SAT Math Lesson Plan 22: Modeling with Functions

Lesson Summary: In this lesson, we will explore the process of modeling real-world situations using functions. Students will learn to identify and apply linear, quadratic, and exponential functions to model various situations. This lesson will cover: - The use of linear functions for constant rates of change. - Quadratic functions for modeling situations involving acceleration, area, and projectile motion. - Exponential functions for modeling growth and decay in various contexts. Understanding these types of functions is vital for solving problems in the SAT Math section, where 15-20% of questions involve functions and their applications.

Lesson Objectives:

• Identify linear, quadratic, and exponential functions.
• Apply these functions to solve real-world problems.
• Understand how the structure of a function relates to its graph and real-world meaning.
• Analyze and interpret functions to make predictions or solve problems.

Common Core Standards:

• CCSS.MATH.CONTENT.HSA.CED.A.2 – Create equations in two variables to represent relationships between quantities.
• CCSS.MATH.CONTENT.HSA.CED.A.3 – Represent constraints by equations and inequalities, and by systems of equations and inequalities.
• CCSS.MATH.CONTENT.HSF.IF.C.7 – Graph functions expressed symbolically and show key features of the graph.

Prerequisite Skills:

• Understanding of basic algebraic operations.
• Ability to solve linear and quadratic equations.
• Familiarity with graphing basic functions.

Key Vocabulary:

• Linear Function: A function of the form \( f(x) = mx + b \), where \( m \) is the slope and \( b \) is the y-intercept.
• Quadratic Function: A function of the form \( f(x) = ax^2 + bx + c \), where \( a \), \( b \), and \( c \) are constants.
• Exponential Function: A function of the form \( f(x) = a \cdot b^x \), where \( a \) is the initial value and \( b \) is the base of the exponential function.
• Domain: The set of all possible input values (x-values) for which a function is defined.
• Range: The set of all possible output values (y-values) of a function.

Warm-Up

Activity 1: Identifying Linear Functions

A linear function is one that has a constant rate of change, and its graph is a straight line. Let’s review the following linear function:

\( f(x) = 2x + 5 \)

What do you notice about this function? The slope is 2, which means that for every unit increase in \( x \), the value of \( f(x) \) increases by 2. The y-intercept is 5, so the graph crosses the y-axis at \( (0, 5) \).

Activity 2: Identifying Quadratic Functions

Quadratic functions model situations with an acceleration or change in rate. Here is an example:

\( f(x) = x^2 - 4x + 3 \)

In this quadratic function, the highest power of \( x \) is 2. The graph of this function is a parabola, and we expect the vertex to represent either a maximum or minimum value. This function will have a range starting from the vertex value and extending infinitely.

Activity 3: Identifying Exponential Functions

Exponential functions are useful for modeling situations like population growth or decay. Here is an example:

\( f(x) = 3 \cdot 2^x \)

This function grows exponentially because the base (2) is greater than 1. It models rapid growth, such as the population of a city or the spread of a disease. The graph of this function starts at \( (0, 3) \) and increases rapidly as \( x \) increases.

Linear	Quadratic	Exponential

Teach

In this section, we will explore how different types of functions—linear, quadratic, and exponential—are used to model real-world situations. We will walk through detailed examples to analyze their graphs, interpret their behavior, and solve problems that might appear on the SAT.

• Linear functions: Analyzing straight-line relationships
• Quadratic functions: Solving for max/min values and interpreting their graphs
• Exponential functions: Understanding growth and decay

Step-by-Step Solution:

1. Domain: 
   The domain of the function represents the number of units that can be sold. In the context of this problem, x represents units sold, so the domain is all non-negative integers (i.e., \(x \geq 0\)), because negative sales don’t make sense in this scenario.
2. Range: 
   The range of the function represents the possible profit values. Since the profit increases as more units are sold, the range starts from \(P(0) = -2000\) (which is the fixed cost) and increases indefinitely. Thus, the range is all values greater than or equal to -2000, i.e., \(P(x) \geq -2000\).

Example 1: Modeling Profit for a Business

Consider the function \( P(x) = 500x - 2000 \), where \( P(x) \) represents the profit (in dollars) and \( x \) represents the number of units sold. The business incurs a fixed cost of \$2000, and for each unit sold, the business earns \$500.

Step-by-Step Solution:

1. Domain: 
   The domain of the function represents the number of units that can be sold. In the context of this problem, x represents units sold, so the domain is all non-negative integers (i.e., \(x \geq 0\)), because negative sales don’t make sense in this scenario.
2. Range: 
   The range of the function represents the possible profit values. Since the profit increases as more units are sold, the range starts from \(P(0) = -2000\) (which is the fixed cost) and increases indefinitely. Thus, the range is all values greater than or equal to -2000, i.e., \(P(x) \geq -2000\).
3. Graph: 
   The graph of the function represents the relationship between the number of units sold (\(x\)) and the profit (\(P(x)\)). For \(x = 0\), the profit is \(P(0) = -2000\). As \(x\) increases, the profit increases, showing a linear relationship with a slope of 500. The graph is a straight line with a positive slope, starting at \(P(0) = -2000\).

Conclusion:

This example shows how to use a linear function to model a real-world business scenario, determining both the domain and range and interpreting the graph. The function \( P(x) = 500x - 2000 \) represents a business with a fixed cost and a linear profit growth per unit sold.

Example 2: Modeling with Quadratic Functions

Quadratic functions are useful for modeling situations where there is acceleration or deceleration. Let’s model the path of a projectile. Consider the function:

A ball is thrown from a height of 5 feet, and its height \( h \) (in feet) after \( t \) seconds is given by the function \( h(t) = -16t^2 + 24t + 5 \). This is a quadratic function because the highest exponent of \( t \) is 2.

Step 1: Identify the features of the function: - The negative coefficient of \( t^2 \) indicates that the parabola opens downward, meaning the ball reaches a maximum height before falling back to the ground. - The constant term 5 represents the initial height of the ball. - The linear coefficient (24) represents the initial velocity of the ball.

Step 2: Analyze the graph: - The vertex of the parabola gives the maximum height the ball will reach. - To find when the ball hits the ground, set \( h(t) = 0 \) and solve for \( t \) (this gives the roots of the quadratic equation).

Step 3: Solve for the roots. The quadratic equation \( -16t^2 + 24t + 5 = 0 \) can be solved using the quadratic formula, factoring, or completing the square. The roots will give the time(s) when the ball hits the ground (i.e., when \( h(t) = 0 \)).

Step-by-Step Solution to Finding Roots:

1. Set the height equal to zero: 
   To find when the ball hits the ground, set \( h(t) = 0 \): \[ -16t^2 + 24t + 5 = 0 \]
2. Use the quadratic formula: 
   The quadratic formula is given by: \[ t = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \] For the equation \( -16t^2 + 24t + 5 = 0 \), we have \( a = -16 \), \( b = 24 \), and \( c = 5 \). Plugging in these values: \[ t = \frac{-24 \pm \sqrt{24^2 - 4(-16)(5)}}{2(-16)} \] Simplifying: \[ t = \frac{-24 \pm \sqrt{576 + 320}}{-32} \] \[ t = \frac{-24 \pm \sqrt{896}}{-32} \] \[ t = \frac{-24 \pm 29.93}{-32} \]
3. Calculate the two possible roots: 
   The two possible solutions for \( t \) are: \[ t_1 = \frac{-24 + 29.93}{-32} = \frac{5.93}{-32} \approx -0.185 \] \[ t_2 = \frac{-24 - 29.93}{-32} = \frac{-53.93}{-32} \approx 1.69 \]
4. Interpret the roots: 
   - The first root \( t_1 \approx -0.185 \) is not physically meaningful because time cannot be negative. Therefore, we discard it. - The second root \( t_2 \approx 1.69 \) represents the time when the ball hits the ground, approximately 1.69 seconds after it is thrown.

Conclusion:

The ball hits the ground approximately 1.69 seconds after it is thrown. The negative root was discarded because negative time is not possible in this context.

Example 3: Solving Exponential Growth Problems

A population of bacteria doubles every 4 hours. Initially, there are 100 bacteria. The population after \( t \) hours is given by the exponential function:

\[ P(t) = 100 \times 2^{\frac{t}{4}} \]

The problem is to determine how long it will take for the population to reach 800 bacteria. This is an example of exponential growth.

1. Set up the equation: 
   We are given the equation \( P(t) = 100 \times 2^{\frac{t}{4}} \), and we want to know when \( P(t) = 800 \). So we set the equation equal to 800:

   \[ 800 = 100 \times 2^{\frac{t}{4}} \]

2. Isolate the exponential expression: 
   Divide both sides by 100:

   \[ 8 = 2^{\frac{t}{4}} \]

3. Take the logarithm of both sides: 
   To solve for \( t \), take the natural logarithm (ln) of both sides:

   \[ \ln(8) = \ln\left(2^{\frac{t}{4}}\right) \]

   Apply the logarithmic identity \( \ln(a^b) = b \ln(a) \):

   \[ \ln(8) = \frac{t}{4} \ln(2) \]

4. Solve for \( t \): 
   We know that \( \ln(8) = \ln(2^3) = 3 \ln(2) \), so:

   \[ 3 \ln(2) = \frac{t}{4} \ln(2) \]

   Cancel out \( \ln(2) \) from both sides (assuming \( \ln(2) \neq 0 \)):

   \[ 3 = \frac{t}{4} \]

   Multiply both sides by 4 to solve for \( t \):

   \[ t = 12 \]

5. Interpret the result: 
   Therefore, it will take exactly 12 hours for the population to reach 800 bacteria.

Here is the graph:

This is an example of how exponential growth works over time, and how we can use logarithms to solve for time when the population reaches a specific value.

Example 4: Building a Linear Model from Data

Problem: The following table shows the number of units sold by a company over five months. You are tasked with modeling the data with a linear function and predicting the number of units sold in the 24th month.

Step 1: Analyze the Data

The data shows a consistent increase in the number of units sold each month. This suggests a linear relationship between the month and units sold. To model this data, we will use a linear equation of the form:

Linear equation format:

\[ y = mx + b \]

Where \( m \) is the slope (rate of change) and \( b \) is the y-intercept (the starting value). We can use the slope formula to find \( m \), which is the change in \( y \) (units sold) divided by the change in \( x \) (months). The slope formula is:

Slope formula:

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

In the given data, the change in \( y \) is the difference between the units sold in month 5 (45 units) and month 1 (25 units), and the change in \( x \) is the difference in months (5 - 1 = 4). Thus:

\[ m = \frac{{45 - 25}}{{5 - 1}} = \frac{{20}}{{4}} = 5 \]

Step 2: Build the Mathematical Model

Now that we know the slope is 5, we can use one of the points to solve for \( b \). Let's use the point (1, 25) and substitute it into the point-slope form of the equation:

Point-slope form:

\[ y - y_1 = m(x - x_1) \]

Substituting the values \( m = 5 \), \( x_1 = 1 \), and \( y_1 = 25 \) into the point-slope form:

\[ y - 25 = 5(x - 1) \]

Now, simplify the equation:

\[ y - 25 = 5x - 5 \]

\[ y = 5x + 20 \]

Step 3: Solve the Problem

Now that we have the equation \( y = 5x + 20 \), we can use this to predict the number of units sold after 24 months. Substitute \( x = 24 \) into the equation:

\[ y = 5(24) + 20 \]

\[ y = 120 + 20 \]

\[ y = 140 \]

Answer: The company is predicted to sell 140 units in the 24th month.

Step 4: Graphing the Data

Here’s the graph of the data and the linear equation. The graph shows how the number of units sold increases linearly over time.

Example 5: Pyrotechnic Display - Finding the Roots of a Parabolic Path

Problem: A firework is launched in a pyrotechnic display, and its trajectory follows a parabolic path. The highest point of the trajectory, where the firework explodes, occurs at a height of 60 meters and at a time of 4 seconds. We will assume that this point is the vertex of the parabola. Find the quadratic model that represents the path of the firework and determine the time it takes for the firework to reach the ground (i.e., find the roots of the quadratic equation).

Step 1: Write the vertex form of the quadratic equation

Since the highest point of the firework's trajectory is the vertex, we will use the vertex form of a quadratic equation:

\[ y = a(x - h)^2 + k \]

Where: 

• \( (h, k) \) is the vertex of the parabola, and
• \( a \) is a constant that determines the shape and direction of the parabola.

From the problem, we know that the vertex occurs at \( (h, k) = (4, 60) \), so the equation becomes:

\[ y = a(x - 4)^2 + 60 \]

Step 2: Use a known point to solve for \( a \)

We also know that at \( t = 0 \), the firework is at the ground level, meaning \( y = 0 \). We can use this point to solve for \( a \). Substituting \( x = 0 \) and \( y = 0 \) into the equation:

\[ 0 = a(0 - 4)^2 + 60 \]

Simplifying the equation:

\[ 0 = a(16) + 60 \] \[ -60 = 16a \] \[ a = \frac{-60}{16} \] \[ a = -\frac{15}{4} \]

Step 3: Write the quadratic equation in vertex form

Now that we know \( a = -\frac{15}{4} \), we can substitute this value back into the vertex form equation:

\[ y = -\frac{15}{4}(x - 4)^2 + 60 \]

Step 4: Convert the equation to standard form

To convert the equation to standard form, expand the vertex form equation:

\[ y = -\frac{15}{4}(x^2 - 8x + 16) + 60 \]

Distribute \( -\frac{15}{4} \):

\[ y = -\frac{15}{4}x^2 + 30x - 60 + 60 \]

Simplifying:

\[ y = -\frac{15}{4}x^2 + 30x \]

Step 5: Find the roots (when \( y = 0 \))

To find the roots, we set \( y = 0 \) and solve for \( x \):

\[ 0 = -\frac{15}{4}x^2 + 30x \]

First, factor out \( x \):

\[ 0 = x\left(-\frac{15}{4}x + 30\right) \]

Now, set each factor equal to zero:

1. \( x = 0 \) is one of the roots, which corresponds to the time when the firework is at the ground.

2. Solve for the second root by setting \( -\frac{15}{4}x + 30 = 0 \):

\[ -\frac{15}{4}x = -30 \quad \Rightarrow \quad x = \frac{4(30)}{15} = 8 \]

Step 6: Interpretation of the roots

The roots are \( x = 0 \) and \( x = 8 \). This means that the firework reaches the ground at \( t = 0 \) (when it is initially launched) and again at \( t = 8 \) seconds (when the firework falls back to the ground after exploding at the peak). The firework reaches its maximum height at \( t = 4 \) seconds, as stated in the problem.

Example 6: Exponential Decay Model (Radioactive Decay)

Context: The decay of a radioactive material, such as Carbon-14, follows an exponential decay model. In this example, we'll use data to derive the exponential decay model and solve a problem.

Suppose the mass of a radioactive substance is measured over time, and the following table shows the remaining mass at various times:

Using this data, we will create a model for the decay of the substance and find out how much of it remains after 4000 years.

Step 1: Analyze the Data and Identify the Type of Model

Notice that the mass remaining is decreasing over time. Since the data is decreasing in a way that suggests constant fractional decay, we can assume that this follows an exponential decay model:

\[ A(t) = A_0 \cdot e^{-kt} \]

Where:

• A(t) is the mass remaining at time t in kilograms
• A₀ is the initial mass (100 kg in this case)
• k is the decay constant we need to find
• t is the time in years

Step 2: Use Data to Find the Decay Constant k

We can use the data point at t = 1000 years, where A = 80 kg. Substituting these values into the exponential decay model:

\[ 80 = 100 \cdot e^{-k \cdot 1000} \]

Now, solve for k:

\[ \frac{80}{100} = e^{-1000k} \] \[ 0.8 = e^{-1000k} \]

Take the natural logarithm of both sides:

\[ \ln(0.8) = -1000k \]

Solve for \( k \):

\[ k = -\frac{\ln(0.8)}{1000} \approx 0.00022314 \]

Step 3: Write the Exponential Decay Model

Now that we have the decay constant k\, the model becomes:

\[ A(t) = 100 \cdot e^{-0.00022314 \cdot t} \]

Step 4: Solve for the Remaining Mass After 4000 Years

Now that we have the model, we can use it to find the remaining mass of the substance after 4000 years.

\[ A(4000) = 100 \cdot e^{-0.00022314 \cdot 4000} \]

This simplifies to:

\[ A(4000) = 100 \cdot e^{-0.89256} \]

Using a calculator:

\[ A(4000) = 100 \cdot 0.409 \]

\[ A(4000) = 40.9 \]

So, after 4000 years, there will be approximately 40.9 kg of the substance left.

Review

In this lesson, we have learned how to apply mathematical models to real-world scenarios. Below are a few examples that reinforce the key concepts we’ve covered:

Example 1: Loan Repayment Model

Context:
Suppose a loan of \$1000 is being paid back in equal installments over time. Each month, the borrower repays \$50. We want to model the remaining balance over time and use this model to determine how much of the loan remains after a certain number of months.

Data Table:
(represents the remaining balance of the loan over time)

Steps for Students:

1. Recognize the pattern: The remaining balance is decreasing by \$50 each month, which suggests a linear relationship.
2. Build the model: The slope is -50 (the rate at which the loan balance decreases) and the y-intercept is 1000 (the original loan amount).
   The linear model can be written as:
   \[ B(t) = 1000 - 50t \]
3. Solve the problem: Use the model to determine the remaining balance after 6 months.
   Substituting \( t = 6 \) into the equation: \[ B(6) = 1000 - 50(6) = 1000 - 300 = 700 \]
   So, after 6 months, the remaining balance will be \$700.

Conclusion:
After 6 months of payments, \$700 of the original loan remains to be paid off.

Example 2: Quadratic Model – Child's Height Growth

Context:

A child’s height is often modeled as growing faster at first, then gradually leveling off. Over the first few years, the growth curve follows a quadratic pattern. The height of a child (in centimeters) at various ages is known to follow this type of model.

Data Table (Child Growth):

Step 1: Graph the Data

Plot the data points with age on the x-axis and height on the y-axis. The resulting graph should show a parabolic curve, suggesting that the child’s height grows rapidly in the first few years but slows down after that. 

Step 2: Build the Quadratic Model

We know the value of \( c \) based on the data. At \( t = 0 \), the initial height is 50 cm, so we have \( c = 50 \).

Since the height increases and then plateaus, the parabola should open downwards. Therefore, the value of \( a \) must be negative.

The linear coefficient \( b \) and the quadratic coefficient \( a \) can be determined using curve fitting based on the data. Using a graphing calculator or curve-fitting software, we can determine the values for \( a \), \( b \), and \( c \), ensuring that the model best fits the observed growth pattern over time.

Step 3: Solve the problem using the quadratic model.

We now have the quadratic model:

\[ h(x) = -0.0172x^2 + 1.4167x + 50.1191 \]

Using this model, we can predict the height of the child at any given age. For example, to find the height of the child at 33 months:

\[ h(12) = -0.0172(33)^2 + 1.4167(33) + 50.1191 \]

\[ h(12) = -0.0172(1089) + 1.4167(33) + 50.1191 \]

\[ h(12) = -18.7308 + 46.7511 + 50.1191 \]

\[ h(12) \approx 78.1394 \, \text{cm} \]

This result suggests that the child's height at 33 months is approximately 78.14 cm.

Note: The constant \( c \) in the quadratic model is slightly different from the value of 50 in the data set. This discrepancy is due to the curve fitting process, which adjusts the model to better match the overall trend of the data rather than exactly matching each individual data point.

Step 4: Domain and Range

The domain of this quadratic model is limited by the age range for which the data is relevant. Since the data points represent the height of a child from birth (0 months) to 36 months, the domain of this model is:

Domain: \( 0 \leq x \leq 36 \), where \( x \) is the age in months. The domain does not extend beyond 36 months because the model is only valid for the age range provided by the data.

The range is the set of possible height values, which are derived from the output of the quadratic model. Based on the data, the minimum height occurs at 0 months, and the maximum height occurs around 36 months.

Range: The height of the child starts at 50 cm at birth and increases until it peaks. The range is approximately \( 50 \leq y \leq 79 \), where \( y \) is the height of the child in centimeters.

Example 3: Exponential Growth – Town Population Model

Context:
A small town has experienced consistent population growth. The table below shows the population every five years, starting in the year 2000.

Step 1: Recognize the pattern.
The population increases by a common ratio rather than a constant amount, which suggests exponential growth. To confirm this, divide each population by the previous population to find the common ratio:

• \( \frac{12,800}{10,000} = 1.28 \)
• \( \frac{16,384}{12,800} = 1.28 \)
• \( \frac{20,971}{16,384} \approx 1.28 \)
• \( \frac{26,843}{20,971} \approx 1.28 \)

The common ratio is approximately 1.28 every 5 years.

Step 2: Build the exponential model.
We use the formula for exponential growth:

\[ P(t) = P_0 \cdot r^t \]

Where:

• \( P_0 = 10,000 \), the initial population in the year 2000
• \( r = 1.28 \), the common ratio (growth factor every 5 years)
• \( t \), the number of 5-year intervals since the year 2000

So, the exponential model is:

\[ P(t) = 10,000 \cdot (1.28)^t \]

Step 3: Use the model to predict the population in 10 years.
Ten years after 2020 is 2030, which is six 5-year intervals after 2000. So, \( t = 6 \):

\[ P(6) = 10,000 \cdot (1.28)^6 \]

\[ P(6) \approx 10,000 \cdot 4.7684 = 47,684 \]

Conclusion:
Using the exponential model, the projected population of the town in 2030 is approximately 47,684 people.

Multimedia Resources

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Quiz

Directions: Solve each problem. Choose the correct answer from the options provided.

1. A business makes a profit described by the function \( P(x) = 500x - 2000 \), where \( x \) is the number of units sold and \( P(x) \) is the profit in dollars. What is the break-even point for this business?
   a) 0 units
   b) 4 units
   c) 5 units
   d) 10 units
    
2. The height of a ball thrown into the air is modeled by the function \( h(t) = -16t^2 + 32t + 5 \), where \( t \) is the time in seconds and \( h(t) \) is the height in feet. How long will it take for the ball to hit the ground?
   a) 1 second
   b) 2 seconds
   c) 3 seconds
   d) 4 seconds
    
3. A car travels on a highway with a speed given by the function \( s(t) = 60t - 3 \), where \( s(t) \) is the speed in miles per hour and \( t \) is the time in hours. What is the car's speed at \( t = 2 \)?
   a) 54 mph
   b) 57 mph
   c) 60 mph
   d) 117 mph
    
4. A company sells a product at a price \( p(x) = 30 - 0.5x \), where \( x \) is the number of units sold and \( p(x) \) is the price in dollars. If the company wants to sell 40 units, what will be the price of each unit?
   a) 5 dollars
   b) 10 dollars
   c) 15 dollars
   d) 20 dollars
    
5. The population \( P \) of a town is modeled by the function \( P(t) = 5000e^{0.03t} \), where \( t \) is the time in years. What is the population after 10 years?
   a) 6000
   b) 6500
   c) 7000
   d) 7500
    
6. A company's revenue is modeled by the quadratic function \( R(x) = -x^2 + 100x \), where \( x \) is the number of units sold and \( R(x) \) is the revenue in dollars. How many units must be sold to achieve maximum revenue?
   a) 25 units
   b) 50 units
   c) 75 units
   d) 100 units
    
7. A factory's production cost is modeled by \( C(x) = 200 + 25x \), where \( x \) is the number of items produced. What is the production cost for producing 50 items?
   a) 600 dollars
   b) 700 dollars
   c) 900 dollars
   d) 1450 dollars
    
8. The growth of a population is modeled by the exponential function \( P(t) = 3000e^{0.02t} \), where \( t \) is the time in years. What is the population at \( t = 5 \)?
   a) 3300
   b) 3500
   c) 3700
   d) 3900
    
9. A restaurant’s revenue is modeled by the function \( R(x) = 100x - 2x^2 \), where \( x \) is the number of customers and \( R(x) \) is the revenue. What is the maximum revenue the restaurant can generate?
   a) 800 dollars
   b) 1000 dollars
   c) 1200 dollars
   d) 1250 dollars
    
10. The equation \( f(x) = 2x + 10 \) models the cost in dollars of renting a bike for \( x \) hours. If the rental charge includes a fixed cost of $10 plus $2 per hour, what is the cost for renting the bike for 4 hours?
    a) 12 dollars
    b) 14 dollars
    c) 18 dollars
    d) 20 dollars
     

Answer Key

1. Answer: b) 4 units
   Explanation: Set \( P(x) = 0 \): \( 500x - 2000 = 0 \Rightarrow x = 4 \)
2. Answer: b) 2 seconds
   Explanation: Solve \( -16t^2 + 32t + 5 = 0 \); the positive root is approximately 2 seconds.
3. Answer: d) 117 mph
   Explanation: \( s(2) = 60(2) - 3 = 117 \)
4. Answer: b) 10 dollars
   Explanation: \( p(40) = 30 - 0.5(40) = 10 \)
5. Answer: c) 7000
   Explanation: \( P(10) = 5000e^{0.3} \approx 5000 \cdot 1.3499 \approx 6749.5 \)
6. Answer: b) 50 units
   Explanation: Maximum of a quadratic occurs at \( x = -\frac{b}{2a} = 50 \)
7. Answer: d) 1450 dollars
   Explanation: \( C(50) = 200 + 25(50) = 1450 \)
8. Answer: a) 3300
   Explanation: \( P(5) = 3000e^{0.1} \approx 3000 \cdot 1.1052 \approx 3315.6 \)
9. Answer: d) 1250 dollars
   Explanation: Vertex of \( R(x) = 100x - 2x^2 \) is at \( x = 25 \); \( R(25) = 1250 \)
10. Answer: c) 18 dollars
    Explanation: \( f(4) = 2(4) + 10 = 18 \)