4.1 Linear and Quadratic Functions

4.1 Linear and Quadratic Functions

Polynomial functions form the foundation of algebraic analysis in Precalculus. The simplest non-trivial polynomials are linear functions (degree 1) and quadratic functions (degree 2). Mastering their structural forms, graphical behaviors, and analytical properties is essential for CLEP Precalculus success.

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1. Linear Functions: Rates of Change and Structural Forms

A linear function is any function that can be expressed in the form: $f(x) = mx + b$ where $m$ and $b$ are real constants. The parameter $m$ represents the slope (average rate of change), while $b$ represents the $y$-intercept $(0, b)$.

Slope Definition and Properties

Given any two distinct points $(x_1, y_1)$ and $(x_2, y_2)$ on a non-vertical line, the slope $m$ is defined as: $m = \frac{\Delta y}{\Delta x} = \frac{y_2 - y_1}{x_2 - x_1}, \quad \text{where } x_1 \neq x_2$

• Positive Slope ($m > 0$): The line rises from left to right; the function is strictly increasing.
• Negative Slope ($m < 0$): The line falls from left to right; the function is strictly decreasing.
• Zero Slope ($m = 0$): The line is horizontal ($y = b$); the function is constant.
• Undefined Slope: A vertical line ($x = c$) has no defined slope and is not a function of $x$.

Common Forms of Linear Equations

Depending on the given information, linear equations can be written in three primary algebraic forms:

Parallel and Perpendicular Lines

Let Line 1 have slope $m_1$ and Line 2 have slope $m_2$:

• Parallel Lines ($\parallel$): Two non-vertical lines are parallel if and only if their slopes are equal: $m_1 = m_2$
• Perpendicular Lines ($\perp$): Two non-vertical lines are perpendicular if and only if their slopes are negative reciprocals: $m_1 \cdot m_2 = -1 \implies m_2 = -\frac{1}{m_1}$

Key Concept: Horizontal lines ($m = 0$) are perpendicular to vertical lines (undefined slope).

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Step-by-Step Worked Example: Constructing Perpendicular Linear Equations

Problem: Find the equation of the line passing through the point $P(-4, 3)$ that is perpendicular to the line $3x - 2y = 8$. Express your answer in slope-intercept form.

Solution:

1. Find the slope of the given line: Convert $3x - 2y = 8$ to slope-intercept form: $-2y = -3x + 8 \implies y = \frac{3}{2}x - 4$ The slope of the given line is $m_1 = \frac{3}{2}$.

2. Determine the perpendicular slope: Taking the negative reciprocal: $m_\perp = -\frac{1}{m_1} = -\frac{2}{3}$

3. Apply point-slope form with $P(-4, 3)$: $y - 3 = -\frac{2}{3}(x - (-4))$ $y - 3 = -\frac{2}{3}(x + 4)$ $y - 3 = -\frac{2}{3}x - \frac{8}{3}$

4. Convert to slope-intercept form: $y = -\frac{2}{3}x - \frac{8}{3} + \frac{9}{3} \implies y = -\frac{2}{3}x + \frac{1}{3}$

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2. Quadratic Functions: Forms, Graphs, and Characteristics

A quadratic function is a polynomial function of degree 2: $f(x) = ax^2 + bx + c \quad (a \neq 0)$ The graph of a quadratic function is a U-shaped curve called a parabola.

Standard Form vs. Vertex Form

While standard form $f(x) = ax^2 + bx + c$ explicitly exposes the $y$-intercept $(0, c)$, vertex form exposes key graphical features: $f(x) = a(x - h)^2 + k$ where $(h, k)$ is the vertex of the parabola.

• Axis of Symmetry: The vertical line $x = h$.
• Direction of Opening:
  • If $a > 0$, the parabola opens upward ($\cup$), and the vertex $(h, k)$ is the absolute minimum.
  • If $a < 0$, the parabola opens downward ($\cap$), and the vertex $(h, k)$ is the absolute maximum.

Converting Standard Form to Vertex Form (Completing the Square)

From standard form $f(x) = ax^2 + bx + c$, the vertex coordinates $(h, k)$ can be calculated directly: $h = -\frac{b}{2a}, \quad k = f(h) = c - \frac{b^2}{4a}$

Alternatively, complete the square algebraically:

1. Factor $a$ out of the first two terms: $f(x) = a\left(x^2 + \frac{b}{a}x\right) + c$.
2. Add and subtract $\left(\frac{b}{2a}\right)^2$ inside the parentheses.
3. Factor the perfect square trinomial to obtain $f(x) = a(x - h)^2 + k$.

Zeros and the Discriminant

The $x$-intercepts (zeros) of $f(x) = ax^2 + bx + c$ are found by solving $ax^2 + bx + c = 0$ using the Quadratic Formula: $x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$

The expression under the radical is the discriminant, denoted $\Delta = b^2 - 4ac$:

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Step-by-Step Worked Example: Vertex Form and Intercept Analysis

Problem: Given $f(x) = -2x^2 + 12x - 10$, (a) rewrite $f(x)$ in vertex form, (b) identify the vertex, axis of symmetry, domain, and range, and (c) find all $x$- and $y$-intercepts.

Solution:

1. Rewrite in vertex form via completing the square: $f(x) = -2(x^2 - 6x) - 10$ Half of $-6$ is $-3$, and $(-3)^2 = 9$. Add and subtract $9$ inside: $f(x) = -2(x^2 - 6x + 9 - 9) - 10$ $f(x) = -2(x^2 - 6x + 9) + 18 - 10$ $f(x) = -2(x - 3)^2 + 8$

2. Identify key properties:

   • Vertex: $(h, k) = (3, 8)$.
   • Axis of Symmetry: $x = 3$.
   • Opening: Since $a = -2 < 0$, the parabola opens downward.
   • Extremum: Absolute maximum of $y = 8$ at $x = 3$.
   • Domain: $(-\infty, \infty)$.
   • Range: $(-\infty, 8]$.

3. Find Intercepts:

   • $y$-intercept: Evaluate $f(0) = -2(0)^2 + 12(0) - 10 = -10 \implies (0, -10)$.
   • $x$-intercepts: Set $f(x) = 0$: $-2(x - 3)^2 + 8 = 0 \implies -2(x - 3)^2 = -8 \implies (x - 3)^2 = 4$ $x - 3 = \pm 2 \implies x = 3 \pm 2$ The $x$-intercepts are $(1, 0)$ and $(5, 0)$.

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3. Real-World Optimization with Quadratic Functions

Quadratic optimization models real-world problems where a quantity reaches a maximum or minimum value. Because the absolute extremum of a parabola occurs strictly at its vertex $(h, k)$, optimization problems reduce to evaluating $h = -\frac{b}{2a}$ and computing $k = f(h)$.

Optimization Steps:

1. Identify the quantity to be optimized and set up an equation in terms of one variable.
2. Ensure the resulting function is quadratic: $f(x) = ax^2 + bx + c$.
3. Check the leading coefficient $a$:
   • If $a < 0$, the vertex yields the maximum value $k$ at $x = h$.
   • If $a > 0$, the vertex yields the minimum value $k$ at $x = h$.
4. Interpret the solution within the physical domain constraints.

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Step-by-Step Worked Example: Revenue Optimization

Problem: A theater company determines that at a ticket price of $$20$, they sell $400$ tickets per performance. For each $$1$ increase in ticket price, they lose $10$ attendees. What ticket price maximizes total revenue, and what is the maximum revenue?

Solution:

1. Define the variable: Let $x$ be the number of $$1$ price increases.

   • New price per ticket: $P(x) = 20 + x$
   • Number of tickets sold: $N(x) = 400 - 10x$

2. Formulate Revenue function $R(x) = P(x) \cdot N(x)$: $R(x) = (20 + x)(400 - 10x)$ $R(x) = 8000 - 200x + 400x - 10x^2$ $R(x) = -10x^2 + 200x + 8000$

3. Find the vertex of $R(x)$: Here $a = -10$, $b = 200$, $c = 8000$. $h = -\frac{b}{2a} = -\frac{200}{2(-10)} = 10$

4. Calculate optimal price and maximum revenue:

   • Optimal number of price increases: $x = 10$.
   • Optimal ticket price: $P(10) = 20 + 10 = $30$.
   • Maximum revenue: $R(10) = -10(10)^2 + 200(10) + 8000 = -1000 + 2000 + 8000 = $9,000$.

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4. Common CLEP Pitfalls & Exam Traps

Trap 1: Confusing "Where the extremum occurs" with "What the extremum is". If a question asks for the maximum value of $f(x) = -3(x - 4)^2 + 15$, the answer is $y = 15$ ($k$), NOT $x = 4$ ($h$). $x = 4$ is where it occurs.

Trap 2: Sign Errors in Completing the Square. When factoring out a negative leading coefficient, ensure you flip the sign of the linear coefficient. For $f(x) = -2x^2 + 12x$, factor out $-2$ to get $-2(x^2 - 6x)$, NOT $-2(x^2 + 6x)$.

Trap 3: Misinterpreting Perpendicular Slopes. The perpendicular slope of $m = -\frac{1}{4}$ is $+4$, not $+\frac{1}{4}$ (must negate AND invert).