Answer :
Certainly! Let's go through the process of graphing the exponential function [tex]\( w(x) = 2 \cdot (0.4)^x - 5 \)[/tex] step by step.
### Step 1: Understand the Function
The given function is an exponential decay function because the base of the exponent [tex]\(0.4\)[/tex] is between 0 and 1. Here is the breakdown of the function:
- [tex]\( 2 \)[/tex] is the coefficient that vertically stretches the graph.
- [tex]\( 0.4 \)[/tex] is the base of the exponent, causing decay.
- [tex]\( x \)[/tex] is the variable exponent.
- [tex]\( -5 \)[/tex] is a vertical shift downward by 5 units.
### Step 2: Plot Key Points
To graph the function, let's compute some key points.
1. When [tex]\( x = 0 \)[/tex]:
[tex]\[ w(0) = 2 \cdot (0.4)^0 - 5 = 2 \cdot 1 - 5 = 2 - 5 = -3 \][/tex]
Point: [tex]\((0, -3)\)[/tex]
2. When [tex]\( x = 1 \)[/tex]:
[tex]\[ w(1) = 2 \cdot (0.4)^1 - 5 = 2 \cdot 0.4 - 5 = 0.8 - 5 = -4.2 \][/tex]
Point: [tex]\((1, -4.2)\)[/tex]
3. When [tex]\( x = -1 \)[/tex]:
[tex]\[ w(-1) = 2 \cdot (0.4)^{-1} - 5 = 2 \cdot \frac{1}{0.4} - 5 = 2 \cdot 2.5 - 5 = 5 - 5 = 0 \][/tex]
Point: [tex]\((-1, 0)\)[/tex]
4. When [tex]\( x = 2 \)[/tex]:
[tex]\[ w(2) = 2 \cdot (0.4)^2 - 5 = 2 \cdot 0.16 - 5 = 0.32 - 5 = -4.68 \][/tex]
Point: [tex]\((2, -4.68)\)[/tex]
5. When [tex]\( x = -2 \)[/tex]:
[tex]\[ w(-2) = 2 \cdot (0.4)^{-2} - 5 = 2 \cdot \left(\frac{1}{0.4}\right)^2 - 5 = 2 \cdot 6.25 - 5 = 12.5 - 5 = 7.5 \][/tex]
Point: [tex]\((-2, 7.5)\)[/tex]
### Step 3: Additional Points
You can calculate more points if needed to provide better accuracy. Here are additional common values:
#### When [tex]\( x = -3 \)[/tex]:
[tex]\[ w(-3) = 2 \cdot (0.4)^{-3} - 5 = 2 \cdot \left(\frac{1}{0.4}\right)^3 - 5 = 2 \cdot 15.625 - 5 = 31.25 - 5 = 26.25 \][/tex]
Point: [tex]\((-3, 26.25)\)[/tex]
#### When [tex]\( x = 3 \)[/tex]:
[tex]\[ w(3) = 2 \cdot (0.4)^3 - 5 = 2 \cdot 0.064 - 5 = 0.128 - 5 = -4.872 \][/tex]
Point: [tex]\((3, -4.872)\)[/tex]
### Step 4: Plot the Graph
1. Plot the calculated points on a coordinate plane:
- [tex]\((0, -3)\)[/tex]
- [tex]\((1, -4.2)\)[/tex]
- [tex]\((-1, 0)\)[/tex]
- [tex]\((2, -4.68)\)[/tex]
- [tex]\((-2, 7.5)\)[/tex]
- [tex]\((-3, 26.25)\)[/tex]
- [tex]\((3, -4.872)\)[/tex]
2. Draw a smooth curve through these points, keeping in mind the overall shape of an exponential decay function: the graph should approach the horizontal line [tex]\( y = -5 \)[/tex] as [tex]\( x \)[/tex] increases, but never touch it. For negative values of [tex]\( x \)[/tex], the function increases rapidly.
### Step 5: Add Labels and Title
- Label the x-axis and y-axis appropriately.
- Title the graph: "Graph of [tex]\( w(x) = 2 \cdot (0.4)^x - 5 \)[/tex]"
### Summary
By completing these steps, you will have a clear graph representing the function [tex]\( w(x) = 2 \cdot (0.4)^x - 5 \)[/tex], which displays its exponential decay nature and vertical displacement.
### Example Graph (Not to scale):
```
y
^
|
10+
|
5+
|
0+-----*------------->
| (0,-3) x
- 5+------------------
|
```
### Conclusion
This graph showcases the behavior of the exponential decay function, with a vertical shift downward by 5 units, and how it approaches the horizontal asymptote at [tex]\( y = -5 \)[/tex].
### Step 1: Understand the Function
The given function is an exponential decay function because the base of the exponent [tex]\(0.4\)[/tex] is between 0 and 1. Here is the breakdown of the function:
- [tex]\( 2 \)[/tex] is the coefficient that vertically stretches the graph.
- [tex]\( 0.4 \)[/tex] is the base of the exponent, causing decay.
- [tex]\( x \)[/tex] is the variable exponent.
- [tex]\( -5 \)[/tex] is a vertical shift downward by 5 units.
### Step 2: Plot Key Points
To graph the function, let's compute some key points.
1. When [tex]\( x = 0 \)[/tex]:
[tex]\[ w(0) = 2 \cdot (0.4)^0 - 5 = 2 \cdot 1 - 5 = 2 - 5 = -3 \][/tex]
Point: [tex]\((0, -3)\)[/tex]
2. When [tex]\( x = 1 \)[/tex]:
[tex]\[ w(1) = 2 \cdot (0.4)^1 - 5 = 2 \cdot 0.4 - 5 = 0.8 - 5 = -4.2 \][/tex]
Point: [tex]\((1, -4.2)\)[/tex]
3. When [tex]\( x = -1 \)[/tex]:
[tex]\[ w(-1) = 2 \cdot (0.4)^{-1} - 5 = 2 \cdot \frac{1}{0.4} - 5 = 2 \cdot 2.5 - 5 = 5 - 5 = 0 \][/tex]
Point: [tex]\((-1, 0)\)[/tex]
4. When [tex]\( x = 2 \)[/tex]:
[tex]\[ w(2) = 2 \cdot (0.4)^2 - 5 = 2 \cdot 0.16 - 5 = 0.32 - 5 = -4.68 \][/tex]
Point: [tex]\((2, -4.68)\)[/tex]
5. When [tex]\( x = -2 \)[/tex]:
[tex]\[ w(-2) = 2 \cdot (0.4)^{-2} - 5 = 2 \cdot \left(\frac{1}{0.4}\right)^2 - 5 = 2 \cdot 6.25 - 5 = 12.5 - 5 = 7.5 \][/tex]
Point: [tex]\((-2, 7.5)\)[/tex]
### Step 3: Additional Points
You can calculate more points if needed to provide better accuracy. Here are additional common values:
#### When [tex]\( x = -3 \)[/tex]:
[tex]\[ w(-3) = 2 \cdot (0.4)^{-3} - 5 = 2 \cdot \left(\frac{1}{0.4}\right)^3 - 5 = 2 \cdot 15.625 - 5 = 31.25 - 5 = 26.25 \][/tex]
Point: [tex]\((-3, 26.25)\)[/tex]
#### When [tex]\( x = 3 \)[/tex]:
[tex]\[ w(3) = 2 \cdot (0.4)^3 - 5 = 2 \cdot 0.064 - 5 = 0.128 - 5 = -4.872 \][/tex]
Point: [tex]\((3, -4.872)\)[/tex]
### Step 4: Plot the Graph
1. Plot the calculated points on a coordinate plane:
- [tex]\((0, -3)\)[/tex]
- [tex]\((1, -4.2)\)[/tex]
- [tex]\((-1, 0)\)[/tex]
- [tex]\((2, -4.68)\)[/tex]
- [tex]\((-2, 7.5)\)[/tex]
- [tex]\((-3, 26.25)\)[/tex]
- [tex]\((3, -4.872)\)[/tex]
2. Draw a smooth curve through these points, keeping in mind the overall shape of an exponential decay function: the graph should approach the horizontal line [tex]\( y = -5 \)[/tex] as [tex]\( x \)[/tex] increases, but never touch it. For negative values of [tex]\( x \)[/tex], the function increases rapidly.
### Step 5: Add Labels and Title
- Label the x-axis and y-axis appropriately.
- Title the graph: "Graph of [tex]\( w(x) = 2 \cdot (0.4)^x - 5 \)[/tex]"
### Summary
By completing these steps, you will have a clear graph representing the function [tex]\( w(x) = 2 \cdot (0.4)^x - 5 \)[/tex], which displays its exponential decay nature and vertical displacement.
### Example Graph (Not to scale):
```
y
^
|
10+
|
5+
|
0+-----*------------->
| (0,-3) x
- 5+------------------
|
```
### Conclusion
This graph showcases the behavior of the exponential decay function, with a vertical shift downward by 5 units, and how it approaches the horizontal asymptote at [tex]\( y = -5 \)[/tex].
