The problem asks us to graph the equation $y = -\frac{1}{2}x + 3$ by creating a table of values and plotting the points. We are specifically instructed not to use slope-intercept form as a shortcut.

1. Problem Description

The problem asks us to graph the equation $y = -\frac{1}{2}x + 3$ by creating a table of values and plotting the points. We are specifically instructed not to use slope-intercept form as a shortcut.

2. Solution Steps

First, create a table of values by choosing several $x$ values and computing the corresponding $y$ values. Choose easy values for $x$, and since we have a fraction with a denominator of 2, choosing even values for $x$ will prevent fractions for $y$.

* Let $x = -4$. Then $y = -\frac{1}{2}(-4) + 3 = 2 + 3 = 5$.

* Let $x = -2$. Then $y = -\frac{1}{2}(-2) + 3 = 1 + 3 = 4$.

* Let $x = 0$. Then $y = -\frac{1}{2}(0) + 3 = 0 + 3 = 3$.

* Let $x = 2$. Then $y = -\frac{1}{2}(2) + 3 = -1 + 3 = 2$.

* Let $x = 4$. Then $y = -\frac{1}{2}(4) + 3 = -2 + 3 = 1$.

* Let $x = 6$. Then $y = -\frac{1}{2}(6) + 3 = -3 + 3 = 0$.

Now, we can write these points in a table:

| $x$ | $y$ |

|---|---|

| -4 | 5 |

| -2 | 4 |

| 0 | 3 |

| 2 | 2 |

| 4 | 1 |

| 6 | 0 |

Plot the points on the graph and connect them with a straight line.

3. Final Answer

The solution is the graph of the line $y = -\frac{1}{2}x + 3$, plotted point-by-point using the table of values described above. The graph would show a straight line passing through the points (-4, 5), (-2, 4), (0, 3), (2, 2), (4, 1), and (6, 0).