Answer :
Certainly! To solve the given problem, we will break it down comprehensively.
First, understand the problem:
1. The exponential function [tex]\( p(x) \)[/tex] increases at a rate of 25%. This implies that the function has a base growth factor greater than 1.
2. The function passes through the ordered pair (0, 10), which means that [tex]\( p(0) = 10 \)[/tex].
3. The function is shifted down by 5 units.
Step-by-step solution:
### Step 1: Define the base exponential function
Given that [tex]\( p(x) \)[/tex] increases at a rate of 25%, we recognize the exponential function will be of the form:
[tex]\[ p(x) = A \cdot e^{kx} \][/tex]
where [tex]\( e \)[/tex] is the base of the natural logarithm, and [tex]\( k \)[/tex] is the rate constant.
### Step 2: Determine the growth rate
A 25% growth rate per interval can be mathematically expressed with the factor:
[tex]\[ e^{k} = 1.25 \][/tex]
### Step 3: Express k
To express [tex]\( k \)[/tex], we take the natural logarithm of both sides:
[tex]\[ k = \ln(1.25) \][/tex]
### Step 4: Apply the initial condition
The condition [tex]\( p(0) = 10 \)[/tex] provides:
[tex]\[ p(x) = 10 \cdot e^{kx} \][/tex]
### Step 5: Shifting the function downward by 5 units
The function is shifted down by 5 units, resulting in:
[tex]\[ p(x) = 10 \cdot e^{kx} - 5 \][/tex]
### Step 6: Understand the horizontal asymptote
When [tex]\( x \)[/tex] approaches [tex]\(\infty\)[/tex], the term [tex]\( 10 \cdot e^{kx} \)[/tex] grows very large, and the function [tex]\( p(x) \approx \infty - 5\)[/tex] increases indefinitely. Therefore, the vertical shift downward by 5 indicates the function’s horizontal asymptote, which is [tex]\(-5\)[/tex]. This leads to two important observations.
### Step 7: Identify intervals
Given the nature of exponential growth and the shift:
- [tex]\( p(x) \)[/tex] will become very large as [tex]\( x \to \infty \)[/tex].
- That makes [tex]\( p(x) \)[/tex] greater than some constant for large positive [tex]\( x \)[/tex].
Comparing the options provided:
- [tex]\( (5, \infty) \)[/tex]
- [tex]\( (-\infty, 5) \)[/tex]
- [tex]\( (-\infty,-5) \)[/tex]
- [tex]\( (-5, \infty) \)[/tex]
### Step 8: Match the intervals to the function behavior:
The exponential function [tex]\( p(x) \)[/tex] with a downward shift of 5 will mean for [tex]\( x \)[/tex] approaching positive values, [tex]\( p(x) \)[/tex] will continue to grow from the horizontal asymptote [tex]\(-5 \)[/tex] upwards.
### Step 9: Conclusion
For exponential growth (positively):
The domain of our function as [tex]\( p(x) \to \infty \)[/tex] is captured by the interval:
[tex]\[ (-5, \infty) \][/tex]
Therefore, the appropriate interval for the exponential function given the conditions is:
[tex]\[ \boxed{(-5, \infty)} \][/tex]
First, understand the problem:
1. The exponential function [tex]\( p(x) \)[/tex] increases at a rate of 25%. This implies that the function has a base growth factor greater than 1.
2. The function passes through the ordered pair (0, 10), which means that [tex]\( p(0) = 10 \)[/tex].
3. The function is shifted down by 5 units.
Step-by-step solution:
### Step 1: Define the base exponential function
Given that [tex]\( p(x) \)[/tex] increases at a rate of 25%, we recognize the exponential function will be of the form:
[tex]\[ p(x) = A \cdot e^{kx} \][/tex]
where [tex]\( e \)[/tex] is the base of the natural logarithm, and [tex]\( k \)[/tex] is the rate constant.
### Step 2: Determine the growth rate
A 25% growth rate per interval can be mathematically expressed with the factor:
[tex]\[ e^{k} = 1.25 \][/tex]
### Step 3: Express k
To express [tex]\( k \)[/tex], we take the natural logarithm of both sides:
[tex]\[ k = \ln(1.25) \][/tex]
### Step 4: Apply the initial condition
The condition [tex]\( p(0) = 10 \)[/tex] provides:
[tex]\[ p(x) = 10 \cdot e^{kx} \][/tex]
### Step 5: Shifting the function downward by 5 units
The function is shifted down by 5 units, resulting in:
[tex]\[ p(x) = 10 \cdot e^{kx} - 5 \][/tex]
### Step 6: Understand the horizontal asymptote
When [tex]\( x \)[/tex] approaches [tex]\(\infty\)[/tex], the term [tex]\( 10 \cdot e^{kx} \)[/tex] grows very large, and the function [tex]\( p(x) \approx \infty - 5\)[/tex] increases indefinitely. Therefore, the vertical shift downward by 5 indicates the function’s horizontal asymptote, which is [tex]\(-5\)[/tex]. This leads to two important observations.
### Step 7: Identify intervals
Given the nature of exponential growth and the shift:
- [tex]\( p(x) \)[/tex] will become very large as [tex]\( x \to \infty \)[/tex].
- That makes [tex]\( p(x) \)[/tex] greater than some constant for large positive [tex]\( x \)[/tex].
Comparing the options provided:
- [tex]\( (5, \infty) \)[/tex]
- [tex]\( (-\infty, 5) \)[/tex]
- [tex]\( (-\infty,-5) \)[/tex]
- [tex]\( (-5, \infty) \)[/tex]
### Step 8: Match the intervals to the function behavior:
The exponential function [tex]\( p(x) \)[/tex] with a downward shift of 5 will mean for [tex]\( x \)[/tex] approaching positive values, [tex]\( p(x) \)[/tex] will continue to grow from the horizontal asymptote [tex]\(-5 \)[/tex] upwards.
### Step 9: Conclusion
For exponential growth (positively):
The domain of our function as [tex]\( p(x) \to \infty \)[/tex] is captured by the interval:
[tex]\[ (-5, \infty) \][/tex]
Therefore, the appropriate interval for the exponential function given the conditions is:
[tex]\[ \boxed{(-5, \infty)} \][/tex]
