Problem 29 - Precalculus: Functions and Graphs

Question

Average low temperatures Three average monthly low temperatures for Detroit are listed in the table. Let \(x=1\) correspond to January, \(x=2\) to February, \(\ldots,\) and \(x=12\) to December. Determine a quadratic function \(f(x)=a x^{2}+b x+c\) that interpolatesthe data - that is, determine the constants \(a, b,\) and \(c\) such that \(f(2)=19, f(8)=59,\) and \(f(11)=26\)(b) Graph \(f\) in the viewing rectangle \([1,12]\) by \([-15,70,5]\) (c) Use \(f\) to approximate the average monthly low temperatures in June and October. Compare your predictions to the actual temperatures of \(58^{\circ} \mathrm{F}\) and \(41^{\circ} \mathrm{F}\), respectively. (IMAGE CAN'T COPY)

Step-by-Step Solution

Answer

The quadratic function is \( f(x) = -1.96x^2 + 26.3x - 25.76 \). June's predicted temperature is \( 57.44^{\circ}F \), and October's is \( 43.24^{\circ}F \), both near the actual values.

Step 1: Set Up the System of Equations

To find the quadratic function, we need the coefficient values \( a \), \( b \), and \( c \). Given the function \( f(x) = ax^2 + bx + c \), we plug the known values into this equation: \( f(2) = 19 \), \( f(8) = 59 \), and \( f(11) = 26 \). This gives us the following system of equations: - \( 4a + 2b + c = 19 \) (for \( x = 2 \)) - \( 64a + 8b + c = 59 \) (for \( x = 8 \)) - \( 121a + 11b + c = 26 \) (for \( x = 11 \)).

Step 2: Solve the System of Equations

We will use the elimination or substitution method to solve the equations for \( a \), \( b \), and \( c \). First, subtract the first equation from the second and third to eliminate \( c \): - \( 60a + 6b = 40 \) from \( (2) - (1) \)- \( 117a + 9b = 7 \) from \( (3) - (1) \)Now solve these two equations for \( a \) and \( b \): From the first equation, \( 60a + 6b = 40 \) or \( 10a + b = 6.67 \).From the second equation, \( 117a + 9b = 7 \). These two new equations can be solved simultaneously to find values of \( a \) and \( b \).

Step 3: Find Coefficients a and b

By solving \( 10a + b = 6.67 \) and \( 117a + 9b = 7 \), multiply the first equation by 9: - \( 90a + 9b = 60 \). Now subtract this from the second equation:- \( 27a = -53 \)Solve for \( a \):- \( a = -\frac{53}{27} \approx -1.96 \). Substitute \( a \) back into \( 10a + b = 6.67 \) to find \( b \):- \( 10(-1.96) + b = 6.67 \)- \( -19.6 + b = 6.67 \)- \( b = 26.27 \approx 26.3 \).

Step 4: Find Coefficient c

Now substitute \( a = -1.96 \) and \( b = 26.3 \) into one of the original equations to solve for \( c \): - \( 4(-1.96) + 2(26.3) + c = 19 \)- \( -7.84 + 52.6 + c = 19 \)Simplify and solve for \( c \):- \( c = 19 - 44.76 = -25.76 \).

Step 5: Write the Quadratic Function

Now, we can write the quadratic function based on the calculated coefficients:\[ f(x) = -1.96x^2 + 26.3x - 25.76 \]

Step 6: Graph the Function

Graph the function \( f(x) = -1.96x^2 + 26.3x - 25.76 \) in the viewing rectangle \([1,12]\) by \([-15,70,5]\) to visually confirm it represents the average low temperatures accurately.

Step 7: Calculate June and October Temperatures

Use the function to calculate the approximated temperatures for June and October. Recall: - For June \((x=6)\): \( f(6) = -1.96(6)^2 + 26.3(6) - 25.76 \)- For October \((x=10)\): \( f(10) = -1.96(10)^2 + 26.3(10) - 25.76 \).Calculate these values:- June \( f(6) \approx 57.44^{\circ}F \)- October \( f(10) \approx 43.24^{\circ}F \).

Step 8: Compare with Actual Temperatures

- June: The calculated temperature \( 57.44^{\circ}F \) is close to the actual temperature of \( 58^{\circ}F \).- October: The calculated temperature \( 43.24^{\circ}F \) is close to the actual temperature of \( 41^{\circ}F \).

Key Concepts

System of Equations

A system of equations is a collection of two or more equations with the same set of variables. In this exercise, we are tasked with finding the quadratic function that best interpolates given temperature data. The quadratic function is represented as:

• \( f(x) = ax^2 + bx + c \)

The problem provides three specific values, creating a system of equations that we need to solve:

• \( 4a + 2b + c = 19 \) (for \( x = 2 \))
• \( 64a + 8b + c = 59 \) (for \( x = 8 \))
• \( 121a + 11b + c = 26 \) (for \( x = 11 \))

The trick is to find the values of \( a \), \( b \), and \( c \) that satisfy all equations simultaneously.
To do this, one can use methods like substitution or elimination. Here, we simplified the system by subtracting equations from one another to eliminate \( c \) and solve for \( a \) and \( b \). This allows us to break it down into simpler linear equations that are easier to solve.

Interpolation

Interpolation involves estimating values between two known values. In the context of this problem, it concerns using a quadratic function to estimate the monthly temperatures at any given month of the year.
The given data points for January, August, and November act as anchors for our interpolation. By finding the quadratic function that passes precisely through these points, we are essentially interpolating the temperature data to understand it more smoothly across the entire interval from January to December.
This function is not only used for known months but is also key in estimating temperatures for June and October, which weren’t among the initially provided data points. This provides an approximation that helps in identifying trends and analyzing the data more efficiently.
While interpolation gives approximate results, it reduces the error margin and offers a sensible estimation within the range of the known data points.

Temperature Approximation

Temperature approximation via a quadratic function allows us to predict temperatures for months that were not initially provided with exact data.
Within the quadratic function \( f(x) = -1.96x^2 + 26.3x - 25.76 \), substituting values like June (\( x=6 \)) and October (\( x=10 \)) helps find these approximations. For instance:

• June: \( f(6) \approx 57.44^{\circ}F \)
• October: \( f(10) \approx 43.24^{\circ}F \)

These calculations provide us with an estimation framework that can then be juxtaposed with the actual recorded temperatures (58°F for June and 41°F for October) for validation.
Although the function yields close approximations, slight discrepancies are typical due to natural variations and the limitations of using a quadratic model to account for potentially more complex data trends. Nonetheless, this method is highly valuable for quick and understandable approximations.

Graphical Representation

Visualizing the quadratic function through a graph can offer a clear and immediate understanding of how well the function approximates the actual temperature data.
Plotting \( f(x) = -1.96x^2 + 26.3x - 25.76 \) in a specified viewing window of \( [1,12] \) by \( [-15,70,5] \), gives a comprehensive look at the function's behavior over the year. This graphical representation displays the smooth curve fitting through the points for January, August, and November — and extends beyond these to predict other months.
By comparing the curve of this function with the temperature data, students can visually assess how well the quadratic function fits the observed temperatures and where slight deviations may occur.
Graphs are particularly beneficial for enhancing comprehension, allowing for analyses such as detecting months with higher rates of temperature change or verifying the accuracy of interpolations visually.