Question

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?

Answer 1

To compare the rate of change between years 9 and 10 for both investment options, we first need to evaluate the functions \( p(x) \) and \( q(x) \) at those specific years.

### Option #1: \( p(x) = 5,000(1.06)^x \)
Calculate \( p(9) \) and \( p(10) \):

1. Calculate \( p(9) \):
\[
p(9) = 5,000(1.06)^9
\]
Using the calculator:
\[
(1.06)^9 \approx 1.689478
\]
So,
\[
p(9) \approx 5,000 \times 1.689478 \approx 8,447.39
\]

2. Calculate \( p(10) \):
\[
p(10) = 5,000(1.06)^{10}
\]
Using the calculator:
\[
(1.06)^{10} \approx 1.790847
\]
So,
\[
p(10) \approx 5,000 \times 1.790847 \approx 8,954.24
\]

3. Calculate the rate of change from year 9 to year 10:
\[
\text{Rate of change for } p = p(10) - p(9) \approx 8,954.24 - 8,447.39 \approx 506.85
\]

### Option #2: \( q(x) = 500x + 5,000 \)
Calculate \( q(9) \) and \( q(10) \):

1. Calculate \( q(9) \):
\[
q(9) = 500 \times 9 + 5,000 = 4,500 + 5,000 = 9,500
\]

2. Calculate \( q(10) \):
\[
q(10) = 500 \times 10 + 5,000 = 5,000 + 5,000 = 10,000
\]

3. Calculate the rate of change from year 9 to year 10:
\[
\text{Rate of change for } q = q(10) - q(9) = 10,000 - 9,500 = 500
\]

### Comparing the Rates of Change
- Rate of change for Option #1 (\( p \)) = 506.85
- Rate of change for Option #2 (\( q \)) = 500

### Conclusion
Between years 9 and 10, Option #1 has a greater rate of change (506.85) compared to Option #2 (500).

#### Recommendation:
Miguel should choose **Option #1**, as it offers a higher rate of growth for his investment.