Graphing Linear Function Worksheets - Free Printable

The image you provided is a worksheet titled "Function Table," which involves completing tables for various functions and plotting points on a graph. Below, I will explain how to solve each problem step by step.

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Problem 1: \( y = \frac{1}{2}x + 1 \)

#### Step 1: Complete the table
We are given the function \( y = \frac{1}{2}x + 1 \). We need to calculate \( y \) for the given values of \( x \).

- For \( x = -4 \):
\[
y = \frac{1}{2}(-4) + 1 = -2 + 1 = -1
\]
- For \( x = -2 \):
\[
y = \frac{1}{2}(-2) + 1 = -1 + 1 = 0
\]
- For \( x = 0 \):
\[
y = \frac{1}{2}(0) + 1 = 0 + 1 = 1
\]
- For \( x = 2 \):
\[
y = \frac{1}{2}(2) + 1 = 1 + 1 = 2
\]
- For \( x = 4 \):
\[
y = \frac{1}{2}(4) + 1 = 2 + 1 = 3
\]

So, the completed table is:
\[
\begin{array}{c|c}
x & y \\
\hline
-4 & -1 \\
-2 & 0 \\
0 & 1 \\
2 & 2 \\
4 & 3 \\
\end{array}
\]

#### Step 2: Plot the points
Plot the points \((-4, -1)\), \((-2, 0)\), \((0, 1)\), \((2, 2)\), and \((4, 3)\) on the graph.

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Problem 2: \( y = -\frac{1}{2}x - 2 \)

#### Step 1: Complete the table
We are given the function \( y = -\frac{1}{2}x - 2 \). We need to calculate \( y \) for the given values of \( x \).

- For \( x = -50 \):
\[
y = -\frac{1}{2}(-50) - 2 = 25 - 2 = 23
\]
- For \( x = -10 \):
\[
y = -\frac{1}{2}(-10) - 2 = 5 - 2 = 3
\]
- For \( x = 0 \):
\[
y = -\frac{1}{2}(0) - 2 = 0 - 2 = -2
\]
- For \( x = 10 \):
\[
y = -\frac{1}{2}(10) - 2 = -5 - 2 = -7
\]
- For \( x = 50 \):
\[
y = -\frac{1}{2}(50) - 2 = -25 - 2 = -27
\]

So, the completed table is:
\[
\begin{array}{c|c}
x & y \\
\hline
-50 & 23 \\
-10 & 3 \\
0 & -2 \\
10 & -7 \\
50 & -27 \\
\end{array}
\]

#### Step 2: Plot the points
Plot the points \((-50, 23)\), \((-10, 3)\), \((0, -2)\), \((10, -7)\), and \((50, -27)\) on the graph.

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Problem 3: \( y = 3x + 3 \)

#### Step 1: Complete the table
We are given the function \( y = 3x + 3 \). We need to calculate \( y \) for the given values of \( x \).

- For \( x = -3 \):
\[
y = 3(-3) + 3 = -9 + 3 = -6
\]
- For \( x = -2 \):
\[
y = 3(-2) + 3 = -6 + 3 = -3
\]
- For \( x = -1 \):
\[
y = 3(-1) + 3 = -3 + 3 = 0
\]
- For \( x = 0 \):
\[
y = 3(0) + 3 = 0 + 3 = 3
\]
- For \( x = 1 \):
\[
y = 3(1) + 3 = 3 + 3 = 6
\]

So, the completed table is:
\[
\begin{array}{c|c}
x & y \\
\hline
-3 & -6 \\
-2 & -3 \\
-1 & 0 \\
0 & 3 \\
1 & 6 \\
\end{array}
\]

#### Step 2: Plot the points
Plot the points \((-3, -6)\), \((-2, -3)\), \((-1, 0)\), \((0, 3)\), and \((1, 6)\) on the graph.

---

Problem 4: \( y = -3x - 3 \)

#### Step 1: Complete the table
We are given the function \( y = -3x - 3 \). We need to calculate \( y \) for the given values of \( x \).

- For \( x = -1 \):
\[
y = -3(-1) - 3 = 3 - 3 = 0
\]
- For \( x = -\frac{1}{2} \):
\[
y = -3\left(-\frac{1}{2}\right) - 3 = \frac{3}{2} - 3 = \frac{3}{2} - \frac{6}{2} = -\frac{3}{2}
\]
- For \( x = 0 \):
\[
y = -3(0) - 3 = 0 - 3 = -3
\]
- For \( x = \frac{1}{2} \):
\[
y = -3\left(\frac{1}{2}\right) - 3 = -\frac{3}{2} - 3 = -\frac{3}{2} - \frac{6}{2} = -\frac{9}{2}
\]
- For \( x = 1 \):
\[
y = -3(1) - 3 = -3 - 3 = -6
\]

So, the completed table is:
\[
\begin{array}{c|c}
x & y \\
\hline
-1 & 0 \\
-\frac{1}{2} & -\frac{3}{2} \\
0 & -3 \\
\frac{1}{2} & -\frac{9}{2} \\
1 & -6 \\
\end{array}
\]

#### Step 2: Plot the points
Plot the points \((-1, 0)\), \(\left(-\frac{1}{2}, -\frac{3}{2}\right)\), \((0, -3)\), \(\left(\frac{1}{2}, -\frac{9}{2}\right)\), and \((1, -6)\) on the graph.

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Problem 5: \( y = -4 + \frac{1}{2}x \)

#### Step 1: Complete the table
We are given the function \( y = -4 + \frac{1}{2}x \). We need to calculate \( y \) for the given values of \( x \).

- For \( x = -2 \):
\[
y = -4 + \frac{1}{2}(-2) = -4 - 1 = -5
\]
- For \( x = 0 \):
\[
y = -4 + \frac{1}{2}(0) = -4 + 0 = -4
\]
- For \( x = 2 \):
\[
y = -4 + \frac{1}{2}(2) = -4 + 1 = -3
\]
- For \( x = 4 \):
\[
y = -4 + \frac{1}{2}(4) = -4 + 2 = -2
\]
- For \( x = 6 \):
\[
y = -4 + \frac{1}{2}(6) = -4 + 3 = -1
\]

So, the completed table is:
\[
\begin{array}{c|c}
x & y \\
\hline
-2 & -5 \\
0 & -4 \\
2 & -3 \\
4 & -2 \\
6 & -1 \\
\end{array}
\]

#### Step 2: Plot the points
Plot the points \((-2, -5)\), \((0, -4)\), \((2, -3)\), \((4, -2)\), and \((6, -1)\) on the graph.

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Problem 6: \( y = -2 + \frac{1}{2}x \)

#### Step 1: Complete the table
We are given the function \( y = -2 + \frac{1}{2}x \). We need to calculate \( y \) for the given values of \( x \).

- For \( x = -10 \):
\[
y = -2 + \frac{1}{2}(-10) = -2 - 5 = -7
\]
- For \( x = -5 \):
\[
y = -2 + \frac{1}{2}(-5) = -2 - \frac{5}{2} = -2 - 2.5 = -4.5
\]
- For \( x = 0 \):
\[
y = -2 + \frac{1}{2}(0) = -2 + 0 = -2
\]
- For \( x = 5 \):
\[
y = -2 + \frac{1}{2}(5) = -2 + \frac{5}{2} = -2 + 2.5 = 0.5
\]
- For \( x = 10 \):
\[
y = -2 + \frac{1}{2}(10) = -2 + 5 = 3
\]

So, the completed table is:
\[
\begin{array}{c|c}
x & y \\
\hline
-10 & -7 \\
-5 & -4.5 \\
0 & -2 \\
5 & 0.5 \\
10 & 3 \\
\end{array}
\]

#### Step 2: Plot the points
Plot the points \((-10, -7)\), \((-5, -4.5)\), \((0, -2)\), \((5, 0.5)\), and \((10, 3)\) on the graph.

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Final Answer

The final answers are the completed tables and the plotted points for each function. The boxed answer is:

\[
\boxed{
\begin{array}{c}
\text{Tables and graphs for all functions are completed as explained above.}
\end{array}
}
\]