Question
Use technology to find points and then graph the function y, equals, 2, to the power x , minus, 8, commay=2
x
−8, following the instructions below.
Equation of asymptote:
y, equalsy=
x
−8, following the instructions below.
Equation of asymptote:
y, equalsy=
Answers
Answered by
GPT-4o mini
To analyze the function \( y = 2^x - 8 \), we first need to identify its properties, including the asymptote.
### 1. Finding the Asymptote:
The function \( y = 2^x - 8 \) is an exponential function that is shifted down by 8 units. As the value of \( x \) approaches negative infinity, \( 2^x \) approaches 0. Therefore:
\[
y = 2^x \rightarrow 0 \quad \text{as} \quad x \rightarrow -\infty
\]
Thus, the horizontal asymptote of this function is:
\[
y = -8
\]
### 2. Finding Points:
To plot this function, we can find several points by selecting different values for \( x \):
- **For \( x = -3 \):**
\[
y = 2^{-3} - 8 = \frac{1}{8} - 8 = \frac{1}{8} - \frac{64}{8} = -\frac{63}{8} \approx -7.875
\]
- **For \( x = -2 \):**
\[
y = 2^{-2} - 8 = \frac{1}{4} - 8 = \frac{1}{4} - \frac{32}{4} = -\frac{31}{4} = -7.75
\]
- **For \( x = -1 \):**
\[
y = 2^{-1} - 8 = \frac{1}{2} - 8 = \frac{1}{2} - \frac{16}{2} = -\frac{15}{2} = -7.5
\]
- **For \( x = 0 \):**
\[
y = 2^0 - 8 = 1 - 8 = -7
\]
- **For \( x = 1 \):**
\[
y = 2^1 - 8 = 2 - 8 = -6
\]
- **For \( x = 2 \):**
\[
y = 2^2 - 8 = 4 - 8 = -4
\]
- **For \( x = 3 \):**
\[
y = 2^3 - 8 = 8 - 8 = 0
\]
- **For \( x = 4 \):**
\[
y = 2^4 - 8 = 16 - 8 = 8
\]
### 3. Points to Plot:
Now, we have the following points:
- \( (-3, -7.875) \)
- \( (-2, -7.75) \)
- \( (-1, -7.5) \)
- \( (0, -7) \)
- \( (1, -6) \)
- \( (2, -4) \)
- \( (3, 0) \)
- \( (4, 8) \)
### 4. Graphing the Function:
To graph the function \( y = 2^x - 8 \):
1. Draw the horizontal asymptote at \( y = -8 \).
2. Plot the points listed above on a coordinate system.
3. Connect these points smoothly, keeping in mind that as \( x \) increases, the function grows rapidly, and as \( x \) decreases, it approaches the asymptote \( y = -8 \).
This graph will rise steeply after crossing the x-axis and approach the horizontal asymptote \( y = -8 \) as \( x \) goes to negative infinity.
### 1. Finding the Asymptote:
The function \( y = 2^x - 8 \) is an exponential function that is shifted down by 8 units. As the value of \( x \) approaches negative infinity, \( 2^x \) approaches 0. Therefore:
\[
y = 2^x \rightarrow 0 \quad \text{as} \quad x \rightarrow -\infty
\]
Thus, the horizontal asymptote of this function is:
\[
y = -8
\]
### 2. Finding Points:
To plot this function, we can find several points by selecting different values for \( x \):
- **For \( x = -3 \):**
\[
y = 2^{-3} - 8 = \frac{1}{8} - 8 = \frac{1}{8} - \frac{64}{8} = -\frac{63}{8} \approx -7.875
\]
- **For \( x = -2 \):**
\[
y = 2^{-2} - 8 = \frac{1}{4} - 8 = \frac{1}{4} - \frac{32}{4} = -\frac{31}{4} = -7.75
\]
- **For \( x = -1 \):**
\[
y = 2^{-1} - 8 = \frac{1}{2} - 8 = \frac{1}{2} - \frac{16}{2} = -\frac{15}{2} = -7.5
\]
- **For \( x = 0 \):**
\[
y = 2^0 - 8 = 1 - 8 = -7
\]
- **For \( x = 1 \):**
\[
y = 2^1 - 8 = 2 - 8 = -6
\]
- **For \( x = 2 \):**
\[
y = 2^2 - 8 = 4 - 8 = -4
\]
- **For \( x = 3 \):**
\[
y = 2^3 - 8 = 8 - 8 = 0
\]
- **For \( x = 4 \):**
\[
y = 2^4 - 8 = 16 - 8 = 8
\]
### 3. Points to Plot:
Now, we have the following points:
- \( (-3, -7.875) \)
- \( (-2, -7.75) \)
- \( (-1, -7.5) \)
- \( (0, -7) \)
- \( (1, -6) \)
- \( (2, -4) \)
- \( (3, 0) \)
- \( (4, 8) \)
### 4. Graphing the Function:
To graph the function \( y = 2^x - 8 \):
1. Draw the horizontal asymptote at \( y = -8 \).
2. Plot the points listed above on a coordinate system.
3. Connect these points smoothly, keeping in mind that as \( x \) increases, the function grows rapidly, and as \( x \) decreases, it approaches the asymptote \( y = -8 \).
This graph will rise steeply after crossing the x-axis and approach the horizontal asymptote \( y = -8 \) as \( x \) goes to negative infinity.