Let's break down the problem step by step:
A.) Write an equation in slope-intercept form.
To represent the costs associated with putting Tyco in the kennel, we can define the variables:
- Let \( y \) be the total cost in dollars.
- Let \( x \) be the number of days Tyco stays at the kennel.
The kennel charges a one-time fee of $15 for the flea bath, and they also charge $5 per day.
The total cost can be expressed as:
\[ y = mx + b \]
Where:
- \( m \) is the slope (the cost per day)
- \( b \) is the y-intercept (the initial cost)
In this case:
- The slope \( m = 5 \) (the daily charge)
- The y-intercept \( b = 15 \) (the initial fee for the flea bath)
So, the equation representing the total cost is:
\[ y = 5x + 15 \]
B.) Graph the equation.
To graph the equation \( y = 5x + 15 \):
- Start with the y-intercept at \( b = 15 \). This means your graph will start at the point (0, 15) on the y-axis.
- The slope is \( 5 \), which means for every day (1 unit increase in \( x \)), the cost increases by $5 (1 unit increase in \( y \)). So from (0, 15), if \( x = 1 \), then \( y = 5(1) + 15 = 20 \). Plot the point (1, 20).
- Continuing this, if \( x = 2 \), then \( y = 5(2) + 15 = 25 \). Plot the point (2, 25).
- Draw a straight line through these points extending to the appropriate range of \( x \).
C.) Explain why only one quadrant is needed to graph this situation.
This situation only needs the first quadrant (Quadrant I) because:
- In this scenario, both variables \( x \) (number of days) and \( y \) (total cost) must be non-negative.
- The number of days \( x \) cannot be negative (you can't have a negative number of days).
- The total cost \( y \) cannot be negative (costs are always non-negative).
Therefore, both \( x \) and \( y \) will always be zero or positive, which confines the graph to the first quadrant where \( x \geq 0 \) and \( y \geq 0 \).