8.2 Linear Functions: Slope, Rate of Change, & Linear Equations

8.2 Linear Functions: Slope, Rate of Change, & Linear Equations

Linear functions represent the most widely used mathematical models on the TABE 13&14 Mathematics assessment and across technical trades. A linear relationship describes any scenario where a dependent variable changes at a constant, steady rate relative to an independent variable—such as steady vehicle fuel consumption, constant hourly wage accumulation, or linear thermal expansion in piping.

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1. Slope as a Constant Rate of Change

The defining characteristic of any linear function is that its rate of change is constant. On a coordinate plane, this constant rate of change is called the slope ($m$) of the line. Geometrically, slope quantifies the steepness and direction of a line, defined as the ratio of vertical change ("rise") to horizontal change ("run"):

$$m = \frac{\text{Vertical Change (Rise)}}{\text{Horizontal Change (Run)}} = \frac{\Delta y}{\Delta x} = \frac{y_2 - y_1}{x_2 - x_1} \quad (x_2 \neq x_1)$$

Step-by-Step Algebraic Slope Calculation

To calculate the slope between two points $(x_1, y_1)$ and $(x_2, y_2)$:

1. Assign coordinates carefully: let $(x_1, y_1) = (-4, 7)$ and $(x_2, y_2) = (6, -8)$.
2. Substitute into the slope formula: $$m = \frac{-8 - 7}{6 - (-4)} = \frac{-15}{6 + 4} = \frac{-15}{10} = -\frac{3}{2} = -1.5$$
3. Workplace Interpretation: For every increase of $2$ units in $x$, the quantity $y$ decreases by $3$ units.

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2. The Four Classifications of Slope

Every straight line in coordinate geometry falls into one of four distinct slope categories:

Four Geometric Slope Classifications:

  1. POSITIVE (m > 0)        2. NEGATIVE (m < 0)        3. ZERO (m = 0)            4. UNDEFINED (m = undefined)
        /                          \                                                      |
       /                            \                     ───────────────────             |
      /                              \                                                    |
   Rises L -> R                 Falls L -> R              Horizontal (y = c)         Vertical (x = c)

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3. The Three Forms of Linear Equations

Depending on the given information (slope, points, intercepts), linear equations can be written in three algebraic forms:

A. Slope-Intercept Form: $y = mx + b$

• Components: $m = \text{slope}$, $b = y\text{-intercept}$ (the point $(0, b)$ where the graph crosses the vertical $y$-axis).
• Best used when: The rate of change and starting value are known, or when graphing directly.

B. Point-Slope Form: $y - y_1 = m(x - x_1)$

• Components: $m = \text{slope}$, $(x_1, y_1) = \text{any specific point on the line}$.
• Best used when: Given a slope and a single point, or when calculating a line through two given points.

C. Standard Form: $Ax + By = C$

• Components: $A, B, C$ are integers (with $A \ge 0$, and $A, B$ not both zero).
• Best used when: Finding coordinate intercepts quickly ($x$-intercept and $y$-intercept) or modeling combination constraints (e.g., $5x + 3y = 60$).

Quick Method: Intercepts from Standard Form

To graph $Ax + By = C$ using intercepts:

1. $x$-intercept (set $y = 0$): $Ax + B(0) = C \implies x = \frac{C}{A} \implies \left(\frac{C}{A}, 0\right)$.
2. $y$-intercept (set $x = 0$): $A(0) + By = C \implies y = \frac{C}{B} \implies \left(0, \frac{C}{B}\right)$.

Example: Find the intercepts of $4x - 5y = 40$.

• Set $y = 0$: $4x = 40 \implies x = 10 \implies (10, 0)$.
• Set $x = 0$: $-5y = 40 \implies y = -8 \implies (0, -8)$.

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4. Converting Among Linear Forms (Step-by-Step)

Mastering algebraic transitions between forms is a core TABE skill:

Worked Demonstration: Find the equation of the line passing through $(2, -3)$ and $(6, 5)$ in slope-intercept form and standard form.

1. Calculate Slope ($m$): $$m = \frac{5 - (-3)}{6 - 2} = \frac{5 + 3}{4} = \frac{8}{4} = 2$$
2. Apply Point-Slope Form using $(2, -3)$: $$y - (-3) = 2(x - 2) \implies y + 3 = 2x - 4$$
3. Convert to Slope-Intercept Form ($y = mx + b$): $$y = 2x - 4 - 3 \implies y = 2x - 7$$
4. Convert to Standard Form ($Ax + By = C$): $$-2x + y = -7 \implies 2x - y = 7 \quad (\text{Multiply by } -1 \text{ so } A > 0)$$

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5. Parallel vs. Perpendicular Lines

The geometric relationship between two straight lines is governed entirely by their slopes:

Line Relationships:
├── Parallel Lines ─────────> Identical Slopes: m1 = m2  (Different intercepts b1 ≠ b2)
└── Perpendicular Lines ────> Negative Reciprocal Slopes: m1 · m2 = -1  (m2 = -1 / m1)

Worked Example: Finding Perpendicular Equations

Problem: Find the slope-intercept equation of the line that passes through $(4, -1)$ and is perpendicular to the line $2x - 3y = 9$.

1. Find slope of the given line: $$-3y = -2x + 9 \implies y = \frac{2}{3}x - 3 \implies m_1 = \frac{2}{3}$$
2. Determine perpendicular slope ($m_2$): $$m_2 = -\frac{1}{m_1} = -\frac{3}{2}$$
3. Use Point-Slope Form with $(4, -1)$: $$\begin{aligned} y - (-1) &= -\frac{3}{2}(x - 4) \\ y + 1 &= -\frac{3}{2}x + \left(-\frac{3}{2}\right)(-4) \\ y + 1 &= -\frac{3}{2}x + 6 \\ y &= -\frac{3}{2}x + 5 \end{aligned}$$

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6. Applied Linear Modeling: Workplace Word Problems & Tables

In practical applications, linear functions model scenarios with a fixed initial amount (the $y$-intercept $b$) plus a variable rate per unit (the slope $m$):

$$\text{Total Cost / Output: } f(x) = (\text{Rate of Change}) \cdot x + (\text{Fixed Starting Value}) = mx + b$$

Worked Workplace Problem: Industrial Tank Drainage

Problem: An industrial water tank is drained at a constant rate. A telemetry sensor records the remaining volume: at $t = 3\text{ hours}$, volume is $4,100\text{ gallons}$; at $t = 8\text{ hours}$, volume is $2,750\text{ gallons}$.

1. Find Rate of Drainage (Slope): $$m = \frac{2750 - 4100}{8 - 3} = \frac{-1350}{5} = -270\text{ gallons per hour}$$
2. Find Initial Tank Capacity ($b$): $$V(t) = -270t + b \implies 4100 = -270(3) + b \implies 4100 = -810 + b \implies b = 4910\text{ gallons}$$
3. Complete Linear Model: $V(t) = -270t + 4910$.
4. Time to Empty Tank ($V(t) = 0$): $$0 = -270t + 4910 \implies 270t = 4910 \implies t = \frac{4910}{270} \approx 18.19\text{ hours}$$