Supply a valid technology formula for the given function.
s(x) = 2^x − 2
s(x) =
Compute the missing values as indicated in the table below.
x	−3	−2	−1	0	1	2	3
s(x)
Show My Work (Opti

Let's start by calculating the missing values of \( s(x) = 2^x - 2 \) for the given \( x \)-values. 

### The function is:
\[
s(x) = 2^x - 2
\]

### Compute the values for each given \( x \):
1. **For \( x = -3 \):**
   \[
   s(-3) = 2^{-3} - 2 = \frac{1}{8} - 2 = \frac{1}{8} - \frac{16}{8} = -\frac{15}{8}
   \]

2. **For \( x = -2 \):**
   \[
   s(-2) = 2^{-2} - 2 = \frac{1}{4} - 2 = \frac{1}{4} - \frac{8}{4} = -\frac{7}{4}
   \]

3. **For \( x = -1 \):**
   \[
   s(-1) = 2^{-1} - 2 = \frac{1}{2} - 2 = \frac{1}{2} - \frac{4}{2} = -\frac{3}{2}
   \]

4. **For \( x = 0 \):**
   \[
   s(0) = 2^0 - 2 = 1 - 2 = -1
   \]

5. **For \( x = 1 \):**
   \[
   s(1) = 2^1 - 2 = 2 - 2 = 0
   \]

6. **For \( x = 2 \):**
   \[
   s(2) = 2^2 - 2 = 4 - 2 = 2
   \]

7. **For \( x = 3 \):**
   \[
   s(3) = 2^3 - 2 = 8 - 2 = 6
   \]

### The computed table is:
\[
\begin{array}{|c|c|c|c|c|c|c|c|}
\hline
x & -3 & -2 & -1 & 0 & 1 & 2 & 3 \\
\hline
s(x) & -\frac{15}{8} & -\frac{7}{4} & -\frac{3}{2} & -1 & 0 & 2 & 6 \\
\hline
\end{array}
\]

Let me know if you'd like further clarification or detailed work on any of these steps!

---

### Related Questions:
1. What happens to \( s(x) \) as \( x \to -\infty \)?
2. How would you describe the behavior of \( s(x) \) for large positive \( x \)?
3. Can we find the roots of the function \( s(x) = 0 \)?
4. What is the range of the function \( s(x) \)?
5. How does the transformation \( 2^x - 2 \) affect the graph of \( 2^x \)?

### Tip:
Remember, for exponential functions like \( 2^x \), negative exponents represent fractions, and positive exponents represent growth!

Supply a valid technology formula for the given function. s(x) = 2^x − 2. Compute the missing values as indicated in the table below.
x  -3   -2   -1   0   1   2   3
s(x)

Explore how to evaluate the exponential function s(x) = 2^x - 2 for a set of x-values including negative exponents. This problem involves calculating function outputs for x-values ranging from -3 to 3. Detailed step-by-step solutions are provided, suitable for students in Grades 9-12.

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