Westonci.ca is the best place to get answers to your questions, provided by a community of experienced and knowledgeable experts. Get quick and reliable solutions to your questions from a community of seasoned experts on our user-friendly platform. Explore comprehensive solutions to your questions from knowledgeable professionals across various fields on our platform.
Sagot :
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]
We appreciate your visit. Our platform is always here to offer accurate and reliable answers. Return anytime. We appreciate your visit. Our platform is always here to offer accurate and reliable answers. Return anytime. Discover more at Westonci.ca. Return for the latest expert answers and updates on various topics.
