We are given a linear equation $y = \frac{1}{2}x - 3$ and a table with some $x$ and $y$ values. We need to find the values of $y$ when $x = -1$ (which will be A) and when $x = 0$ (which will be B). Then, we need to draw the line $y = \frac{1}{2}x - 3$ on the provided axes.

1. Problem Description

We are given a linear equation $y = \frac{1}{2}x - 3$ and a table with some $x$ and $y$ values. We need to find the values of $y$ when $x = -1$ (which will be A) and when $x = 0$ (which will be B). Then, we need to draw the line $y = \frac{1}{2}x - 3$ on the provided axes.

2. Solution Steps

First, let's find the value of A by substituting $x = -1$ into the equation:

$y = \frac{1}{2}x - 3$

$A = \frac{1}{2}(-1) - 3$

$A = -\frac{1}{2} - 3$

$A = -\frac{1}{2} - \frac{6}{2}$

$A = -\frac{7}{2} = -3.5$

Next, let's find the value of B by substituting $x = 0$ into the equation:

$y = \frac{1}{2}x - 3$

$B = \frac{1}{2}(0) - 3$

$B = 0 - 3$

$B = -3$

To draw the line, we can use the points we already have in the table: $(-2, -4)$, $(-1, -3.5)$, $(0, -3)$, $(1, -2.5)$, and $(2, -2)$. Plot these points on the axes and draw a line through them.

3. Final Answer

$A = -3.5$

$B = -3$

The line $y = \frac{1}{2}x - 3$ can be plotted by using the points $(-2, -4)$, $(-1, -3.5)$, $(0, -3)$, $(1, -2.5)$, and $(2, -2)$. Plot these points on the provided coordinate system and connect them to form a straight line.