Problem 12 - A Graphical Approach to Precalculus with Limits

Question

Work each problem related to linear functions. (a) Evaluate \(f(-2)\) and \(f(4)\) (b) Graph \(f\). How can the graph of \(f\) be used to determine the zero of \(f ?\) (c) Find the zero of \(f\) $$f(x)=\frac{1}{4} x+\frac{1}{2}$$

Step-by-Step Solution

Answer

The zero of \(f\) is \(x = -2\).

Step 1: Evaluate f(-2)

To evaluate \(f(-2)\), substitute \(x = -2\) into the function \(f(x) = \frac{1}{4}x + \frac{1}{2}\). This gives us: \[ f(-2) = \frac{1}{4}(-2) + \frac{1}{2} = -\frac{1}{2} + \frac{1}{2} = 0. \] Therefore, \(f(-2) = 0.\)

Step 2: Evaluate f(4)

Substitute \(x = 4\) into the function \(f(x) = \frac{1}{4}x + \frac{1}{2}\). This gives us: \[ f(4) = \frac{1}{4}(4) + \frac{1}{2} = 1 + \frac{1}{2} = \frac{3}{2}. \] Therefore, \(f(4) = \frac{3}{2}.\)

Step 3: Graph the Function

To graph \(f(x) = \frac{1}{4}x + \frac{1}{2}\), identify the y-intercept at \(\left(0, \frac{1}{2}\right)\) and use the slope \(\frac{1}{4}\). The slope indicates that for every 4 units increase horizontally (x-direction), there is a 1 unit increase vertically (y-direction). Plot more points or use these properties to sketch the line.

Step 4: Determine Zero from Graph

The zero of the function is where the graph crosses the x-axis. From the graph of \(f(x)\), identify the x-coordinate at this intersection point, which should be consistent with our earlier calculations.

Step 5: Find the Zero Algebraically

Set \(f(x) = 0\) to find when the function equals zero: \[ \frac{1}{4}x + \frac{1}{2} = 0. \] Solve for \(x\): \[ \frac{1}{4}x = -\frac{1}{2}, \quad x = -2. \] Thus, the zero of the function is \(x = -2\).

Key Concepts

Function Evaluation

Function evaluation is the process of calculating the value of a function for specific input values. In our exercise, we evaluated the function \(f(x)=\frac{1}{4}x+\frac{1}{2}\) at \(x = -2\) and \(x = 4\). By substituting \(-2\) into the function, we solved:

• \(f(-2) = \frac{1}{4}(-2) + \frac{1}{2} = -\frac{1}{2} + \frac{1}{2} = 0\)

Similarly, substituting \(4\) we found:

• \(f(4) = \frac{1}{4}(4) + \frac{1}{2} = 1 + \frac{1}{2} = \frac{3}{2}\)

Function evaluation is important because it tells us the output of a function for any given input. This process is essential for understanding the behavior of functions in different situations.

Graphing Functions

Graphing a linear function involves plotting the line on a coordinate plane based on its equation. For the function \(f(x) = \frac{1}{4}x + \frac{1}{2}\), we first identify the y-intercept, which is \(\frac{1}{2}\). This means the line crosses the y-axis at the point \((0, \frac{1}{2})\). Next, consider the slope, \(\frac{1}{4}\), which indicates that for every 4 units of horizontal movement, the function value increases by 1 unit vertically.
By marking the y-intercept, then using the slope to plot a second point, we can draw the line of the function. These two components, the y-intercept and the slope, fully define the line in the graph. Graphing helps visually depict how changes in \(x\) affect \(f(x)\) and is essential to understand a function's overall behavior.

Finding Zeros

Finding the zeros of a function means determining the points where the function value equals zero. For a linear function like \(f(x) = \frac{1}{4}x + \frac{1}{2}\), setting the function equal to zero helps find these points algebraically.

• \[\frac{1}{4}x + \frac{1}{2} = 0\]
• Solving this equation involves isolating \(x\):\(\frac{1}{4}x = -\frac{1}{2}\)
• Thus, \(x = -2\)

This result shows that the function changes from negative to positive or vice versa as it crosses the x-axis. Knowing where the function's zero is located allows us to understand crucial points on a graph and is key for analyzing the function's behavior.

Slope and Intercept

The slope and intercept are fundamental concepts in understanding linear functions. The slope reflects the rate of change of the function. It indicates how steep the line is. For the function \(f(x) = \frac{1}{4}x + \frac{1}{2}\), the slope is \(\frac{1}{4}\). This means for each increase of 4 units in \(x\), \(f(x)\) rises by 1 unit.
The y-intercept is where the function crosses the y-axis. This value is \(\frac{1}{2}\). It's the output of the function when \(x\) is zero. Understanding these two values helps quickly determine the function graph's shape without having to plot several points.

• These parameter values help when predicting the function's response to input changes and understanding its real-world implications.

In essence, the slope and intercept are keys to unlocking the characteristics and applications of linear functions.