Problem 86 - Algebra A Combined Function

Question

Notice that the shape of the temperature graph for Exercises 81 through 84 is similar to a parabola (see Section 8.5). In fact, this graph can be approximated by the quadratic function \(f(x)=3 x^{2}-18 x+57,\) where \(f(x)\) is the temperature in degrees Fahrenheit and \(x\) is the number of days from Sunday. Use this function to answer. Use the given function and the quadratic formula to find when the low temperature was \(35^{\circ} \mathrm{F}\). [Hint: Let \(f(x)=35\) and solve for \(x .]\) Round your answer to one decimal place and interpret your result. Does your answer agree with the graph on the previous page?

Step-by-Step Solution

Answer

The low temperature of 35°F occurred roughly 1.7 and 4.6 days after Sunday.

Step 1: Understanding the Problem

We need to find when the temperature equaled 35°F. The quadratic function approximating temperature is given by \(f(x) = 3x^2 - 18x + 57\). To find when the temperature was 35°F, we need to set \(f(x) = 35\) and solve for \(x\).

Step 2: Setting Up the Equation

We set up the equation by substituting 35 into the quadratic function: \(3x^2 - 18x + 57 = 35\). Simplifying, we get \(3x^2 - 18x + 22 = 0\).

Step 3: Using the Quadratic Formula

The quadratic formula is \(x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\), where \(a = 3\), \(b = -18\), and \(c = 22\). Substitute these values into the formula to solve for \(x\).

Step 4: Calculating the Discriminant

Calculate the discriminant \(b^2 - 4ac\): \((-18)^2 - 4 \times 3 \times 22\). This equals \(324 - 264 = 60\). Since the discriminant is positive, there are two real solutions.

Step 5: Solving for x

Substitute back into the quadratic formula: \(x = \frac{-(-18) \pm \sqrt{60}}{6}\), which simplifies to \(x = \frac{18 \pm \sqrt{60}}{6}\). Calculating further, \(x_1 = \frac{18 + 7.75}{6} = 4.6\) and \(x_2 = \frac{18 - 7.75}{6} = 1.7\).

Step 6: Interpreting the Result

The solutions \(x = 1.7\) and \(x = 4.6\) represent the number of days from Sunday when the temperature was 35°F. Since the solutions are within the week range, they agree with the temperature graph mentioned in context.

Key Concepts

Temperature Modeling

Temperature modeling involves using mathematical functions to represent temperature changes over time. In this exercise, we use a quadratic function to model the temperature throughout a week. The given quadratic expression is:

• \(f(x) = 3x^2 - 18x + 57\)

This function models how the temperature in degrees Fahrenheit changes based on \(x\), the number of days from Sunday.
By approximating temperature changes with a quadratic equation, we can predict temperatures for specific days. This approach is useful because many natural phenomena, like temperature variations, can be cycled and graphed as parabolic curves.
The function here allows you to find specific temperatures, like when it hits 35°F, by setting \(f(x) = 35\) and solving for \(x\). Understanding temperature modeling is crucial for interpreting such data, especially in terms of forecasts and historical temperature tracking.

Quadratic Formula

The quadratic formula is a powerful tool for solving quadratic equations, which are equations in the form \(ax^2 + bx + c = 0\). It provides a way to find the values of \(x\) by using the coefficients \(a\), \(b\), and \(c\) in the formula:

• \(x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\)

This formula is derived from completing the square of a quadratic equation, and it can solve any quadratic equation, regardless of whether its roots are real or complex.
In our problem, we use the quadratic formula to solve \(3x^2 - 18x + 22 = 0\), which is obtained by setting the temperature function equal to 35°F. By substituting the values \(a = 3\), \(b = -18\), and \(c = 22\) into the formula, we can find the precise days when the temperature reached 35°F.

Discriminant

The discriminant is an important part of the quadratic formula that helps determine the nature of the roots of a quadratic equation. It is calculated as:

• \(b^2 - 4ac\)

The discriminant provides insight into the roots without solving the entire equation. Based on its value, we can easily understand the type of solutions:

• If the discriminant is positive, there are two distinct real solutions.
• If it is zero, there is exactly one real solution (a repeated root).
• If it is negative, there are no real solutions, but two complex solutions.

In this exercise, the discriminant is \(60\), showing that our quadratic equation has two real solutions. This means the temperature reached 35°F on two different days, illustrating the practical impact of the discriminant in understanding the solution's nature.

Real Solutions

Real solutions refer to the actual numerical answers we get when solving an equation. These solutions correspond to real numbers as opposed to complex numbers. In the context of a quadratic equation, real solutions indicate the points where the curve intersects the x-axis.
In this temperature modeling problem, after applying the quadratic formula, we found the real solutions:

• \(x_1 = 4.6\)
• \(x_2 = 1.7\)

These solutions tell us the days (from Sunday) on which the temperature dropped to 35°F. Real solutions are important as they represent actual, feasible answers in real-world problems, such as predicting temperatures on specific days in this scenario.