Lesson 06: Rational Functions — Zeros, Vertical Asymptotes & Holes

Warm-Up

Three functions that look nearly identical:

f(x) = (x − 2)/(x − 2)²     g(x) = (x − 2)²/(x − 2)     h(x) = (x − 2)/(x + 2)

At x = 2: f blows up (vertical asymptote), g has a tiny puncture (hole), and h calmly equals 0 (a zero). Three completely different behaviors, controlled entirely by where the factor (x − 2) lives and how many times. Today's job: never confuse these three again.

Core Concept

The domain comes first

A rational function r(x) = p(x)/q(x) is undefined wherever q(x) = 0. Every x-value that zeroes the denominator is out of the domain — the only question is what the graph does near it: asymptote or hole.

The three-way sort

For each x-value a that zeroes the numerator or denominator, compare multiplicities. Write m = multiplicity of a in the numerator, n = multiplicity in the denominator (either can be 0):

The slogan: denominator wins → asymptote; numerator ties or wins → hole.

(If m > n, the hole sits on the x-axis at (a, 0) — the simplified function still has a zero there, but r itself is undefined at a, so it's a hole, not an x-intercept.)

Finding a hole's y-coordinate

Cancel the common factors completely; evaluate what's left at a:

r(x) = (x − 3)(x + 3) / ((x − 3)(x + 2))   →   simplified: (x + 3)/(x + 2)
hole at x = 3:  y = (3 + 3)/(3 + 2) = 6/5   →   hole at (3, 6/5)

In limit notation: lim x→3 r(x) = 6/5, even though r(3) is undefined. That sentence — finite limit, undefined value — is the definition of a hole on this exam.

Behavior near a vertical asymptote: one-sided limits

Near a vertical asymptote the outputs run to +∞ or −∞, and the two sides can disagree. Determine each side with a sign check: plug in a test value (or reason about signs of each factor) just left and just right of a.

For r(x) = (x + 1)/((x − 2)(x + 3)) near x = 2:

• x → 2⁺ (e.g., x = 2.01): numerator ≈ 3 (+), (x − 2) is tiny (+), (x + 3) ≈ 5 (+) → overall + and huge: lim x→2⁺ r(x) = +∞
• x → 2⁻ (e.g., x = 1.99): (x − 2) is tiny (−) → overall −: lim x→2⁻ r(x) = −∞

[GRAPH: r(x) = (x+1)/((x−2)(x+3)). Vertical asymptotes x = −3 and x = 2 (dashed). Zero at (−1, 0). Horizontal asymptote y = 0. Near x = 2: left branch plunges to −∞, right branch rises from +∞. Near x = −3: left branch falls to −∞ (as x→−3⁻, numerator ≈ −2 and denominator is a tiny positive), right branch rises to +∞ (as x→−3⁺, denominator is a tiny negative). Middle branch connects across the zero at (−1, 0).]

A useful parity shortcut: if the asymptote factor's net multiplicity (n − m) is odd, the two sides point opposite directions; if even, both sides point the same direction (like 1/x² at 0).

Why your calculator can't be trusted here 📱

Graphing calculators plot pixels; a hole is a single missing point — usually invisible. And in "connected" mode, some calculators draw a fake near-vertical line at asymptotes. The exam knows this: hole/asymptote questions live in the no-calculator sections, and FRQ justifications must cite multiplicities, not screens.

Worked Examples

Example 1 (easy) — Full inventory 🚫 No-Calc

Problem: For r(x) = (x² − 9)/(x² − x − 6), find all zeros, vertical asymptotes, and holes.

Solution: Factor:

r(x) = (x − 3)(x + 3) / ((x − 3)(x + 2))

• x = 3: multiplicity 1 up top, 1 below → m ≥ n → hole; simplified (x+3)/(x+2) gives y = 6/5. Hole at (3, 6/5).
• x = −3: numerator only → zero; x-intercept (−3, 0).
• x = −2: denominator only → vertical asymptote x = −2.

Interpretation: Factor completely, then sort each special x-value with the multiplicity comparison. Three candidates, three different fates.

Example 2 (medium) — Hole on the axis 🚫 No-Calc

Problem: Analyze x = 1 for r(x) = (x − 1)²/((x − 1)(x + 5)).

Solution: m = 2 (numerator), n = 1 (denominator): m ≥ n → hole. Simplify: (x − 1)/(x + 5); at x = 1: y = 0/6 = 0. Hole at (1, 0) — a puncture sitting exactly on the x-axis.

Is x = 1 a zero of r? No. r(1) is undefined (original denominator is 0), so (1, 0) is not an x-intercept — the graph approaches the axis there and skips the point itself. In limit language: lim x→1 r(x) = 0 but r(1) does not exist.

Interpretation: "Zero" requires the function to actually equal zero there. Domain first, always.

Example 3 (medium) — One-sided limits at an asymptote 🚫 No-Calc

Problem: For r(x) = (2x + 6)/((x − 4)²), describe the behavior near x = 4 with limit notation.

Solution: At x = 4: numerator = 14 ≠ 0, denominator has multiplicity 2 → vertical asymptote. Sign check: (x − 4)² is positive on both sides; numerator ≈ 14 > 0 near 4. So:

lim x→4⁻ r(x) = +∞     and     lim x→4⁺ r(x) = +∞

Both branches shoot up — the even net multiplicity means the sides agree.

Interpretation: Odd net multiplicity → branches split (one up, one down). Even → branches match. Check the sign of the rest of the function to know which direction they take.

Example 4 (AP-style) — Build to order 🚫 No-Calc

Problem: Write a rational function with: a zero at x = 2, a vertical asymptote at x = 5, a hole at x = −1, and horizontal asymptote y = 3.

Strategy: Place factors by job, then tune leading coefficients for the HA.

Solution: Zero → (x − 2) up top. VA → (x − 5) below. Hole → (x + 1) in both. Draft:

r(x) = (x − 2)(x + 1) / ((x − 5)(x + 1))

Check HA: degrees 2/2, ratio 1/1 = 1. Need 3 → multiply the numerator by 3:

r(x) = 3(x − 2)(x + 1) / ((x − 5)(x + 1))

Verify: zero at 2 ✓ (numerator only); VA at 5 ✓ (denominator only); hole at −1 ✓ (multiplicity 1 = 1); HA: leading ratio 3x²/x² = 3 ✓.

Interpretation: Each feature is a factor-placement instruction; the horizontal asymptote is a knob you turn last with the leading constant. This construction task is a recurring MC and FRQ item.

Common Mistakes

1. Calling a hole a zero. If the x-value kills the denominator too, it's out of the domain — no x-intercept, even when the hole sits at height 0 (Example 2). Fix: check the denominator before declaring any zero.
2. Canceling and forgetting. After simplifying, students treat the reduced function as the original. The cancelled factor leaves a scar: the hole. Fix: record excluded x-values before canceling.
3. Assuming every denominator zero is an asymptote. Only when the denominator's multiplicity exceeds the numerator's. Fix: compare m vs. n every time.
4. Skipping the sign check at asymptotes. "It goes to infinity" costs the point when the exam asks which infinity on which side. Fix: test values just left and right, or use the parity shortcut + overall sign.
5. Trusting the graphing window. No hole will ever show up as a gap in a standard plot. Fix: algebra decides; the calculator merely decorates.

Practice Problems

Problems 1–3 refer to r(x) = (x² − 9)/(x² − x − 6).

12. (FRQ-style) 🚫 Let r(x) = (2(x − 1)(x + 4)) / ((x − 1)(x − 3)). (i) Identify the domain of r, and classify the graph's behavior at each excluded x-value (hole or vertical asymptote), with coordinates for any hole. (ii) Write one-sided limit statements describing the behavior of r on each side of any vertical asymptote (include a sign check). (iii) Find the horizontal asymptote and the intercepts of the graph.

FRQ Practice — Task Model: Communicating about Functions (FRQ 4 style) 🚫 No-Calc

Let f(x) = ((x + 2)²(x − 3)) / ((x + 2)(x − 3)²).

(a) (i) State the domain of f. (ii) Classify the behavior of the graph of f at x = −2 and at x = 3 (zero, hole, or vertical asymptote), using multiplicities to justify each classification.

(b) (i) Find the coordinates of any hole. (ii) Write a limit statement that expresses what the hole means.

(c) (i) Using one-sided limit notation, describe the behavior of f on both sides of any vertical asymptote. (ii) Justify the signs with a sign analysis of the simplified expression.

Model Response & Rubric (6 points)

(a) [2 pts] (i) [1 pt] Domain: all real x with x ≠ −2 and x ≠ 3. (ii) [1 pt] At x = −2: numerator multiplicity 2 ≥ denominator multiplicity 1 → hole. At x = 3: numerator multiplicity 1 < denominator multiplicity 2 → vertical asymptote.

(b) [2 pts] (i) [1 pt] Simplify: f(x) = (x + 2)/(x − 3) for x ≠ −2, 3. At x = −2: y = (−2 + 2)/(−2 − 3) = 0/(−5) = 0. Hole at (−2, 0). (ii) [1 pt] lim x→−2 f(x) = 0, although f(−2) is undefined — the outputs approach 0 near x = −2, but there is no point on the graph at x = −2.

(c) [2 pts] (i) [1 pt] lim x→3⁻ f(x) = −∞ and lim x→3⁺ f(x) = +∞. (ii) [1 pt] Using the simplified form (x + 2)/(x − 3): near x = 3 the numerator ≈ 5 > 0. For x → 3⁻, (x − 3) is a small negative → quotient → −∞; for x → 3⁺, (x − 3) is a small positive → quotient → +∞. (Note the net multiplicity at x = 3 is 2 − 1 = 1, odd — consistent with the two sides pointing opposite directions.)