1. **State the problem:** We need to evaluate the function $y = \left(\frac{1}{2}\right)^x$ at the given values of $x$: $-3, -2, -1, 0, 1, 2$. 2. **Recall the formula:** The function is an exponential function where the base is $\frac{1}{2}$ and the exponent is $x$. For any real number $x$, $$y = \left(\frac{1}{2}\right)^x = 2^{-x}$$ This means when $x$ is negative, the function becomes a positive power of 2. 3. **Calculate each value:** - For $x = -3$: $$y = \left(\frac{1}{2}\right)^{-3} = 2^{3} = 8$$ - For $x = -2$: $$y = \left(\frac{1}{2}\right)^{-2} = 2^{2} = 4$$ - For $x = -1$: $$y = \left(\frac{1}{2}\right)^{-1} = 2^{1} = 2$$ - For $x = 0$: $$y = \left(\frac{1}{2}\right)^{0} = 1$$ - For $x = 1$: $$y = \left(\frac{1}{2}\right)^{1} = \frac{1}{2}$$ - For $x = 2$: $$y = \left(\frac{1}{2}\right)^{2} = \frac{1}{4}$$ 4. **Summary:** The function values are: $$\begin{aligned} x & : -3 & -2 & -1 & 0 & 1 & 2 \\ y & : 8 & 4 & 2 & 1 & \frac{1}{2} & \frac{1}{4} \end{aligned}$$ This shows the exponential decay behavior for positive $x$ and growth for negative $x$. **Final answer:** The values of $y$ at the given $x$ are $8, 4, 2, 1, \frac{1}{2}, \frac{1}{4}$ respectively.