Use the graph for both parts of the question. (The graph is a ray, so it's not all real numbers here. It starts at 0,-2.)

Question

Part 1:

Identify the domain of the function.

Part 2:

Identify the range of the function.

Answer 1

To provide accurate answers for the domain and range of the function represented by the ray that starts at the point (0, -2), I will describe the characteristics of a ray.

Part 1: Domain of the Function

Since the ray starts at the point (0, -2) and extends indefinitely in one direction, the domain will consist of all \( x \)-values that correspond to the \( y \)-values of the ray. If the ray stretches to the right (which is typical for a ray starting at a given point), the domain would be all \( x \)-values starting from \( 0 \) and going to positive infinity.

So, the domain of the function is: \[ [0, \infty) \]

Part 2: Range of the Function

To find the range, we look at the \( y \)-values that the ray attains. Since the ray starts at the point (0, -2) and extends infinitely in the upward direction, the minimum \( y \)-value is \( -2 \), and it will extend upward indefinitely.

Thus, the range of the function is: \[ [-2, \infty) \]

In summary:

• Domain: \([0, \infty)\)
• Range: \([-2, \infty)\)

Answer 2

The ray DOES end somewhere. At (5,0).

Answer 3

Thank you for the clarification! The fact that the ray has a definite endpoint at (5, 0) changes the analysis of both the domain and the range. Let's revise the answers accordingly.

Part 1: Domain of the Function

The ray starts at (0, -2) and ends at (5, 0). The domain consists of all \( x \)-values from the starting point to the endpoint of the ray, which means it covers the interval from \( 0 \) to \( 5 \).

So, the domain of the function is: \[ [0, 5] \]

Part 2: Range of the Function

The ray starts at \( y = -2 \) when \( x = 0 \) and ends at \( y = 0 \) when \( x = 5 \). The \( y \)-values increase from \( -2 \) to \( 0 \) as \( x \) goes from \( 0 \) to \( 5 \).

Thus, the range of the function is: \[ [-2, 0] \]

In summary:

• Domain: \([0, 5]\)
• Range: \([-2, 0]\)

Answer 4

Posh Pups charges $35 for a shampoo and hair cut. The salon charges an additional $10 per hour for each hour a dog stays at the salon.

Part 1:

Write an equation in Slope-Intercept Form to represent the relationship between the hours (x) the dog is at the salon and total cost (y).

Part 2:

Write the equation in Standard Form.

Answer 5

To determine the equations for the cost of taking a dog to Posh Pups, we start with the information provided.

The base cost for a shampoo and haircut is $35, and there is an additional charge of $10 per hour that the dog stays at the salon. We can use this information to write the equations.

Part 1: Slope-Intercept Form

In slope-intercept form, the equation is given as:

\[ y = mx + b \]

Where:

• \( m \) is the slope (the rate of change in cost per hour),
• \( b \) is the y-intercept (the starting cost).

For this scenario:

• The slope \( m \) (cost per hour) is $10.
• The y-intercept \( b \) (starting cost for the haircut) is $35.

So the equation representing the total cost \( y \) based on the hours \( x \) the dog stays at the salon is:

\[ y = 10x + 35 \]

Part 2: Standard Form

To write the equation in standard form, we need it in the format \( Ax + By = C \) where \( A \), \( B \), and \( C \) are integers. Starting from the slope-intercept form \( y = 10x + 35 \):

1. Rearrange to get \( -10x + y = 35 \).
2. To convert it into standard form, we typically want \( A \) to be positive. Hence, multiply the entire equation by \(-1\):

\[ 10x - y = -35 \]

Thus, the standard form of the equation is:

\[ 10x - y = -35 \]

In summary:

• Slope-Intercept Form: \( y = 10x + 35 \)
• Standard Form: \( 10x - y = -35 \)

Answer 6

Write the equation for the line that passes through (3, -4) and is perpendicular to the line that passes through (5, 6) and (8, 15).

Part 1:

Write the equation in Point-Slope Form.

Part 2:

Write the equation in Slope-Intercept Form.

Answer 7

To find the equation of the line that passes through the point (3, -4) and is perpendicular to the line passing through the points (5, 6) and (8, 15), we will follow these steps:

Part 1: Find the slope of the line through (5, 6) and (8, 15)

1. Calculate the slope \( m \) of the line through points \( (5, 6) \) and \( (8, 15) \): \[ m = \frac{y_2 - y_1}{x_2 - x_1} = \frac{15 - 6}{8 - 5} = \frac{9}{3} = 3 \]

2. Find the slope of the perpendicular line. The slope of a line that is perpendicular to another is the negative reciprocal of the original slope. Thus, if the slope of the original line is \( 3 \): \[ m_{perpendicular} = -\frac{1}{3} \]

3. Use the point-slope form of the equation for a line, which is given by: \[ y - y_1 = m(x - x_1) \] Here, \( (x_1, y_1) = (3, -4) \) and \( m = -\frac{1}{3} \).

   Substituting in the values: \[ y - (-4) = -\frac{1}{3}(x - 3) \] This simplifies to: \[ y + 4 = -\frac{1}{3}(x - 3) \]

Part 2: Write in Slope-Intercept Form

To convert the equation from point-slope form to slope-intercept form (\( y = mx + b \)), we need to distribute and isolate \( y \):

1. Distributing the right side: \[ y + 4 = -\frac{1}{3}x + 1 \]

2. Subtracting 4 from both sides: \[ y = -\frac{1}{3}x + 1 - 4 \] \[ y = -\frac{1}{3}x - 3 \]

Final Results

• Part 1: Point-Slope Form: \[ y + 4 = -\frac{1}{3}(x - 3) \]

• Part 2: Slope-Intercept Form: \[ y = -\frac{1}{3}x - 3 \]

These are the equations for the line passing through the point (3, -4) and perpendicular to the line connecting (5, 6) and (8, 15).

Answer 8

Dennis is attending the Plant City Strawberry Festival. He plans to park his car in a lot from 6:00pm to 11:00pm. The lot charges $5 for the first hour and $2 for each additional hour.

Part 1:

Create a graph that represents the relationship between the price (y) and hours parked (x).

Part 2:

Identify the Domain of the function.

Part 3:

Identify the Range of the function.

Answer 9

To solve this problem, let’s first break down the costs associated with Dennis parking his car at the Plant City Strawberry Festival.

Part 1: Create a Graph

Calculating the Costs

• Time Period: Dennis parks from 6:00 PM to 11:00 PM, which is a total of 5 hours.
• Cost Structure:
  • The first hour (6:00 PM to 7:00 PM) costs $5.
  • The next four hours (7:00 PM to 11:00 PM) cost $2 each, which totals $8 for the additional hours.

Total Cost Calculation:

• Total cost after 5 hours = $5 (first hour) + $8 (next four hours) = $13.

Breaking down the hours and corresponding costs:

• 0 hours: $0
• 1 hour: $5
• 2 hours: $7 ($5 + $2)
• 3 hours: $9 ($5 + $2 + $2)
• 4 hours: $11 ($5 + $2 + $2 + $2)
• 5 hours: $13 ($5 + $2 + $2 + $2 + $2)

Thus, the relationship can be summarized as:

• ( \begin{align*} (0, 0), \ (1, 5), \ (2, 7), \ (3, 9), \ (4, 11), \ (5, 13) \end{align*} )

Graphing: You can plot these points on a Cartesian plane where the \( x \)-axis represents the number of hours parked (from 0 to 5) and the \( y \)-axis represents the cost in dollars.

The graph will be a piecewise function with:

• A constant increase: from (0, 0) to (1, 5), then a slope of 1 for the increase from (1, 5) to (5, 13) where it becomes linear.

Part 2: Identify the Domain of the Function

The domain represents the possible values of \( x \), which in this context, refers to the number of hours Dennis can park.

• Since Dennis plans to park for a minimum of 0 hours (not parked) to a maximum of 5 hours, the domain of the function is: \[ [0, 5] \]

Part 3: Identify the Range of the Function

The range represents the possible values of \( y \), which corresponds to the total cost incurred from parking.

• The minimum cost is $0 when \( x = 0 \) hours, and the maximum cost is $13 when \( x = 5 \) hours, so the range of the function is: \[ [0, 13] \]

Summary

• Graph: Plot the points (0, 0), (1, 5), (2, 7), (3, 9), (4, 11), (5, 13) on a Cartesian plane with a straight line connecting these points.
• Domain: \( [0, 5] \)
• Range: \( [0, 13] \)

If you need a visual representation of the graph, tools like graphing calculators or software can help create that based on the plotted points.

Answer 10

Please do inequalities for the domain and range.

Answer 11

Certainly! To represent the domain and range using inequalities, we will express each one accordingly.

Domain

The domain refers to the possible values of \( x \), which represents the number of hours parked.

• From the previous discussion, the domain is \( [0, 5] \).
• Using inequalities, this can be expressed as: \[ 0 \leq x \leq 5 \]

Range

The range refers to the possible values of \( y \), which represents the total cost.

• From the previous discussion, the range is \( [0, 13] \).
• Using inequalities, this can be expressed as: \[ 0 \leq y \leq 13 \]

Summary of Inequalities

• Domain: \( 0 \leq x \leq 5 \)
• Range: \( 0 \leq y \leq 13 \)

These inequalities clearly state the valid intervals for the number of hours parked and the total cost associated with those hours.