MATH 110 — Module 10 Exam (Respondus LockDown + Webcam) 2026 Update | 40 Questions | 100 pts
MATH 110 — MODULE 10 EXAM
REQUIRES RESPONDUS LOCKDOWN BROWSER + WEBCAM | 2026 UPDATE | 100% CORRECT
College Algebra & Precalculus • Conic Sections · Sequences/Series · Binomial/Combinatorics · Modeling
Aligned with 2026 | 2027 academic standards • Proctored final module — high academic integrity required
Questions 40 multiple-choice (4 sections) Time (suggested) 90 minutes
Points 100 (2.5 pts per question) Passing ≥ 70 (Proficient)
Cognitive Mix 30% recall · 50% application · 20% analysis Format 75% scenario / 25% direct
Proctoring Respondus LockDown + Webcam Calculator Approved scientific/graphing
Examination Instructions
This Module 10 final-module exam is administered under Respondus LockDown Browser with webcam
monitoring to ensure academic integrity. Questions test deep conceptual understanding, multi-step cognitive
application, and synthesis of end-of-course topics. Rationales include step-by-step algebraic manipulations,
series summations, and combinatorial calculations, and explicitly identify the common student errors that
produce each distractor (e.g., confusing permutations with combinations, misidentifying the conic type,
forgetting the factorial in a binomial coefficient, or using the wrong common ratio in a geometric series). Section
4 integrates 2026|2027 modern computational contexts such as graphing-utility conic intersections,
contemporary financial modeling with infinite series, and modern algorithmic applications of combinatorics.
Section 1: Conic Sections (Circles, Parabolas, Ellipses, Hyperbolas) (Q1–Q10)
Identifying, graphing, and writing equations of conic sections in standard and general forms; completing the square;
identifying centers, vertices, foci, and asymptotes; and using the conic discriminant for classification.
Q1. Identify the center and radius of the circle x² + y² − 6x + 8y = 0 by completing the square.
A. Center (3, −4), radius 5 [CORRECT]
B. Center (−3, 4), radius 5
C. Center (3, −4), radius 25
D. Center (3, 4), radius 5
Correct Answer: A — Center (3, −4), radius 5
Rationale: Complete the square: (x²−6x+9) + (y²+8y+16) = 25 → (x−3)² + (y+4)² = 25. Center = (3, −4), r = 5. B
reverses both signs; C uses r² as the radius; D reverses the y-coordinate sign.
Q2. Write y = x² − 4x + 7 in vertex form and identify the vertex.
A. y = (x − 2)² + 3; vertex (2, 3) [CORRECT]
B. y = (x + 2)² + 3; vertex (−2, 3)
C. y = (x − 2)² + 7; vertex (2, 7)
D. y = (x − 2)² − 3; vertex (2, −3)
Correct Answer: A — y = (x − 2)² + 3; vertex (2, 3)
Rationale: y = (x²−4x+4) + 3 = (x−2)² + 3. Vertex = (2, 3). B reverses the horizontal shift sign; C forgets to subtract 4
from the constant; D subtracts instead of adds.
Q3. For the ellipse x²/25 + y²/9 = 1, find the foci.
A. (±4, 0) [CORRECT]
Proctored Final Module Exam • Conics · Sequences · Binomial · Modeling Page 1
, MATH 110 — Module 10 Exam (Respondus LockDown + Webcam) 2026 Update | 40 Questions | 100 pts
B. (0, ±4)
C. (±5, 0)
D. (±3, 0)
Correct Answer: A — (±4, 0)
Rationale: a² = 25, b² = 9, so c² = a² − b² = 16, c = 4. Major axis is horizontal (a > b), so foci = (±4, 0). B places foci on
the wrong axis; C/D use a or b instead of c.
Q4. Find the equations of the asymptotes of the hyperbola x²/9 − y²/16 = 1.
A. y = ±(4/3)x [CORRECT]
B. y = ±(3/4)x
C. y = ±(16/9)x
D. y = ±(9/16)x
Correct Answer: A — y = ±(4/3)x
Rationale: For x²/a² − y²/b² = 1, asymptotes are y = ±(b/a)x with a = 3, b = 4 → y = ±(4/3)x. B inverts the ratio; C uses
b²/a; D uses a/b².
Q5. Identify the conic section: 9x² + 16y² = 144.
A. Ellipse [CORRECT]
B. Hyperbola
C. Parabola
D. Circle
Correct Answer: A — Ellipse
Rationale: Divide by 144: x²/16 + y²/9 = 1. Both x² and y² terms are present with the same sign and different
denominators → ellipse. B requires opposite signs; C lacks an xy or single squared term; D requires equal
denominators.
Q6. For the parabola x² = 12y, find the focus and directrix.
A. Focus (0, 3); directrix y = −3 [CORRECT]
B. Focus (0, 3); directrix y = 3
C. Focus (3, 0); directrix x = −3
D. Focus (0, 6); directrix y = −6
Correct Answer: A — Focus (0, 3); directrix y = −3
Rationale: x² = 4py → 4p = 12 → p = 3. Focus = (0, 3); directrix y = −3. B uses the wrong sign on the directrix; C swaps
axes; D uses p = 6.
Q7. Find the equation of the ellipse with foci (±3, 0) and vertices (±5, 0).
A. x²/25 + y²/16 = 1 [CORRECT]
B. x²/25 + y²/9 = 1
C. x²/16 + y²/25 = 1
D. x²/9 + y²/25 = 1
Correct Answer: A — x²/25 + y²/16 = 1
Rationale: a = 5, c = 3, so b² = a² − c² = 25 − 9 = 16. Major axis horizontal: x²/25 + y²/16 = 1. B uses c² as b²; C swaps
a and b; D places the major axis on the wrong axis.
Q8. Find the center and asymptotes of the hyperbola (x − 1)²/16 − (y + 2)²/9 = 1.
A. Center (1, −2); asymptotes y + 2 = ±(3/4)(x − 1) [CORRECT]
B. Center (1, −2); asymptotes y + 2 = ±(4/3)(x − 1)
C. Center (−1, 2); asymptotes y − 2 = ±(3/4)(x + 1)
Proctored Final Module Exam • Conics · Sequences · Binomial · Modeling Page 2
MATH 110 — MODULE 10 EXAM
REQUIRES RESPONDUS LOCKDOWN BROWSER + WEBCAM | 2026 UPDATE | 100% CORRECT
College Algebra & Precalculus • Conic Sections · Sequences/Series · Binomial/Combinatorics · Modeling
Aligned with 2026 | 2027 academic standards • Proctored final module — high academic integrity required
Questions 40 multiple-choice (4 sections) Time (suggested) 90 minutes
Points 100 (2.5 pts per question) Passing ≥ 70 (Proficient)
Cognitive Mix 30% recall · 50% application · 20% analysis Format 75% scenario / 25% direct
Proctoring Respondus LockDown + Webcam Calculator Approved scientific/graphing
Examination Instructions
This Module 10 final-module exam is administered under Respondus LockDown Browser with webcam
monitoring to ensure academic integrity. Questions test deep conceptual understanding, multi-step cognitive
application, and synthesis of end-of-course topics. Rationales include step-by-step algebraic manipulations,
series summations, and combinatorial calculations, and explicitly identify the common student errors that
produce each distractor (e.g., confusing permutations with combinations, misidentifying the conic type,
forgetting the factorial in a binomial coefficient, or using the wrong common ratio in a geometric series). Section
4 integrates 2026|2027 modern computational contexts such as graphing-utility conic intersections,
contemporary financial modeling with infinite series, and modern algorithmic applications of combinatorics.
Section 1: Conic Sections (Circles, Parabolas, Ellipses, Hyperbolas) (Q1–Q10)
Identifying, graphing, and writing equations of conic sections in standard and general forms; completing the square;
identifying centers, vertices, foci, and asymptotes; and using the conic discriminant for classification.
Q1. Identify the center and radius of the circle x² + y² − 6x + 8y = 0 by completing the square.
A. Center (3, −4), radius 5 [CORRECT]
B. Center (−3, 4), radius 5
C. Center (3, −4), radius 25
D. Center (3, 4), radius 5
Correct Answer: A — Center (3, −4), radius 5
Rationale: Complete the square: (x²−6x+9) + (y²+8y+16) = 25 → (x−3)² + (y+4)² = 25. Center = (3, −4), r = 5. B
reverses both signs; C uses r² as the radius; D reverses the y-coordinate sign.
Q2. Write y = x² − 4x + 7 in vertex form and identify the vertex.
A. y = (x − 2)² + 3; vertex (2, 3) [CORRECT]
B. y = (x + 2)² + 3; vertex (−2, 3)
C. y = (x − 2)² + 7; vertex (2, 7)
D. y = (x − 2)² − 3; vertex (2, −3)
Correct Answer: A — y = (x − 2)² + 3; vertex (2, 3)
Rationale: y = (x²−4x+4) + 3 = (x−2)² + 3. Vertex = (2, 3). B reverses the horizontal shift sign; C forgets to subtract 4
from the constant; D subtracts instead of adds.
Q3. For the ellipse x²/25 + y²/9 = 1, find the foci.
A. (±4, 0) [CORRECT]
Proctored Final Module Exam • Conics · Sequences · Binomial · Modeling Page 1
, MATH 110 — Module 10 Exam (Respondus LockDown + Webcam) 2026 Update | 40 Questions | 100 pts
B. (0, ±4)
C. (±5, 0)
D. (±3, 0)
Correct Answer: A — (±4, 0)
Rationale: a² = 25, b² = 9, so c² = a² − b² = 16, c = 4. Major axis is horizontal (a > b), so foci = (±4, 0). B places foci on
the wrong axis; C/D use a or b instead of c.
Q4. Find the equations of the asymptotes of the hyperbola x²/9 − y²/16 = 1.
A. y = ±(4/3)x [CORRECT]
B. y = ±(3/4)x
C. y = ±(16/9)x
D. y = ±(9/16)x
Correct Answer: A — y = ±(4/3)x
Rationale: For x²/a² − y²/b² = 1, asymptotes are y = ±(b/a)x with a = 3, b = 4 → y = ±(4/3)x. B inverts the ratio; C uses
b²/a; D uses a/b².
Q5. Identify the conic section: 9x² + 16y² = 144.
A. Ellipse [CORRECT]
B. Hyperbola
C. Parabola
D. Circle
Correct Answer: A — Ellipse
Rationale: Divide by 144: x²/16 + y²/9 = 1. Both x² and y² terms are present with the same sign and different
denominators → ellipse. B requires opposite signs; C lacks an xy or single squared term; D requires equal
denominators.
Q6. For the parabola x² = 12y, find the focus and directrix.
A. Focus (0, 3); directrix y = −3 [CORRECT]
B. Focus (0, 3); directrix y = 3
C. Focus (3, 0); directrix x = −3
D. Focus (0, 6); directrix y = −6
Correct Answer: A — Focus (0, 3); directrix y = −3
Rationale: x² = 4py → 4p = 12 → p = 3. Focus = (0, 3); directrix y = −3. B uses the wrong sign on the directrix; C swaps
axes; D uses p = 6.
Q7. Find the equation of the ellipse with foci (±3, 0) and vertices (±5, 0).
A. x²/25 + y²/16 = 1 [CORRECT]
B. x²/25 + y²/9 = 1
C. x²/16 + y²/25 = 1
D. x²/9 + y²/25 = 1
Correct Answer: A — x²/25 + y²/16 = 1
Rationale: a = 5, c = 3, so b² = a² − c² = 25 − 9 = 16. Major axis horizontal: x²/25 + y²/16 = 1. B uses c² as b²; C swaps
a and b; D places the major axis on the wrong axis.
Q8. Find the center and asymptotes of the hyperbola (x − 1)²/16 − (y + 2)²/9 = 1.
A. Center (1, −2); asymptotes y + 2 = ±(3/4)(x − 1) [CORRECT]
B. Center (1, −2); asymptotes y + 2 = ±(4/3)(x − 1)
C. Center (−1, 2); asymptotes y − 2 = ±(3/4)(x + 1)
Proctored Final Module Exam • Conics · Sequences · Binomial · Modeling Page 2
