Answer :
Sure, let's examine the function [tex]\( g(x) = 2^{x-1} + 3 \)[/tex] in detail in order to find its graph:
1. Understanding the Function:
This function involves an exponential term [tex]\( 2^{x-1} \)[/tex] and a vertical shift of 3 units upward.
2. Key Points Calculation:
To understand the shape of the graph, it's useful to calculate several key points.
- When [tex]\( x = 0 \)[/tex]:
[tex]\[ g(0) = 2^{0-1} + 3 = 2^{-1} + 3 = \frac{1}{2} + 3 = 3.5 \][/tex]
So, [tex]\( (0, 3.5) \)[/tex] is a point on the graph.
- When [tex]\( x = 1 \)[/tex]:
[tex]\[ g(1) = 2^{1-1} + 3 = 2^{0} + 3 = 1 + 3 = 4 \][/tex]
So, [tex]\( (1, 4) \)[/tex] is a point on the graph.
- When [tex]\( x = 2 \)[/tex]:
[tex]\[ g(2) = 2^{2-1} + 3 = 2^{1} + 3 = 2 + 3 = 5 \][/tex]
So, [tex]\( (2, 5) \)[/tex] is a point on the graph.
3. Behavior as [tex]\( x \)[/tex] Changes:
- As [tex]\( x \to \infty \)[/tex], [tex]\( 2^{x-1} \)[/tex] grows exponentially large, making [tex]\( g(x) \)[/tex] grow without bound.
- As [tex]\( x \to -\infty \)[/tex], [tex]\( 2^{x-1} \)[/tex] approaches 0 (but never quite reaches it), meaning [tex]\( g(x) \)[/tex] approaches 3 from above. Thus, [tex]\( y = 3 \)[/tex] is a horizontal asymptote.
4. Plotting the Points & Graph:
Plot the points (0, 3.5), (1, 4), and (2, 5). Since [tex]\( g(x) \)[/tex] is an exponential function shifted upwards, its shape will be that of a typical exponential growth curve:
- Starting just above the asymptote [tex]\( y = 3 \)[/tex] as [tex]\( x \to -\infty \)[/tex]
- Increasing rapidly as [tex]\( x \)[/tex] increases, through the points calculated.
5. Graph Description:
The graph will look like an upward-sloping curve starting close to [tex]\( y = 3 \)[/tex] for very negative [tex]\( x \)[/tex], passing through the calculated points, and then rising steeply for positive [tex]\( x \)[/tex].
This is how we derive the graph of [tex]\( g(x) = 2^{x-1} + 3 \)[/tex].
1. Understanding the Function:
This function involves an exponential term [tex]\( 2^{x-1} \)[/tex] and a vertical shift of 3 units upward.
2. Key Points Calculation:
To understand the shape of the graph, it's useful to calculate several key points.
- When [tex]\( x = 0 \)[/tex]:
[tex]\[ g(0) = 2^{0-1} + 3 = 2^{-1} + 3 = \frac{1}{2} + 3 = 3.5 \][/tex]
So, [tex]\( (0, 3.5) \)[/tex] is a point on the graph.
- When [tex]\( x = 1 \)[/tex]:
[tex]\[ g(1) = 2^{1-1} + 3 = 2^{0} + 3 = 1 + 3 = 4 \][/tex]
So, [tex]\( (1, 4) \)[/tex] is a point on the graph.
- When [tex]\( x = 2 \)[/tex]:
[tex]\[ g(2) = 2^{2-1} + 3 = 2^{1} + 3 = 2 + 3 = 5 \][/tex]
So, [tex]\( (2, 5) \)[/tex] is a point on the graph.
3. Behavior as [tex]\( x \)[/tex] Changes:
- As [tex]\( x \to \infty \)[/tex], [tex]\( 2^{x-1} \)[/tex] grows exponentially large, making [tex]\( g(x) \)[/tex] grow without bound.
- As [tex]\( x \to -\infty \)[/tex], [tex]\( 2^{x-1} \)[/tex] approaches 0 (but never quite reaches it), meaning [tex]\( g(x) \)[/tex] approaches 3 from above. Thus, [tex]\( y = 3 \)[/tex] is a horizontal asymptote.
4. Plotting the Points & Graph:
Plot the points (0, 3.5), (1, 4), and (2, 5). Since [tex]\( g(x) \)[/tex] is an exponential function shifted upwards, its shape will be that of a typical exponential growth curve:
- Starting just above the asymptote [tex]\( y = 3 \)[/tex] as [tex]\( x \to -\infty \)[/tex]
- Increasing rapidly as [tex]\( x \)[/tex] increases, through the points calculated.
5. Graph Description:
The graph will look like an upward-sloping curve starting close to [tex]\( y = 3 \)[/tex] for very negative [tex]\( x \)[/tex], passing through the calculated points, and then rising steeply for positive [tex]\( x \)[/tex].
This is how we derive the graph of [tex]\( g(x) = 2^{x-1} + 3 \)[/tex].
