MTH 123 - College Algebra and Trigonometry Final Exam 2026/2027 | Complete Questions & Verified Answers | College-Level Precalculus Assessment

MTH 123 - College Algebra and Trigonometry Final
Exam (2026/2027) | QUESTIONS AND ANSWERS
MTH 123: College Algebra and Trigonometry Comprehensive Examination | Core Domains:
Algebraic Functions & Their Graphs (Polynomial, Rational, Exponential, Logarithmic),
Trigonometric Functions & Identities, Analytic Trigonometry, Systems of Equations &
Matrices, Conic Sections, Complex Numbers, Sequences & Series, and Applications & Modeling
| College-Level Precalculus Focus | Comprehensive Course Final Exam Format


Exam Structure

The MTH 123 College Algebra and Trigonometry Final Exam for the 2026/2027 academic cycle
is a 90-question, multiple-choice question (MCQ) and free-response problem-solving
examination.

Introduction​
This MTH 123 College Algebra and Trigonometry Final Exam guide for the 2026/2027 cycle
assesses mastery of advanced algebraic and trigonometric concepts essential for calculus and
STEM fields. The content emphasizes function analysis, equation solving, graphical
interpretation, and the application of trigonometric identities and laws to solve real-world
problems.

Answer Format​
All correct answers and mathematical solutions must be presented in bold and green,
followed by step-by-step rationales that demonstrate algebraic manipulation, function
transformations, application of trigonometric identities (e.g., Pythagorean, sum/difference), and
clear graphical or analytical reasoning.


Question 1: Solve for \( x \): \( \log_2(x) + \log_2(x - 2) = 3 \).



(A) \( x = 4 \)



(B) \( x = -2 \)



(C) \( x = 2 \)



(D) \( x = 1 \)



(E) No real solution

,Correct Answer: (A) \( x = 4 \)


Rationale: Use the product rule for logarithms: \( \log_b(M) + \log_b(N) = \log_b(MN) \).
So, \( \log_2[x(x - 2)] = 3 \Rightarrow \log_2(x^2 - 2x) = 3 \). Convert to exponential form: \(
x^2 - 2x = 2^3 = 8 \). Rearrange: \( x^2 - 2x - 8 = 0 \). Factor: \( (x - 4)(x + 2) = 0 \). Solutions:
\( x = 4 \) or \( x = -2 \). Check domain: arguments of logs must be positive. For \( x = -2 \), \(
\log_2(-2) \) undefined. For \( x = 4 \), \( \log_2(4) + \log_2(2) = 2 + 1 = 3 \). Valid. Thus, only
\( x = 4 \) is acceptable.

Question 2: Find the exact value of \( \sin\left(\frac{5\pi}{12}\right) \).



(A) \( \frac{\sqrt{6} + \sqrt{2}}{4} \)



(B) \( \frac{\sqrt{6} - \sqrt{2}}{4} \)



(C) \( \frac{\sqrt{3}}{2} \)



(D) \( \frac{1}{2} \)



(E) \( \frac{\sqrt{2}}{2} \)


Correct Answer: (A) \( \frac{\sqrt{6} + \sqrt{2}}{4} \)


Rationale: Note that \( \frac{5\pi}{12} = \frac{\pi}{4} + \frac{\pi}{6} \). Use the sine sum
identity: \( \sin(A + B) = \sin A \cos B + \cos A \sin B \). So, \( \sin\left(\frac{\pi}{4} +
\frac{\pi}{6}\right) = \sin\frac{\pi}{4}\cos\frac{\pi}{6} + \cos\frac{\pi}{4}\sin\frac{\pi}{6}
\). Substitute known values: \( = \left(\frac{\sqrt{2}}{2}\right)\left(\frac{\sqrt{3}}{2}\right) +
\left(\frac{\sqrt{2}}{2}\right)\left(\frac{1}{2}\right) = \frac{\sqrt{6}}{4} + \frac{\sqrt{2}}{4}
= \frac{\sqrt{6} + \sqrt{2}}{4} \).

Question 3: What is the horizontal asymptote of the rational function \( f(x) = \frac{3x^2 - 2x
+ 1}{x^2 + 4} \)?



(A) \( y = 0 \)



(B) \( y = 3 \)

,(C) \( y = -2 \)



(D) \( y = \frac{1}{4} \)



(E) No horizontal asymptote


Correct Answer: (B) \( y = 3 \)


Rationale: For a rational function \( \frac{P(x)}{Q(x)} \), if degrees of numerator and
denominator are equal, the horizontal asymptote is the ratio of leading coefficients. Here, both
numerator and denominator are degree 2. Leading coefficient of numerator is 3, denominator is
1. So, horizontal asymptote is \( y = \frac{3}{1} = 3 \).

Question 4: Solve the system:

\( x + y + z = 6 \)

\( 2x - y + z = 3 \)

\( x + 2y - z = 2 \)



(A) \( (1, 2, 3) \)



(B) \( (2, 1, 3) \)



(C) \( (3, 2, 1) \)



(D) \( (1, 3, 2) \)



(E) No solution


Correct Answer: (A) \( (1, 2, 3) \)

, Rationale: Use elimination. Add equations (1) and (3): \( (x + y + z) + (x + 2y - z) = 6 + 2
\Rightarrow 2x + 3y = 8 \) → Eq (4). Add equations (2) and (3): \( (2x - y + z) + (x + 2y - z) = 3
+ 2 \Rightarrow 3x + y = 5 \) → Eq (5). Now solve (4) and (5): From (5), \( y = 5 - 3x \).
Substitute into (4): \( 2x + 3(5 - 3x) = 8 \Rightarrow 2x + 15 - 9x = 8 \Rightarrow -7x = -7
\Rightarrow x = 1 \). Then \( y = 5 - 3(1) = 2 \). Plug into (1): \( 1 + 2 + z = 6 \Rightarrow z = 3
\). Solution: \( (1, 2, 3) \). Verify in all equations—valid.

Question 5: Write the complex number \( z = 1 - i \) in polar form.



(A) \( \sqrt{2} \left( \cos\frac{\pi}{4} + i\sin\frac{\pi}{4} \right) \)



(B) \( \sqrt{2} \left( \cos\frac{7\pi}{4} + i\sin\frac{7\pi}{4} \right) \)



(C) \( 2 \left( \cos\frac{5\pi}{4} + i\sin\frac{5\pi}{4} \right) \)



(D) \( \sqrt{2} \left( \cos\frac{3\pi}{4} + i\sin\frac{3\pi}{4} \right) \)



(E) \( 1 \left( \cos\frac{\pi}{2} + i\sin\frac{\pi}{2} \right) \)


Correct Answer: (B) \( \sqrt{2} \left( \cos\frac{7\pi}{4} + i\sin\frac{7\pi}{4}
\right) \)


Rationale: For \( z = a + bi = 1 - i \), modulus \( r = \sqrt{a^2 + b^2} = \sqrt{1^2 + (-1)^2} =
\sqrt{2} \). Argument \( \theta \): point is in quadrant IV. \( \tan\theta = \frac{-1}{1} = -1 \), so
reference angle is \( \frac{\pi}{4} \), thus \( \theta = 2\pi - \frac{\pi}{4} = \frac{7\pi}{4} \).
Polar form: \( z = r(\cos\theta + i\sin\theta) = \sqrt{2} \left( \cos\frac{7\pi}{4} +
i\sin\frac{7\pi}{4} \right) \).

Question 6: Find the vertex of the parabola \( y = -2x^2 + 8x - 5 \).



(A) \( (2, 3) \)



(B) \( (-2, -29) \)



(C) \( (0, -5) \)