Question 3
Chosen Amount for Carly's Babysitting:
Let's assume Carly earns $10 each week for babysitting.
Explanation of Linear Change:
Carly’s savings represent a linear change because she is saving a fixed amount each week. This means that as time progresses (in weeks), her savings increase by a consistent amount ($10) at regular intervals (each week).
Sequence of Savings Over 20 Weeks:
The sequence of her savings over 20 weeks can be described as:
- Week 1: $10
- Week 2: $20
- Week 3: $30
- Week 4: $40
- ...
- Week 20: $200
This can be expressed as:
\[ S(n) = 10n \]
where \( S(n) \) is the savings at week \( n \).
Graphing Carly's Savings:
On graph paper, plot the number of weeks (x-axis) against the savings (y-axis).
Key Feature Indicating Linear Change:
The graph will show a straight line, indicating a constant rate of increase.
Question 4
Chosen Amount for Nico's Savings Account:
Let’s assume Nico's grandparents put in $1000 when he was born.
Explanation of Exponential Growth:
The amount in the savings account grows exponentially because the account earns 8% interest annually on the total amount, meaning that the growth each year is a percentage of the current amount, not a fixed amount. Thus, the growth compounds over time.
Sequence of Savings Over 20 Years:
Using compound interest, the amount in the account can be described as follows:
- Year 1: $1000 * (1 + 0.08) = $1080
- Year 2: $1000 * (1 + 0.08)² = $1166.40
- Year 3: $1000 * (1 + 0.08)³ = $1259.71
- ...
- Year 20: $1000 * (1 + 0.08)²⁰ ≈ $4660.67
This can be expressed as:
\[ A(t) = 1000(1.08)^t \]
where \( A(t) \) is the amount in the account after \( t \) years.
Graphing Nico’s Savings:
On graph paper, plot the number of years (x-axis) against the savings (y-axis).
Key Feature Indicating Exponential Growth:
The graph will show a curve that starts off slow but increases rapidly, indicating accelerating growth.
Question 5
Chosen Amount for Aria's Car:
Let’s assume Aria buys her new car for $20,000.
Explanation of Exponential Decay:
The value of Aria’s car depreciates by 15% each year. This means that each year, the car's value is multiplied by a constant factor (0.85), which results in exponential decay since the amount lost each year increases in absolute terms as the car's value decreases.
Sequence of Car Value Over 10 Years:
- Year 1: $20000 * 0.85 = $17,000
- Year 2: $20000 * 0.85² = $14,450
- Year 3: $20000 * 0.85³ = $12,282.50
- ...
- Year 10: $20000 * 0.85^{10} ≈ $3,415.24
This can be expressed as:
\[ V(t) = 20000(0.85)^t \]
where \( V(t) \) is the value of the car after \( t \) years.
Graphing Aria’s Car Value:
On graph paper, plot the number of years (x-axis) against the car's value (y-axis).
Key Feature Indicating Exponential Decay:
The graph will show a curve that decreases rapidly at first and then level off, indicating decreasing value over time.
Question 6
Linear vs. Exponential Growth:
- Linear functions grow at a constant rate. Regardless of the consistent increment, the growth remains proportional and creates a straight line when graphed.
- Exponential functions, however, increase at an accelerating rate. The growth increases multiplicatively, meaning larger increments occur as the value grows.
Statement Explanation:
The statement “No matter what the growth rate of a linear function is, that of an exponential function will always exceed it” implies that over time, the effects of compounding in exponential growth will outpace the fixed increments of linear growth.
Demonstration of Statement Using Carly and Nico’s Functions:
- In Carly’s case, if she saves $10 weekly, after 20 weeks, she’ll have $200.
- In Nico’s case, with an initial deposit of $1000 growing at 8%, he’ll have approximately $4660.67 after 20 years.
This stark difference between $200 and $4660.67 illustrates how exponential growth can quickly surpass linear growth, even over shorter time frames in the context of different applications. The final amounts clearly demonstrate that Nico’s savings grow faster than Carly’s linear savings trajectory.