1. The problem gives three linear functions: $$y = \frac{1}{2}x$$ $$y = \frac{1}{2}x + 2$$ $$y = \frac{1}{2}x - 3$$ and a table with values of $x$ as 0 and 2, asking to find corresponding $y$ values. 2. The formula for each function is linear: $$y = mx + b$$ where $m$ is the slope and $b$ is the y-intercept. 3. Calculate $y$ for each function at $x=0$ and $x=2$: For $$y = \frac{1}{2}x$$: - At $x=0$: $$y = \frac{1}{2} \times 0 = 0$$ - At $x=2$: $$y = \frac{1}{2} \times 2 = 1$$ For $$y = \frac{1}{2}x + 2$$: - At $x=0$: $$y = \frac{1}{2} \times 0 + 2 = 2$$ - At $x=2$: $$y = \frac{1}{2} \times 2 + 2 = 1 + 2 = 3$$ For $$y = \frac{1}{2}x - 3$$: - At $x=0$: $$y = \frac{1}{2} \times 0 - 3 = -3$$ - At $x=2$: $$y = \frac{1}{2} \times 2 - 3 = 1 - 3 = -2$$ 4. Fill the table with these values: | x | 0 | 2 | |---|---|---| | y (for $$y=\frac{1}{2}x$$) | 0 | 1 | | y (for $$y=\frac{1}{2}x + 2$$) | 2 | 3 | | y (for $$y=\frac{1}{2}x - 3$$) | -3 | -2 | This completes the evaluation of $y$ values for the given $x$ values in each function.