Week 10 — Quiz (auto-graded) · Polynomial & Rational Functions
Course: College Algebra (MATH 120) · Silver Oak University (fictional sample) · Prof. Calloway
Objective tested: Objective 7 — polynomial end behavior, zeros and multiplicity, rational-function domain, vertical asymptotes, horizontal asymptotes.
Points: 10 (1 each) · Assignment group: Quizzes (15% of grade) · Due: end of Module 10.
This is the human-readable quiz with its vetted answer key and feedback. The import-ready Classic QTI is in
F-quiz-week-10-qti.xml. AI is not permitted on quizzes (course AI policy). Every numeric answer below is pre-computed and independently re-verified (Pythonw10_verify.py, PASS).
Blueprint
| # | Type | Concept | Objective |
|---|---|---|---|
| 1 | Multiple choice | End behavior — odd degree, positive leading coefficient | 7 |
| 2 | Multiple choice | End behavior — even degree, negative leading coefficient | 7 |
| 3 | Multiple choice | Zeros of a factored polynomial | 7 |
| 4 | Multiple choice | Multiplicity — touch vs. cross | 7 |
| 5 | Multiple choice | Domain of a rational function | 7 |
| 6 | Multiple choice | Vertical asymptotes of a rational function | 7 |
| 7 | Multiple choice | Horizontal asymptote — equal degrees | 7 |
| 8 | Multiple choice | Horizontal asymptote — numerator degree less | 7 |
| 9 | Multiple choice | Horizontal asymptote — numerator degree greater | 7 |
| 10 | Matching | Degree comparison ↔ horizontal-asymptote rule | 7 |
No trick questions; distractors target the Week 10 misconceptions named in the lecture outline (even vs. odd end behavior, leading-coefficient sign, multiplicity touch vs. cross, degree-comparison rules for horizontal asymptotes, hole vs. vertical asymptote).
Questions, key, and feedback
Q1 (MC). Describe the end behavior of f(x) = x³.
- A. rises left, rises right
- B. falls left, rises right ✅
- C. falls left, falls right
- D. rises left, falls right
Feedback: The leading term x³ has odd degree (3) and positive leading coefficient (+1). Odd degree → opposite tails; positive leading coefficient → right tail rises. Therefore: falls left, rises right. (A would be even degree, positive; C would be even degree, negative; D would be odd degree, negative.)
Q2 (MC). Describe the end behavior of f(x) = −2x⁴.
- A. falls left, falls right ✅
- B. rises left, rises right
- C. falls left, rises right
- D. rises left, falls right
Feedback: The leading term −2x⁴ has even degree (4) and negative leading coefficient (−2). Even degree → same-direction tails; negative leading coefficient → both tails fall. (B would be even/positive; C would be odd/positive; D would be odd/negative.)
Q3 (MC). Find the zeros of f(x) = (x − 2)(x + 3).
- A. x = 2 only
- B. x = −2 and x = 3
- C. x = 2 and x = −3 ✅
- D. x = −2 and x = −3
Feedback: Set each factor to zero: x − 2 = 0 gives x = 2; x + 3 = 0 gives x = −3. (B reverses the signs of both zeros — a common sign error.)
Q4 (MC). For f(x) = (x − 1)²(x + 4), at which zero does the graph touch the x-axis without crossing?
- A. x = −4
- B. x = 1 ✅
- C. Both x = 1 and x = −4
- D. Neither — both zeros are crossed
Feedback: The factor (x − 1)² has multiplicity 2 (even) → the graph touches and turns back at x = 1. The factor (x + 4) has multiplicity 1 (odd) → the graph crosses at x = −4. (C confuses all zeros with "touch"; D ignores multiplicity entirely.)
Q5 (MC). What is the domain of f(x) = (x + 1) / (x² − 4)?
- A. all real numbers
- B. x ≠ 4
- C. x ≠ 2 and x ≠ −2 ✅
- D. x ≠ 0
Feedback: Factor the denominator: x² − 4 = (x − 2)(x + 2). Setting each factor to zero: x = 2 and x = −2. The domain excludes both values: x ≠ 2 and x ≠ −2. (B forgets to take the square root; D incorrectly sets the numerator to zero.)
Q6 (MC). Identify the vertical asymptote(s) of f(x) = (x + 1) / (x² − 4).
- A. x = 4 only
- B. x = −1 only
- C. x = 1 and x = −1
- D. x = 2 and x = −2 ✅
Feedback: Factor the denominator: (x − 2)(x + 2). Check the numerator at each zero: at x = 2, numerator = 3 ≠ 0 → vertical asymptote; at x = −2, numerator = −1 ≠ 0 → vertical asymptote. Neither factor cancels, so both give true vertical asymptotes. (B sets the numerator to zero instead; C gives the wrong values.)
Q7 (MC). Find the horizontal asymptote of f(x) = (3x + 1) / (x − 5).
- A. y = 0
- B. y = 3 ✅
- C. y = 1
- D. no horizontal asymptote
Feedback: The numerator and denominator have equal degree (both 1). The horizontal asymptote is the ratio of leading coefficients: 3 / 1 = 3, so y = 3. (A is the rule for deg numerator < deg denominator; C wrongly gives 1 (coefficient of x in denominator) or equates leading coefficients to 1; D applies when deg numerator > deg denominator.)
Q8 (MC). Find the horizontal asymptote of f(x) = 2x / (x² + 1).
- A. y = 0 ✅
- B. y = 2
- C. y = 1
- D. no horizontal asymptote
Feedback: The numerator has degree 1 and the denominator has degree 2. Since deg(numerator) < deg(denominator), the horizontal asymptote is y = 0 — the denominator grows much faster than the numerator, so the fraction shrinks to zero. (B uses the leading coefficient of the numerator instead; D applies when numerator degree is greater.)
Q9 (MC). What is the horizontal asymptote of f(x) = (x² + 1) / (x − 1)?
- A. y = 0
- B. y = 1
- C. y = −1
- D. no horizontal asymptote ✅
Feedback: The numerator has degree 2 and the denominator has degree 1. Since deg(numerator) > deg(denominator), there is no horizontal asymptote — the function grows without bound as x → ±∞. (A is the rule for deg numerator < denominator; B and C confuse the coefficients with the asymptote.)
Q10 (Matching). Match each degree comparison to the correct horizontal-asymptote rule.
| Degree comparison | Rule |
|---|---|
| deg(numerator) < deg(denominator) | y = 0 |
| deg(numerator) = deg(denominator) | y = (leading coeff of numerator) / (leading coeff of denominator) |
| deg(numerator) > deg(denominator) | no horizontal asymptote |
Feedback: When the denominator dominates, the fraction shrinks to zero (y = 0). When they're equal, the leading coefficients determine the level the function approaches. When the numerator dominates, the function grows without bound — no horizontal floor or ceiling exists.
Answer key (quick reference)
| Q | Answer |
|---|---|
| 1 | B (falls left, rises right) |
| 2 | A (falls left, falls right) |
| 3 | C (x = 2 and x = −3) |
| 4 | B (x = 1) |
| 5 | C (x ≠ 2 and x ≠ −2) |
| 6 | D (x = 2 and x = −2) |
| 7 | B (y = 3) |
| 8 | A (y = 0) |
| 9 | D (no horizontal asymptote) |
| 10 | less-than→y=0 / equal→ratio of leading coefficients / greater-than→no HA |
Quality gate (self-checked, computer-verified): each single-answer item (Q1–Q9) has exactly one correct option; the matching item (Q10) pairs all three comparisons one-to-one with no leftover options. Arithmetic pre-computed and independently re-verified (w10_verify.py, PASS): Q3 zeros of (x−2)(x+3) = {2,−3} ✓; Q5 zeros of x²−4 = {2,−2} ✓; Q6 numerator at x=2 is 3≠0 and at x=−2 is −1≠0, both VAs ✓; Q7 ratio 3/1=3 ✓; Q8 deg 1 < deg 2 → y=0 ✓; Q9 deg 2 > deg 1 → no HA ✓. All checks PASS. QTI parse confirmation: F-quiz-week-10-qti.xml generated via qtigen.py — parses OK, 10 items.
Item-bank entries (for variants + the final)
All ten items are tagged course=MATH120 · week=10 · objective=7 · topic=polynomial-rational-functions and deposited in Item Bank: Week 10 — Polynomial & Rational Functions. The final (Week 16) draws fresh items from this bank. (Tags: q1 end-behavior-odd, q2 end-behavior-even, q3 zeros-factored, q4 multiplicity, q5 rational-domain, q6 vertical-asymptote, q7 ha-equal-degree, q8 ha-less-degree, q9 ha-greater-degree, q10 ha-matching.)
Canvas placement block
canvas_object = Quizzes::Quiz
title = "Week 10 Quiz — Polynomial & Rational Functions"
assignment_group = "Quizzes"
points_possible = 10
grading_type = points
due_offset_days = 6 # 6 days after module start (Sun Nov 8)
published = true
shuffle_answers = true
provenance = "~ Prof. Calloway's edition · Fall 2026 · built with thecoursemaker.com"
F-quiz-week-10-qti.xml) ships inside the course's .imscc package — it lands in the Canvas gradebook on import.~ Prof. Calloway's edition · Fall 2026 · built with thecoursemaker.com