Problem 56 - College Algebra with Modeling and Visualization

Question

Compute the average rate of change of \(f\) from \(x_{1}\) to \(x_{2}\). Round your answer to two decimal places when appropriate. Interpret your result graphically. $$ f(x)=0.5 x^{2}-5, x_{1}=-1, \text { and } x_{2}=4 $$

Step-by-Step Solution

Answer

The average rate of change is 1.5.

Step 1: Understand the formula for average rate of change

The average rate of change of a function \(f\) from \(x_1\) to \(x_2\) is given by \(\frac{f(x_2) - f(x_1)}{x_2 - x_1}\). This formula represents the slope of the secant line that passes through the points \((x_1, f(x_1))\) and \((x_2, f(x_2))\) on the graph of \(f\).

Step 2: Compute f(x_1) and f(x_2)

First, substitute \(x_1\) into the function to find \(f(-1)\). We have:\[f(-1) = 0.5(-1)^2 - 5 = 0.5(1) - 5 = 0.5 - 5 = -4.5\]Now substitute \(x_2\) into the function to find \(f(4)\). We have:\[f(4) = 0.5(4)^2 - 5 = 0.5(16) - 5 = 8 - 5 = 3\]

Step 3: Calculate the numerator of the average rate of change

Subtract \(f(x_1)\) from \(f(x_2)\):\[ f(4) - f(-1) = 3 - (-4.5) = 3 + 4.5 = 7.5\]

Step 4: Calculate the denominator of the average rate of change

Subtract \(x_1\) from \(x_2\):\[x_2 - x_1 = 4 - (-1) = 4 + 1 = 5\]

Step 5: Compute the average rate of change

Divide the numerator by the denominator:\[\frac{f(4) - f(-1)}{x_2 - x_1} = \frac{7.5}{5} = 1.5\]This result represents how much the function's output (\(f(x)\)) changes on average for each unit increase in \(x\) between \(x_1\) and \(x_2\).

Step 6: Interpret the result graphically

The average rate of change, 1.5, represents the slope of the secant line connecting the points \((-1, -4.5)\) and \((4, 3)\) on the graph of \(f(x) = 0.5x^2 - 5\). Graphically, this means that for every unit increase in \(x\) from \(-1\) to \(4\), the function's value increases by 1.5 on average.

Key Concepts

Understanding the Secant Line

A secant line is a straight line that intersects a curve at two or more points.
This line is crucial when discussing the average rate of change in functions.
Essentially, it gives us a visual representation of the change in the function between two points.

• The secant line represents the average rate of change over an interval.
• For a function like a quadratic, which curves, it provides a linear estimation between these points.
• In the context of the exercise, the secant line connects the points \((-1, -4.5)\) and \(4, 3 \) on the graph.

Seeing the secant line helps understand how the function behaves across these points.
It shows how the function averages out from one point to another, despite the actual curve being nonlinear.

The Quadratic Function

Quadratic functions are a type of polynomial function that have the general form
\(f(x) = ax^2 + bx + c\), where \(a\), \(b\), and \(c\) are constants.
They are characterized by their U-shaped graphs, known as parabolas.

• In the formula provided, \(a = 0.5, b = 0\), and \(c = -5\).
• The parabola opens upwards because the coefficient \(a\) is positive.
• The vertex or the turning point of this parabola gives the minimum value of the function.

These functions are important in algebra due to their symmetry and predictable shapes.
In our exercise, we are looking at how this specific quadratic function behaves from \(-1\) to \4\.
This quadratic describes a scenario where the function first decreases and then increases.

Calculating the Slope

When calculating the slope of a line, we essentially assess how steep that line is.
For the secant line, the slope is determined based on how rapidly the quadratic function changes between two points.
This slope is also the average rate of change and is computed using the formula: \[\text{slope} = \frac{f(x_2) - f(x_1)}{x_2 - x_1}\]

• Plugging in values from the exercise gives \(\text{slope} = \frac{3 - (-4.5)}{4 - (-1)} = \frac{7.5}{5} = 1.5\).
• This slope tells us that for each increase of 1 unit in \(x\), the function value increases by 1.5 units.

This slope not only helps in understanding the average change but also provides insight into how much the function's output alters with input.
This concept is vital to understand changes over an interval, especially in nonlinear functions like quadratics.