Question
Given that graph q(x) is linear and p(x) is exponential solve the problem
Juan is planning to invest $5000 With option 1 , his money will grow according to the function p(x) =5,000*1.06^x. With option 2 , his money will grow according to the function q(x)= 509x+5,000. He decides to graph both functions. What does he observe from the graph
For the first 10 years, Juan’s money will grow faster with option ___ but after that it will grow faster with option ___
Juan is planning to invest $5000 With option 1 , his money will grow according to the function p(x) =5,000*1.06^x. With option 2 , his money will grow according to the function q(x)= 509x+5,000. He decides to graph both functions. What does he observe from the graph
For the first 10 years, Juan’s money will grow faster with option ___ but after that it will grow faster with option ___
Answers
GPT-4o mini
To determine which investment option grows faster for Juan over time, we can analyze the two functions given:
1. **Exponential Growth (Option 1)**:
\( p(x) = 5000 \cdot 1.06^x \)
2. **Linear Growth (Option 2)**:
\( q(x) = 509x + 5000 \)
### Step 1: Calculate the Values for Both Options
We can compute the values of \( p(x) \) and \( q(x) \) for the first ten years (from \( x = 0 \) to \( x = 10 \)).
#### For Option 1 (Exponential Growth):
- \( p(0) = 5000 \cdot 1.06^0 = 5000 \)
- \( p(1) = 5000 \cdot 1.06^1 = 5000 \cdot 1.06 \approx 5300 \)
- \( p(2) = 5000 \cdot 1.06^2 \approx 5618 \)
- \( p(3) = 5000 \cdot 1.06^3 \approx 5951 \)
- \( p(4) = 5000 \cdot 1.06^4 \approx 6308 \)
- \( p(5) = 5000 \cdot 1.06^5 \approx 6681 \)
- \( p(6) = 5000 \cdot 1.06^6 \approx 7079 \)
- \( p(7) = 5000 \cdot 1.06^7 \approx 7495 \)
- \( p(8) = 5000 \cdot 1.06^8 \approx 7930 \)
- \( p(9) = 5000 \cdot 1.06^9 \approx 8386 \)
- \( p(10) = 5000 \cdot 1.06^{10} \approx 8862 \)
#### For Option 2 (Linear Growth):
- \( q(0) = 509 \cdot 0 + 5000 = 5000 \)
- \( q(1) = 509 \cdot 1 + 5000 = 5509 \)
- \( q(2) = 509 \cdot 2 + 5000 = 6018 \)
- \( q(3) = 509 \cdot 3 + 5000 = 6527 \)
- \( q(4) = 509 \cdot 4 + 5000 = 7036 \)
- \( q(5) = 509 \cdot 5 + 5000 = 7545 \)
- \( q(6) = 509 \cdot 6 + 5000 = 8054 \)
- \( q(7) = 509 \cdot 7 + 5000 = 8563 \)
- \( q(8) = 509 \cdot 8 + 5000 = 9072 \)
- \( q(9) = 509 \cdot 9 + 5000 = 9581 \)
- \( q(10) = 509 \cdot 10 + 5000 = 10090 \)
### Step 2: Compare Values for \( x=0 \) to \( x=10 \)
Now we can summarize the computations:
- **Year 0**: \( p(0) = 5000 \) and \( q(0) = 5000 \) (equal)
- **Year 1**: \( p(1) \approx 5300 \) and \( q(1) = 5509 \) (option 2 is larger)
- **Year 2**: \( p(2) \approx 5618 \) and \( q(2) = 6018 \) (option 2 is larger)
- **Year 3**: \( p(3) \approx 5951 \) and \( q(3) = 6527 \) (option 2 is larger)
- **Year 4**: \( p(4) \approx 6308 \) and \( q(4) = 7036 \) (option 2 is larger)
- **Year 5**: \( p(5) \approx 6681 \) and \( q(5) = 7545 \) (option 2 is larger)
- **Year 6**: \( p(6) \approx 7079 \) and \( q(6) = 8054 \) (option 2 is larger)
- **Year 7**: \( p(7) \approx 7495 \) and \( q(7) = 8563 \) (option 2 is larger)
- **Year 8**: \( p(8) \approx 7930 \) and \( q(8) = 9072 \) (option 2 is larger)
- **Year 9**: \( p(9) \approx 8386 \) and \( q(9) = 9581 \) (option 2 is larger)
- **Year 10**: \( p(10) \approx 8862 \) and \( q(10) = 10090 \) (option 2 is larger)
### Conclusion
Based on this analysis, for the first **10 years**, Juan’s money grows faster with **option 2** (linear growth).
To find out when option 1 begins to exceed option 2, we can set the two equations equal:
\( 5000 \cdot 1.06^x = 509x + 5000 \)
This equation does not lend itself easily to algebraic manipulation, so numerical methods or graphing could be used to solve for \( x \) beyond 10 years. However, given that option 1 is exponential and option 2 is linear, option 1 will eventually grow faster for a larger value of \( x \).
Thus, the final observation:
- **For the first 10 years**: Juan’s money will grow faster with option **2**.
- **After that**: Juan’s money will grow faster with option **1** (exponential).
1. **Exponential Growth (Option 1)**:
\( p(x) = 5000 \cdot 1.06^x \)
2. **Linear Growth (Option 2)**:
\( q(x) = 509x + 5000 \)
### Step 1: Calculate the Values for Both Options
We can compute the values of \( p(x) \) and \( q(x) \) for the first ten years (from \( x = 0 \) to \( x = 10 \)).
#### For Option 1 (Exponential Growth):
- \( p(0) = 5000 \cdot 1.06^0 = 5000 \)
- \( p(1) = 5000 \cdot 1.06^1 = 5000 \cdot 1.06 \approx 5300 \)
- \( p(2) = 5000 \cdot 1.06^2 \approx 5618 \)
- \( p(3) = 5000 \cdot 1.06^3 \approx 5951 \)
- \( p(4) = 5000 \cdot 1.06^4 \approx 6308 \)
- \( p(5) = 5000 \cdot 1.06^5 \approx 6681 \)
- \( p(6) = 5000 \cdot 1.06^6 \approx 7079 \)
- \( p(7) = 5000 \cdot 1.06^7 \approx 7495 \)
- \( p(8) = 5000 \cdot 1.06^8 \approx 7930 \)
- \( p(9) = 5000 \cdot 1.06^9 \approx 8386 \)
- \( p(10) = 5000 \cdot 1.06^{10} \approx 8862 \)
#### For Option 2 (Linear Growth):
- \( q(0) = 509 \cdot 0 + 5000 = 5000 \)
- \( q(1) = 509 \cdot 1 + 5000 = 5509 \)
- \( q(2) = 509 \cdot 2 + 5000 = 6018 \)
- \( q(3) = 509 \cdot 3 + 5000 = 6527 \)
- \( q(4) = 509 \cdot 4 + 5000 = 7036 \)
- \( q(5) = 509 \cdot 5 + 5000 = 7545 \)
- \( q(6) = 509 \cdot 6 + 5000 = 8054 \)
- \( q(7) = 509 \cdot 7 + 5000 = 8563 \)
- \( q(8) = 509 \cdot 8 + 5000 = 9072 \)
- \( q(9) = 509 \cdot 9 + 5000 = 9581 \)
- \( q(10) = 509 \cdot 10 + 5000 = 10090 \)
### Step 2: Compare Values for \( x=0 \) to \( x=10 \)
Now we can summarize the computations:
- **Year 0**: \( p(0) = 5000 \) and \( q(0) = 5000 \) (equal)
- **Year 1**: \( p(1) \approx 5300 \) and \( q(1) = 5509 \) (option 2 is larger)
- **Year 2**: \( p(2) \approx 5618 \) and \( q(2) = 6018 \) (option 2 is larger)
- **Year 3**: \( p(3) \approx 5951 \) and \( q(3) = 6527 \) (option 2 is larger)
- **Year 4**: \( p(4) \approx 6308 \) and \( q(4) = 7036 \) (option 2 is larger)
- **Year 5**: \( p(5) \approx 6681 \) and \( q(5) = 7545 \) (option 2 is larger)
- **Year 6**: \( p(6) \approx 7079 \) and \( q(6) = 8054 \) (option 2 is larger)
- **Year 7**: \( p(7) \approx 7495 \) and \( q(7) = 8563 \) (option 2 is larger)
- **Year 8**: \( p(8) \approx 7930 \) and \( q(8) = 9072 \) (option 2 is larger)
- **Year 9**: \( p(9) \approx 8386 \) and \( q(9) = 9581 \) (option 2 is larger)
- **Year 10**: \( p(10) \approx 8862 \) and \( q(10) = 10090 \) (option 2 is larger)
### Conclusion
Based on this analysis, for the first **10 years**, Juan’s money grows faster with **option 2** (linear growth).
To find out when option 1 begins to exceed option 2, we can set the two equations equal:
\( 5000 \cdot 1.06^x = 509x + 5000 \)
This equation does not lend itself easily to algebraic manipulation, so numerical methods or graphing could be used to solve for \( x \) beyond 10 years. However, given that option 1 is exponential and option 2 is linear, option 1 will eventually grow faster for a larger value of \( x \).
Thus, the final observation:
- **For the first 10 years**: Juan’s money will grow faster with option **2**.
- **After that**: Juan’s money will grow faster with option **1** (exponential).