Rational Functions and Their Graphs

〽️ Rational Functions and Their Graphs

Part 1 of 5 — What Is a Rational Function?

Topics in This Part

🔑 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

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

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.

Notice two things:

• 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.

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:

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

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

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.

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 $x = \,?$ 2) $g(x) = \dfrac{4}{x^2 - 9}$. Smaller excluded $x = \,?$ 3) $g(x) = \dfrac{4}{x^2 - 9}$. Larger excluded $x = \,?$

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.
• 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).

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.

〽️ 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: $f(x) = \dfrac{1}{x - 4}$

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 $x = 4$.

⚠️ 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 $(3, 6)$.

🔑 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:

1. The $x$-coordinate is the value that made the canceled factor zero.
2. 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 $x \ne 2$.

• Hole's $x$: the canceled factor $x - 2 = 0 \Rightarrow x = 2$.
• Hole's $y$: plug into $x + 3$: $\;2 + 3 = 5$.

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 $x+2$) $= \,?$

〽️ 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 ($d$):

Examples

• $\dfrac{3x+1}{x^2 - 5}$: degree $1 < 2$, so $y = 0$.
• $\dfrac{2x^2 + 1}{5x^2 - x}$: degrees equal, so $y = \dfrac{2}{5}$.
• $\dfrac{x^2 + 1}{x - 1}$: 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 $x \to \pm\infty$, the remainder term $\frac{6}{x-1} \to 0$, 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) $f(x) = \dfrac{8}{x^2 + 1}$. $y = \,?$

〽️ 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 $P(x) = 0$ and solve.

$y$-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 $f(x) = \dfrac{P(x)}{Q(x)}$:

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

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

💡 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: $x = \,?$

〽️ 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

⚠️ 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.