Question
1) when the Bryant’s leave town for a vacation, they put their dog Tyco in a kennel. The kennel charges $15 for an immediate flea bath and $5 per day.
A.) write and equation in slope intercept form to represent the situation.
B.) graph the equation.
C.) explain why only one quadrant is needed to graph this situation.
A.) write and equation in slope intercept form to represent the situation.
B.) graph the equation.
C.) explain why only one quadrant is needed to graph this situation.
Answers
GPT-4o mini
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 \( y = 5x + 15 \):
1. Start with the y-intercept at \( b = 15 \). This means your graph will start at the point (0, 15) on the y-axis.
2. 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).
3. Continuing this, if \( x = 2 \), then \( y = 5(2) + 15 = 25 \). Plot the point (2, 25).
4. 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 \( y \geq 0 \).
### 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 \( y = 5x + 15 \):
1. Start with the y-intercept at \( b = 15 \). This means your graph will start at the point (0, 15) on the y-axis.
2. 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).
3. Continuing this, if \( x = 2 \), then \( y = 5(2) + 15 = 25 \). Plot the point (2, 25).
4. 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 \( y \geq 0 \).