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To solve the function table for \( h(x) = 2x + 5 \), substitute each \( x \) value into the function:

1. \( x = -5 \)
   \[
   h(-5) = 2(-5) + 5 = -10 + 5 = -5
   \]

2. \( x = -4 \)
   \[
   h(-4) = 2(-4) + 5 = -8 + 5 = -3
   \]

3. \( x = 0 \)
   \[
   h(0) = 2(0) + 5 = 0 + 5 = 5
   \]

4. \( x = 1 \)
   \[
   h(1) = 2(1) + 5 = 2 + 5 = 7
   \]

5. \( x = 4 \)
   \[
   h(4) = 2(4) + 5 = 8 + 5 = 13
   \]

The completed function table is:

\[
\begin{array}{|c|c|}
\hline
x & h(x) \\
\hline
-5 & -5 \\
-4 & -3 \\
0 & 5 \\
1 & 7 \\
4 & 13 \\
\hline
\end{array}
\]

Complete the function table for h(x) = 2x + 5.

Explore how to complete the function table for the linear function h(x) = 2x + 5. Learn the step-by-step substitution of x-values to find h(x) for each entry in the table. Suitable for students in Grades 7-8.

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