Comprehensive Guide to Graphing and Applying Linear Functions

Core Competencies in Linear Functions

The fundamental objectives when studying linear functions involve several key analytical skills. Students must be able to graph a linear function and precisely determine its essential components. These components include the domain, which represents the set of all possible input values for the variable $x$, and the range, which represents the set of all possible output values for the variable $y$. Additionally, students are expected to identify the intercepts, specifically the $x$-intercept (where the line crosses the horizontal axis at $y = 0$) and the $y$-intercept (where the line crosses the vertical axis at $x = 0$). Understanding and calculating the slope ($m$), which defines the steepness and direction of the line, is also required. Beyond theoretical graphing, the competencies include representing real-life relationships using various formats—such as verbal descriptions, tables, graphs, and equations—and solving complex word problems that involve linear models.

Methodologies for Graphing Linear Functions

There are several distinct methods to graph a linear function, depending on the information provided or the form of the equation. Each method offers a systematic way to represent the linear relationship on a Cartesian plane.

The first method is using intercepts. This approach is highly efficient for equations written in standard form ($Ax + By = C$). The procedure involves three steps: first, set $y = 0$ to solve for the $x$-intercept; second, set $x = 0$ to solve for the $y$-intercept; and third, plot these two points on the respective axes and draw a straight line through them. For example, in the equation $2x + y = 4$, setting $y = 0$ yields $2x = 4$, giving an $x$-intercept of $(2, 0)$. Setting $x = 0$ yields $y = 4$, giving a $y$-intercept of $(0, 4)$. The resulting graph has a domain and range of all real numbers, expressed as $\{x | x \in \mathbb{R}\}$ or $(-\infty, \infty)$, and a slope of $m = -2$.

The second method utilizes the slope-intercept form, defined by the equation $y = mx + b$. In this form, $m$ represents the slope (calculated as $\frac{\text{rise}}{\text{run}}$) and $b$ represents the $y$-intercept $(0, b)$. To graph using this method, identify the $y$-intercept and plot it first. Then, use the slope's rise and run to find a second point. Finally, connect the points. In Example 2 ($y = \frac{1}{2}x + 1$), the slope is $m = \frac{1}{2}$ and the $y$-intercept is $(0, 1)$. Starting at $(0, 1)$, a rise of $1$ and a run of $2$ leads to the point $(2, 2)$. The $x$-intercept is identified at $(-2, 0)$.

The third method involves using a table of values. This requires choosing convenient $x$-values to calculate corresponding $y$-values. In Example 3, given $2x + 3y = 6$, specific points found in the table include $(-3, 4)$, $(0, 2)$, and $(3, 0)$. Plotting these points reveals a line with a slope of $m = -\frac{2}{3}$, an $x$-intercept of $(3, 0)$, and a $y$-intercept of $(0, 2)$.

The fourth method involves using two given points. If a linear function passes through two specific coordinates, such as $(2, 3)$ and $(-2, -1)$, the line can be drawn simply by connecting them. In this specific case, the slope is calculated as $m = 1$. The line crosses the axes at the $x$-intercept of $(-1, 0)$ and the $y$-intercept of $(0, 1)$.

The fifth method involves using a slope and a single point. For example, a function with a slope of $m = -\frac{1}{4}$ passing through the point $(4, -2)$ can be graphed by starting at the given point and applying the negative slope. This specific function results in an $x$-intercept at $(-4, 0)$ and a $y$-intercept at $(0, -1)$.

Linear Relationships in Real-Life Contexts

A linear relationship occurs whenever one quantity changes at a constant rate relative to another quantity. Real-world scenarios are modeled using four primary representations: verbal descriptions, tables of values, graphs, and algebraic equations. To effectively model these situations, one must identify the rate of change (slope) and the starting value (the constant or $y$-intercept).

A classic example is transportation fares. A modern jeepney might charge a base fare of $\text{‑}17$ for the first $4\,\text{km}$ and $\text{‑}2.30$ for every additional kilometer. If we let $x$ represent the kilometers traveled beyond the first $4\,\text{km}$ and $y$ represent the total fare, the linear function form is $y = 2.30x + 17$. This demonstrates a starting value of $17$ and a constant rate of increase.

Additional examples of algebraic modeling include:

1. Total earnings for a worker who makes $\text{‑}400$ per day: $y = 400x$, where $x$ is the number of days worked.
2. Distance traveled by a bicycle at a constant speed of $12\,\text{km/h}$: $y = 12x$, where $x$ is the number of hours.
3. Laundry service costs with a fixed fee of $\text{‑}30$ plus $\text{‑}45$ per kilogram: $y = 45x + 30$, where $x$ is the weight in kilograms.
4. Printing shop charges with a $\text{‑}15$ layout fee and $\text{‑}3$ per page: $y = 3x + 15$, where $x$ is the number of pages.

Practical Problem-Solving Applications

Applying linear functions allows for the prediction of costs, earnings, and quantities in various service-based and commercial scenarios. Below are specific problem cases derived from real-world operations:

In retail and services: A school canteen selling sandwiches at $\text{‑}35$ each can be modeled as $y = 35x$. For a sari-sari store selling bottled water at $\text{‑}15$ each, the total receipt for selling $24$ bottles is calculated as $15 \times 24 = 360$.

In transportation and logistics: A delivery service charging a fixed booking fee of $\text{‑}40$ plus $\text{‑}12$ per kilometer yields the equation $y = 12x + 40$. A regional tricycle fare system charging $\text{‑}16$ for the first $3\,\text{km}$ and $\text{‑}2.50$ for each additional kilometer can be used to find the fare for a $9\,\text{km}$ trip. Here, the additional distance is $9 - 3 = 6\,\text{km}$, making the total fare $16 + (2.50 \times 6) = 16 + 15 = 31$.

In employment and commissions: A student tutor earning $\text{‑}150$ per hour, working $4$ hours a day for $3$ days, calculates total earnings as $150 \times 4 \times 3 = 1800$. In sales, a salesperson earning a base salary of $\text{‑}9500$ and a $\text{‑}400$ commission per phone sold would earn $9500 + (400 \times 7) = 12300$ for selling $7$ phones in a month.

Questions & Discussion

Pre-test and Post-test Evaluation

Question 1: In a certain municipality, the Rural Health Unit recorded the number of Persons Under Investigation (PUI) for COVID-19 starting on the first month of community quarantine. They observed a constant increase. If the pattern continues, can you predict the number of PUIs in the 7th month? Answer Options: A. No, because data is insufficient. B. No, because it is not stipulated. C. Yes, the number is 85. D. Yes, the number is 97.

Question 2: A delivery service charges $\text{‑}50$ base fee plus $\text{‑}15$ per kilometer. Which equation represents the total cost ($y$)? Answer Options: A. $y = 15x$; B. $y = 50x + 15$; C. $y = 15x + 50$; D. $y = 50 + x$. Analysis: The correct equation is C, as $\text{‑}15$ represents the rate ($m$) and $\text{‑}50$ the fixed value ($b$).

Question 3: A graph shows a line decreasing from $(0, 10)$ to $(5, 0)$. Which equation represents this relationship? Answer Options: A. $y = 2x + 10$; B. $y = -2x + 10$; C. $y = -10x + 2$; D. $y = x - 10$. Analysis: The slope $m = \frac{0 - 10}{5 - 0} = -2$. The $y$-intercept is $10$. Therefore, B is correct.

Question 4 to 6 Context: A small online delivery business charges a base fee plus an additional charge by distance. The table shows: $1\,\text{km}$ costs $\text{‑}60$, $2\,\text{km}$ costs $\text{‑}80$, $3\,\text{km}$ costs $\text{‑}100$, and $4\,\text{km}$ costs $\text{‑}120$.

Question 4: Which statement best describes the relationship? Answer Choice: C. The cost increases at a constant rate.

Question 5: Which equation represents the relationship? Analysis: The cost increases by $\text{‑}20$ per km. If $x = 1, y = 60$. Using $y = mx + b \rightarrow 60 = 20(1) + b \rightarrow b = 40$. The equation is $y = 20x + 40$. (Option B).

Question 6: If a customer travels $10\,\text{km}$, what is the total delivery cost? Analysis: $y = 20(10) + 40 = 200 + 40 = 240$. (Option C).