Question
Leticia invests $200 at 5% interest. If y represents the amount of money after x time periods, which describes the graph of the exponential function relating time and money?
The initial value of the graph is 200. The graph increases by a factor of 1.05 per 1 unit increase in time.
The initial value of the graph is 200. The graph increases by a factor of 5 per 1 unit increase in time.
The initial value of the graph is 500. The graph increases by a factor of 2 per 1 unit increase in time.
The initial value of the graph is 500. The graph increases by a factor of 1.02 per 1 unit increase in time.
The initial value of the graph is 200. The graph increases by a factor of 1.05 per 1 unit increase in time.
The initial value of the graph is 200. The graph increases by a factor of 5 per 1 unit increase in time.
The initial value of the graph is 500. The graph increases by a factor of 2 per 1 unit increase in time.
The initial value of the graph is 500. The graph increases by a factor of 1.02 per 1 unit increase in time.
Answers
GPT-4o mini
To determine the correct description of the exponential function relating time and money for Leticia's investment, we first identify the parameters of exponential growth formula, which is generally expressed as:
\[ y = a \cdot (1 + r)^x \]
Where:
- \( y \) represents the amount of money after \( x \) time periods,
- \( a \) is the initial amount (the principal),
- \( r \) is the interest rate (in decimal form),
- \( x \) is the number of time periods.
Given the problem:
- Leticia invests $200, which becomes \( a = 200 \).
- The interest rate is 5%, so \( r = 0.05 \).
Putting these values into the formula:
1. The initial value \( a = 200 \).
2. The amount increases by a factor of \( 1 + r = 1 + 0.05 = 1.05 \).
Thus, the function relating time and money is:
\[ y = 200 \cdot (1.05)^x \]
Now, examining the options provided:
1. **The initial value of the graph is 200. The graph increases by a factor of 1.05 per 1 unit increase in time.** - This is consistent with our analysis.
2. The initial value of the graph is 200. The graph increases by a factor of 5 per 1 unit increase in time. - This is incorrect as the factor is 1.05, not 5.
3. The initial value of the graph is 500. The graph increases by a factor of 2 per 1 unit increase in time. - This is incorrect on both initial value and growth factor.
4. The initial value of the graph is 500. The graph increases by a factor of 1.02 per 1 unit increase in time. - This is incorrect as the initial value is not 500 and the growth factor is not 1.02.
Thus, the correct description is:
**The initial value of the graph is 200. The graph increases by a factor of 1.05 per 1 unit increase in time.**
\[ y = a \cdot (1 + r)^x \]
Where:
- \( y \) represents the amount of money after \( x \) time periods,
- \( a \) is the initial amount (the principal),
- \( r \) is the interest rate (in decimal form),
- \( x \) is the number of time periods.
Given the problem:
- Leticia invests $200, which becomes \( a = 200 \).
- The interest rate is 5%, so \( r = 0.05 \).
Putting these values into the formula:
1. The initial value \( a = 200 \).
2. The amount increases by a factor of \( 1 + r = 1 + 0.05 = 1.05 \).
Thus, the function relating time and money is:
\[ y = 200 \cdot (1.05)^x \]
Now, examining the options provided:
1. **The initial value of the graph is 200. The graph increases by a factor of 1.05 per 1 unit increase in time.** - This is consistent with our analysis.
2. The initial value of the graph is 200. The graph increases by a factor of 5 per 1 unit increase in time. - This is incorrect as the factor is 1.05, not 5.
3. The initial value of the graph is 500. The graph increases by a factor of 2 per 1 unit increase in time. - This is incorrect on both initial value and growth factor.
4. The initial value of the graph is 500. The graph increases by a factor of 1.02 per 1 unit increase in time. - This is incorrect as the initial value is not 500 and the growth factor is not 1.02.
Thus, the correct description is:
**The initial value of the graph is 200. The graph increases by a factor of 1.05 per 1 unit increase in time.**