Answer :
To graph the function [tex]\( f(x) = 2^{x-1} \)[/tex], we will plot several key points and then draw a smooth curve through those points. Let's plot the points at [tex]\( x = -2, 0, 2, \)[/tex] and [tex]\( 4 \)[/tex].
### Step-by-Step Solution:
1. Identify Key Points:
- For [tex]\( x = -2 \)[/tex]:
[tex]\[ f(-2) = 2^{-2 - 1} = 2^{-3} = \frac{1}{2^3} = \frac{1}{8} = 0.125 \][/tex]
Therefore, the point is [tex]\((-2, 0.125)\)[/tex].
- For [tex]\( x = 0 \)[/tex]:
[tex]\[ f(0) = 2^{0 - 1} = 2^{-1} = \frac{1}{2} = 0.5 \][/tex]
Therefore, the point is [tex]\((0, 0.5)\)[/tex].
- For [tex]\( x = 2 \)[/tex]:
[tex]\[ f(2) = 2^{2 - 1} = 2^{1} = 2 \][/tex]
Therefore, the point is [tex]\((2, 2)\)[/tex].
- For [tex]\( x = 4 \)[/tex]:
[tex]\[ f(4) = 2^{4 - 1} = 2^{3} = 8 \][/tex]
Therefore, the point is [tex]\((4, 8)\)[/tex].
2. Plot the Key Points:
Plot the points [tex]\((-2, 0.125)\)[/tex], [tex]\((0, 0.5)\)[/tex], [tex]\((2, 2)\)[/tex], and [tex]\((4, 8)\)[/tex] on a coordinate plane.
3. Draw the Curve:
After plotting these points, you can visually ascertain the behavior of the exponential function.
- Recognize that for negative values of [tex]\(x\)[/tex], the function [tex]\( f(x) = 2^{x-1} \)[/tex] approaches zero but never quite reaches it (as [tex]\( x \to -\infty \)[/tex]).
- The function increases rapidly for positive values of [tex]\( x \)[/tex].
Using these points, draw a smooth, continuous curve that passes through each of these plotted points. The curve should show an increasing trend, steeply rising as [tex]\( x \)[/tex] becomes larger.
Graph:
Here is a simple schematic for how the graph looks:
```
9 |
8 | • (4,8)
7 |
6 |
5 |
4 |
3 |
2 | • (2,2)
1 |
0.5 | • (0,0.5)
0.125|• (-2,0.125)
0 +-----------------------------------------------------------
-3 -2 -1 0 1 2 3 4 5
```
Ensure your paper graph shows a smooth exponential curve passing accurately through these key points.
### Step-by-Step Solution:
1. Identify Key Points:
- For [tex]\( x = -2 \)[/tex]:
[tex]\[ f(-2) = 2^{-2 - 1} = 2^{-3} = \frac{1}{2^3} = \frac{1}{8} = 0.125 \][/tex]
Therefore, the point is [tex]\((-2, 0.125)\)[/tex].
- For [tex]\( x = 0 \)[/tex]:
[tex]\[ f(0) = 2^{0 - 1} = 2^{-1} = \frac{1}{2} = 0.5 \][/tex]
Therefore, the point is [tex]\((0, 0.5)\)[/tex].
- For [tex]\( x = 2 \)[/tex]:
[tex]\[ f(2) = 2^{2 - 1} = 2^{1} = 2 \][/tex]
Therefore, the point is [tex]\((2, 2)\)[/tex].
- For [tex]\( x = 4 \)[/tex]:
[tex]\[ f(4) = 2^{4 - 1} = 2^{3} = 8 \][/tex]
Therefore, the point is [tex]\((4, 8)\)[/tex].
2. Plot the Key Points:
Plot the points [tex]\((-2, 0.125)\)[/tex], [tex]\((0, 0.5)\)[/tex], [tex]\((2, 2)\)[/tex], and [tex]\((4, 8)\)[/tex] on a coordinate plane.
3. Draw the Curve:
After plotting these points, you can visually ascertain the behavior of the exponential function.
- Recognize that for negative values of [tex]\(x\)[/tex], the function [tex]\( f(x) = 2^{x-1} \)[/tex] approaches zero but never quite reaches it (as [tex]\( x \to -\infty \)[/tex]).
- The function increases rapidly for positive values of [tex]\( x \)[/tex].
Using these points, draw a smooth, continuous curve that passes through each of these plotted points. The curve should show an increasing trend, steeply rising as [tex]\( x \)[/tex] becomes larger.
Graph:
Here is a simple schematic for how the graph looks:
```
9 |
8 | • (4,8)
7 |
6 |
5 |
4 |
3 |
2 | • (2,2)
1 |
0.5 | • (0,0.5)
0.125|• (-2,0.125)
0 +-----------------------------------------------------------
-3 -2 -1 0 1 2 3 4 5
```
Ensure your paper graph shows a smooth exponential curve passing accurately through these key points.