To represent Kelly's savings account situation:
Table Representation
We will calculate the amount in the savings account after each specified month:
- Initial Amount: $30
- Monthly Addition: $15
| Time in Months | Dollars in the Savings Account | |----------------|-------------------------------| | 0 | 30 | | 1 | 30 + 15 × 1 = 45 | | 2 | 30 + 15 × 2 = 60 | | 3 | 30 + 15 × 3 = 75 |
Completed Table
| Time in Months | Dollars in the Savings Account | |----------------|-------------------------------| | 0 | 30 | | 1 | 45 | | 2 | 60 | | 3 | 75 |
Points to Plot
The points to plot on the coordinate plane are:
- (0, 30)
- (1, 45)
- (2, 60)
- (3, 75)
Equation Representation
To formulate the equation of the linear function, we will use the slope-intercept form of a linear equation:
\[ y = mx + b \]
Where:
- \( m \) is the rate of change (slope).
- \( b \) is the initial amount (y-intercept).
Finding the Rate of Change (m):
- The monthly increase is $15. Therefore,
\[ m = 15 \]
Finding the Initial Amount (b):
- The amount when \( t = 0 \) is $30.
Thus, the equation becomes:
\[ y = 15t + 30 \]
Summary
- Rate of Change: $15 per month
- Initial Amount: $30
- Equation: \( y = 15t + 30 \)
Kelly's savings can be represented as a table, plotted as points on a graph, and expressed with this linear equation.