Problem 48 - A Graphical Approach to Precalculus with Limits

Question

Solve each problem. The table shows a person's heart rate during the first 4 minutes after exercise has stopped. $$\begin{array}{|l|c|c|c|} \hline \text { Time (min) } & 0 & 2 & 4 \\ \hline \text { Heart rate (bpm) } & 154 & 106 & 90 \\ \hline \end{array}$$ (a) Find a formula \(f(x)=a(x-h)^{2}+k\) that models the data, where \(x\) represents time and \(0 \leq x \leq 4 .\) Use \((4,90)\) as the vertex. (b) Evaluate \(f(1)\) and interpret the result. (c) Estimate the times when the heart rate was from 115 to 125 beats per minute.

Step-by-Step Solution

Answer

Formula: \(f(x) = 4(x-4)^2 + 90\). At 1 min, heart rate is 126 bpm. Heart rate 115-125 bpm from 1.042 to 1.5 min.

Step 1: Set up the Vertex Form Equation

The problem provides the vertex of the parabola as the point \((4, 90)\). Using this, we can write the vertex form of the quadratic function \(f(x) = a(x-4)^2 + 90\). Our task now is to find the value of \(a\).

Step 2: Use Known Point for Equation

Use the point \((0, 154)\) to solve for \(a\). Substitute \(x = 0\) and \(f(x) = 154\) into the vertex form equation: \(154 = a(0-4)^2 + 90\). Simplifying this gives \(154 = 16a + 90\).

Step 3: Solve for 'a'

Subtract 90 from both sides: \(154 - 90 = 16a\), resulting in \(64 = 16a\). Divide both sides by 16 to get \(a = 4\). So, the formula is \(f(x) = 4(x-4)^2 + 90\).

Step 4: Evaluate f(1)

Substitute \(x = 1\) into the formula: \(f(1) = 4(1-4)^2 + 90\). Calculate the value: \(f(1) = 4(9) + 90 = 36 + 90 = 126\). The heart rate after 1 minute is 126 bpm.

Step 5: Set Up Inequality for Estimation

To find when the heart rate was between 115 and 125 bpm, set up the inequality \(115 \leq 4(x-4)^2 + 90 \leq 125\).

Step 6: Solve the Inequality for Time

Separate into two inequalities and solve. First solve \(4(x-4)^2 + 90 = 115\): 1. Subtract 90: \(4(x-4)^2 = 25\) 2. Divide by 4: \((x-4)^2 = 6.25\) 3. Solve \(x - 4 = \pm 2.5\) 4. So, \(x = 6.5, 1.5\).Next, solve \(4(x-4)^2 + 90 = 125\):1. Subtract 90: \(4(x-4)^2 = 35\)2. Divide by 4: \((x-4)^2 = 8.75\)3. Solve \(x - 4 = \pm 2.958\)4. So, \(x = 6.958, 1.042\).

Step 7: Conclude the Time Interval

The heart rate is between 115 and 125 bpm when \(1.042 \leq x \leq 1.5\). This means the heart rate was in that range approximately between 1.042 and 1.5 minutes after exercise stopped.

Key Concepts

Vertex Form and Its Importance in Quadratic Functions

In order to understand quadratic functions more effectively, it is helpful to use the vertex form of a quadratic equation. This form is particularly useful when you know the vertex of the parabola. A quadratic function can be expressed as:

• Vertex Form: \( f(x) = a(x-h)^2 + k \)

The vertex of the parabola is represented by the point \((h, k)\). This point gives us valuable information about the graph itself, such as its highest or lowest point depending on the orientation of the parabola. The value of "a" affects the width and direction of the parabola's opening. If "a" is positive, the parabola opens upwards, and if "a" is negative, it opens downwards.
To solve the original problem, we used the given vertex \((4, 90)\) to set up the equation \(f(x) = a(x-4)^2 + 90\). Then, by substituting the known point \((0, 154)\) into the equation, we solved for "a". Once "a" is found, the exact model of this quadratic function can be used to further explore and predict heart rates post-exercise.

Modeling Heart Rate After Exercise

Heart rate can vary significantly with exercise and during the recovery period. When analyzing heart rate data, particularly during recovery, a quadratic model can be a good tool. In the given problem, heart rate was measured over time to establish a model for the data. A suitable quadratic equation was needed to predict heart rate at any given time after exercise.
After figuring out the function \(f(x) = 4(x-4)^2 + 90\), we evaluated \(f(1)\) to interpret the heart rate one minute post-exercise. The calculation showed that the heart rate was 126 bpm after one minute, which represents a visible decrease from the initial rate of 154 bpm.
This kind of modeling helps in understanding how quickly a heart rate returns to a normal state after the intensities of exercise have subsided.

Solving Inequalities for Time Estimation

Solving inequalities is a key technique used to find when events occur within specific limits. In this scenario, the aim was to estimate the timeframe during which the heart rate fell between 115 and 125 beats per minute (bpm) after exercise.
First, the inequality \(115 \leq 4(x-4)^2 + 90 \leq 125\) was set up. This inequality splits into two separate inequalities:

• \(\; 4(x-4)^2 + 90 = 115 \)
• \(\; 4(x-4)^2 + 90 = 125 \)

After solving these inequalities, the solutions indicated that the heart rate was within the desired range between approximately 1.042 minutes and 1.5 minutes.
By solving inequalities, we not only defined a specific model for heart rate reduction but also captured the precise moments when the rate passed through this interval. This technique can be quite useful in medical and fitness-related fields where understanding recovery phases is crucial.