Rational Functions

Understand functions with variables in the denominator

You have seen fractions before. You have divided numbers, simplified expressions, and worked with ratios. A rational function is simply the next step: instead of dividing one number by another, you divide one polynomial by another. The word “rational” comes from “ratio,” and that is exactly what these functions are - ratios of polynomials.

Think about calculating your average speed on a road trip. If you drive 180 miles in $t$ hours, your average speed is $\frac{180}{t}$ miles per hour. That is a rational function - and notice something interesting happens as $t$ gets smaller. Drive those 180 miles in 3 hours, and your average speed is 60 mph. Drive them in 2 hours, and it jumps to 90 mph. What if you tried to drive them in 0 hours? The calculation breaks. You cannot divide by zero, and this fundamental limitation shapes everything about rational functions.

Core Concepts

What Is a Rational Function?

A rational function is any function that can be written as the ratio of two polynomials:

$$f(x) = \frac{p(x)}{q(x)}$$

where $p(x)$ and $q(x)$ are polynomials, and crucially, $q(x) \neq 0$. The polynomial on top is the numerator, and the polynomial on the bottom is the denominator.

Here are some examples of rational functions:

  • $f(x) = \frac{1}{x}$ (the simplest rational function)
  • $g(x) = \frac{x + 3}{x - 2}$
  • $h(x) = \frac{x^2 - 4}{x^2 + x - 6}$
  • $r(x) = \frac{2x^3 + x - 1}{x + 5}$

And here are some things that are NOT rational functions:

  • $f(x) = \frac{\sqrt{x}}{x + 1}$ (square roots are not polynomials)
  • $g(x) = \frac{2^x}{x}$ (exponential expressions are not polynomials)

Every polynomial is technically a rational function too - just put 1 in the denominator. So $f(x) = x^2 - 3x + 2$ is the same as $f(x) = \frac{x^2 - 3x + 2}{1}$. But when we talk about rational functions, we usually mean the interesting cases where the denominator has a variable in it.

Domain Restrictions: Where the Function Breaks

Here is the central challenge with rational functions: you cannot divide by zero. Ever. This means any $x$-value that makes the denominator equal to zero is forbidden - it is not in the domain.

To find the domain of a rational function:

  1. Set the denominator equal to zero
  2. Solve for $x$
  3. Exclude those values from the domain

For example, consider $f(x) = \frac{3}{x + 5}$. The denominator is $x + 5$. Setting it equal to zero: $x + 5 = 0$, so $x = -5$. The domain is all real numbers except $x = -5$.

In interval notation, you write this as: $(-\infty, -5) \cup (-5, \infty)$

Or in set-builder notation: ${x \mid x \neq -5}$

These excluded values are where the interesting behavior happens - they often become vertical asymptotes or holes in the graph.

Vertical Asymptotes: Walls the Graph Cannot Cross

A vertical asymptote is a vertical line that the graph approaches but never touches or crosses. As the $x$-values get closer and closer to the asymptote, the $y$-values shoot off toward positive or negative infinity.

To find vertical asymptotes:

  1. Factor both the numerator and denominator completely
  2. Cancel any common factors (these become holes, not asymptotes)
  3. Set the remaining denominator factors equal to zero
  4. Solve - these $x$-values are where the vertical asymptotes occur

For $f(x) = \frac{2}{x - 3}$, the denominator is zero when $x = 3$. Since there is nothing to cancel with the numerator, there is a vertical asymptote at $x = 3$.

The graph will approach this vertical line but never touch it. On one side, the function values will head toward $+\infty$; on the other side, toward $-\infty$.

Think of vertical asymptotes as invisible walls. The function can get arbitrarily close to the wall, but it can never reach it or pass through it.

Horizontal Asymptotes: Where the Graph Levels Off

A horizontal asymptote is a horizontal line that the graph approaches as $x$ goes to positive or negative infinity. Unlike vertical asymptotes, the graph CAN cross a horizontal asymptote - it just settles toward it in the long run.

The rule for finding horizontal asymptotes depends on comparing the degree of the numerator and denominator:

Let $n$ = degree of numerator and $m$ = degree of denominator.

Comparison Horizontal Asymptote Why
$n < m$ $y = 0$ (the x-axis) Denominator grows faster, fraction shrinks to 0
$n = m$ $y = \frac{\text{leading coefficient of numerator}}{\text{leading coefficient of denominator}}$ Terms of same degree dominate
$n > m$ No horizontal asymptote Numerator grows faster (may have oblique asymptote)

Example: For $f(x) = \frac{3x^2 + 1}{2x^2 - 5}$:

  • Numerator degree: 2
  • Denominator degree: 2
  • Since they are equal, the horizontal asymptote is $y = \frac{3}{2}$ (ratio of leading coefficients)

Example: For $g(x) = \frac{x + 1}{x^2 - 4}$:

  • Numerator degree: 1
  • Denominator degree: 2
  • Since $1 < 2$, the horizontal asymptote is $y = 0$

Oblique (Slant) Asymptotes: When Horizontal Is Not Enough

When the degree of the numerator is exactly one more than the degree of the denominator, the graph does not level off horizontally - instead, it approaches a slanted line called an oblique asymptote or slant asymptote.

To find an oblique asymptote, perform polynomial long division. The quotient (ignoring the remainder) gives you the equation of the slant asymptote.

For $f(x) = \frac{x^2 + 3x + 2}{x - 1}$:

  • Numerator degree: 2
  • Denominator degree: 1
  • Since $2 = 1 + 1$, there is an oblique asymptote

Divide $x^2 + 3x + 2$ by $x - 1$:

  • $x^2 \div x = x$
  • $x(x - 1) = x^2 - x$
  • $(x^2 + 3x + 2) - (x^2 - x) = 4x + 2$
  • $4x \div x = 4$
  • $4(x - 1) = 4x - 4$
  • $(4x + 2) - (4x - 4) = 6$

The quotient is $x + 4$ with remainder 6, so: $$f(x) = x + 4 + \frac{6}{x - 1}$$

As $x \to \pm\infty$, the fraction $\frac{6}{x-1} \to 0$, so the function approaches the line $y = x + 4$.

The oblique asymptote is $y = x + 4$.

Holes: The Invisible Points

A hole (also called a removable discontinuity) occurs when a factor cancels from both the numerator and denominator. The function is still undefined at that point, but instead of the graph shooting off to infinity, there is just a “hole” - a single missing point.

To find holes:

  1. Factor both numerator and denominator completely
  2. Identify any common factors
  3. The $x$-value that makes the common factor zero is where the hole occurs
  4. To find the $y$-coordinate of the hole, cancel the common factor and evaluate at that $x$-value

Example: Consider $f(x) = \frac{x^2 - 4}{x - 2}$.

Factor the numerator: $\frac{(x+2)(x-2)}{x-2}$

The factor $(x - 2)$ appears in both numerator and denominator. Cancel it:

$$f(x) = x + 2 \quad \text{(for } x \neq 2\text{)}$$

There is a hole at $x = 2$. To find the $y$-coordinate, evaluate the simplified function: $y = 2 + 2 = 4$.

The hole is at the point $(2, 4)$.

Key distinction: If a factor cancels, you get a hole. If it does not cancel, you get a vertical asymptote. Both represent values excluded from the domain, but they look very different on the graph.

X-Intercepts and Y-Intercept

X-intercepts occur where the function equals zero. For a fraction to equal zero, the numerator must be zero (while the denominator is not).

To find x-intercepts:

  1. Set the numerator equal to zero
  2. Solve for $x$
  3. Check that these values do not also make the denominator zero

Y-intercept is the function value when $x = 0$ (if 0 is in the domain).

To find the y-intercept:

  1. Substitute $x = 0$ into the function
  2. Simplify

Example: For $f(x) = \frac{x + 3}{x - 2}$:

  • X-intercept: Set numerator to zero. $x + 3 = 0$, so $x = -3$. Check: denominator at $x = -3$ is $-3 - 2 = -5 \neq 0$. X-intercept is $(-3, 0)$.

  • Y-intercept: $f(0) = \frac{0 + 3}{0 - 2} = \frac{3}{-2} = -\frac{3}{2}$. Y-intercept is $(0, -\frac{3}{2})$.

Graphing Rational Functions

To graph a rational function, gather all the key features systematically:

  1. Find the domain - identify all values where the denominator is zero
  2. Factor completely - this reveals holes vs. asymptotes
  3. Find holes - where common factors cancel
  4. Find vertical asymptotes - where remaining denominator factors equal zero
  5. Find horizontal or oblique asymptotes - compare degrees
  6. Find intercepts - where graph crosses the axes
  7. Plot additional points - especially near asymptotes
  8. Sketch the graph - draw curves approaching asymptotes correctly

The asymptotes act as guidelines. The graph will hug them but (for vertical) never touch them.

Solving Rational Equations

A rational equation contains at least one rational expression with a variable in the denominator. The strategy is to eliminate denominators by multiplying both sides by the LCD (least common denominator).

Steps to solve:

  1. Find the LCD of all rational expressions
  2. Multiply every term by the LCD
  3. Simplify (denominators should cancel)
  4. Solve the resulting equation
  5. Check for extraneous solutions - any answer that makes a denominator zero is invalid

That last step is crucial. When you multiply by an expression containing the variable, you might introduce false solutions. Always verify your answers in the original equation.

Notation and Terminology

Term Meaning Example
Rational function $f(x) = \frac{p(x)}{q(x)}$ where $p$ and $q$ are polynomials $f(x) = \frac{x + 1}{x - 2}$
Vertical asymptote Line $x = a$ that the graph approaches but never touches $x = 2$ for $f(x) = \frac{1}{x-2}$
Horizontal asymptote Line $y = b$ that the graph approaches as $x \to \pm\infty$ $y = 0$ for $f(x) = \frac{1}{x}$
Oblique asymptote Slanted line the graph approaches When numerator degree = denominator degree + 1
Hole Removable discontinuity; missing point When a factor cancels from top and bottom
Domain restriction Value(s) excluded because denominator = 0 $x \neq 2$

Examples

Example 1: Finding the Domain

Find the domain of $f(x) = \frac{3}{x + 5}$.

Solution:

The domain includes all real numbers except those that make the denominator zero.

Set the denominator equal to zero: $$x + 5 = 0$$ $$x = -5$$

The function is undefined at $x = -5$ because that would create division by zero.

Domain: All real numbers except $x = -5$.

In interval notation: $(-\infty, -5) \cup (-5, \infty)$

Think about it this way: If you tried to evaluate $f(-5)$, you would get $\frac{3}{0}$, which is undefined. Every other real number works just fine.

Example 2: Identifying Vertical and Horizontal Asymptotes

Identify the vertical and horizontal asymptotes of $f(x) = \frac{2}{x - 3}$.

Solution:

Vertical asymptote:

Set the denominator equal to zero: $$x - 3 = 0$$ $$x = 3$$

There is no common factor with the numerator (the numerator is just 2), so $x = 3$ is a vertical asymptote.

Horizontal asymptote:

Compare degrees:

  • Numerator degree: 0 (it is a constant, 2)
  • Denominator degree: 1

Since the numerator degree (0) is less than the denominator degree (1), the horizontal asymptote is $y = 0$.

Answer:

  • Vertical asymptote: $x = 3$
  • Horizontal asymptote: $y = 0$

Interpretation: As $x$ approaches 3, the function values explode toward $\pm\infty$. As $x$ goes to $\pm\infty$, the function values shrink toward 0.

Example 3: Finding All Asymptotes with a Hole

Find all asymptotes of $f(x) = \frac{2x^2 - 8}{x^2 - 4}$.

Solution:

Step 1: Factor completely.

Numerator: $2x^2 - 8 = 2(x^2 - 4) = 2(x + 2)(x - 2)$

Denominator: $x^2 - 4 = (x + 2)(x - 2)$

So: $f(x) = \frac{2(x + 2)(x - 2)}{(x + 2)(x - 2)}$

Step 2: Identify common factors.

Both $(x + 2)$ and $(x - 2)$ appear in numerator and denominator. They cancel:

$$f(x) = 2 \quad \text{(for } x \neq 2 \text{ and } x \neq -2\text{)}$$

Step 3: Determine asymptotes and holes.

Since all factors cancelled, there are no remaining denominator factors. This means:

  • No vertical asymptotes
  • Holes at $x = 2$ and $x = -2$ (where the cancelled factors equal zero)

Step 4: Find horizontal asymptote.

The original function had equal degrees (both degree 2). But since the function simplifies to a constant, another way to see it: the simplified function $f(x) = 2$ is a horizontal line.

Horizontal asymptote: $y = 2$

Actually, in this case, the function IS the line $y = 2$, just with two holes punched out of it.

Answer:

  • No vertical asymptotes
  • Horizontal asymptote: $y = 2$
  • Holes at $x = 2$ and $x = -2$ (both at height $y = 2$)
Example 4: Graphing a Rational Function

Graph $f(x) = \frac{x + 1}{x - 2}$.

Solution:

Step 1: Find the domain.

Denominator zero when $x - 2 = 0$, so $x = 2$. Domain: all real numbers except $x = 2$.

Step 2: Check for holes.

Factor: numerator is $x + 1$, denominator is $x - 2$. No common factors, so no holes.

Step 3: Find vertical asymptote.

Since $x = 2$ makes the denominator zero and there is no cancellation, there is a vertical asymptote at $x = 2$.

Step 4: Find horizontal asymptote.

  • Numerator degree: 1
  • Denominator degree: 1

Degrees are equal, so horizontal asymptote is: $$y = \frac{\text{leading coefficient of numerator}}{\text{leading coefficient of denominator}} = \frac{1}{1} = 1$$

Horizontal asymptote: $y = 1$

Step 5: Find intercepts.

X-intercept: Set numerator to zero. $$x + 1 = 0 \Rightarrow x = -1$$ X-intercept: $(-1, 0)$

Y-intercept: Evaluate at $x = 0$. $$f(0) = \frac{0 + 1}{0 - 2} = \frac{1}{-2} = -\frac{1}{2}$$ Y-intercept: $(0, -\frac{1}{2})$

Step 6: Plot additional points and sketch.

Choose points on each side of the vertical asymptote:

$x$ $f(x)$
$-2$ $\frac{-1}{-4} = \frac{1}{4}$
$1$ $\frac{2}{-1} = -2$
$3$ $\frac{4}{1} = 4$
$5$ $\frac{6}{3} = 2$

Graph description:

  • Draw dashed lines for asymptotes at $x = 2$ (vertical) and $y = 1$ (horizontal)
  • Plot intercepts and additional points
  • The curve approaches the vertical asymptote from both sides
  • The curve approaches the horizontal asymptote as $x \to \pm\infty$
  • To the left of $x = 2$: curve goes from below $y = 1$, through the intercepts, down toward $-\infty$ as it nears $x = 2$
  • To the right of $x = 2$: curve comes from $+\infty$, decreases, and levels off toward $y = 1$
Example 5: Finding Holes and Asymptotes

Find all holes and asymptotes of $f(x) = \frac{x^2 - 9}{x^2 + x - 6}$.

Solution:

Step 1: Factor completely.

Numerator: $x^2 - 9 = (x + 3)(x - 3)$

Denominator: $x^2 + x - 6 = (x + 3)(x - 2)$

So: $f(x) = \frac{(x + 3)(x - 3)}{(x + 3)(x - 2)}$

Step 2: Identify common factors.

The factor $(x + 3)$ appears in both numerator and denominator. This means there is a hole at $x = -3$.

Step 3: Simplify.

$$f(x) = \frac{x - 3}{x - 2} \quad \text{(for } x \neq -3\text{)}$$

Step 4: Find the hole’s coordinates.

The hole is at $x = -3$. To find the $y$-coordinate, plug $x = -3$ into the simplified function: $$y = \frac{-3 - 3}{-3 - 2} = \frac{-6}{-5} = \frac{6}{5}$$

Hole: $\left(-3, \frac{6}{5}\right)$

Step 5: Find vertical asymptote.

After cancellation, the denominator is $x - 2$. Setting it to zero: $$x - 2 = 0 \Rightarrow x = 2$$

Vertical asymptote: $x = 2$

Step 6: Find horizontal asymptote.

Original function had degree 2 in both numerator and denominator. Leading coefficients are both 1.

Horizontal asymptote: $y = \frac{1}{1} = 1$

Summary:

  • Hole: $\left(-3, \frac{6}{5}\right)$
  • Vertical asymptote: $x = 2$
  • Horizontal asymptote: $y = 1$
Example 6: Solving a Rational Equation

Solve $\frac{2}{x - 1} + \frac{3}{x + 2} = 1$.

Solution:

Step 1: Identify domain restrictions.

The denominators are $x - 1$ and $x + 2$.

  • $x - 1 = 0 \Rightarrow x = 1$ (excluded)
  • $x + 2 = 0 \Rightarrow x = -2$ (excluded)

Any solution must avoid $x = 1$ and $x = -2$.

Step 2: Find the LCD.

The LCD of $(x - 1)$ and $(x + 2)$ is $(x - 1)(x + 2)$.

Step 3: Multiply every term by the LCD.

$$(x - 1)(x + 2) \cdot \frac{2}{x - 1} + (x - 1)(x + 2) \cdot \frac{3}{x + 2} = (x - 1)(x + 2) \cdot 1$$

Step 4: Simplify.

$$2(x + 2) + 3(x - 1) = (x - 1)(x + 2)$$

Step 5: Expand.

Left side: $$2x + 4 + 3x - 3 = 5x + 1$$

Right side: $$x^2 + 2x - x - 2 = x^2 + x - 2$$

Step 6: Set up and solve the equation.

$$5x + 1 = x^2 + x - 2$$

Move everything to one side: $$0 = x^2 + x - 2 - 5x - 1$$ $$0 = x^2 - 4x - 3$$

This does not factor nicely, so use the quadratic formula with $a = 1$, $b = -4$, $c = -3$:

$$x = \frac{-(-4) \pm \sqrt{(-4)^2 - 4(1)(-3)}}{2(1)}$$ $$x = \frac{4 \pm \sqrt{16 + 12}}{2}$$ $$x = \frac{4 \pm \sqrt{28}}{2}$$ $$x = \frac{4 \pm 2\sqrt{7}}{2}$$ $$x = 2 \pm \sqrt{7}$$

Step 7: Check for extraneous solutions.

Our solutions are $x = 2 + \sqrt{7} \approx 4.65$ and $x = 2 - \sqrt{7} \approx -0.65$.

Neither of these equals 1 or $-2$, so both are valid solutions.

Solution: $x = 2 + \sqrt{7}$ and $x = 2 - \sqrt{7}$

Verification tip: You can plug these back into the original equation to confirm they work. The arithmetic is messy but the equation will balance.

Key Properties and Rules

Asymptote Summary

Feature How to Find It What It Means
Vertical asymptote Set remaining denominator factors = 0 Graph goes to $\pm\infty$
Horizontal asymptote Compare degrees of numerator and denominator Graph levels off
Oblique asymptote Polynomial division (when num. degree = denom. degree + 1) Graph approaches a slant line
Hole Factor that cancels from both top and bottom Missing point on graph

Horizontal Asymptote Rules

For $f(x) = \frac{a_n x^n + \ldots}{b_m x^m + \ldots}$:

  • If $n < m$: horizontal asymptote is $y = 0$
  • If $n = m$: horizontal asymptote is $y = \frac{a_n}{b_m}$
  • If $n > m$: no horizontal asymptote (check for oblique if $n = m + 1$)

Solving Rational Equations: Key Steps

  1. Find domain restrictions FIRST
  2. Multiply all terms by the LCD
  3. Solve the resulting polynomial equation
  4. Check EVERY answer against domain restrictions
  5. Reject any extraneous solutions

Common Mistakes to Avoid

  • Forgetting domain restrictions - Always check what values make the denominator zero, both before and after solving
  • Confusing holes and asymptotes - Holes occur when factors cancel; asymptotes occur when they do not
  • Wrong horizontal asymptote - Remember: compare degrees first, then look at leading coefficients only if degrees are equal
  • Not checking solutions - Multiplying by the LCD can introduce extraneous solutions
  • Crossing vertical asymptotes - The graph can cross horizontal asymptotes but NEVER crosses vertical asymptotes

Real-World Applications

Average Cost Per Unit (Economics)

A company has fixed costs of $10,000 (rent, equipment, etc.) plus variable costs of $5 per unit produced. The total cost for producing $x$ units is:

$$C(x) = 10000 + 5x$$

The average cost per unit is:

$$A(x) = \frac{10000 + 5x}{x} = \frac{10000}{x} + 5$$

This is a rational function. Notice:

  • When $x$ is small (few units), average cost is high because fixed costs spread over few items
  • As $x$ increases, average cost approaches $5 (the horizontal asymptote)
  • You can never reach exactly $5 per unit, but you can get arbitrarily close

This explains economies of scale: producing more units lowers the average cost per unit.

Concentration of Medication Over Time

When medication enters your bloodstream, its concentration changes over time. A typical model:

$$C(t) = \frac{50t}{t^2 + 4}$$

where $C$ is concentration (mg/L) and $t$ is time (hours) after taking the medication.

  • At $t = 0$: $C(0) = 0$ (no medication yet in blood)
  • The concentration rises, reaches a peak, then gradually decreases
  • The horizontal asymptote is $y = 0$ (medication eventually leaves the system)

Doctors use these models to determine dosing schedules that keep medication in the therapeutic range.

Electrical Resistance in Parallel

When two resistors with resistances $R_1$ and $R_2$ are connected in parallel, the total resistance $R$ is given by:

$$\frac{1}{R} = \frac{1}{R_1} + \frac{1}{R_2}$$

Solving for $R$:

$$R = \frac{R_1 \cdot R_2}{R_1 + R_2}$$

This is a rational function of $R_1$ (or $R_2$). Notice that the total resistance is always less than either individual resistance - current can flow through both paths.

Inverse Relationships in Science

Many physical relationships are inverse: when one quantity doubles, another halves. These are modeled by rational functions.

Boyle’s Law: For a gas at constant temperature, pressure and volume are inversely related:

$$P = \frac{k}{V}$$

Squeeze a gas into half the volume, and the pressure doubles. This is why compressed air tanks can be dangerous.

Light intensity: Light intensity decreases with the square of distance:

$$I = \frac{k}{d^2}$$

Stand twice as far from a light source, and you receive one-fourth the light. This inverse-square law appears throughout physics.

Self-Test Problems

Problem 1: Find the domain of $f(x) = \frac{x + 2}{x^2 - 9}$.

Show Answer

Set the denominator equal to zero: $$x^2 - 9 = 0$$ $$(x + 3)(x - 3) = 0$$ $$x = -3 \text{ or } x = 3$$

Domain: All real numbers except $x = -3$ and $x = 3$.

In interval notation: $(-\infty, -3) \cup (-3, 3) \cup (3, \infty)$

Problem 2: Find the vertical and horizontal asymptotes of $f(x) = \frac{4x - 1}{2x + 6}$.

Show Answer

Vertical asymptote: $$2x + 6 = 0 \Rightarrow x = -3$$ Vertical asymptote: $x = -3$

Horizontal asymptote:

  • Numerator degree: 1
  • Denominator degree: 1
  • Degrees are equal, so $y = \frac{4}{2} = 2$

Horizontal asymptote: $y = 2$

Problem 3: Identify any holes in $f(x) = \frac{x^2 - 1}{x^2 - 3x + 2}$.

Show Answer

Factor both:

  • Numerator: $x^2 - 1 = (x + 1)(x - 1)$
  • Denominator: $x^2 - 3x + 2 = (x - 1)(x - 2)$

$$f(x) = \frac{(x + 1)(x - 1)}{(x - 1)(x - 2)}$$

The factor $(x - 1)$ cancels, so there is a hole at $x = 1$.

To find the $y$-coordinate, use the simplified function: $$f(x) = \frac{x + 1}{x - 2}$$ $$f(1) = \frac{1 + 1}{1 - 2} = \frac{2}{-1} = -2$$

Hole: $(1, -2)$

Note: There is also a vertical asymptote at $x = 2$ (from the remaining denominator factor).

Problem 4: Find all asymptotes of $f(x) = \frac{x^2 + 2x - 3}{x - 1}$.

Show Answer

First, check for common factors:

  • Numerator: $x^2 + 2x - 3 = (x + 3)(x - 1)$
  • Denominator: $x - 1$

The factor $(x - 1)$ cancels, so there is a hole at $x = 1$, not a vertical asymptote.

Simplified: $f(x) = x + 3$ (for $x \neq 1$)

Since the function simplifies to a line, there are no vertical or horizontal asymptotes. The graph is just the line $y = x + 3$ with a hole at $(1, 4)$.

No asymptotes (just a hole at $(1, 4)$).

Problem 5: Solve $\frac{5}{x} = \frac{3}{x - 2}$.

Show Answer

Domain restrictions: $x \neq 0$ and $x \neq 2$

Cross-multiply: $$5(x - 2) = 3x$$ $$5x - 10 = 3x$$ $$2x = 10$$ $$x = 5$$

Check: $x = 5$ is not 0 or 2, so it is valid.

Verify: $\frac{5}{5} = 1$ and $\frac{3}{5-2} = \frac{3}{3} = 1$ ✓

Solution: $x = 5$

Problem 6: Find the x-intercept and y-intercept of $f(x) = \frac{2x - 6}{x + 4}$.

Show Answer

X-intercept: Set numerator equal to zero. $$2x - 6 = 0$$ $$x = 3$$ Check: $x = 3$ does not make denominator zero. ✓

X-intercept: $(3, 0)$

Y-intercept: Evaluate at $x = 0$. $$f(0) = \frac{2(0) - 6}{0 + 4} = \frac{-6}{4} = -\frac{3}{2}$$

Y-intercept: $\left(0, -\frac{3}{2}\right)$

Problem 7: Determine whether $f(x) = \frac{3x^2 + 1}{x^3 - x}$ has a horizontal or oblique asymptote. What is it?

Show Answer

Compare degrees:

  • Numerator degree: 2
  • Denominator degree: 3

Since numerator degree (2) < denominator degree (3), there is a horizontal asymptote at $y = 0$.

Horizontal asymptote: $y = 0$

There is no oblique asymptote because the numerator degree is not exactly one more than the denominator degree.

Summary

  • A rational function is a ratio of two polynomials: $f(x) = \frac{p(x)}{q(x)}$. The key restriction is that the denominator cannot equal zero.

  • Domain restrictions come from setting the denominator equal to zero and excluding those $x$-values.

  • Vertical asymptotes occur at $x$-values where the denominator is zero but the factor does not cancel with the numerator. The graph shoots toward $\pm\infty$ near these invisible walls.

  • Horizontal asymptotes depend on comparing degrees: if numerator degree < denominator degree, the asymptote is $y = 0$; if they are equal, it is $y =$ ratio of leading coefficients; if numerator degree > denominator degree, there is no horizontal asymptote.

  • Oblique (slant) asymptotes appear when the numerator degree is exactly one more than the denominator degree. Find them using polynomial long division.

  • Holes (removable discontinuities) occur when a factor cancels from both numerator and denominator. The function is still undefined there, but the graph has a missing point rather than an asymptote.

  • X-intercepts are found by setting the numerator equal to zero (and verifying the denominator is not also zero). The y-intercept is found by evaluating $f(0)$, if 0 is in the domain.

  • To solve rational equations, multiply by the LCD to clear denominators, solve the resulting polynomial equation, and always check for extraneous solutions that would make an original denominator zero.

  • Rational functions appear everywhere in the real world: average costs in economics, medication concentrations in pharmacology, electrical resistance in physics, and any situation involving inverse relationships.

You have now learned to work with functions that have variables in the denominator. The key insight is that division by zero creates boundaries - places where the function cannot go. Understanding these boundaries through asymptotes and holes gives you a complete picture of how these functions behave. These skills will serve you well in calculus and beyond, where rational functions appear constantly.