Question

Use the table to answer the question. (1 point) Miguel is planning to invest $5,000. He plans to withdraw his investment and earnings in approximately 10 years. 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)=500x+5,000. He creates sequences for both functions. Which option has the greater rate of change between years 9 and 10? Which option should he choose for his money? Option # 2 has the greater rate of change between years 9 and 10. If Miguel plans to invest his money for only 10 years, he should choose Option # 1

Answer 1

To determine which option has a greater rate of change between years 9 and 10, let’s analyze both functions.

**Option #1: Function p(x)**
The function for option #1 is given by:
\[ p(x) = 5,000(1.06)^x \]

To find the values at years 9 and 10:
- \( p(9) = 5,000(1.06)^9 \)
- \( p(10) = 5,000(1.06)^{10} \)

Calculating these values:
- \( p(9) = 5,000(1.06)^9 \approx 5,000(1.6895) \approx 8,447.50 \)
- \( p(10) = 5,000(1.06)^{10} \approx 5,000(1.7908) \approx 8,954.00 \)

Now calculate the rate of change between years 9 and 10 for option #1:
\[
\text{Rate of Change} = p(10) - p(9) = 8,954.00 - 8,447.50 = 506.50
\]

**Option #2: Function q(x)**
The function for option #2 is given by:
\[ q(x) = 500x + 5,000 \]

To find the values at years 9 and 10:
- \( q(9) = 500(9) + 5,000 = 4,500 + 5,000 = 9,500 \)
- \( q(10) = 500(10) + 5,000 = 5,000 + 5,000 = 10,000 \)

Now calculate the rate of change between years 9 and 10 for option #2:
\[
\text{Rate of Change} = q(10) - q(9) = 10,000 - 9,500 = 500
\]

**Comparison of Rate of Change**
- Rate of change for Option #1: 506.50
- Rate of change for Option #2: 500

Between years 9 and 10, **Option #1** has the greater rate of change (506.50 > 500).

**Conclusion:**
- The correct statement is: Option #1 has the greater rate of change between years 9 and 10.
- Since Miguel plans to withdraw after 10 years, he should choose **Option #1** as it provides more growth in that time frame.