Question

Use technology to find points and then graph the function y, equals, left parenthesis, one half, right parenthesis, start superscript, x, plus, 2, end superscript, minus, 9, commay=(
2
1
​
)
x+2
−9, following the instructions below.
Equation of asymptote:
y, equalsy=

Plot at least five points with integer coordinates

Answer 1

To analyze and graph the function \( y = \left( \frac{1}{2} \right)^{x+2} - 9 \), we can start by identifying the key features of the function, such as its asymptote and points for plotting.

### Step 1: Determine the Asymptote
The asymptote of exponential functions is generally determined by the constant term in the equation. Here, the function approaches the value of \(-9\) as \(x\) approaches infinity. Therefore, the equation of the asymptote is:
\[
y = -9
\]

### Step 2: Find Points with Integer Coordinates
To find points on the graph, we'll calculate the value of \(y\) for a range of \(x\) values (let's choose \(x = -5\) to \(x = 3\)):

1. **For \(x = -5\):**
\[
y = \left( \frac{1}{2} \right)^{-5+2} - 9 = \left( \frac{1}{2} \right)^{-3} - 9 = 2^3 - 9 = 8 - 9 = -1
\]
So the point is \((-5, -1)\).

2. **For \(x = -4\):**
\[
y = \left( \frac{1}{2} \right)^{-4+2} - 9 = \left( \frac{1}{2} \right)^{-2} - 9 = 2^2 - 9 = 4 - 9 = -5
\]
So the point is \((-4, -5)\).

3. **For \(x = -3\):**
\[
y = \left( \frac{1}{2} \right)^{-3+2} - 9 = \left( \frac{1}{2} \right)^{-1} - 9 = 2^1 - 9 = 2 - 9 = -7
\]
So the point is \((-3, -7)\).

4. **For \(x = -2\):**
\[
y = \left( \frac{1}{2} \right)^{-2+2} - 9 = \left( \frac{1}{2} \right)^{0} - 9 = 1 - 9 = -8
\]
So the point is \((-2, -8)\).

5. **For \(x = -1\):**
\[
y = \left( \frac{1}{2} \right)^{-1+2} - 9 = \left( \frac{1}{2} \right)^{1} - 9 = \frac{1}{2} - 9 = -8.5
\]
(not an integer point)

6. **For \(x = 0\):**
\[
y = \left( \frac{1}{2} \right)^{0+2} - 9 = \left( \frac{1}{2} \right)^{2} - 9 = \frac{1}{4} - 9 = -8.75
\]
(not an integer point)

7. **For \(x = 1\):**
\[
y = \left( \frac{1}{2} \right)^{1+2} - 9 = \left( \frac{1}{2} \right)^{3} - 9 = \frac{1}{8} - 9 = -8.875
\]
(not an integer point)

8. **For \(x = 2\):**
\[
y = \left( \frac{1}{2} \right)^{2+2} - 9 = \left( \frac{1}{2} \right)^{4} - 9 = \frac{1}{16} - 9 = -8.9375
\]
(not an integer point)

9. **For \(x = 3\):**
\[
y = \left( \frac{1}{2} \right)^{3+2} - 9 = \left( \frac{1}{2} \right)^{5} - 9 = \frac{1}{32} - 9 = -8.96875
\]
(not an integer point)

### Summary of Points with Integer Coordinates
- \((-5, -1)\)
- \((-4, -5)\)
- \((-3, -7)\)
- \((-2, -8)\)

### Graphing
You would plot these points on a graph, drawing the horizontal line \(y = -9\) for the asymptote. As \(x\) increases, the function descends toward the asymptote without ever crossing or touching it.

### Conclusion
These procedures allow us to analyze and graph the function \(y = \left( \frac{1}{2} \right)^{x+2} - 9\) while identifying key features like the asymptote and specific coordinates for plotting points. The points can be represented on a coordinate system for visualization.