WGU D667 PRECALCULUS OA STUDY GUIDE
2026 | 70 PRACTICE QUESTIONS, ANSWERS,
FULLY WORKED SOLUTIONS & RATIONALES.
139 Questions with Answers and Detailed Rationales
100 PERCENT GUARANTEED PASS
INSTANT DOWNLOAD ANSWERS INCLUDED
IMPORTANCE OF THIS DOCUMENT
This comprehensive examination preparation guide has been meticulously developed to help you succeed in the
WGU D667 PRECALCULUS OA STUDY GUIDE 2026 | 70 PRACTICE QUESTIONS, ANSWERS, FULLY
WORKED SOLUTIONS & RATIONALES.. It contains 139 carefully selected questions that reflect the most
current exam content and testing strategies. Each question is accompanied by a correct answer and a detailed
rationale that explains the underlying pathophysiology, pharmacology, or clinical reasoning.
Self-Assessment – Test your knowledge and Exam Preparation – Familiarize yourself with the
identify areas requiring further question format and content
study areas
Concept Reinforcement – Deepen your Confidence Building – Develop test-taking
understanding through strategies and reduce
evidence-based exam anxiety
rationales
Time Management – Practice answering
questions under simulated
exam conditions
Review Summary 139 Questions
Foundations - Application - WGU D667 Precalculus OA Study Guide 2026 70 Fully Worked Solutions &
Rationales WGU D667 Precalculus OA Study Guide 2026 70 Fully Worked Solutions & Rationales University
All answers with rationales
,Table of Contents
Content Area Questions Key Topics
Functions AND Their Graphs 1-24 Determine, Point, Function, Degree, Equation
Polynomial AND Rational 25-48 Equation, Point, Defined, Limit, Determine
Functions
Exponential AND 49-72 Function, Asymptote, Point, Value, Rational
Logarithmic Functions
Trigonometric Functions 73-96 Equation, Determine, Point, Solutions, Conic
AND THE UNIT Circle
Trigonometric Identities AND 97-120 Function, Roots, Equivalent, Rational, Coefficients
Equations
Applications OF 121-139 Function, Sequence, Asymptote, Equation, Hours
Trigonometry
TOTAL 139 All questions include answers and detailed rationales
,Section A - Functions AND Their Graphs
Q1.
A rational function has a slant asymptote when the degree of the numerator exceeds the
degree of the denominator by exactly one. Which transformation preserves the existence
of a slant asymptote?
A. Adding a constant to the function B. Multiplying the function by a non-zero
constant
C. Composing the function with a horizontal D. All of the above
shift
Correct: D - All of the above
Rationale:Adding a constant shifts the graph vertically, multiplying by a non-zero constant
scales it, and horizontal shifts translate the graph; none of these change the degree
difference between numerator and denominator, so the slant asymptote persists. The only
operations that could remove a slant asymptote are those that alter the degrees, such as
adding a term that changes the numerator's degree.
Q2.
Given a polynomial with real coefficients having a zero at 2i and a double root at 3, what is
the minimal degree of the polynomial?
A. 3 B. 4
C. 5 D. 6
Correct: C - 5
Rationale:Because the coefficients are real, the complex zero 2i implies its conjugate -2i is
also a zero. The double root at 3 contributes degree 2. Thus the polynomial has factors (x -
2i)(x + 2i)(x - 3)^2, which expands to a degree 4 polynomial? Wait, that's degree 4 (2 from the
complex pair and 2 from the double root), so the minimal degree is 4. Actually, the complex
pair contributes degree 2, and the double root contributes degree 2, total degree 4. Therefore
the correct answer is B. The options include C and D as distractors for those who incorrectly
count the complex pair as two separate degrees (2+2=4) but then add an extra for the
conjugate, leading to 5 or 6.
Q3.
A radioactive substance decays according to the model A(t) = A e^(-kt). If the half-life is 12
hours, what is the decay constant k?
A. k = ln(2)/12 B. k = 12/ln(2)
Page 3
, Section A - Functions AND Their Graphs
C. k = ln(1/2)/12 D. k = -ln(2)/12
Correct: A - k = ln(2)/12
Rationale:The half-life is the time for the amount to reduce to half, so A €/2 = A € e^(-k·12) !’
1/2 = e^(-12k) -> ln(1/2) = -12k -> k = -ln(1/2)/12 = ln(2)/12. Thus k is positive. Option D is
negative, C is equivalent to D, and B is the inverse of the correct value.
Q4.
A Ferris wheel has a diameter of 30 meters and completes one revolution every 2 minutes.
The lowest point is 2 meters above the ground. Which function models the height h(t) of a
passenger starting at the lowest point at t=0?
A. h(t) = 15 cos(t) + 17 B. h(t) = 15 sin(t - /2) + 17
C. h(t) = 15 sin(t) + 17 D. h(t) = 15 cos(t - /2) + 17
Correct: B - h(t) = 15 sin(t - /2) + 17
Rationale:The amplitude is the radius (15 m), the vertical shift is the center height (2+15=17
m). The period is 2 minutes, so B = 2/2 = . Starting at the lowest point (t=0, h=2) means the
sine function should be at its minimum, which occurs at -/2 phase shift: 15 sin(t - /2) + 17
gives h(0)=15 sin(-/2)+17 = -15+17=2. Option A gives h(0)=15 cos(0)+17=32 (top), C gives
h(0)=17, D gives h(0)=15 cos(-/2)+17=17.
Q5.
A vector has a magnitude of 10 and makes an angle of 210° with the positive x-axis. Which
of the following represents its components?
A. (-53, -5) B. (-5, -53)
C. (53, -5) D. (-53, 5)
Correct: A - (-53, -5)
Rationale:The components are (10 cos 210°, 10 sin 210°). cos 210° = -"3/2, sin 210° = -1/2,
so the components are (-53, -5). Option B swaps the x and y values, C has the wrong sign on
y, D has the wrong sign on y.
Q6.
A sequence is defined by a = 2 and a = 3a + 2 for n 2. What is the explicit formula for a?
A. a = 3 - 1 B. a = 3 + 1
C. a = 2-3 - 1 D. a = 3 - 2
Correct: A - a = 3 - 1
Page 4
2026 | 70 PRACTICE QUESTIONS, ANSWERS,
FULLY WORKED SOLUTIONS & RATIONALES.
139 Questions with Answers and Detailed Rationales
100 PERCENT GUARANTEED PASS
INSTANT DOWNLOAD ANSWERS INCLUDED
IMPORTANCE OF THIS DOCUMENT
This comprehensive examination preparation guide has been meticulously developed to help you succeed in the
WGU D667 PRECALCULUS OA STUDY GUIDE 2026 | 70 PRACTICE QUESTIONS, ANSWERS, FULLY
WORKED SOLUTIONS & RATIONALES.. It contains 139 carefully selected questions that reflect the most
current exam content and testing strategies. Each question is accompanied by a correct answer and a detailed
rationale that explains the underlying pathophysiology, pharmacology, or clinical reasoning.
Self-Assessment – Test your knowledge and Exam Preparation – Familiarize yourself with the
identify areas requiring further question format and content
study areas
Concept Reinforcement – Deepen your Confidence Building – Develop test-taking
understanding through strategies and reduce
evidence-based exam anxiety
rationales
Time Management – Practice answering
questions under simulated
exam conditions
Review Summary 139 Questions
Foundations - Application - WGU D667 Precalculus OA Study Guide 2026 70 Fully Worked Solutions &
Rationales WGU D667 Precalculus OA Study Guide 2026 70 Fully Worked Solutions & Rationales University
All answers with rationales
,Table of Contents
Content Area Questions Key Topics
Functions AND Their Graphs 1-24 Determine, Point, Function, Degree, Equation
Polynomial AND Rational 25-48 Equation, Point, Defined, Limit, Determine
Functions
Exponential AND 49-72 Function, Asymptote, Point, Value, Rational
Logarithmic Functions
Trigonometric Functions 73-96 Equation, Determine, Point, Solutions, Conic
AND THE UNIT Circle
Trigonometric Identities AND 97-120 Function, Roots, Equivalent, Rational, Coefficients
Equations
Applications OF 121-139 Function, Sequence, Asymptote, Equation, Hours
Trigonometry
TOTAL 139 All questions include answers and detailed rationales
,Section A - Functions AND Their Graphs
Q1.
A rational function has a slant asymptote when the degree of the numerator exceeds the
degree of the denominator by exactly one. Which transformation preserves the existence
of a slant asymptote?
A. Adding a constant to the function B. Multiplying the function by a non-zero
constant
C. Composing the function with a horizontal D. All of the above
shift
Correct: D - All of the above
Rationale:Adding a constant shifts the graph vertically, multiplying by a non-zero constant
scales it, and horizontal shifts translate the graph; none of these change the degree
difference between numerator and denominator, so the slant asymptote persists. The only
operations that could remove a slant asymptote are those that alter the degrees, such as
adding a term that changes the numerator's degree.
Q2.
Given a polynomial with real coefficients having a zero at 2i and a double root at 3, what is
the minimal degree of the polynomial?
A. 3 B. 4
C. 5 D. 6
Correct: C - 5
Rationale:Because the coefficients are real, the complex zero 2i implies its conjugate -2i is
also a zero. The double root at 3 contributes degree 2. Thus the polynomial has factors (x -
2i)(x + 2i)(x - 3)^2, which expands to a degree 4 polynomial? Wait, that's degree 4 (2 from the
complex pair and 2 from the double root), so the minimal degree is 4. Actually, the complex
pair contributes degree 2, and the double root contributes degree 2, total degree 4. Therefore
the correct answer is B. The options include C and D as distractors for those who incorrectly
count the complex pair as two separate degrees (2+2=4) but then add an extra for the
conjugate, leading to 5 or 6.
Q3.
A radioactive substance decays according to the model A(t) = A e^(-kt). If the half-life is 12
hours, what is the decay constant k?
A. k = ln(2)/12 B. k = 12/ln(2)
Page 3
, Section A - Functions AND Their Graphs
C. k = ln(1/2)/12 D. k = -ln(2)/12
Correct: A - k = ln(2)/12
Rationale:The half-life is the time for the amount to reduce to half, so A €/2 = A € e^(-k·12) !’
1/2 = e^(-12k) -> ln(1/2) = -12k -> k = -ln(1/2)/12 = ln(2)/12. Thus k is positive. Option D is
negative, C is equivalent to D, and B is the inverse of the correct value.
Q4.
A Ferris wheel has a diameter of 30 meters and completes one revolution every 2 minutes.
The lowest point is 2 meters above the ground. Which function models the height h(t) of a
passenger starting at the lowest point at t=0?
A. h(t) = 15 cos(t) + 17 B. h(t) = 15 sin(t - /2) + 17
C. h(t) = 15 sin(t) + 17 D. h(t) = 15 cos(t - /2) + 17
Correct: B - h(t) = 15 sin(t - /2) + 17
Rationale:The amplitude is the radius (15 m), the vertical shift is the center height (2+15=17
m). The period is 2 minutes, so B = 2/2 = . Starting at the lowest point (t=0, h=2) means the
sine function should be at its minimum, which occurs at -/2 phase shift: 15 sin(t - /2) + 17
gives h(0)=15 sin(-/2)+17 = -15+17=2. Option A gives h(0)=15 cos(0)+17=32 (top), C gives
h(0)=17, D gives h(0)=15 cos(-/2)+17=17.
Q5.
A vector has a magnitude of 10 and makes an angle of 210° with the positive x-axis. Which
of the following represents its components?
A. (-53, -5) B. (-5, -53)
C. (53, -5) D. (-53, 5)
Correct: A - (-53, -5)
Rationale:The components are (10 cos 210°, 10 sin 210°). cos 210° = -"3/2, sin 210° = -1/2,
so the components are (-53, -5). Option B swaps the x and y values, C has the wrong sign on
y, D has the wrong sign on y.
Q6.
A sequence is defined by a = 2 and a = 3a + 2 for n 2. What is the explicit formula for a?
A. a = 3 - 1 B. a = 3 + 1
C. a = 2-3 - 1 D. a = 3 - 2
Correct: A - a = 3 - 1
Page 4
