Rational Functions and Asymptotes

➗ Rational Functions and Asymptotes

Part 1 of 7 — Meet the Rational Function

Topics in This Part

🔑 Key Concept: A rational function is a ratio of two polynomials, $f(x) = \dfrac{P(x)}{Q(x)}$. Everything interesting about its graph — gaps, towers shooting to infinity, and flat lines it hugs — traces back to where $Q(x) = 0$ and to the degrees of $P$ and $Q$.

What Is a Rational Function?

A rational function has the form

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

where $P(x)$ and $Q(x)$ are polynomials and $Q(x)$ is not the zero polynomial.

Examples and Non-Examples

💡 The word "rational" comes from ratio — a rational function is literally a ratio of polynomials, just as a rational number is a ratio of integers.

Concept Check 🎯

Finding the Domain

The domain of a rational function is all real numbers except where the denominator equals zero.

To find the domain:

1. Set the denominator equal to $0$.
2. Solve for $x$ — these are the excluded values.
3. The domain is every other real number.

Worked Example: $f(x) = \dfrac{x+4}{x^2 - 9}$

Set the denominator to zero:

$x^2 - 9 = 0 \;\Rightarrow\; (x-3)(x+3) = 0 \;\Rightarrow\; x = 3 \text{ or } x = -3$

So the domain is all real numbers except $x = 3$ and $x = -3$, written $x \neq \pm 3$.

⚠️ Look only at the denominator for the domain. The numerator can be zero with no problem — that just makes the whole fraction equal $0$.

Find the Excluded Values 🧮

Enter the $x$-value(s) excluded from the domain. If there are two, enter them smallest first.

1) $f(x) = \dfrac{x}{x - 7}$, excluded value $= \,?$ 2) $g(x) = \dfrac{x+1}{x^2 - 25}$, smaller excluded value $= \,?$ 3) Same $g(x)$: larger excluded value $= \,?$

Where the Graph "Breaks"

At an excluded $x$-value, the graph cannot have a point — something special happens there. There are exactly two possibilities, and which one occurs is the heart of this whole lesson:

Plus, far out to the left and right, the graph often levels off toward a horizontal asymptote.

🔑 The plan: Part 2 handles vertical asymptotes and holes. Part 3 handles horizontal asymptotes. Part 4 handles slant asymptotes. By Part 6 you'll sketch the whole graph from the equation alone.

Quick Preview 🔽

You don't need to compute anything yet — just match each phrase to the idea it describes.

➗ Rational Functions and Asymptotes

Part 2 of 7 — Vertical Asymptotes & Holes

🔑 The Distinction: Both vertical asymptotes and holes happen where the denominator is zero. The difference is whether that factor cancels with the numerator. Always factor and simplify first.

Vertical Asymptotes

A vertical asymptote (VA) is a vertical line $x = a$ that the graph approaches but never touches, racing toward $+\infty$ or $-\infty$.

How to find them:

1. Factor numerator and denominator completely.
2. Cancel any common factors (those become holes — see below).
3. In the simplified function, set the denominator equal to $0$. Each solution gives a vertical asymptote.

Worked Example: $f(x) = \dfrac{x + 2}{x - 5}$

Nothing cancels, so set the denominator to zero:

$x - 5 = 0 \;\Rightarrow\; x = 5$

The graph has a vertical asymptote at $x = 5$.

💡 Why does it blow up? As $x \to 5$, the denominator $x - 5 \to 0$ while the numerator $\to 7$. A fixed number divided by something tiny is enormous — so $|f(x)| \to \infty$.

Locate the Vertical Asymptotes 🧮

After simplifying, enter the vertical asymptote location(s). If there are two, enter them smallest first.

1) $f(x) = \dfrac{x+3}{x-8}$, VA at $x = \,?$ 2) $g(x) = \dfrac{5}{x^2 - x - 6}$, smaller VA at $x = \,?$ 3) Same $g(x)$: larger VA at $x = \,?$

Holes (Removable Discontinuities)

A hole appears when a factor cancels from both the numerator and denominator. The graph looks normal there except for one missing point, drawn as an open circle.

How to find a hole:

1. Factor and cancel the common factor.
2. The $x$-value that made the canceled factor zero is the hole's $x$-coordinate.
3. Plug that $x$ into the simplified function to get the hole's $y$-coordinate.

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

Factor and cancel:

$f(x) = \frac{(x-2)(x+2)}{x-2} = x + 2, \quad x \neq 2$

The factor $(x-2)$ canceled, so there's a hole at $x = 2$. Its height:

$y = (2) + 2 = 4$

So the hole is at the point $(2, 4)$ — the graph is the line $y = x+2$ with one point punched out.

⚠️ Use the simplified function for the hole's height — never plug into the original (you'd get $\frac{0}{0}$, which is undefined).

Concept Check 🎯

The Decision Rule, Summarized

When the denominator is zero at $x = a$, factor everything first and ask one question:

🔑 Always simplify before you classify. A factor that survives in the denominator is an asymptote; a factor that cancels is a hole. The next drill puts this into practice — you'll get both the hole's location and its height.

Find the Hole's Coordinates 🧮

The function $f(x) = \dfrac{x^2 - 9}{x - 3}$ has a hole.

1) $x$-coordinate of the hole $= \,?$ 2) $y$-coordinate of the hole $= \,?$

➗ Rational Functions and Asymptotes

Part 3 of 7 — Horizontal Asymptotes (The Degree Test)

🔑 The End-Behavior Question: A horizontal asymptote (HA) describes what $f(x)$ approaches as $x \to \pm\infty$. You find it by comparing the degree of the numerator to the degree of the denominator — three cases, one rule.

The Degree Test for Horizontal Asymptotes

Let $n = \deg(\text{numerator})$ and $d = \deg(\text{denominator})$.

💡 Memory aids: "Bottom Bigger → $y = 0$" (bigger bottom drags the value to zero). "Same degree → ratio of leaders." "Top bigger → no horizontal asymptote."

Three Quick Examples

Bottom-heavy ($n < d$): $f(x) = \frac{2x + 1}{x^2 - 5} \quad n=1,\ d=2 \;\Rightarrow\; \text{HA: } y = 0$

Equal degrees ($n = d$): $g(x) = \frac{3x^2 - x}{6x^2 + 4} \quad n=2,\ d=2 \;\Rightarrow\; \text{HA: } y = \frac{3}{6} = \frac{1}{2}$

Top-heavy ($n > d$): $h(x) = \frac{x^2 + 1}{x - 3} \quad n=2,\ d=1 \;\Rightarrow\; \text{no HA}$

⚠️ For the equal-degree case, use only the leading coefficients — the numbers in front of the highest-power terms. Ignore everything else.

Concept Check 🎯

Can a Graph Cross Its Horizontal Asymptote?

Yes! Unlike a vertical asymptote, a graph may cross its horizontal asymptote — the HA only describes the long-run end behavior, not the middle.

To find where $f$ crosses its HA $y = c$, just solve $f(x) = c$.

Example: $f(x) = \dfrac{x}{x^2 + 1}$ has HA $y = 0$ (bottom-heavy). Solving $\frac{x}{x^2+1} = 0$ gives $x = 0$, so the graph crosses $y = 0$ at the origin, then approaches it again far out.

💡 This is the big behavioral difference: a curve never crosses a vertical asymptote, but it can cross a horizontal (or slant) asymptote.

Classify Each Function 🔽

For each function, choose its horizontal asymptote.

Watch the Leading Coefficients

For the equal-degree case, the value of the horizontal asymptote is just the ratio of leading coefficients — the constant and lower-degree terms don't matter at infinity.

$\frac{a x^n + \cdots}{b x^n + \cdots} \;\xrightarrow[x \to \pm\infty]{}\; \frac{a}{b}$

💡 So $\dfrac{10x^2 + 1}{5x^2 - 3}$ has HA $y = \frac{10}{5} = 2$, regardless of the $+1$ and $-3$. Compute a couple of these in the drill below.

Horizontal Asymptote Values 🧮

Each function below has a horizontal asymptote $y = c$. Enter the value of $c$.

1) $f(x) = \dfrac{10x^2 + 1}{5x^2 - 3}$, $\;c = \,?$ 2) $g(x) = \dfrac{9x - 4}{6x + 7}$, $\;c = \,?$ (decimal or fraction)

➗ Rational Functions and Asymptotes

Part 4 of 7 — Slant (Oblique) Asymptotes

🔑 One Degree Too Big: When the numerator's degree is exactly one more than the denominator's, there's no horizontal asymptote — instead the graph follows a slanted line. You find it with polynomial long division.

When Does a Slant Asymptote Occur?

A slant (oblique) asymptote occurs exactly when

$\deg(\text{numerator}) = \deg(\text{denominator}) + 1$

That is, the top is one degree higher than the bottom.

⚠️ A function has at most one of {horizontal, slant} asymptote — never both. If you have a slant asymptote, there is no horizontal asymptote.

Concept Check 🎯

Finding It by Long Division

To find the slant asymptote, divide the numerator by the denominator. The quotient (ignoring the remainder) is the line $y = mx + b$.

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

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

$\frac{x^2 + 3x + 2}{x - 1} = x + 4 + \frac{6}{x - 1}$

Let's verify the division:

• $x \cdot (x - 1) = x^2 - x$. Subtract: $(x^2 + 3x) - (x^2 - x) = 4x$.
• Bring down $+2$: $4x + 2$. Then $4 \cdot (x-1) = 4x - 4$. Subtract: $(4x+2)-(4x-4) = 6$.

So the quotient is $x + 4$ with remainder $6$.

As $x \to \pm\infty$, the term $\dfrac{6}{x-1} \to 0$, so the graph approaches the line

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

💡 The remainder term always dies off at infinity — that's why the quotient line is the asymptote.

Find the Slant Asymptote 🧮

Divide and read off the line $y = mx + b$.

For $f(x) = \dfrac{x^2 - 5x + 3}{x - 2}$, the division gives $f(x) = (x - 3) - \dfrac{3}{x-2}$.

1) Slope $m$ of the slant asymptote $= \,?$ 2) Intercept $b$ of the slant asymptote $= \,?$

End-Behavior Decision Tree

Putting Parts 3 and 4 together, the degree comparison alone decides the entire end behavior:

🔑 Check the degrees first, every time. Only when $n = d+1$ do you actually need to perform the long division.

End-Behavior Round-Up 🔽

For each function, choose the correct end-behavior asymptote.

➗ Rational Functions and Asymptotes

Part 5 of 7 — Intercepts & Sign Analysis

🔑 The Remaining Anchors: Asymptotes give the skeleton; intercepts and sign analysis tell you where the curve sits — above or below the $x$-axis in each region. Together they pin the graph down.

Finding Intercepts

$x$-intercepts (zeros): set $f(x) = 0$. A fraction is zero only when its numerator is zero (and the denominator is not zero there).

• Solve $P(x) = 0$, then discard any solution that also makes $Q(x) = 0$ (that's a hole, not a zero).

$y$-intercept: plug in $x = 0$, i.e. compute $f(0)$ — as long as $x = 0$ is in the domain.

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

• $x$-intercept: numerator $x - 3 = 0 \Rightarrow x = 3$. Check: denominator at $x=3$ is $4 \neq 0$. ✓ So $(3, 0)$.
• $y$-intercept: $f(0) = \dfrac{0 - 3}{0 + 1} = \dfrac{-3}{1} = -3$. So $(0, -3)$.

⚠️ A value that makes both top and bottom zero is not an $x$-intercept — it's a hole. Always check the denominator before declaring a zero.

Find the Intercepts 🧮

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

1) $x$-intercept: $x = \,?$ 2) $y$-intercept: $y = \,?$ (decimal or fraction is fine)

Sign Analysis

To know whether the curve is above ($+$) or below ($-$) the $x$-axis, mark every $x$-intercept and every vertical asymptote on a number line. These split the line into intervals. The sign of $f$ can only change at those marks.

Pick a test point in each interval and check the sign of $f$.

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

Critical $x$-values: zero at $x = 1$, vertical asymptote at $x = -2$. Test each region:

So the graph is above the axis for $x < -2$, below it between $-2$ and $1$, and above it again for $x > 1$.

💡 The sign tells you which way the graph escapes along each vertical asymptote: approaching $x = -2$ from the right, $f < 0$, so the branch dives to $-\infty$.

Concept Check 🎯

The Sign-Analysis Routine

Here is the compact procedure you'll use on every graph:

1. Find the $x$-intercepts (zeros of the numerator) and vertical asymptotes (zeros of the denominator).
2. Mark them on a number line — they split it into intervals.
3. Test one value in each interval; the sign of $f$ there is the sign for the whole interval.

💡 You only need the sign of each test value, not its exact size — so just track whether each factor is positive or negative. Try the routine on the function below.

Determine Each Sign 🔽

For $g(x) = \dfrac{x}{x - 4}$ (zero at $x = 0$, vertical asymptote at $x = 4$), pick the sign of $g$ on each interval.

➗ Rational Functions and Asymptotes

Part 6 of 7 — Putting It All Together: Sketching the Graph

🔑 The Full Recipe: You now have every ingredient. Sketching a rational function is just running a checklist in order, then connecting the dots through the regions sign analysis hands you.

The Graphing Checklist

💡 Do the steps in order — each one constrains the next, and by step 6 the shape is basically forced.

Full Worked Example: $f(x) = \dfrac{x^2 - 4}{x^2 - x - 6}$

Step 1 — Factor: $f(x) = \frac{(x-2)(x+2)}{(x-3)(x+2)}$ The $(x+2)$ cancels $\Rightarrow$ hole at $x = -2$. Simplified: $f(x) = \dfrac{x-2}{x-3}$.

Step 2 — Vertical asymptote: simplified denominator $x - 3 = 0 \Rightarrow$ VA at $x = 3$.

Step 3 — End behavior: equal degrees ($1$ and $1$), leading coefficients $\frac{1}{1}$ $\Rightarrow$ HA $y = 1$.

Step 4 — Intercepts:

• $x$-intercept: $x - 2 = 0 \Rightarrow x = 2$, point $(2, 0)$.
• $y$-intercept: $f(0) = \frac{-2}{-3} = \frac{2}{3}$, point $\left(0, \tfrac{2}{3}\right)$.

Step 5 — Hole height: in simplified $f$, at $x = -2$: $\frac{-2-2}{-2-3} = \frac{-4}{-5} = \frac{4}{5}$, so the hole is at $\left(-2, \tfrac{4}{5}\right)$.

✅ The graph is the curve $\frac{x-2}{x-3}$ approaching $y = 1$ at the ends, breaking at $x = 3$, crossing the axis at $(2,0)$, with an open circle at $\left(-2, \tfrac{4}{5}\right)$.

Concept Check 🎯

Based on the worked example $f(x) = \dfrac{x^2 - 4}{x^2 - x - 6}$:

A Simpler Sketch: $h(x) = \dfrac{2x}{x - 1}$

Not every function has a hole. Many are already in lowest terms, so the checklist is even quicker. For $h(x) = \dfrac{2x}{x-1}$:

• Vertical asymptote: $x - 1 = 0 \Rightarrow x = 1$.
• Horizontal asymptote: equal degrees, ratio $\frac{2}{1} \Rightarrow y = 2$.
• $x$-intercept: $2x = 0 \Rightarrow x = 0$ — so the curve passes through the origin $(0,0)$, which is also the $y$-intercept.

💡 Two branches: one in the lower-left, one in the upper-right, each hugging the lines $x = 1$ and $y = 2$. Confirm the features yourself below.

Match the Function to Its Features 🔽

For $h(x) = \dfrac{2x}{x - 1}$, identify each feature.

Your Turn — Run the Whole Checklist

Time to drive the process yourself. For the next function, work through the steps in order: denominator $\to$ vertical asymptote, degree test $\to$ horizontal asymptote, numerator $\to$ $x$-intercept.

🔑 Notice the leading coefficient of the numerator is not $1$ this time — that matters for the horizontal asymptote, since the ratio uses the actual leading coefficients.

Run the Checklist 🧮

For $f(x) = \dfrac{3x - 6}{x + 4}$:

1) Vertical asymptote: $x = \,?$ 2) Horizontal asymptote: $y = \,?$ 3) $x$-intercept: $x = \,?$

➗ Rational Functions and Asymptotes

Part 7 of 7 — Mixed Practice & Mastery Check

You can now find the domain, locate vertical asymptotes & holes, determine horizontal & slant asymptotes, compute intercepts, run a sign analysis, and sketch the whole graph. Let's put it all together.

Quick Reference

⚠️ Three traps to avoid: (1) forgetting to factor and cancel first, (2) mixing up the degree cases, (3) calling a $\frac{0}{0}$ point a zero when it is really a hole.

Mixed Practice 🎯

From Recognizing to Computing

The questions above asked you to recognize features. The next drill asks you to compute them from scratch — multiple asymptotes in a single function. Factor the denominator, check nothing cancels, then apply the degree test.

💡 When the denominator is a difference of squares like $x^2 - 9$, expect two vertical asymptotes — one for each factor.

Mixed Drill 🧮

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

1) Smaller vertical asymptote: $x = \,?$ 2) Larger vertical asymptote: $x = \,?$ 3) Horizontal asymptote: $y = \,?$

The Sneaky One: a Hidden Cancellation

The trickiest functions look like they have an asymptote where they actually have a hole. The only way to be sure is to factor the numerator too and check for a common factor before you classify anything.

⚠️ If a denominator factor cancels, that location is a hole, not a vertical asymptote. If everything in the denominator cancels, the graph is the leftover polynomial with one point removed, so it has no asymptotes at all.

Full Feature Identification 🔽

For $f(x) = \dfrac{x^2 - x - 6}{x + 2}$, work out each feature. (Hint: factor the numerator first.)

You've Got the Full Toolkit

You can now take any rational function and read off its domain, holes, vertical asymptotes, horizontal or slant asymptote, intercepts, and sign behavior — everything you need to sketch it.

🔑 The single most important habit: factor and simplify first. Almost every mistake on this topic comes from skipping that step. One last quiz to lock it in.

Exit Quiz ✅

Answer all three to finish the lesson.