Problem 16 - 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 \(f\) $$f(x)=x+0.5$$

Step-by-Step Solution

Answer

(a) \(f(-2) = -1.5\), \(f(4) = 4.5\); (b) Zero is \(-0.5\); (c) Zero is \(-0.5\).

Step 1: Evaluate f(-2)

To evaluate the function at \(x = -2\), substitute \(-2\) in for \(x\) in the function: \[f(-2) = -2 + 0.5 = -1.5\] Thus, \(f(-2) = -1.5\).

Step 2: Evaluate f(4)

To evaluate the function at \(x = 4\), substitute \(4\) in for \(x\) in the function: \[f(4) = 4 + 0.5 = 4.5\] Thus, \(f(4) = 4.5\).

Step 3: Graph f(x)

The linear function \(f(x) = x + 0.5\) can be graphed as a straight line. This line has a slope of \(1\) (indicating it rises 1 unit for each unit it moves right) and a y-intercept at \(0.5\) (indicating it crosses the y-axis at \(0.5\)). Plot these points and draw the line through them.

Step 4: Determine Zero of f(x) from Graph

The zero of the function \(f(x)\) is the x-value where the graph crosses the x-axis. Looking at the graph, this is where the line intersects the x-axis. Since the y-coordinate is \(0\) at this intersection, it corresponds to solving \(x + 0.5 = 0\). From visual inspection, this appears to occur at \(-0.5\).

Step 5: Calculate Zero of f(x) Algebraically

To find the zero algebraically, solve \(f(x) = 0\): \[x + 0.5 = 0\] Subtract \(0.5\) from both sides: \[x = -0.5\] Thus, the zero of the function is \(x = -0.5\).

Key Concepts

Evaluating Functions

Evaluating functions is a fundamental skill in mathematics, especially when dealing with linear functions like the one given in the problem. When we evaluate a function, we're essentially plugging in specific values for the variable and performing the operations outlined in the function's expression. For example:

\( f(x) = x + 0.5 \)

When evaluating \( f(-2) \), you substitute \(-2\) for \(x\) in the equation:

• \( f(-2) = -2 + 0.5 = -1.5 \)

Similarly, for \( f(4) \), substitute \(4\) for \(x\):

• \( f(4) = 4 + 0.5 = 4.5 \)

Evaluating functions helps us understand how the output (or y-value) changes as the input (or x-value) changes. It's like performing an activity with instructions - just follow the steps and you'll arrive at the answer!

This process is crucial when graphing or finding zeros, setting the stage for deeper analysis of the function.

Graphing Linear Equations

Graphing linear equations is an excellent way to visualize how functions behave. For the function \( f(x) = x + 0.5 \), you are essentially graphing a straight line on a coordinate plane. This line will have a slope, indicating the steepness, and an intercept, indicating where it crosses the y-axis.

With \( f(x) = x + 0.5 \), the slope is \(1\), meaning for each unit increase in \(x\), \(y\) also increases by one unit. The y-intercept is \(0.5\), so the line crosses the y-axis at point (0, 0.5). Here's how you graph it:

• Start at the y-intercept (0, 0.5).
• From that point, use the slope to find another point: move right 1 unit and up 1 unit to plot the next point (1, 1.5).
• Draw a straight line through these points, extending it across the graph.

The graph not only shows the function's general progression but also lets us visually identify key features like intercepts and slope, assisting in understanding the overall behavior of the function.

Finding Zeros of Functions

Finding zeros of functions is akin to solving the puzzle where the goal is to identify points where the function equals zero. These points are often referred to as x-intercepts because that's where the graph crosses the x-axis.

For our linear function \( f(x) = x + 0.5 \), the zero is found by setting the function equal to zero and solving for \(x\):

• \( x + 0.5 = 0 \)

Subtract \(0.5\) from both sides:

• \( x = -0.5 \)

Thus, the function's zero is at \(x = -0.5\). On the graph, this appears where the line toggles the x-axis. Recognizing zeros is crucial as they often mark crucial points in real-world scenarios, like determining break-even points in business or transitions in trends.