🎯⭐ INTERACTIVE LESSON

Rational Functions and Their Graphs

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Rational Functions and Their Graphs - Complete Interactive Lesson

Part 1: What Is a Rational Function? (definition, domain, the 1/x parent graph)

〽️ Rational Functions and Their Graphs

Part 1 of 5 — What Is a Rational Function?

Topics in This Part

Section
Definition & the Parent Function
Domain: Where the Function "Breaks"
Reading the Graph of $\frac{1}{x}$

🔑 Key Concept: A rational function is a fraction of two polynomials, $f(x) = \dfrac{P(x)}{Q(x)}$ . Everything interesting about its graph — the gaps, the invisible walls, the leveling-off — comes from one question: where does the denominator misbehave?

Definition & the Parent Function

A rational function has the form

$f(x) = \frac{P(x)}{Q(x)}, \qquad Q(x) \ne 0$

Concept Check 🎯

Domain: Where the Function "Breaks"

The domain of a rational function is all real numbers except where the denominator equals zero. To find the excluded values:

Set the denominator equal to $0$ .
Solve. Those solutions are the forbidden $x$ -values.

Example: $f(x) = \dfrac{x+5}{x-3}$

Find the Excluded Values 🧮

For each function, enter the $x$ -value(s) that are not in the domain. If there are two, enter the smaller one first.

1) $f(x) = \dfrac{x-1}{x+6}$ . Excluded . Smaller excluded . Larger excluded

Reading the Graph of $\frac{1}{x}$

The parent graph lives in the top-right and bottom-left corners. Two invisible lines guide it:

The vertical line $x = 0$ (the $y$ -axis): the curve shoots up on the right and down on the left, never touching it.

Read the Parent Graph 🔽

Use the behavior of $f(x) = \dfrac{1}{x}$ to fill in each blank.

What the Domain Tells Us

Every excluded $x$ -value is a place where the graph has something missing — either an invisible vertical wall the curve can't cross (a vertical asymptote) or a single missing point (a hole). Part 2 shows you how to tell which is which.

🔑 Takeaway: Finding the domain is always step one. Set the denominator to zero, solve, and you've located every "break" in the graph.

Part 2: Vertical Asymptotes & Holes (factor, cancel, classify)

〽️ Rational Functions and Their Graphs

Part 2 of 5 — Vertical Asymptotes & Holes

🔑 The Idea: Both vertical asymptotes and holes come from zeros of the denominator. The difference: a hole appears where a factor cancels with the numerator; a vertical asymptote appears where it does not.

Vertical Asymptotes

A vertical asymptote is a vertical line $x = a$ that the graph approaches but never touches. After fully simplifying the fraction, you get one wherever the denominator is zero.

Example: $f(x) = \dfrac{1}{x - 4}$

Part 3: Horizontal & Slant Asymptotes (degree rules, polynomial division)

〽️ Rational Functions and Their Graphs

Part 3 of 5 — Horizontal & Slant Asymptotes

🔑 The Idea: Vertical asymptotes describe what happens near a forbidden $x$ . Horizontal and slant asymptotes describe the end behavior — where the graph heads as $x \to \pm\infty$ . The answer depends only on the degrees of the top and bottom.

The Degree Rules

Compare the degree of the numerator ( $n$ ) to the degree of the denominator ():

Part 4: Intercepts & Sketching the Whole Graph (full recipe)

〽️ Rational Functions and Their Graphs

Part 4 of 5 — Intercepts & Sketching the Whole Graph

🔑 Big Payoff: With domain, asymptotes, and intercepts in hand, you can sketch any rational graph. This part adds the last two pieces — the $x$ - and $y$ -intercepts — and assembles a full step-by-step recipe.

Finding Intercepts

$x$ -intercepts (where the graph crosses the $x$ -axis): a fraction equals zero only when its numerator equals zero (and the denominator is nonzero there). So set and solve.

Part 5: Mixed Practice & Mastery Check (Exit Quiz)

〽️ Rational Functions and Their Graphs

Part 5 of 5 — Mixed Practice & Mastery Check

You can now (1) find the domain, (2) locate vertical asymptotes and holes, (3) determine horizontal/slant asymptotes from degrees, and (4) find intercepts and sketch the full graph. Let's put it all together.

Quick Reference

Feature	How to find it
Domain	denominator $\ne 0$
Hole	factor cancels top & bottom
Vertical asymptote	zero of the remaining denominator
Horizontal asymptote	$n < d \Rightarrow y =$ ;

where $P$ and $Q$ are polynomials. The simplest one is the parent function:

$f(x) = \frac{1}{x}$

Its graph is a hyperbola sitting in two opposite corners of the plane.

$x$	$-2$	$-1$	$-\tfrac{1}{2}$	$\tfrac{1}{2}$	$1$	$2$
$\frac{1}{x}$	$-0.5$	$-1$	$-2$

As $x$ gets close to $0$ , $\frac{1}{x}$ blows up toward $\pm\infty$ .
As $x$ gets large, $\frac{1}{x}$ shrinks toward $0$ .

🔑 Key Idea: Division by zero is undefined, so $x = 0$ is not allowed. That single forbidden value creates the most important feature of the graph.

Set $x - 3 = 0 \Rightarrow x = 3$ . So the domain is all reals except $x = 3$ .

Example: $g(x) = \dfrac{x}{x^2 - 4}$

Factor the denominator: $x^2 - 4 = (x-2)(x+2)$ . Set each factor to $0$ :

$x - 2 = 0 \Rightarrow x = 2, \qquad x + 2 = 0 \Rightarrow x = -2$

Domain: all reals except $x = 2$ and $x = -2$ .

💡 The numerator never restricts the domain — only the denominator can make a rational function undefined.

g(x) = \dfrac{4}{x^2 - 9}

g(x) = \dfrac{4}{x^2 - 9}

The horizontal line $y = 0$ (the

x

-axis): far out, both branches flatten toward it.

These guide lines are called asymptotes — lines a graph approaches but never reaches. Almost every rational graph you'll meet is built from this same idea.

💡 As $x \to 0$ , $\frac{1}{x} \to \pm\infty$ (vertical asymptote). As $x \to \pm\infty$ , $\frac{1}{x} \to 0$ (horizontal asymptote).

The denominator is zero at $x = 4$ , and nothing cancels, so there is a vertical asymptote at $x = 4$ .

As $x \to 4$ from either side, the denominator shrinks toward $0$ , so $|f(x)| \to \infty$ . The curve races up or down along the line $x = 4$ .

$x$	$3.9$	$3.99$	$4.01$	$4.1$
$\frac{1}{x-4}$	$-10$	$-100$	$100$	$10$

⚠️ The graph gets arbitrarily close to $x = 4$ but never crosses it — a function can never output a value at a point where it's undefined.

Holes (Removable Discontinuities)

When a factor cancels from both the numerator and denominator, the graph has a hole — a single missing point — instead of an asymptote.

Example: $f(x) = \dfrac{x^2 - 9}{x - 3}$

Factor and cancel:

$f(x) = \frac{(x-3)(x+3)}{x-3} = x + 3, \quad x \ne 3$

The simplified function is the line $y = x + 3$ , but $x = 3$ is still excluded from the original domain. So the graph is that line with a single hole at $x = 3$ .

Find the hole's $y$ -coordinate by plugging $x = 3$ into the simplified form: $y = 3 + 3 = 6$ . The hole is at $(3,$ .

🔑 Cancels → hole. Doesn't cancel → vertical asymptote. Always factor first to find out which.

Concept Check 🎯

Asymptote or Hole? 🔽

Factor each function and decide what happens at the given $x$ -value.

Pinpointing a Hole's Coordinates

When a factor cancels, finding the exact missing point takes two quick steps:

The $x$ -coordinate is the value that made the canceled factor zero.
The $y$ -coordinate comes from plugging that $x$ into the simplified function.

Example: $f(x) = \dfrac{x^2 + x - 6}{x - 2}$

Factor: $\dfrac{(x-2)(x+3)}{x-2} = x + 3$ , with .

Hole's $x$ : the canceled factor $x - 2 = 0 \Rightarrow x = 2$ .
Hole's $y$ : plug into $x + 3$ : .

So the hole is at $(2, 5)$ .

💡 Never plug into the original fraction to find the $y$ -value — it's undefined there. Always use the simplified form.

Locate the Hole 🧮

$f(x) = \dfrac{x^2 - 4}{x - 2}$

1) This simplifies to $y = x + 2$ for $x \ne 2$ . The hole is at $x = \,?$ 2) The hole's $y$ -coordinate (plug $x=2$ into $x+2$ ) $= \,?$

Case	End behavior
$n < d$ (bottom-heavy)	Horizontal asymptote $y = 0$
$n = d$ (equal degrees)	Horizontal asymptote $y = \dfrac{\text{lead coeff of top}}{\text{lead coeff of bottom}}$
$n = d + 1$ (top one higher)	Slant (oblique) asymptote — do polynomial division
$n > d + 1$	No horizontal or slant asymptote

$\dfrac{3x+1}{x^2 - 5}$ : degree $1 < 2$ , so $y = 0$ .
$\dfrac{2x^2 + 1}{5x^2 - x}$ : degrees equal, so .
$\dfrac{x^2 + 1}{x - 1}$ : top degree is one higher → slant asymptote.

💡 A graph can cross a horizontal asymptote in the middle, but it always levels off toward it at the far ends.

Concept Check 🎯

Slant (Oblique) Asymptotes

When the numerator's degree is exactly one more than the denominator's, the graph follows a slanted line at the ends. Find it by polynomial long division — the quotient (ignoring the remainder) is the slant asymptote.

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

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

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

So $f(x) = x + 4 + \dfrac{6}{x - 1}$ . As , the remainder term , and the graph hugs the line

$y = x + 4 \quad (\text{the slant asymptote}).$

⚠️ A graph cannot have both a horizontal and a slant asymptote — it's one or the other (or neither).

Pick the End Behavior 🔽

For each function, choose the correct horizontal/slant asymptote.

Find the Horizontal Asymptote 🧮

Enter the $y$ -value of the horizontal asymptote. (Decimals or fractions are fine.)

1) $f(x) = \dfrac{7x - 2}{x + 5}$ . $y = \,?$ 2) $f(x) = \dfrac{3x^2 + x}{6x^2 - 4}$ . $y = \,?$ (fraction or decimal) 3) $f(x) = \dfrac{8}{x^2 + 1}$ . $y = \,?$

$y$ -intercept (where it crosses the $y$ -axis): plug in $x = 0$ , i.e. compute $f(0)$ — provided $0$ is in the domain.

Example: $f(x) = \dfrac{x - 2}{x + 4}$

$x$ -intercept: numerator $x - 2 = 0 \Rightarrow x = 2$ . Point $(2, 0)$ .
$y$ -intercept: $f(0) = \dfrac{0 - 2}{0 + 4} = \dfrac{-2}{4} = -\dfrac{1}{2}$ . Point $\left(0, -\tfrac{1}{2}\right)$ .

🔑 Numerator zero → $x$ -intercept. Plug in $0$ → $y$ -intercept. Don't mix them up: the denominator's zeros give asymptotes/holes, not intercepts.

Concept Check 🎯

The Full Sketching Recipe

To graph any rational function $f(x) = \dfrac{P(x)}{Q(x)}$ :

Factor both $P$ and $Q$ , and cancel common factors (those give holes).
Vertical asymptotes: zeros of the remaining denominator.
Horizontal/slant asymptote: compare degrees (Part 3 rules).
$x$ -intercepts: zeros of the numerator.
$y$ -intercept: compute $f(0)$ .
Plot asymptotes (dashed), intercepts, and a test point in each region; connect smoothly.

Worked Example: $f(x) = \dfrac{x - 1}{x - 3}$

Feature	Result
Vertical asymptote	$x = 3$
Horizontal asymptote	$y = 1$ (equal degrees, $\frac{1}{1}$ )

💡 Asymptotes are dashed guide lines, not part of the curve. They tell the two branches where to go.

Build the Graph 🔽

Analyze $f(x) = \dfrac{x + 2}{x - 4}$ feature by feature.

Intercept Practice 🧮

Use $f(x) = \dfrac{3x - 12}{x + 2}$ .

1) $x$ -intercept: solve $3x - 12 = 0$ , so $x = \,?$ 2) $y$ -intercept value $f(0) = \,?$ (fraction or decimal) 3) Vertical asymptote: $x = \,?$

n < d \Rightarrow y = 0

\;n=d \Rightarrow y=\frac{\text{lead}}{\text{lead}}

⚠️ Top-three traps: (1) forgetting to cancel first and mislabeling a hole as an asymptote, (2) using leading coefficients only when degrees are equal, and (3) confusing which zero (top vs. bottom) gives intercepts vs. asymptotes.

Mixed Practice 🎯

One Big Analysis 🧮

Analyze $f(x) = \dfrac{(x - 5)(x + 1)}{(x + 1)(x - 2)}$ completely.

1) A factor cancels, creating a hole at $x = \,?$ 2) After canceling, the vertical asymptote is at $x = \,?$ 3) The $x$ -intercept comes from the surviving numerator factor: $x = \,?$

Exit Quiz ✅

Answer all three to finish the lesson.

\frac{6}{x-1} \to 0

Rational Functions and Their Graphs

Rational Functions and Their Graphs - Complete Interactive Lesson

Part 1: What Is a Rational Function? (definition, domain, the 1/x parent graph)

〽️ Rational Functions and Their Graphs

Topics in This Part

Definition & the Parent Function

Domain: Where the Function "Breaks"

Example: f(x)=x+5x−3f(x) = \dfrac{x+5}{x-3}f(x)=

Reading the Graph of 1x\frac{1}{x}x1​

What the Domain Tells Us

Part 2: Vertical Asymptotes & Holes (factor, cancel, classify)

〽️ Rational Functions and Their Graphs

Vertical Asymptotes

Example: f(x)=1x−4f(x) = \dfrac{1}{x - 4}f(x)

Part 3: Horizontal & Slant Asymptotes (degree rules, polynomial division)

〽️ Rational Functions and Their Graphs

The Degree Rules

Part 4: Intercepts & Sketching the Whole Graph (full recipe)

〽️ Rational Functions and Their Graphs

Finding Intercepts

Part 5: Mixed Practice & Mastery Check (Exit Quiz)

〽️ Rational Functions and Their Graphs

Quick Reference

Example: g(x)=xx2−4g(x) = \dfrac{x}{x^2 - 4}g(x)=x2−4x​

Holes (Removable Discontinuities)

Example: f(x)=x2−9x−3f(x) = \dfrac{x^2 - 9}{x - 3}f(x)=x−3x2−9​

Pinpointing a Hole's Coordinates

Example: f(x)=x2+x−6x−2f(x) = \dfrac{x^2 + x - 6}{x - 2}f(x)=x−2x2+x−6​

Examples

Slant (Oblique) Asymptotes

Example: f(x)=x2+3x+2x−1f(x) = \dfrac{x^2 + 3x + 2}{x - 1}f(x)=x−1x2+3x+2​

Example: f(x)=x−2x+4f(x) = \dfrac{x - 2}{x + 4}f(x)=x+4x−2​

The Full Sketching Recipe

Worked Example: f(x)=x−1x−3f(x) = \dfrac{x - 1}{x - 3}f(x)=x−3x−1​

Example: $f(x) = \dfrac{x+5}{x-3}$

Reading the Graph of $\frac{1}{x}$

Example: $f(x) = \dfrac{1}{x - 4}$

Example: $g(x) = \dfrac{x}{x^2 - 4}$

Example: $f(x) = \dfrac{x^2 - 9}{x - 3}$

Example: $f(x) = \dfrac{x^2 + x - 6}{x - 2}$

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

Example: $f(x) = \dfrac{x - 2}{x + 4}$

Worked Example: $f(x) = \dfrac{x - 1}{x - 3}$