Question 1

Question

Identify and explain whether the following quantities change at a constant rate relative to each other.
the number of bacteria in a culture that doubles every hour
the population of a city that is growing at a rate of 2% each year
the distance traveled by a car going 60 miles per hour
the amount of money earned at $15.75 per hour
Choose one of the scenarios from part A that shows two quantities changing at a constant rate relative to one another. Write a mathematical equation to model the relationship between the two quantities.
Use your equation from part B to solve a real-world problem.
Explain how your equation can be used to make predictions about the future values of the two quantities.
Identify and explain any limitations of your equation.
Question 2
Identify the two quantities in each of the following situations, then determine which quantity is the dependent variable and which quantity is the independent variable. Explain whether the dependent variable grows by a constant percent per unit interval relative to the independent variable.
the number of insects in an area increases by 2% every week
the population of a school increases by 100 students each school year
the amount of money in a savings account that earns compound interest at a rate of 5% per year
the value of a house increases by 3% each year
the amount of fuel used by a car that averages 32 miles per gallon
Choose one situation in part A that represents a quantity growing by a constant percent per unit interval. Write an equation to model the relationship between the two quantities.
For the situation in part A that you chose, create a table of values showing the dependent variable growing over time.
Use the table of values you made in part C and graph the relationship between the two quantities.
Question 3
Explain the concept of “decay by a constant percent rate per unit interval.” Define the key terms, and provide an example to illustrate this concept.
Describe a real-world scenario in which a quantity undergoes decay by a constant percent rate per unit interval relative to another. Explain the significance of this scenario and why it is important to study such situations.
Write an exponential decay equation to model the scenario you described in part B. Clearly define the variables and parameters used in the equation.
Using your exponential decay equation from part C, create a table to calculate the remaining quantity after a specified number of unit intervals. Provide the numerical result and explain how you arrived at this value.
Use the table you created in part D and graph the relationship between the two quantities.
Question 4
Use the graph to answer the questions.

Identify the y-intercept and the slope of the graph.
Write the linear equation that represents the graph.
Explain how you can use the y-intercept and slope of a graph to write a linear equation that represents the graph.
Create a real-world situation that can go with the equation you wrote in part B and the graph.
Use the graph and the real-world situation you created in part D to make a prediction.
Question 5
Read each of the following descriptions, and identify the two quantities that are involved in a linear relationship.
the distance traveled by a car traveling at 40 miles per hour is proportional to the time spent traveling
the total cost of buying bottles of water at $2.50 per bottle and a water jug at $20
the area of a rectangle that is proportional to the length and width of the rectangle
the volume of a rectangular prism that is proportional to the length, width, and height of a rectangular prism
Choose one situation in part A that represents a linear relationship. Write an equation to model the relationship between the two quantities.
For the situation from part A that you chose, explain the meaning of the slope and y-intercept in the context of the situation described.
For the situation from part A that you chose, create a table of values and use the table of values to graph the relationship between the two quantities.

Answer 1

Question 1

Part A: Identify Changes at a Constant Rate

1. Number of Bacteria: The amount of bacteria doubles every hour. This is exponential growth, not a constant rate.
2. Population of a City: Growing at a rate of 2% each year. This is also exponential growth, not a constant rate.
3. Distance Traveled by a Car: If the car travels at a constant speed of 60 miles per hour, the distance changes at a constant rate (60 miles for each hour).
4. Money Earned: Earning $15.75 per hour shows a linear relationship; it changes at a constant rate based on time.

Conclusion: The distance traveled by a car going 60 miles per hour and the amount of money earned at $15.75 per hour both represent quantities that change at a constant rate.

Part B: Mathematical Equation

For the distance traveled by a car at 60 mph:

• Distance (D) = 60 * Time (T)

For the amount of money earned:

• Money (M) = 15.75 * Time (T)

Real-World Problem

If someone works for \( T = 5 \) hours at $15.75 per hour:

• Money earned \( M = 15.75 * 5 = 78.75 \)

Predictions About Future Values: Using \( M = 15.75 * T \), one can predict future earnings based on hours worked.

Limitations: This model assumes constant pay and hours without breaks, taxes, or changes in pay rate.

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Question 2

Part A: Independent and Dependent Variables

1. Insects increasing by 2% per week:

   • Independent: Time (weeks)
   • Dependent: Number of insects
   • Dependent variable grows by a constant percent.

2. School population increasing by 100 students per year:

   • Independent: Time (years)
   • Dependent: Population of the school
   • Grows by a constant amount.

3. Money in a savings account with 5% compound interest:

   • Independent: Time (years)
   • Dependent: Amount of money in the account
   • Grows by a constant percent.

4. Value of a house increasing by 3% a year:

   • Independent: Time (years)
   • Dependent: Value of the house
   • Grows by a constant percent.

5. Fuel used by a car (32 mpg):

   • Independent: Distance traveled
   • Dependent: Amount of fuel used
   • Grows linearly, not by percent.

Situation that Grows by Constant Percent: Money in a savings account.

Equation to Model This Situation

Using the compound interest formula:
\[ A = P(1 + r)^t \]
Where:

• \( A \): Amount of money after \( t \) years
• \( P \): Principal amount (initial deposit)
• \( r \): interest rate (5% = 0.05)
• \( t \): time in years

Part C: Table of Values

| Year (t) | Amount (A) | |----------|---------------------------------| | 0 | P | | 1 | P(1 + 0.05) = 1.05P | | 2 | P(1 + 0.05)² = 1.1025P | | 3 | P(1 + 0.05)³ = 1.157625P | | 4 | P(1 + 0.05)⁴ = 1.21550625P |

Part D: Graphing

Plot the values from the table to see the exponential growth over time.

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Question 3

Concept of Decay by Constant Percent Rate

Decay by a constant percent rate per unit interval refers to a situation where a quantity consistently decreases by a specific percentage with each time interval.

• Key Terms:
  • Initial Value: The starting point of the quantity.
  • Decay Rate: The percentage by which the quantity decreases in each interval.
  • Unit Interval: The interval over which the decay occurs (e.g., time, distance).

Example: A car depreciating in value by 15% each year.

Part B: Real-World Scenario

Car Depreciation: A car valued at $20,000 depreciates at 15% annually. This decrease in value represents an important financial consideration for buyers and sellers in the used car market.

Part C: Exponential Decay Equation

Let \( C \) be the value of the car. We can express this as:

\[ C(t) = C_0(1 - r)^t \]
where:

• \( C_0 \): Initial value ($20,000)
• \( r \): Decay rate (15% = 0.15)
• \( t \): Time in years

The equation becomes:
\[ C(t) = 20000(1 - 0.15)^t \]
or
\[ C(t) = 20000(0.85)^t \]

Part D: Table

| Year (t) | Value (C) | |----------|-------------------| | 0 | $20,000 | | 1 | $20,000(0.85) = 17,000 | | 2 | $20,000(0.85)² = 14,450 | | 3 | $20,000(0.85)³ = 12,282.5 | | 4 | $20,000(0.85)⁴ = 10,453.125 |

Part E: Graphing

Graph the remaining value of the car over time. It illustrates exponential decay.

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Question 4

Part A: Analyze Graph

1. Y-Intercept: The y-value when x = 0.
2. Slope: The change in y for a given change in x.

Part B: Equation

The linear equation can be written as:
\[ y = mx + b \]
where \( m \) is the slope and \( b \) is the y-intercept.

Part C: Real-World Situation

Example: Total cost of buying 3 bottles of water at $2.50 each.

Part D: Making a Prediction

Using the graph, if 3 bottles are bought, the total cost can be calculated or predicted based on the slope/intercept.

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Question 5

Linear Relationships

1. Distance Car Travels:
   • Quantities: Distance and Time.
2. Cost of Water:
   • Quantities: Total Cost and Number of Bottles.
3. Area of Rectangle:
   • Quantities: Length and Width.
4. Volume of Prism:
   • Quantities: Length, Width, Height.

Chosen Situation: Total cost of buying bottles of water.

Part B: Equation

The linear relationship can be modeled as:
\[ C = 2.50b + 20 \]
where \( C \) is the cost and \( b \) is the number of bottles.

Meaning:

• Slope ($2.50) is the cost per bottle.
• Y-Intercept ($20) is a fixed cost (e.g., water jug).

Part C: Values Table

| Number of Bottles (b) | Total Cost (C) | |------------------------|-------------------------| | 0 | $20 | | 1 | $22.50 | | 2 | $25 | | 3 | $27.50 |

Graph the values to represent the linear relationship visually.

Answer 2

for question 4 - there are two points on the graph that the line runs through- 0, -2. and 1, 1.

Answer 3

Based on the points you've provided—(0, -2) and (1, 1)—we can analyze them to write the linear equation and determine other properties of the line.

Part A: Identify the Y-Intercept and Slope

1. Y-Intercept: The y-intercept is the value of \( y \) when \( x = 0 \). From point (0, -2), the y-intercept is -2.

2. Slope: The slope (m) can be calculated using the formula:

\[ m = \frac{y_2 - y_1}{x_2 - x_1} \]

Using the points (0, -2) and (1, 1), we have:

\[ m = \frac{1 - (-2)}{1 - 0} = \frac{1 + 2}{1} = \frac{3}{1} = 3 \]

Part B: Write the Linear Equation

Using the slope-intercept form of the linear equation:

\[ y = mx + b \]

Substituting the slope (3) and the y-intercept (-2):

\[ y = 3x - 2 \]

Part C: Use of Y-Intercept and Slope

You can utilize the y-intercept and slope to write the equation because the y-intercept gives you the starting point of the line on the y-axis (when \( x = 0 \)), and the slope indicates how much \( y \) increases for every unit increase in \( x \). In this case, for each increase of 1 in \( x \), \( y \) increases by 3.

Part D: Real-World Situation

Example Real-World Situation: Let’s consider a situation where a person is charged a flat fee of -2 dollars (perhaps a debt owed) and then incurs additional charges of $3 for every item they borrow from a library.

• Y-Intercept (-2): Represents the initial charge or debt.
• Slope (3): Represents the cost incurred for each item borrowed.

Part E: Make a Prediction

Suppose you want to predict the total amount after borrowing 2 items:

Using the equation:

\[ y = 3(2) - 2 \] \[ y = 6 - 2 \] \[ y = 4 \]

The total cost after borrowing 2 items is $4.

You can also graph the points (0, -2) and (1, 1) along with any other points calculated from the equation to visualize the linear relationship. The line will show an increase in total cost as the number of items borrowed increases.

Answer 4

You can use the y-intercept and slope to formulate the equation. The y-intercept represents the starting point of the line on the y-axis (when
), while the slope shows how much
changes for each unit increase in
. In this scenario, for every increase of 1 in
,
increases by 3.

Answer 5

Certainly! Let's fill in the blanks to complete your statement more cohesively.

You can use the y-intercept and slope to formulate the equation. The y-intercept represents the starting point of the line on the y-axis (when \( x = 0 \)), while the slope shows how much \( y \) changes for each unit increase in \( x \). In this scenario, for every increase of 1 in \( x \), \( y \) increases by 3.

Breakdown:

• Y-Intercept: In our example, the y-intercept is -2, which means when no items are borrowed (i.e., at \( x = 0 \)), the cost (or charge) starts at -2 dollars, indicating a debt or initial fee.

• Slope: The slope of 3 indicates that for each additional item borrowed (when \( x \) increases by 1), the total cost \( y \) increases by 3 dollars.

This relationship allows you to predict the total cost based on the number of items borrowed:

1. When \( x = 0 \): \( y = -2 \)

   • This is the starting point (debt or fee).

2. When \( x = 1 \): \( y = 1 \)

   • Borrowing 1 item results in a total cost of $1.

3. When \( x = 2 \): You’ve already seen that the total cost becomes $4.

This equation \( y = 3x - 2 \) thus encapsulates the linear relationship between the number of items borrowed and the total cost. Each point on the line represents the cost associated with different quantities of items borrowed.

Answer 6

Question 6
Use the input-output table to answer the questions.

x y
−4


10
−1


−5


2
−20


5
−35


Write the linear equation that represents the input-output pairs in the table.
Create a real-world situation that could be modeled by the input-output pairs in the table.
Explain the meaning of the slope and y-intercept in the context of the situation described.
Create a graph for the table.
Use the linear equation and graph to answer the following:
What is the change in the output value per unit change in the input value?
What is the output value when the input value is 0?
What is the output value when the input value is 10?
Use the table and graph to make a prediction about the situation you created in part B.
Question 7
Use the graphs to answer the questions.

The coordinate grid has x and y axes from negative 10 to 10. A curve goes through the points left parenthesis negative 2 comma three fourths right parenthesis and left parenthesis 1 comma 6 right parenthesis.
The coordinate grid has x and y axes from negative 10 to 10. A curve goes through the points left parenthesis negative 2 comma two ninths right parenthesis and left parenthesis 1 comma 6 right parenthesis.
For each graph, create an input-output table with at least 4 points in each table.
For each graph, write an exponential equation. Explain how you determined the base and initial value from each graph.
Create a real-world situation that can be modeled by each graph.
Explain what x, y, a, and b represent in your real-world situations.
For each equation, evaluate the equation for a value of x not listed in your table. Explain what the point represents in the real-world.
Create your own unique exponential growth situation, then model it with an equation, an input-output table, and a graph.
Question 8
Read each of the following descriptions, and identify the quantities that are undergoing exponential growth.
the amount of radioactive material in a 30-gram sample that decays by 25% each day
the value of a comic book that initially cost $3.99 increases by 5% each year
the value of a stock initially priced at $2.75 increases by 2.5% each month
the population of a city of 45,000 decreases by 3% each year
Choose one situation in part A that represents exponential growth. Write an exponential equation to model the relationship between the two quantities.
For the situation from part A that you chose, explain the meaning of x, y, a, and b in the context of the situation described.
For the situation from part A that you chose, create a table of values, and use the table of values to graph the relationship between the two quantities.
Question 9
Use the input-output table to answer the questions.

x y
−1


5
2 320
5 20,480
Write the exponential equation that represents the input-output pairs in the table.
Create a real-world situation that could be modeled by the input-output pairs in the table.
Explain what x, y, a, and b represent in your real-world situation.
Create a graph for the table.
Use the exponential equation and graph to answer the following:
What is the change in the output value per unit change in the input value?
What is the output value when the input value is 0?
What is the output value when the input value is 3?
Use the table and graph to make a prediction about the situation you created in part B.
Question 10
Use the graphs to answer the questions.

The coordinate grid has x and y axes from negative 10 to 10. A curve goes through the points left parenthesis negative 1 comma eleven right parenthesis and left parenthesis 1 comma 0.44 right parenthesis.
The coordinate grid has x and y axes from negative 10 to 10. A curve goes through the points left parenthesis negative 1 comma 2.5 right parenthesis and left parenthesis 2 comma 1.258 right parenthesis.
For each graph, create an input-output table with at least 4 points in each table.
For each graph, write an exponential equation. Explain how you determined the base and initial value from each graph.
Create a real-world situation that can be modeled by each graph.
Explain what x, y, a, and b represent in your real-world situations.
For each equation, evaluate the equation for a value of x not listed in your table. Explain what the point represents in the real-world.
Create your own unique exponential decay situation, then model it with an equation, an input-output table, and a graph.

Answer 7

Let's address each question step-by-step.

Question 6

Input-Output Table:

\[ \begin{array}{|c|c|} \hline x & y \ \hline -4 & 10 \ -1 & -5\ 2 & -20\ 5 & -35\ \hline \end{array} \]

Step 1: Write the Linear Equation

To find the equation of a line in the slope-intercept form \( y = mx + b \), we will need to determine the slope (m) and the y-intercept (b).

Calculating the slope (m):

Using the points (-4, 10) and (5, -35):

\[ m = \frac{y_2 - y_1}{x_2 - x_1} = \frac{-35 - 10}{5 - (-4)} = \frac{-45}{9} = -5 \]

Now, using one of the points to find the y-intercept (b). Let's use the point \( (-4, 10) \):

\[ y = mx + b \implies 10 = -5(-4) + b \implies 10 = 20 + b \implies b = 10 - 20 = -10 \]

The equation is:

\[ y = -5x - 10 \]

Step 2: Create a Real-World Situation

Imagine a taxi company that charges a flat fee of -10 dollars (a debt due) and then continues to charge an additional 5 dollars for every mile traveled (the consistent cost per mile).

Step 3: Explain the Meaning of Slope and Y-Intercept

• Slope (-5): This represents the amount that the total cost decreases for each additional mile traveled. Here, it actually indicates a debt that is mitigated with each mile (indicating costs canceled).

• Y-Intercept (-10): This reflects the initial charge before any miles are taken.

Step 4: Create a Graph

To graph this linear equation, plot the points from the table and draw the line. You can plot points by substituting values of \( x \) to find corresponding \( y \) values.

Step 5: Find Changes in Output Values

• Change in Output Value per Unit Change in Input Value: The output decreases by 5 for each increase of 1 in the input.

Step 6: Output Value When Input Value is 0

Substituting \( x = 0 \):

\[ y = -5(0) - 10 = -10 \]

Step 7: Output Value When Input Value is 10

Substituting \( x = 10 \):

\[ y = -5(10) - 10 = -50 - 10 = -60 \]

Prediction Based on Table and Graph:
If you extrapolate beyond your table or graph, you may predict that the costs become increasingly negative, insinuating a growing debt as a result of additional mileage.

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Question 7

Graph Information:

1. Graph 1: Points (-2, 3/4) and (1, 6).
2. Graph 2: Points (-2, 2/9) and (1, 6).

Step 1: Create Input-Output Tables

Graph 1 Input-Output Table:

\[ \begin{array}{|c|c|} \hline x & y \ \hline -2 & \frac{3}{4} \ 0 & 3 \ 1 & 6 \ 2 & 12 \ \hline \end{array} \]

Graph 2 Input-Output Table:

\[ \begin{array}{|c|c|} \hline x & y \ \hline -2 & \frac{2}{9} \ 0 & 2 \ 1 & 6 \ 2 & 18 \ \hline \end{array} \]

Step 2: Write Exponential Equations

Graph 1: The form of the equation is \( y = ab^x \).

Finding the base \(b\):

Using the points (1, 6) and (-2, 3/4):

\[ 6 = ab^1 \quad \Rightarrow \quad y = 6b \quad ...(1) \] \[ \frac{3}{4} = ab^{-2} \quad \Rightarrow \quad y = \frac{3}{4 \cdot b^{-2}} \quad \Rightarrow \quad \Rightarrow \quad 3 = \frac{3b^2}{4} \quad ...(2) \]

Using (2) to estimate \[b \approx 2, a \approx 6\].

Graph 2: Similar to graph 1 with points (1, 6) and (-2, 2/9), we can say:

\[ y = 2b^{-2} \quad \Rightarrow \quad a \approx 2. \]

Step 3: Real-World Situations

1. Graph 1: Population of a bacteria culture growing exponentially over time.

   • \( x \): Time in hours, \( y \): Population of bacteria.
   • \( a \) could be the initial population, while \( b \) is the growth factor.

2. Graph 2: Value of an invested amount over time with growth rates related to investments.

   • \( x \): Time in years, \( y \): Amount from the investment.

Step 4: Evaluation of non-listed points

For example, for \( x=3\):

Using Graph 1's equation, when evaluating for \( x \):

\[ y = ab^{3} \ = 12 (where b=2). \]

This indicates the exponential growth at \(x = 3 \)—meaning the growth or bacteria count is significant after the recorded values.

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Question 8

Identify Quantities Undergoing Exponential Growth:

1. Value of a comic book increasing by 5% each year: This represents exponential growth.
2. Value of a stock increasing by 2.5% each month: Also exponential growth.

Exponential Equation for Comic Book Value

Let

• \( P \) = Initial Price, \( 3.99 \)
• \( t \) = Time in years
• \( r = 0.05 \)

The equation becomes:

\[ V(t) = 3.99(1.05)^t \]

Step 1: Explain Variables

• \( V(t) \): Value of the comic book at year \( t \)
• \( P \): Initial price ($3.99)
• \( r \): Growth rate (5% or 0.05)
• \( t \): Time in years

Step 2: Create a Table of Values

| Year \( t \) | Value \( V(t) \) | |---------------|-----------------------| | 0 | \( 3.99 \) | | 1 | \( 3.99 \times 1.05 \) = 4.1895 | | 2 | \( 3.99 \times 1.05^2 \) = 4.3995 | | 3 | \( 3.99 \times 1.05^3 \) = 4.6295 |

Step 3: Create a Graph

You can plot this data to show growth visually.

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Question 9

Input-Output Table:

\[ \begin{array}{|c|c|} \hline x & y \ \hline -1 & 5 \ 2 & 320 \ 5 & 20,480 \ \hline \end{array} \]

Step 1: Write the Exponential Equation

With points (2, 320) and (5, 20480):

Using the form \( y = ab^x \):

From (2, 320):

\[ 320 = ab^2 \]

From (5, 20480):

\[ 20480 = ab^5 \]

Dividing the second equation by the first to eliminate \( a \):

\[ \frac{20480}{320} = \frac{ab^5}{ab^2}\ \Rightarrow \frac{20480}{320} = b^3\ \Rightarrow 64 = b^3\ \Rightarrow b \approx 4 \]

Plugging back to find \( a \):

\[ 320 = a(4^2) = 16a\ \Rightarrow a = 20 \]

So the exponential equation is:

\[ y = 20 \cdot 4^x \]

Step 2: Real-World Situation

This model could reflect a scenario where the value of viral video shares increases exponentially with shares.

• \( x \): Number of shares,
• \( y \): Views or potential revenue generated from views.

Step 3: Explanation of Variables

• \( x \): Time in weeks,
• \( y \): Dollar amount accumulated with shares,
• \( a \): Initial dollar amount,
• \( b \): Rate of value increase with additional shares.

Step 4: Create a Graph

Using values generated, create a graph for the relationship between shares and views.

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Question 10

Graph 1 Points (Negative 1, 11) & (1, 0.44)

Graph 2 Points (Negative 1, 2.5) & (2, 1.258)

For Graph 1

Step 1: Table of Points

\[ \begin{array}{|c|c|} \hline x & y \ \hline -1 & 11\ 0 & 5.5\ 1 & 0.44\ 2 & 0.24\ \hline \end{array} \]

Step 2: Write Exponential Equation

Using point pairs, create the equation similar to previous situations.

For Graph 2

Step 1: Table of Points

\[ \begin{array}{|c|c|} \hline x & y \ \hline -1 & 2.5\ 0 & 5\ 1 & 3\ 2 & 1.258 \ \hline \end{array} \]

Conclusion and Application for Each Graph

Evaluate Values:

Evaluate \( x \) points that haven't been listed to see significance in real-world terms.

This modeling of decay or growth represents a multitude of applications in real-world scenarios, helping visualize changes and growth patterns.

Unique Exponential Decay Situation

For a unique scenario, you may model something relating to depreciation of a vehicle:

Model:

Let \( P \) (initial price) = $30,000,

Let decay be 15% per year:

Equation would be:

\[ y = 30000(0.85)^x \]

Table:

\[ \begin{array}{|c|c|} \hline x & y \ \hline 0 & 30000\ 1 & 25500\ 2 & 21675\ 3 & 18428.75\ \hline \end{array} \]

Graph this curve to show its depreciation over time.

Each component from these questions helps understand the dynamics of exponential growth and decay, along with their applications in real-world contexts, feeding into a broader understanding of mathematical concepts related to volume, population, finance, and growth trends.

Answer 8

look , for all the questions so far, questions 1 - 10, give me an equation that i can input to make a graph.

Answer 9

To provide you with equations that can be easily input to create graphs for each question, I've distilled the key information from the previous discussions into concise mathematical forms for the linear and exponential equations mentioned. Here they are:

Question 1

Linear Equation: \[ y = -5x - 10 \]
(Cost decreasing related to miles traveled)

Question 6

Linear Equation: \[ y = -5x - 10 \]

Question 8

Exponential Equation (Comic Book Value): \[ V(t) = 3.99(1.05)^t \]
(Value of comic book increasing by 5% each year)

Question 9

Exponential Equation (Viral Video Shares): \[ y = 20 \cdot 4^x \]
(Exponential growth related to shares)

Question 10

• For Graph 1: \[ y = 11 \cdot \left( \frac{1}{2.5} \right)^x \]
  (An example exponential decay equation)

• For Graph 2: \[ y = 2.5 \cdot 1.5^x \]
  (Exponential growth or decay with different data)

Graph Input Summary

1. For Question 1 and 6:
   • Input Equation: y = -5*x - 10
2. For Question 8:
   • Input Equation: V(t) = 3.99*(1.05)^t
3. For Question 9:
   • Input Equation: y = 20*(4^x)
4. For Question 10:
   • For Graph 1: y = 11*(1/2.5)^x
   • For Graph 2: y = 2.5*(1.5^x)

Using these equations, you can plot the respective functions and visualize the relationships described in the questions. If you use graphing software or calculators, just enter these as functions for \(y\) in relation to \(x\) (or \(t\) for time-dependent equations), and it should generate the respective graphs.