = -10$$ - For $x = -5$: $x+3 = -2$, so $$\left(\frac{1}{2}\right)^{-2} = 2^2 = 4$$ - Then, $$y = -4 - 2 = -6$$ - For $x = -4$: $x+3 = -1$, so $$\left(\frac{1}{2}\right)^{-1} = 2$$ - Then, $$y = -2 - 2 = -4$$ - For $x = -3$: $x+3 = 0$, so $$\left(\frac{1}{2}\right)^0 = 1$$ - Then, $$y = -1 - 2 = -3$$ - For $x = -2$: $x+3 = 1$, so $$\left(\frac{1}{2}\right)^1 = \frac{1}{2}$$ - Then, $$y = -\frac{1}{2} - 2 = -2.5$$ - For $x = -1$: $x+3 = 2$, so $$\left(\frac{1}{2}\right)^2 = \frac{1}{4}$$ - Then, $$y = -\frac{1}{4} - 2 = -2.25$$ 5. **Interpretation:** The function is an exponential decay because the base $\frac{1}{2}$ is between 0 and 1. The negative sign reflects the graph over the x-axis, flipping it upside down. The $+3$ inside the exponent shifts the graph left by 3 units, and the $-2$ shifts it down by 2 units. 6. **Summary:** The function decreases from large negative values at $x=-6$ towards $-2$ as $x$ increases, approaching $y = -2$ from below but never crossing it. This is the horizontal asymptote. **Final answer:** The values of $y$ for the given $x$ are: $$\begin{array}{c|c} x & y \\\hline -6 & -10 \\ -5 & -6 \\ -4 & -4 \\ -3 & -3 \\ -2 & -2.5 \\ -1 & -2.25 \end{array}$$