Rewrite the equation with proper formatting:

[tex]\[ y = -\frac{1}{2}x + 3 \][/tex]

Now, organize the table for clarity:

[tex]\[
\begin{array}{c|c}
x & y \\
\hline
0 & \\
4 & \\
-2 & \\
\end{array}
\][/tex]

Fill in the values for [tex]\( y \)[/tex] based on the given equation [tex]\( y = -\frac{1}{2}x + 3 \)[/tex]:

[tex]\[
\begin{array}{c|c}
x & y \\
\hline
0 & 3 \\
4 & 1 \\
-2 & 4 \\
\end{array}
\][/tex]



Answer :

Let's solve the given question step by step by finding the [tex]\( y \)[/tex]-values for each [tex]\( x \)[/tex]-value using the equation [tex]\( y = -\frac{1}{2}x + 3 \)[/tex].

### Step 1: Calculate [tex]\( y \)[/tex] when [tex]\( x = 0 \)[/tex]

Substitute [tex]\( x = 0 \)[/tex] into the equation:
[tex]\[ y = -\frac{1}{2}(0) + 3 \][/tex]

This simplifies to:
[tex]\[ y = 3 \][/tex]

So, the point [tex]\((0, y)\)[/tex] becomes [tex]\((0, 3)\)[/tex].

### Step 2: Calculate [tex]\( y \)[/tex] when [tex]\( x = 4 \)[/tex]

Substitute [tex]\( x = 4 \)[/tex] into the equation:
[tex]\[ y = -\frac{1}{2}(4) + 3 \][/tex]

This simplifies to:
[tex]\[ y = -2 + 3 = 1 \][/tex]

So, the point [tex]\((4, y)\)[/tex] becomes [tex]\((4, 1)\)[/tex].

### Step 3: Calculate [tex]\( y \)[/tex] when [tex]\( x = -2 \)[/tex]

Substitute [tex]\( x = -2 \)[/tex] into the equation:
[tex]\[ y = -\frac{1}{2}(-2) + 3 \][/tex]

This simplifies to:
[tex]\[ y = 1 + 3 = 4 \][/tex]

So, the point [tex]\((-2, y)\)[/tex] becomes [tex]\((-2, 4)\)[/tex].

### Summary of Points

1. When [tex]\( x = 0 \)[/tex], [tex]\( y = 3 \)[/tex]. The point is [tex]\((0, 3)\)[/tex].
2. When [tex]\( x = 4 \)[/tex], [tex]\( y = 1 \)[/tex]. The point is [tex]\((4, 1)\)[/tex].
3. When [tex]\( x = -2 \)[/tex], [tex]\( y = 4 \)[/tex]. The point is [tex]\((-2, 4)\)[/tex].

Therefore, the values of [tex]\( y \)[/tex] corresponding to [tex]\( x \)[/tex] values of [tex]\( 0, 4, \)[/tex] and [tex]\( -2 \)[/tex] are [tex]\( 3.0, 1.0, \)[/tex] and [tex]\( 4.0 \)[/tex], respectively.