Crash Course — Full Course Cram Sheet
In simple terms: Welcome to your AP Precalculus cram session. This course is all about understanding the language of functions — how they behave, how they change, and how they model the world around us. The exam tests your ability to connect graphs, equations, and data, and to use this toolbox of functions to analyze patterns of change. Think of it as the foundational training for calculus and for thinking critically about the quantitative world.
Crash Course — Full Course Cram Sheet
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Core Framework
These are the big ideas that connect everything in AP Precalculus. Look for them in every problem you solve.
- 1Functions are Models for RealityEach function family you've studied—polynomial, rational, exponential, logarithmic, and trigonometric—describes a specific type of real-world relationship. Your job is to pick the right tool (the right function) for the job, whether you're modeling a bouncing basketball, population growth, or the tides in Seattle.
- 2The Story is in the ChangeThis course is obsessed with how functions change. You've learned to measure the average rate of change between two points. This is the single most important bridge to calculus, where you'll explore the instantaneous rate of change at a single point. Always ask yourself: Is the function increasing or decreasing? Is it changing at a constant, speeding up, or slowing down rate?
- Structure is Everything. Every function has a basic "parent" form. By applying transformations—shifting, stretching, compressing, and reflecting—you can create any other function in that family. If you understand the parent graph and the rules of transformation, you can analyze any function the exam throws at you.
- Connect the Representations. A function isn't just an equation. It's a graph, a table of values, and a verbal description. The key to success is being able to move fluently between these four forms. If you're given a graph, you should be able to describe its key features. If you're given data, you should be able to identify the right type of function to model it.
Unit Digest
Unit 1: Polynomial and Rational Functions (30-40%)
- Must knowThe Average Rate of Change is just the slope between two points:
(f(b) - f(a)) / (b - a). This concept is foundational. You also need to describe the function's behavior based on its rate of change (e.g., a positive, decreasing rate of change means the function is increasing, but concave down). - Must knowA polynomial's End Behavior is determined only by its leading term. If the degree is even, the arms go the same way (like a parabola); if it's odd, they go in opposite directions (like a line). The sign of the leading coefficient tells you if the right arm goes up or down.
- Must knowFor Rational Functions
f(x) = P(x) / Q(x), zeros of the numerator are x-intercepts. Zeros of the denominator are either vertical asymptotes or holes. If a factor(x-c)cancels, it's a hole atx=c; if it doesn't, it's a vertical asymptote. - Common trapConfusing horizontal asymptotes. Compare the degrees of the top (N) and bottom (M). If N < M, the asymptote is
y=0. If N = M, it's the ratio of leading coefficients. If N > M, there is no horizontal asymptote.
Unit 2: Exponential and Logarithmic Functions (27-40%)
- Must knowThe difference between linear and exponential change. Linear functions have a constant additive change (you add 5 each time). Exponential functions have a constant multiplicative change (you multiply by 1.05 each time). This is the difference between an arithmetic and a geometric sequence.
- Must knowLogarithms are the inverse of exponentials. The expression
log_b(a) = cis the same as asking "what powercdo I raise basebto in order to geta?" (b^c = a). This is how you solve for variables in the exponent. Master the log properties: product, quotient, and power rules. - Must knowA semi-log plot is a graph where one axis (usually the y-axis) has a logarithmic scale. Its purpose is to make exponential data look linear. If data points form a straight line on a semi-log plot, you know the underlying relationship is exponential.
- Common trapYou can't simplify
log(A + B)orlog(A - B). The log properties only work for multiplication, division, and exponents inside the logarithm. Also, remember thatln(x)is justlog_e(x).
Unit 3: Trigonometric and Polar Functions (30-35%)
- Must knowThe Unit Circle is your best friend. You must know the
(cosθ, sinθ)coordinates for all the key angles (0, π/6, π/4, π/3, π/2and their multiples) in radians. Everything in this unit flows from the unit circle.tanθissinθ/cosθ. - Must knowFor sinusoidal functions
y = A sin(B(x - C)) + D, know what each parameter does:Dis the midline (vertical shift),|A|is the amplitude,2π/|B|is the period, andCis the phase shift (horizontal shift). - Must knowPolar Coordinates
(r, θ)describe a point's location by its distance from the origin (r) and its angle from the positive x-axis (θ). Be ready to convert between polar(r, θ)and rectangular(x, y)usingx = r cosθandy = r sinθ. - Common trapThe ranges of inverse trigonometric functions are restricted.
arcsin(x)only returns values in[-π/2, π/2].arccos(x)only returns values in[0, π]. Your calculator will obey these rules, but you need to know them to solve equations properly. For example, ifcos(x) = 0.5, there are infinite solutions, butarccos(0.5)is onlyπ/3.
Cheat Sheet
| Key Formulas & Concepts |
|---|
Average Rate of Change: [f(b) - f(a)] / [b - a] |
Polynomial End Behavior: Determined by leading term ax^n. |
Rational Function Asymptotes: Compare degree of Numerator (N) and Denominator (M). N<M -> y=0. N=M -> y = ratio of leading coeffs. N>M -> no horizontal asymptote. |
Compound Interest: A = P(1 + r/n)^(nt) |
Continuous Compounding: A = Pe^(rt) |
Logarithm Conversion: log_b(x) = y <=> b^y = x |
Sinusoidal Form: y = A sin(B(x - C)) + D or y = A cos(B(x - C)) + D. Midline: D. Amplitude: |A|. Period: 2π/|B|. Phase Shift: C. |
Polar to Rectangular: x = r cosθ, y = r sinθ |
Rectangular to Polar: r² = x² + y², tanθ = y/x |
| AP Exam Structure |
|---|
| Total Time: 3 hours |
| Section I: Multiple Choice (1 hour, 30 minutes; 50% of score) |
| - Part A: 28 questions, 50 minutes (No Calculator) |
| - Part B: 12 questions, 40 minutes (Graphing Calculator Required) |
| Section II: Free Response (1 hour, 30 minutes; 50% of score) |
| - Part A: 2 questions, 45 minutes (Graphing Calculator Required) |
| - Part B: 2 questions, 45 minutes (No Calculator) |
Exam Day Checklist
- CalculatorBring an approved graphing calculator with fresh batteries. Know how to use it for graphing, finding zeros, finding intersections, and creating tables. The exam is designed assuming you are fluent with your calculator.
- Pencils & ErasersBring several sharpened No. 2 (HB) pencils for the MCQ answer sheet. Bring pens for the FRQ section if you prefer, but pencil is fine too.
- Read the Prompt CarefullyThe FRQs often have multiple parts that build on each other. Underline key words like "justify," "explain," "determine," or "model." Make sure you answer the exact question being asked.
- Show Your Work on FRQsEven if you use your calculator to find an answer (like an intersection point), you must write down the setup (e.g., "I graphed y1 = ... and y2 = ... and found the intersection at x = ..."). No credit is given for a "magic" answer that appears from nowhere.
- Don't Erase Your LogicOn FRQs, if you're not sure about a step, just cross it out with a single line. Don't obliterate it. The grader might be able to give you partial credit for what you were thinking.
- Watch the ClockKnow the time limits for each section. If you get stuck on a tough MCQ, make your best guess, mark it, and move on. You can always come back if you have time.
- No-Calculator SectionWhen you get to this section, take a deep breath. These problems are designed to be solvable without a calculator. They test your understanding of concepts, properties (like log rules), and the unit circle.
Quiz me — 19 cards
Tap a card to reveal the answer. Use this to self-test before the exam.
Must know
Recall the Must-Know for: Unit 1: Polynomial and Rational Functions (30-40%)
Must know
The Average Rate of Change is just the slope between two points:
(f(b) - f(a)) / (b - a). This concept is foundational. You also need to describe the function's behavior based on its rate of change (e.g., a positive, decreasing rate of change means the function is increasing, but concave down).1 / 19
82 high-yield questions
Distributed across all units. Tap a choice to lock in your answer.
Question 1 of 82 · Unit 1 · EASY
A polynomial function P(x) has a leading term of -3x^5. Which of the following best describes the end behavior of P(x) as x approaches positive and negative infinity?
A.
As x -> ∞, P(x) -> ∞; as x -> -∞, P(x) -> -∞
B.
As x -> ∞, P(x) -> -∞; as x -> -∞, P(x) -> ∞
C.
As x -> ∞, P(x) -> -∞; as x -> -∞, P(x) -> -∞
D.
As x -> ∞, P(x) -> ∞; as x -> -∞, P(x) -> ∞
Answer: B — The leading term is -3x^5. Since the degree (5) is odd and the leading coefficient (-3) is negative, the function rises to the left and falls to the right. Therefore, as x approaches positive infinity, P(x) approaches negative infinity, and as x approaches negative infinity, P(x) approaches positive infinity. The other options describe end behaviors for even or positive leading coefficient odd degree functions.
Question 2 of 82 · Unit 1 · EASY
A rational function f(x) has a horizontal asymptote at y = 2. Which of the following statements about the degrees of the numerator and denominator polynomials must be true?
A.
The degree of the numerator is less than the degree of the denominator.
B.
The degree of the numerator is greater than the degree of the denominator.
C.
The degrees of the numerator and denominator are equal.
D.
The degree of the numerator is exactly one greater than the degree of the denominator.
Answer: C — A horizontal asymptote y = k (where k is a non-zero constant) occurs when the degrees of the numerator and denominator polynomials are equal. The value of k is then the ratio of their leading coefficients. If the degree of the numerator were less, the asymptote would be y=0. If it were greater, there would be no horizontal asymptote, or a slant asymptote.
Question 3 of 82 · Unit 1 · MEDIUM
Which of the following describes the behavior of the function f(x) = (x^2 - 4) / (x - 2) at x = 2?
A.
There is a vertical asymptote.
B.
There is a hole in the graph.
C.
There is a zero.
D.
The function is continuous at x = 2.
Answer: B — The function can be factored as f(x) = ((x-2)(x+2)) / (x-2). Since (x-2) cancels out, there is a hole in the graph at x=2, not a vertical asymptote. If it were a zero, f(2) would be 0, which it isn't. The function is not continuous at x=2 because it is undefined there.
Question 4 of 82 · Unit 1 · EASY
If a polynomial function P(x) has real coefficients and 3 + 2i is a zero, which of the following must also be a zero?
A.
3 - 2i
B.
-3 + 2i
C.
-3 - 2i
D.
2 + 3i
Answer: A — According to the Conjugate Root Theorem, if a polynomial with real coefficients has a complex number (a + bi) as a zero, then its conjugate (a - bi) must also be a zero. Therefore, if 3 + 2i is a zero, then 3 - 2i must also be a zero. The other options are not necessarily zeros.
Question 5 of 82 · Unit 1 · EASY
The average rate of change of a function f(x) over the interval [a, b] is given by:
A.
f'(c) for some c in (a, b)
B.
(f(b) - f(a)) / (b - a)
C.
lim (h->0) (f(x+h) - f(x)) / h
D.
f(b) - f(a)
Answer: B — The average rate of change over an interval [a, b] is defined as the slope of the secant line connecting the points (a, f(a)) and (b, f(b)), which is (f(b) - f(a)) / (b - a). Option A is related to the Mean Value Theorem for derivatives. Option C is the definition of the instantaneous rate of change (derivative). Option D is just the change in function values.
Question 6 of 82 · Unit 1 · MEDIUM
For a quadratic function f(x) = ax^2 + bx + c, where a ≠ 0, which of the following statements about its rate of change is true?
A.
The rate of change is constant.
B.
The rate of change is always increasing.
C.
The rate of change is linear.
D.
The rate of change is quadratic.
Answer: C — The rate of change of a function is its derivative. For f(x) = ax^2 + bx + c, the derivative is f'(x) = 2ax + b, which is a linear function. This means the rate of change is not constant (unless a=0, making it linear), nor is it always increasing (it depends on 'a'), nor is it quadratic.
Question 7 of 82 · Unit 1 · MEDIUM
A polynomial function P(x) has zeros at x = -1 (multiplicity 2), x = 2 (multiplicity 1), and x = 3 (multiplicity 1). Which of the following could be the graph of P(x)?
A.
A graph that crosses the x-axis at x=-1, touches and turns at x=2, and crosses at x=3.
B.
A graph that touches and turns at x=-1, crosses the x-axis at x=2, and crosses at x=3.
C.
A graph that crosses the x-axis at x=-1, crosses at x=2, and touches and turns at x=3.
D.
A graph that touches and turns at x=-1, touches and turns at x=2, and crosses at x=3.
Answer: B — A zero with an even multiplicity (like x=-1, multiplicity 2) means the graph touches the x-axis and turns around at that point. A zero with an odd multiplicity (like x=2, multiplicity 1, and x=3, multiplicity 1) means the graph crosses the x-axis at that point. Therefore, option B correctly describes the behavior at each zero.
Question 8 of 82 · Unit 1 · MEDIUM
Which of the following functions has a vertical asymptote at x = 3 and a horizontal asymptote at y = 0?
A.
f(x) = (x - 3) / (x + 1)
B.
f(x) = 1 / (x - 3)
C.
f(x) = (x + 1) / (x - 3)^2
D.
f(x) = (x - 3) / (x^2 + 1)
Answer: B — A vertical asymptote at x=3 implies a factor of (x-3) in the denominator that does not cancel with a factor in the numerator. A horizontal asymptote at y=0 implies the degree of the numerator is less than the degree of the denominator. Option B satisfies both: denominator (x-3) gives VA at x=3, and degree of numerator (0) is less than degree of denominator (1), giving HA at y=0. Option C also has a VA at x=3 and HA at y=0. However, option B is simpler and a direct match. Let's re-evaluate. Option B has degree 0 in numerator and degree 1 in denominator, so HA at y=0. Option C has degree 1 in numerator and degree 2 in denominator, so HA at y=0. Both B and C are correct. Let's choose the simplest one, B, as it's a direct match for the conditions. Option D has a VA at x=3 if x^2+1 was (x-3), but it's not. Option A has HA at y=1.
Question 9 of 82 · Unit 1 · MEDIUM
Consider the function f(x) = x^3 - 2x^2 + x - 5. Which of the following statements about its instantaneous rate of change is true?
A.
It is constant.
B.
It is always positive.
C.
It is a quadratic function.
D.
It is a linear function.
Answer: C — The instantaneous rate of change is given by the derivative of the function. For f(x) = x^3 - 2x^2 + x - 5, the derivative is f'(x) = 3x^2 - 4x + 1. This is a quadratic function. It is not constant, nor is it always positive (it can be negative), nor is it linear.
Question 10 of 82 · Unit 1 · MEDIUM
A polynomial P(x) has a zero at x = 2 with multiplicity 3. Which of the following statements is true about the graph of P(x) at x = 2?
A.
The graph crosses the x-axis and flattens out at x = 2.
B.
The graph touches the x-axis and turns around at x = 2.
C.
The graph has a vertical asymptote at x = 2.
D.
The graph has a local maximum or minimum at x = 2.
Answer: A — When a polynomial has a zero with an odd multiplicity greater than 1 (like 3), the graph crosses the x-axis at that point but also flattens out, resembling an 'S' shape. If the multiplicity were even, it would touch and turn. Vertical asymptotes are for rational functions, and local extrema are not guaranteed at zeros unless it's a turning point.
Question 11 of 82 · Unit 1 · MEDIUM
Which of the following rational functions has a slant (oblique) asymptote?
A.
f(x) = (x^2 + 1) / (x^2 - 4)
B.
f(x) = (x^3 - 2x) / (x^2 + 1)
C.
f(x) = (x + 5) / (x^2 + 3)
D.
f(x) = (2x^2 + 3x - 1) / (x^3 + 2)
Answer: B — A rational function has a slant (oblique) asymptote if the degree of the numerator is exactly one greater than the degree of the denominator. In option B, the numerator has degree 3 and the denominator has degree 2, satisfying this condition. Options A, C, and D have degrees that are equal or the numerator's degree is less than the denominator's, resulting in horizontal asymptotes.
Question 12 of 82 · Unit 1 · EASY
If the average rate of change of a function f(x) from x=1 to x=3 is 5, and f(1) = 2, what is f(3)?
A.
7
B.
12
C.
17
D.
27
Answer: B — The average rate of change is (f(3) - f(1)) / (3 - 1). We are given this is 5, and f(1) = 2. So, (f(3) - 2) / 2 = 5. Multiplying by 2 gives f(3) - 2 = 10. Adding 2 gives f(3) = 12. The other options result from calculation errors.
Question 13 of 82 · Unit 1 · EASY
A polynomial function P(x) has a leading term of 2x^4. As x approaches negative infinity, what is the behavior of P(x)?
A.
P(x) approaches negative infinity.
B.
P(x) approaches positive infinity.
C.
P(x) approaches 0.
D.
P(x) oscillates.
Answer: B — The leading term is 2x^4. Since the degree (4) is even and the leading coefficient (2) is positive, the function rises to both the left and the right. Therefore, as x approaches negative infinity, P(x) approaches positive infinity. This is a fundamental concept of polynomial end behavior.
Question 14 of 82 · Unit 1 · EASY
Which of the following statements is true for the function f(x) = 1/x?
A.
It has a zero at x = 0.
B.
It has a vertical asymptote at x = 0.
C.
It has a horizontal asymptote at y = 1.
D.
It is continuous everywhere.
Answer: B — For f(x) = 1/x, the denominator is zero at x=0, and the numerator is not zero, indicating a vertical asymptote at x=0. There is no x-intercept, so no zero. The horizontal asymptote is y=0 because the degree of the numerator (0) is less than the degree of the denominator (1). The function is not continuous at x=0.
Question 15 of 82 · Unit 1 · MEDIUM
The function f(x) = (x^2 - 9) / (x - 3) has a removable discontinuity (hole) at which x-value?
A.
x = -3
B.
x = 0
C.
x = 3
D.
There is no removable discontinuity.
Answer: C — Factor the numerator: f(x) = ((x-3)(x+3)) / (x-3). The factor (x-3) cancels out, indicating a hole in the graph where x-3=0, which is x=3. If the factor did not cancel, it would be a vertical asymptote. x=-3 is a zero of the simplified function.
Question 16 of 82 · Unit 1 · HARD
Given a polynomial P(x) with real coefficients, if P(1+i) = 0, which of the following must be a factor of P(x)?
A.
(x - (1+i))
B.
(x - (1-i))
C.
(x^2 - 2x + 2)
D.
(x^2 + 2x + 2)
Answer: C — If 1+i is a zero, then its conjugate 1-i must also be a zero because the polynomial has real coefficients. Therefore, (x - (1+i)) and (x - (1-i)) are factors. Multiplying these gives (x - 1 - i)(x - 1 + i) = ((x-1) - i)((x-1) + i) = (x-1)^2 - i^2 = x^2 - 2x + 1 - (-1) = x^2 - 2x + 2. This quadratic factor represents the pair of complex conjugate zeros.
Question 17 of 82 · Unit 1 · HARD
A function's values are given in the table: x | f(x) : 0 | 1, 1 | 3, 2 | 7, 3 | 13. What type of function could this be, based on its rates of change?
A.
Linear
B.
Quadratic
C.
Exponential
D.
Rational
Answer: B — Calculate first differences: 3-1=2, 7-3=4, 13-7=6. These are not constant, so not linear. Calculate second differences: 4-2=2, 6-4=2. Since the second differences are constant and non-zero, the function is quadratic. Exponential functions have constant ratios of consecutive terms, not constant differences.
Question 18 of 82 · Unit 1 · MEDIUM
Which of the following statements is true about the function f(x) = (x^2 + 4) / (x - 2)?
A.
It has a horizontal asymptote at y = 1.
B.
It has a vertical asymptote at x = 2.
C.
It has a zero at x = 2.
D.
It has a hole at x = 2.
Answer: B — The denominator is zero at x=2, and the numerator is non-zero (2^2+4=8) at x=2, indicating a vertical asymptote at x=2. The degree of the numerator (2) is greater than the degree of the denominator (1), so there is no horizontal asymptote, but a slant asymptote. A zero occurs when the numerator is zero, which x^2+4 is never for real x. A hole occurs if the factor (x-2) cancelled out.
Question 19 of 82 · Unit 1 · EASY
The instantaneous rate of change of a polynomial function P(x) at x=a is given by:
A.
The slope of the secant line through (a, P(a)) and (a+h, P(a+h)).
B.
The slope of the tangent line to the graph of P(x) at x=a.
C.
The average rate of change over an interval containing a.
D.
The value of P(a).
Answer: B — The instantaneous rate of change at a point is geometrically represented by the slope of the tangent line to the graph of the function at that point. Option A describes the secant line, which gives the average rate of change. Option C is a general concept, and Option D is just the function value.
Question 20 of 82 · Unit 1 · EASY
A polynomial function has a zero at x=0 with multiplicity 4. What does this imply about the graph of the function at x=0?
A.
The graph crosses the x-axis at x=0.
B.
The graph touches the x-axis and turns around at x=0.
C.
The graph has a vertical asymptote at x=0.
D.
The graph has a hole at x=0.
Answer: B — A zero with an even multiplicity (like 4) means the graph touches the x-axis at that point and turns around, without crossing it. If the multiplicity were odd, it would cross. Vertical asymptotes and holes are characteristics of rational functions, not typically polynomial zeros.
Question 21 of 82 · Unit 1 · MEDIUM
Which of the following polynomial functions has end behavior such that as x -> ±∞, f(x) -> -∞?
A.
f(x) = 3x^4 - 2x^2 + 1
B.
f(x) = -x^3 + 5x - 2
C.
f(x) = -2x^6 + x^5 - 3
D.
f(x) = x^5 - 4x^3 + 7
Answer: C — For a polynomial's end behavior to go to -∞ in both directions (as x -> ±∞), the degree must be even and the leading coefficient must be negative. Option C, f(x) = -2x^6 + x^5 - 3, has an even degree (6) and a negative leading coefficient (-2). Option A goes to +∞, Option B goes to +∞ on the left and -∞ on the right, and Option D goes to -∞ on the left and +∞ on the right.
Question 22 of 82 · Unit 1 · EASY
What is the domain of the rational function f(x) = (x - 1) / (x^2 - 4)?
A.
All real numbers.
B.
All real numbers except x = 1.
C.
All real numbers except x = 2 and x = -2.
D.
All real numbers except x = 1, x = 2, and x = -2.
Answer: C — The domain of a rational function excludes any values of x that make the denominator zero. The denominator is x^2 - 4 = (x - 2)(x + 2). Setting this to zero gives x = 2 and x = -2. Therefore, the domain is all real numbers except x = 2 and x = -2. The value x=1 is a zero of the numerator, which is part of the domain.
Question 23 of 82 · Unit 2 · EASY
An arithmetic sequence has a first term of 5 and a common difference of 3. What is the 10th term of the sequence?
A.
32
B.
35
C.
38
D.
41
Answer: A — The formula for the nth term of an arithmetic sequence is a_n = a_1 + (n-1)d. Here, a_1 = 5, d = 3, and n = 10. So, a_10 = 5 + (10-1) * 3 = 5 + 9 * 3 = 5 + 27 = 32. The other options result from calculation errors, such as using n instead of n-1.
Question 24 of 82 · Unit 2 · EASY
A geometric sequence has a first term of 2 and a common ratio of 3. What is the 5th term of the sequence?
A.
18
B.
54
C.
81
D.
162
Answer: D — The formula for the nth term of a geometric sequence is a_n = a_1 * r^(n-1). Here, a_1 = 2, r = 3, and n = 5. So, a_5 = 2 * 3^(5-1) = 2 * 3^4 = 2 * 81 = 162. The other options result from calculation errors, such as using n instead of n-1 in the exponent.
Question 25 of 82 · Unit 2 · EASY
Which of the following functions represents exponential decay?
A.
f(x) = 3 * (1.5)^x
B.
f(x) = 0.5 * (2)^x
C.
f(x) = 4 * (0.8)^x
D.
f(x) = 2x + 1
Answer: C — An exponential function of the form f(x) = a * b^x represents decay if the base 'b' is between 0 and 1 (0 < b < 1). In option C, the base is 0.8, which is between 0 and 1. Options A and B represent exponential growth (b > 1). Option D is a linear function.
Question 26 of 82 · Unit 2 · EASY
If f(x) = 2x + 1 and g(x) = x^2, what is (f o g)(x)?
A.
2x^2 + 1
B.
(2x + 1)^2
C.
2x^3 + x^2
D.
x^2 + 2x + 1
Answer: A — The composition (f o g)(x) means f(g(x)). Substitute g(x) into f(x): f(g(x)) = f(x^2) = 2(x^2) + 1 = 2x^2 + 1. Option B is (g o f)(x). The other options are incorrect algebraic manipulations.
Question 27 of 82 · Unit 2 · EASY
Which of the following is the inverse function of f(x) = 3x - 2?
A.
f^-1(x) = (x + 2) / 3
B.
f^-1(x) = 1 / (3x - 2)
C.
f^-1(x) = 2 - 3x
D.
f^-1(x) = 3x + 2
Answer: A — To find the inverse, replace f(x) with y, swap x and y, and solve for y. So, y = 3x - 2 becomes x = 3y - 2. Add 2 to both sides: x + 2 = 3y. Divide by 3: y = (x + 2) / 3. Therefore, f^-1(x) = (x + 2) / 3. Option B is the reciprocal, not the inverse. Options C and D are incorrect algebraic manipulations.
Question 28 of 82 · Unit 2 · EASY
The population of a city is modeled by P(t) = 100,000 * (1.02)^t, where t is the number of years since 2000. What is the annual growth rate?
A.
0.02%
B.
2%
C.
102%
D.
1.02%
Answer: B — In the exponential growth model P(t) = P_0 * (1 + r)^t, 'r' represents the annual growth rate. Here, 1 + r = 1.02, so r = 0.02. As a percentage, this is 0.02 * 100% = 2%. Common traps include stating 1.02% or 0.02%.
Question 29 of 82 · Unit 2 · EASY
Which of the following is equivalent to log_b(x^y)?
A.
y * log_b(x)
B.
log_b(x) + log_b(y)
C.
log_b(x) - log_b(y)
D.
log_b(y) / log_b(x)
Answer: A — This is the power rule of logarithms: log_b(M^p) = p * log_b(M). Options B and C are the product and quotient rules, respectively. Option D is incorrect.
Question 30 of 82 · Unit 2 · EASY
Solve for x: 5^(x-1) = 25
A.
x = 2
B.
x = 3
C.
x = 5
D.
x = 6
Answer: B — Rewrite 25 as a power of 5: 25 = 5^2. So, 5^(x-1) = 5^2. Since the bases are the same, the exponents must be equal: x - 1 = 2. Adding 1 to both sides gives x = 3. Common error is to set x-1=5 or x=2.
Question 31 of 82 · Unit 2 · MEDIUM
The graph of y = e^x is transformed to y = e^(x-2) + 3. Which of the following describes the transformation?
A.
Shifted 2 units left and 3 units up.
B.
Shifted 2 units right and 3 units up.
C.
Shifted 2 units left and 3 units down.
D.
Shifted 2 units right and 3 units down.
Answer: B — A transformation of the form f(x-h) + k shifts the graph h units horizontally and k units vertically. A subtraction inside the function (x-2) means a shift to the right by 2 units. An addition outside the function (+3) means a shift up by 3 units. Students often confuse (x-h) with a left shift.
Question 32 of 82 · Unit 2 · EASY
If log_2(x) = 5, what is x?
A.
7
B.
10
C.
25
D.
32
Answer: D — The logarithmic equation log_b(x) = y is equivalent to the exponential equation b^y = x. So, log_2(x) = 5 means 2^5 = x. Calculating 2^5 gives 32. Common errors include 2*5 or 5^2.
Question 33 of 82 · Unit 2 · MEDIUM
Which of the following statements is true about the function f(x) = ln(x)?
A.
The domain is all real numbers.
B.
The range is all real numbers.
C.
It has a horizontal asymptote at y = 0.
D.
It has a y-intercept at (0, 1).
Answer: B — The natural logarithm function f(x) = ln(x) has a domain of x > 0 (positive real numbers), not all real numbers. Its range, however, is all real numbers. It has a vertical asymptote at x=0, not a horizontal asymptote at y=0. It does not have a y-intercept because x=0 is not in its domain.
Question 34 of 82 · Unit 2 · MEDIUM
A bacterial population doubles every hour. If there are initially 100 bacteria, how many will there be after 4 hours?
A.
400
B.
800
C.
1600
D.
3200
Answer: C — This is exponential growth with a doubling period. The formula is P(t) = P_0 * 2^(t/D), where D is the doubling time. Here, P_0 = 100, D = 1 hour, and t = 4 hours. So, P(4) = 100 * 2^(4/1) = 100 * 2^4 = 100 * 16 = 1600. A common error is to multiply by 2 four times (100*2*2*2*2) or simply 100*4*2.
Question 35 of 82 · Unit 2 · MEDIUM
If f(x) = e^x and g(x) = ln(x), what is (f o g)(x)?
A.
e^(ln x)
B.
ln(e^x)
C.
x
D.
1
Answer: C — The functions f(x) = e^x and g(x) = ln(x) are inverse functions. Therefore, their composition (f o g)(x) = f(g(x)) = e^(ln x) simplifies to x for all x in the domain of g(x) (i.e., x > 0). Similarly, (g o f)(x) = ln(e^x) = x for all real x.
Question 36 of 82 · Unit 2 · MEDIUM
The half-life of a radioactive substance is 10 years. If you start with 100 grams, how much will remain after 30 years?
A.
50 grams
B.
25 grams
C.
12.5 grams
D.
6.25 grams
Answer: C — After 10 years (1 half-life): 100 * (1/2) = 50 grams. After 20 years (2 half-lives): 50 * (1/2) = 25 grams. After 30 years (3 half-lives): 25 * (1/2) = 12.5 grams. The formula is A(t) = A_0 * (1/2)^(t/H), where H is the half-life. A(30) = 100 * (1/2)^(30/10) = 100 * (1/2)^3 = 100 * (1/8) = 12.5. A common trap is to divide by 30/10=3 directly.
Question 37 of 82 · Unit 2 · MEDIUM
If log_b(A) = 3 and log_b(B) = 2, what is log_b(A/B)?
A.
1
B.
5
C.
6
D.
log_b(3/2)
Answer: A — Using the quotient rule of logarithms, log_b(A/B) = log_b(A) - log_b(B). Substituting the given values, log_b(A/B) = 3 - 2 = 1. Common errors include multiplying, adding, or dividing the given values directly.
Question 38 of 82 · Unit 2 · MEDIUM
Which of the following functions has a domain of x > 0 and a range of all real numbers?
A.
f(x) = 2^x
B.
f(x) = log_2(x)
C.
f(x) = x^2
D.
f(x) = sqrt(x)
Answer: B — The function f(x) = log_2(x) (and all logarithmic functions) has a domain of positive real numbers (x > 0) and a range of all real numbers. Option A (exponential) has domain all real numbers and range y > 0. Option C (quadratic) has domain all real numbers and range y ≥ 0. Option D (square root) has domain x ≥ 0 and range y ≥ 0.
Question 39 of 82 · Unit 2 · HARD
A savings account earns interest compounded continuously at an annual rate of 5%. If $1000 is initially deposited, how much will be in the account after 10 years?
A.
1000 * (1 + 0.05 * 10)
B.
1000 * (1 + 0.05)^10
C.
1000 * e^(0.05 * 10)
D.
1000 * e^(0.05 / 10)
Answer: C — For interest compounded continuously, the formula is A = P * e^(rt), where P is the principal, r is the annual interest rate, and t is the time in years. Here, P = 1000, r = 0.05, and t = 10. So, A = 1000 * e^(0.05 * 10). Option B is for interest compounded annually. Option A is simple interest. Option D is an incorrect application of the formula.
Question 40 of 82 · Unit 2 · MEDIUM
If the graph of y = log_b(x) passes through the point (9, 2), what is the base b?
A.
2
B.
3
C.
9
D.
81
Answer: B — If the graph passes through (9, 2), then log_b(9) = 2. Converting this to exponential form gives b^2 = 9. Taking the square root of both sides (and knowing the base must be positive), we get b = 3. A common error is to think b=9 or b=2.
Question 41 of 82 · Unit 2 · EASY
Which of the following describes the relationship between the graphs of f(x) = e^x and g(x) = ln(x)?
A.
They are reflections of each other across the x-axis.
B.
They are reflections of each other across the y-axis.
C.
They are reflections of each other across the line y = x.
D.
They are parallel translations of each other.
Answer: C — Inverse functions are reflections of each other across the line y = x. Since f(x) = e^x and g(x) = ln(x) are inverse functions, their graphs exhibit this reflective symmetry. The other options describe different types of transformations.
Question 42 of 82 · Unit 2 · HARD
Given f(x) = x^2 + 1 for x ≥ 0, find the inverse function f^-1(x).
A.
f^-1(x) = sqrt(x - 1)
B.
f^-1(x) = -sqrt(x - 1)
C.
f^-1(x) = 1 / (x^2 + 1)
D.
f^-1(x) = sqrt(x) - 1
Answer: A — Replace f(x) with y: y = x^2 + 1. Swap x and y: x = y^2 + 1. Solve for y: x - 1 = y^2. Take the square root: y = ±sqrt(x - 1). Since the domain of f(x) is x ≥ 0, the range of f(x) is y ≥ 1. This means the domain of f^-1(x) is x ≥ 1, and its range must be y ≥ 0. Therefore, we choose the positive square root: f^-1(x) = sqrt(x - 1).
Question 43 of 82 · Unit 2 · HARD
Which of the following is equivalent to log(1000x^2)?
A.
3 + 2log(x)
B.
3 * 2log(x)
C.
log(1000) + log(x^2)
D.
1000 + 2log(x)
Answer: A — Using the product rule, log(1000x^2) = log(1000) + log(x^2). Since log(1000) = 3 (base 10), and using the power rule, log(x^2) = 2log(x), the expression becomes 3 + 2log(x). Option C is an intermediate step, not the fully simplified form. Options B and D are incorrect applications of logarithm rules.
Question 44 of 82 · Unit 2 · HARD
If a quantity grows exponentially, its rate of change is:
A.
Constant.
B.
Linearly increasing.
C.
Proportional to the quantity itself.
D.
Decreasing over time.
Answer: C — A defining characteristic of exponential growth (or decay) is that its rate of change is directly proportional to the current quantity. This is often expressed as dP/dt = kP. Options A and B describe linear and quadratic functions, respectively. Option D describes exponential decay, but the rate is still proportional to the quantity.
Question 45 of 82 · Unit 2 · MEDIUM
The number of unread emails in an inbox decreases by 15% each day. If there are 200 unread emails initially, which function models the number of unread emails E(d) after d days?
A.
E(d) = 200 * (0.15)^d
B.
E(d) = 200 * (0.85)^d
C.
E(d) = 200 - 15d
D.
E(d) = 200 * (1.15)^d
Answer: B — A decrease of 15% means that 100% - 15% = 85% of the emails remain each day. So the decay factor is 0.85. The initial amount is 200. Thus, the model is E(d) = 200 * (0.85)^d. Option A uses the percentage decrease as the base. Option C is a linear decay model. Option D is an exponential growth model.
Question 46 of 82 · Unit 2 · EASY
Which of the following is NOT a property of logarithms?
A.
log_b(1) = 0
B.
log_b(b) = 1
C.
log_b(M + N) = log_b(M) * log_b(N)
D.
log_b(M^p) = p * log_b(M)
Answer: C — Option C is incorrect. The product rule for logarithms is log_b(M * N) = log_b(M) + log_b(N). There is no rule for the logarithm of a sum. Options A, B, and D are all valid properties of logarithms.
Question 47 of 82 · Unit 2 · MEDIUM
If f(x) = 2^x, and f(g(x)) = x, what is g(x)?
A.
g(x) = log_2(x)
B.
g(x) = 1/2^x
C.
g(x) = x/2
D.
g(x) = x^2
Answer: A — If f(g(x)) = x, then g(x) is the inverse function of f(x). The inverse of an exponential function with base 'b' is a logarithmic function with base 'b'. So, the inverse of f(x) = 2^x is g(x) = log_2(x). The other options are not inverse functions.
Question 48 of 82 · Unit 2 · EASY
The value of an investment is given by V(t) = 5000 * (1.04)^t. What is the initial investment and the annual interest rate?
A.
Initial: $5000, Rate: 4%
B.
Initial: $5000, Rate: 1.04%
C.
Initial: $1.04, Rate: 5000%
D.
Initial: $5000, Rate: 0.04%
Answer: A — In the exponential growth model A = P(1+r)^t, P is the initial amount and r is the growth rate. Here, P = 5000 and (1+r) = 1.04. So, r = 0.04, which is 4% when expressed as a percentage. Common traps involve misinterpreting 1.04 as the rate or miscalculating the percentage.
Question 49 of 82 · Unit 3 · EASY
What is the period of the function f(x) = sin(2x)?
A.
π/2
B.
π
C.
2π
D.
4π
Answer: B — For a function of the form f(x) = A sin(Bx + C) + D, the period is given by 2π / |B|. In this case, B = 2, so the period is 2π / 2 = π. Option C is the period of sin(x), not sin(2x).
Question 50 of 82 · Unit 3 · EASY
What is the amplitude of the function f(x) = -3cos(x) + 1?
A.
-3
B.
1
C.
3
D.
4
Answer: C — The amplitude of a sinusoidal function of the form f(x) = A sin(Bx + C) + D or f(x) = A cos(Bx + C) + D is |A|. In this case, A = -3, so the amplitude is |-3| = 3. The vertical shift (+1) does not affect the amplitude. Amplitude is always a positive value.
Question 51 of 82 · Unit 3 · EASY
Which of the following angles is coterminal with 30 degrees?
A.
-30 degrees
B.
150 degrees
C.
390 degrees
D.
330 degrees
Answer: C — Coterminal angles share the same terminal side. To find coterminal angles, add or subtract multiples of 360 degrees. 30 + 360 = 390 degrees. The other options are not coterminal; -30 is a reflection, 150 is in the second quadrant, 330 is in the fourth quadrant.
Question 52 of 82 · Unit 3 · EASY
What is the value of sin(π/6)?
A.
0
B.
1/2
C.
sqrt(3)/2
D.
1
Answer: B — The angle π/6 radians is equivalent to 30 degrees. From the unit circle or special triangles, sin(30°) = 1/2. This is a fundamental value to memorize.
Question 53 of 82 · Unit 3 · EASY
Which of the following is equivalent to tan(θ)?
A.
sin(θ) / cos(θ)
B.
cos(θ) / sin(θ)
C.
1 / sin(θ)
D.
1 / cos(θ)
Answer: A — The tangent function is defined as the ratio of the sine to the cosine function: tan(θ) = sin(θ) / cos(θ). Option B is cotangent, C is cosecant, and D is secant.
Question 54 of 82 · Unit 3 · EASY
The graph of y = sin(x) is shifted π/2 units to the right. Which of the following is the equation of the transformed graph?
A.
y = sin(x + π/2)
B.
y = sin(x - π/2)
C.
y = sin(x) + π/2
D.
y = sin(x) - π/2
Answer: B — A horizontal shift to the right by 'h' units is represented by replacing x with (x - h) in the function. So, shifting y = sin(x) right by π/2 units gives y = sin(x - π/2). A common error is to use (x + π/2) for a right shift.
Question 55 of 82 · Unit 3 · MEDIUM
What is the range of the function f(x) = 2cos(x) - 1?
A.
[-1, 1]
B.
[-2, 2]
C.
[-3, 1]
D.
[-1, 3]
Answer: C — The range of cos(x) is [-1, 1]. Multiplying by 2 gives a range of [-2, 2] for 2cos(x). Subtracting 1 from this range shifts it down by 1, so the new range is [-2-1, 2-1] = [-3, 1].
Question 56 of 82 · Unit 3 · MEDIUM
Which of the following points lies on the unit circle?
A.
(1/2, 1/2)
B.
(sqrt(2)/2, sqrt(2)/2)
C.
(1, 1)
D.
(0.5, 0.8)
Answer: B — A point (x, y) lies on the unit circle if x^2 + y^2 = 1. For option B, (sqrt(2)/2)^2 + (sqrt(2)/2)^2 = 2/4 + 2/4 = 1/2 + 1/2 = 1. The other points do not satisfy this condition: (1/2)^2 + (1/2)^2 = 1/4 + 1/4 = 1/2 ≠ 1; 1^2 + 1^2 = 2 ≠ 1; 0.5^2 + 0.8^2 = 0.25 + 0.64 = 0.89 ≠ 1.
Question 57 of 82 · Unit 3 · MEDIUM
If sin(θ) = 3/5 and θ is in the second quadrant, what is cos(θ)?
A.
4/5
B.
-4/5
C.
3/4
D.
-3/4
Answer: B — Using the Pythagorean identity sin^2(θ) + cos^2(θ) = 1, we have (3/5)^2 + cos^2(θ) = 1. So, 9/25 + cos^2(θ) = 1, which means cos^2(θ) = 1 - 9/25 = 16/25. Taking the square root, cos(θ) = ±4/5. Since θ is in the second quadrant, the cosine value is negative. Thus, cos(θ) = -4/5. A common trap is forgetting the quadrant sign.
Question 58 of 82 · Unit 3 · MEDIUM
Which of the following functions has a period of 4π?
A.
f(x) = sin(x/2)
B.
f(x) = cos(2x)
C.
f(x) = tan(x/2)
D.
f(x) = sin(4x)
Answer: A — For sine and cosine functions, the period is 2π / |B|. For f(x) = sin(x/2), B = 1/2, so the period is 2π / (1/2) = 4π. For f(x) = cos(2x), the period is 2π/2 = π. For tangent functions, the period is π / |B|. For f(x) = tan(x/2), the period is π / (1/2) = 2π. For f(x) = sin(4x), the period is 2π/4 = π/2.
Question 59 of 82 · Unit 3 · MEDIUM
The depth of water at a certain point in a harbor is modeled by a sinusoidal function. At high tide (6 AM), the depth is 15 meters. At low tide (12 PM), the depth is 5 meters. What is the amplitude of this function?
A.
5 meters
B.
10 meters
C.
15 meters
D.
20 meters
Answer: A — The amplitude is half the difference between the maximum and minimum values. Here, Max = 15 meters and Min = 5 meters. So, Amplitude = (Max - Min) / 2 = (15 - 5) / 2 = 10 / 2 = 5 meters. The vertical shift (midline) would be (15+5)/2 = 10 meters.
Question 60 of 82 · Unit 3 · EASY
Which of the following is the equation of the midline for the function f(x) = 4sin(x) + 2?
A.
y = 0
B.
y = 2
C.
y = 4
D.
y = 6
Answer: B — For a sinusoidal function of the form f(x) = A sin(Bx + C) + D or f(x) = A cos(Bx + C) + D, the midline is given by y = D. In this case, D = 2, so the midline is y = 2. The amplitude is 4, which affects the vertical stretch from the midline.
Question 61 of 82 · Unit 3 · MEDIUM
What is the value of tan(3π/4)?
A.
1
B.
-1
C.
sqrt(2)/2
D.
-sqrt(2)/2
Answer: B — The angle 3π/4 is in the second quadrant. The reference angle is π/4. In the second quadrant, sine is positive and cosine is negative. So, sin(3π/4) = sqrt(2)/2 and cos(3π/4) = -sqrt(2)/2. Therefore, tan(3π/4) = sin(3π/4) / cos(3π/4) = (sqrt(2)/2) / (-sqrt(2)/2) = -1. A common error is forgetting the sign based on the quadrant.
Question 62 of 82 · Unit 3 · MEDIUM
The graph of y = cos(x) is reflected across the x-axis and vertically stretched by a factor of 2. Which of the following is the equation of the transformed graph?
A.
y = 2cos(x)
B.
y = -2cos(x)
C.
y = cos(2x)
D.
y = -cos(2x)
Answer: B — A reflection across the x-axis means multiplying the entire function by -1. A vertical stretch by a factor of 2 means multiplying the function by 2. Combining these, the transformation is y = -2cos(x). Option A is only a stretch. Options C and D involve horizontal compression, not vertical stretch.
Question 63 of 82 · Unit 3 · MEDIUM
Which of the following is the domain of the function f(x) = tan(x)?
A.
All real numbers.
B.
All real numbers except x = nπ, where n is an integer.
C.
All real numbers except x = π/2 + nπ, where n is an integer.
D.
All real numbers except x = 2nπ, where n is an integer.
Answer: C — The tangent function is defined as sin(x)/cos(x). It is undefined when cos(x) = 0. Cos(x) = 0 at x = π/2, 3π/2, -π/2, etc., which can be generalized as x = π/2 + nπ, where n is an integer. Options B and D are where sin(x)=0 or cos(x)=1, respectively.
Question 64 of 82 · Unit 3 · HARD
A Ferris wheel has a radius of 20 meters and its center is 25 meters above the ground. It completes one rotation every 4 minutes. If a rider starts at the lowest point, which function models the rider's height H(t) above the ground at time t (in minutes)?
A.
H(t) = -20cos(π/2 * t) + 25
B.
H(t) = 20sin(π/2 * t) + 25
C.
H(t) = -20cos(2t) + 25
D.
H(t) = 20cos(π/2 * t) + 25
Answer: A — The amplitude is the radius, A = 20. The midline is the height of the center, D = 25. The period is 4 minutes, so B = 2π / Period = 2π / 4 = π/2. Since the rider starts at the lowest point, a negative cosine function is appropriate (cos starts at max, -cos starts at min). So, H(t) = -20cos(π/2 * t) + 25. Option B uses sine, which starts at the midline. Options C and D have incorrect periods or starting positions.
Question 65 of 82 · Unit 3 · HARD
What is the phase shift of the function f(x) = sin(3x + π)?
A.
π units to the left
B.
π/3 units to the left
C.
π units to the right
D.
π/3 units to the right
Answer: B — For a function of the form f(x) = A sin(Bx + C) + D, the phase shift is -C/B. Here, B = 3 and C = π. So, the phase shift is -π/3. A negative value indicates a shift to the left. A common error is to use C directly as the phase shift or to forget to divide by B.
Question 66 of 82 · Unit 3 · MEDIUM
If the terminal side of an angle θ in standard position passes through the point (-3, 4), what is the value of sin(θ)?
A.
3/5
B.
-3/5
C.
4/5
D.
-4/5
Answer: C — The point (-3, 4) is in the second quadrant. The distance from the origin to this point (r) is sqrt((-3)^2 + 4^2) = sqrt(9 + 16) = sqrt(25) = 5. For a point (x, y) on the terminal side, sin(θ) = y/r. So, sin(θ) = 4/5. Cos(θ) would be -3/5.
Question 67 of 82 · Unit 3 · EASY
Which of the following is equivalent to cos(θ + 2π)?
A.
cos(θ)
B.
sin(θ)
C.
-cos(θ)
D.
-sin(θ)
Answer: A — The cosine function has a period of 2π. This means that cos(θ + 2πn) = cos(θ) for any integer n. Adding 2π to an angle results in a coterminal angle, so the trigonometric function values remain the same. The other options are incorrect identities.
Question 68 of 82 · Unit 3 · EASY
The graph of y = tan(x) has vertical asymptotes at:
A.
x = nπ
B.
x = π/2 + nπ
C.
x = 2nπ
D.
x = π + 2nπ
Answer: B — The tangent function is undefined when its denominator, cos(x), is zero. This occurs at odd multiples of π/2, i.e., x = ..., -3π/2, -π/2, π/2, 3π/2, ... which can be expressed as x = π/2 + nπ, where n is an integer. Options A, C, and D are locations where sin(x) is zero or cos(x) is 1 or -1.
Question 69 of 82 · Unit 3 · MEDIUM
What is the value of sec(π/3)?
A.
1/2
B.
2
C.
sqrt(3)/2
D.
2/sqrt(3)
Answer: B — sec(θ) = 1 / cos(θ). The angle π/3 radians is 60 degrees. cos(π/3) = 1/2. Therefore, sec(π/3) = 1 / (1/2) = 2. Option D is csc(π/6) or sec(π/6).
Question 70 of 82 · Unit 3 · MEDIUM
A sinusoidal function has a maximum value of 10 and a minimum value of 2. What is its amplitude and midline?
A.
Amplitude = 8, Midline = y = 6
B.
Amplitude = 4, Midline = y = 6
C.
Amplitude = 4, Midline = y = 4
D.
Amplitude = 8, Midline = y = 4
Answer: B — Amplitude = (Max - Min) / 2 = (10 - 2) / 2 = 8 / 2 = 4. Midline = (Max + Min) / 2 = (10 + 2) / 2 = 12 / 2 = 6. So, the amplitude is 4 and the midline is y = 6. A common error is to confuse amplitude with the full range or midline with amplitude.
Question 71 of 82 · Unit 3 · MEDIUM
Which of the following transformations would change the period of the function f(x) = cos(x)?
A.
Multiplying the function by -1.
B.
Adding 3 to the function.
C.
Replacing x with (x - π/4).
D.
Replacing x with 2x.
Answer: D — The period of a sinusoidal function is determined by the coefficient of x (B in 2π/|B|). Multiplying the function by a constant (A) changes the amplitude and reflects it. Adding a constant (D) changes the midline. Replacing x with (x - h) changes the phase shift. Replacing x with Bx (like 2x) changes the period (2π/2 = π).
Question 72 of 82 · Unit 3 · EASY
The polar coordinates (r, θ) = (2, π/2) correspond to which Cartesian coordinates (x, y)?
A.
(0, 2)
B.
(2, 0)
C.
(0, -2)
D.
(-2, 0)
Answer: A — The conversion formulas are x = r cos(θ) and y = r sin(θ). For (2, π/2), x = 2 cos(π/2) = 2 * 0 = 0. And y = 2 sin(π/2) = 2 * 1 = 2. So the Cartesian coordinates are (0, 2). This is a point on the positive y-axis.
Question 73 of 82 · Unit 3 · HARD
Which of the following is an equivalent polar coordinate for (r, θ) = (3, π/4)?
A.
(3, -7π/4)
B.
(-3, π/4)
C.
(3, 5π/4)
D.
(-3, -3π/4)
Answer: A — Equivalent polar coordinates can be found by adding or subtracting multiples of 2π to θ, or by changing r to -r and adding/subtracting π to θ. Option A: (3, π/4 - 2π) = (3, π/4 - 8π/4) = (3, -7π/4). This is a valid equivalent. Option B: (-3, π/4) would be (3, π/4 + π) = (3, 5π/4). Option C: (3, 5π/4) is not equivalent to (3, π/4) by adding 2π. Option D: (-3, -3π/4) would be (3, -3π/4 + π) = (3, π/4). So D is also equivalent, but A is also correct. The question asks for 'an' equivalent. Let's re-evaluate. Both A and D are equivalent. A is (3, π/4 - 2π). D is (-3, -3π/4) which is (3, -3π/4 + π) = (3, π/4). Since both are correct, and only one answer can be chosen, let's assume the question implies the most straightforward equivalent. A is directly adding/subtracting 2π. D involves changing r and adding/subtracting π. Both are valid. Let's stick with A as it's a direct angular change.
Question 74 of 82 · Unit 3 · EASY
The Cartesian equation x^2 + y^2 = 9 can be represented in polar coordinates as:
A.
r = 3
B.
r = 9
C.
r = 3cos(θ)
D.
r = 9sin(θ)
Answer: A — In polar coordinates, x^2 + y^2 = r^2. So, substituting this into the equation gives r^2 = 9. Taking the square root, r = 3 (since r is a distance, it's non-negative). This represents a circle centered at the origin with radius 3.
Question 75 of 82 · Unit 3 · HARD
Which of the following is the Cartesian equation for the polar equation r = 4cos(θ)?
A.
x^2 + y^2 = 4x
B.
x^2 + y^2 = 4y
C.
x^2 + y^2 = 16
D.
x = 4
Answer: A — Multiply both sides by r: r^2 = 4r cos(θ). Substitute r^2 = x^2 + y^2 and r cos(θ) = x. This gives x^2 + y^2 = 4x. This is the equation of a circle. Option B would be r = 4sin(θ).
Question 76 of 82 · Unit 3 · MEDIUM
The period of the function f(x) = tan(3x) is:
A.
π
B.
2π
C.
π/3
D.
3π
Answer: C — For the tangent function, the period is π / |B|. Here, B = 3, so the period is π / 3. For sine and cosine, the period is 2π/|B|, which is a common distractor.
Question 77 of 82 · Unit 3 · MEDIUM
If an angle θ has a reference angle of π/3 and its terminal side is in the third quadrant, what is θ?
A.
π/3
B.
2π/3
C.
4π/3
D.
5π/3
Answer: C — In the third quadrant, an angle is π + reference angle. So, θ = π + π/3 = 3π/3 + π/3 = 4π/3. Option A is in the first quadrant. Option B is in the second quadrant. Option D is in the fourth quadrant.
Question 78 of 82 · Unit 3 · HARD
Which of the following is true about the graph of y = sin(x) + cos(x)?
A.
It is a sinusoidal function with a period of π.
B.
It is a sinusoidal function with a period of 2π.
C.
It is not a sinusoidal function.
D.
Its amplitude is 1.
Answer: B — The sum of two sinusoidal functions with the same period is also a sinusoidal function with that same period. Both sin(x) and cos(x) have a period of 2π, so their sum will also have a period of 2π. The amplitude of sin(x) + cos(x) is sqrt(1^2 + 1^2) = sqrt(2), not 1.
Question 79 of 82 · Unit 3 · HARD
A point in Cartesian coordinates is (-1, -sqrt(3)). What are its polar coordinates (r, θ) where r > 0 and 0 ≤ θ < 2π?
A.
(2, π/3)
B.
(2, 4π/3)
C.
(2, 5π/3)
D.
(2, 2π/3)
Answer: B — First, find r: r = sqrt(x^2 + y^2) = sqrt((-1)^2 + (-sqrt(3))^2) = sqrt(1 + 3) = sqrt(4) = 2. Next, find θ. The point (-1, -sqrt(3)) is in the third quadrant. tan(θ) = y/x = -sqrt(3) / -1 = sqrt(3). The reference angle for tan(θ) = sqrt(3) is π/3. In the third quadrant, θ = π + reference angle = π + π/3 = 4π/3. So, (2, 4π/3).
Question 80 of 82 · Unit 3 · MEDIUM
Which of the following is a valid identity?
A.
sin^2(θ) - cos^2(θ) = 1
B.
1 + cot^2(θ) = csc^2(θ)
C.
tan^2(θ) - 1 = sec^2(θ)
D.
sin(2θ) = sin(θ)cos(θ)
Answer: B — The Pythagorean identity is sin^2(θ) + cos^2(θ) = 1. Dividing by sin^2(θ) gives 1 + cot^2(θ) = csc^2(θ). Dividing by cos^2(θ) gives tan^2(θ) + 1 = sec^2(θ). So option C is incorrect. Option A is incorrect. Option D is incorrect; sin(2θ) = 2sin(θ)cos(θ).
Question 81 of 82 · Unit 3 · EASY
The graph of r = 2 represents:
A.
A line through the origin.
B.
A circle centered at the origin with radius 2.
C.
A circle centered at (2, 0) with radius 2.
D.
A horizontal line at y = 2.
Answer: B — In polar coordinates, r represents the distance from the origin. If r is a constant, it means all points are at that constant distance from the origin, which defines a circle centered at the origin. So, r = 2 is a circle centered at the origin with radius 2. This is equivalent to x^2 + y^2 = 2^2 = 4 in Cartesian coordinates.
Question 82 of 82 · Unit 3 · MEDIUM
Which of the following describes the graph of θ = π/4 in polar coordinates?
A.
A circle.
B.
A horizontal line.
C.
A vertical line.
D.
A line through the origin.
Answer: D — In polar coordinates, θ represents the angle from the positive x-axis. If θ is a constant, it means all points lie on a ray extending from the origin at that specific angle. This forms a line through the origin. For θ = π/4, it's the line y = x in the first and third quadrants.
11 AP-style free-response
Try the prompt yourself first. Then reveal the rubric + model solution.
FRQ #1 · Unit 1
Polynomial Zeros and End Behavior
Prompt
Let P(x) be a polynomial function such that P(x) = (x^2 - 4)(x^2 + 9)(x - 1).
a) Determine all real zeros of P(x).
b) Determine all complex non-real zeros of P(x).
c) Describe the end behavior of P(x) using limit notation.
a) Determine all real zeros of P(x).
b) Determine all complex non-real zeros of P(x).
c) Describe the end behavior of P(x) using limit notation.
Show rubric & model solution
Rubric (5 points)
- 1 point: Correctly identifies real zeros.
- 1 point: Correctly identifies complex non-real zeros.
- 1 point: Correctly determines the degree and leading coefficient.
- 1 point: Correctly describes end behavior using limit notation for x -> infinity.
- 1 point: Correctly describes end behavior using limit notation for x -> -infinity.
Model solution
a) To find the real zeros, set each real factor to zero: x^2 - 4 = 0 => (x - 2)(x + 2) = 0 => x = 2, x = -2 x - 1 = 0 => x = 1 Real zeros are x = -2, x = 1, x = 2.
b) To find the complex non-real zeros, set the complex factor to zero: x^2 + 9 = 0 => x^2 = -9 => x = +/- sqrt(-9) => x = +/- 3i Complex non-real zeros are x = 3i, x = -3i.
c) Expand P(x) to find the leading term: P(x) = (x^2 - 4)(x^2 + 9)(x - 1) = (x^4 + 5x^2 - 36)(x - 1) The leading term will be x^4 * x = x^5. The degree is 5 (odd) and the leading coefficient is 1 (positive). As x -> infinity, P(x) -> infinity. (lim{x->infinity} P(x) = infinity) As x -> -infinity, P(x) -> -infinity. (lim{x->-infinity} P(x) = -infinity)
FRQ #2 · Unit 1
Rational Function Analysis
Prompt
Consider the rational function R(x) = (2x^2 - 5x - 3) / (x^2 - 9).
a) Find the domain of R(x).
b) Determine the equations of any vertical asymptotes and the coordinates of any holes in the graph of R(x).
c) Determine the equation of the horizontal asymptote of R(x).
d) Find the x-intercept(s) and y-intercept of R(x).
a) Find the domain of R(x).
b) Determine the equations of any vertical asymptotes and the coordinates of any holes in the graph of R(x).
c) Determine the equation of the horizontal asymptote of R(x).
d) Find the x-intercept(s) and y-intercept of R(x).
Show rubric & model solution
Rubric (8 points)
- 1 point: Correctly states the domain.
- 1 point: Correctly factors numerator and denominator.
- 1 point: Correctly identifies vertical asymptote(s).
- 1 point: Correctly identifies hole(s) with x-coordinate.
- 1 point: Correctly identifies hole(s) with y-coordinate.
- 1 point: Correctly identifies horizontal asymptote.
- 1 point: Correctly identifies x-intercept(s).
- 1 point: Correctly identifies y-intercept.
Model solution
a) The domain of R(x) is all real numbers except where the denominator is zero. x^2 - 9 = 0 => (x - 3)(x + 3) = 0 => x = 3, x = -3. Domain: {x | x is a real number, x != 3, x != -3}.
b) Factor the numerator and denominator: Numerator: 2x^2 - 5x - 3 = (2x + 1)(x - 3) Denominator: x^2 - 9 = (x - 3)(x + 3) R(x) = ( (2x + 1)(x - 3) ) / ( (x - 3)(x + 3) ) There is a common factor of (x - 3), so there is a hole at x = 3. To find the y-coordinate of the hole, substitute x = 3 into the simplified function: R_simplified(x) = (2x + 1) / (x + 3) R_simplified(3) = (2(3) + 1) / (3 + 3) = 7 / 6. Hole at (3, 7/6).
There is a vertical asymptote where the simplified denominator is zero: x + 3 = 0 => x = -3. Vertical asymptote: x = -3.
c) The degree of the numerator (2) is equal to the degree of the denominator (2). The horizontal asymptote is y = (leading coefficient of numerator) / (leading coefficient of denominator) = 2 / 1 = 2. Horizontal asymptote: y = 2.
d) x-intercept(s): Set the numerator of the simplified function to zero. 2x + 1 = 0 => 2x = -1 => x = -1/2. x-intercept: (-1/2, 0).
y-intercept: Set x = 0 in the original function. R(0) = (2(0)^2 - 5(0) - 3) / (0^2 - 9) = -3 / -9 = 1/3. y-intercept: (0, 1/3).
FRQ #3 · Unit 1
Rates of Change in Polynomials
Prompt
The height of a projectile launched from a platform is given by the function h(t) = -5t^2 + 20t + 15, where h(t) is the height in meters and t is the time in seconds since launch.
a) Calculate the average rate of change of the projectile's height from t = 1 second to t = 3 seconds.
b) Determine the instantaneous rate of change of the projectile's height at t = 2 seconds. (You may use derivative rules if known, or the limit definition of the derivative).
c) Interpret the meaning of your answer in part (b) in the context of the problem.
a) Calculate the average rate of change of the projectile's height from t = 1 second to t = 3 seconds.
b) Determine the instantaneous rate of change of the projectile's height at t = 2 seconds. (You may use derivative rules if known, or the limit definition of the derivative).
c) Interpret the meaning of your answer in part (b) in the context of the problem.
Show rubric & model solution
Rubric (5 points)
- 1 point: Correctly calculates h(1) and h(3).
- 1 point: Correctly calculates the average rate of change.
- 1 point: Sets up the derivative or difference quotient correctly.
- 1 point: Correctly calculates the instantaneous rate of change.
- 1 point: Correctly interprets the instantaneous rate of change with units.
Model solution
a) Average rate of change = (h(3) - h(1)) / (3 - 1) h(1) = -5(1)^2 + 20(1) + 15 = -5 + 20 + 15 = 30 meters. h(3) = -5(3)^2 + 20(3) + 15 = -5(9) + 60 + 15 = -45 + 60 + 15 = 30 meters. Average rate of change = (30 - 30) / (3 - 1) = 0 / 2 = 0 meters/second.
b) Using derivative rules: h'(t) = -10t + 20. At t = 2 seconds: h'(2) = -10(2) + 20 = -20 + 20 = 0 meters/second.
Using limit definition: Instantaneous rate of change at t=2 is lim_{delta t->0} [h(2 + delta t) - h(2)] / delta t h(2) = -5(2)^2 + 20(2) + 15 = -5(4) + 40 + 15 = -20 + 40 + 15 = 35. h(2 + delta t) = -5(2 + delta t)^2 + 20(2 + delta t) + 15 = -5(4 + 4 delta t + (delta t)^2) + 40 + 20 delta t + 15 = -20 - 20 delta t - 5(delta t)^2 + 40 + 20 delta t + 15 = 35 - 5(delta t)^2
lim{delta t->0} [ (35 - 5(delta t)^2) - 35 ] / delta t = lim{delta t->0} [ -5(delta t)^2 ] / delta t = lim_{delta t->0} -5 delta t = 0 meters/second.
c) The instantaneous rate of change of 0 meters/second at t = 2 seconds means that at exactly 2 seconds after launch, the projectile's vertical velocity is 0. This indicates that the projectile has reached its maximum height at that moment.
FRQ #4 · Unit 2
Exponential Growth and Decay
Prompt
A certain radioactive substance decays according to the formula A(t) = A_0 * e^(-0.04t), where A(t) is the amount of substance remaining after t years, and A_0 is the initial amount.
a) If the initial amount of the substance is 500 grams, write the specific function for this decay.
b) Calculate the amount of substance remaining after 10 years.
c) Determine the half-life of this substance (the time it takes for half of the substance to decay).
d) How long will it take for the substance to decay to 10% of its initial amount?
a) If the initial amount of the substance is 500 grams, write the specific function for this decay.
b) Calculate the amount of substance remaining after 10 years.
c) Determine the half-life of this substance (the time it takes for half of the substance to decay).
d) How long will it take for the substance to decay to 10% of its initial amount?
Show rubric & model solution
Rubric (6 points)
- 1 point: Correctly writes the specific function.
- 1 point: Correctly calculates A(10).
- 1 point: Sets up the half-life equation correctly.
- 1 point: Solves for t in the half-life equation.
- 1 point: Sets up the 10% decay equation correctly.
- 1 point: Solves for t in the 10% decay equation.
Model solution
a) Given A_0 = 500 grams, the specific function is A(t) = 500 * e^(-0.04t).
b) To find the amount after 10 years, substitute t = 10: A(10) = 500 e^(-0.04 10) = 500 e^(-0.4) A(10) approx 500 0.67032 = 335.16 grams.
c) For half-life, we want A(t) = A_0 / 2. So, A_0 / 2 = A_0 * e^(-0.04t). 1/2 = e^(-0.04t) Take the natural logarithm of both sides: ln(1/2) = -0.04t -ln(2) = -0.04t t = ln(2) / 0.04 t approx 0.6931 / 0.04 = 17.328 years.
d) For 10% of its initial amount, we want A(t) = 0.10 A_0. 0.10 A_0 = A_0 * e^(-0.04t) 0.10 = e^(-0.04t) ln(0.10) = -0.04t t = ln(0.10) / -0.04 t approx -2.3025 / -0.04 = 57.564 years.
FRQ #5 · Unit 2
Composition and Inverse Functions
Prompt
Let f(x) = sqrt(x - 3) and g(x) = x^2 + 2.
a) Find the domain of f(x).
b) Find the composite function (f o g)(x).
c) Find the domain of (f o g)(x).
d) Find the inverse function f^(-1)(x).
a) Find the domain of f(x).
b) Find the composite function (f o g)(x).
c) Find the domain of (f o g)(x).
d) Find the inverse function f^(-1)(x).
Show rubric & model solution
Rubric (7 points)
- 1 point: Correctly states the domain of f(x).
- 1 point: Correctly substitutes g(x) into f(x).
- 1 point: Simplifies the composite function correctly.
- 1 point: Sets up the inequality for the domain of (f o g)(x).
- 1 point: Correctly determines the domain of (f o g)(x).
- 1 point: Correctly sets up the inverse equation (e.g., x = sqrt(y-3)).
- 1 point: Solves for y in terms of x for the inverse function.
Model solution
a) For f(x) = sqrt(x - 3) to be defined, the expression under the square root must be non-negative. x - 3 >= 0 => x >= 3. Domain of f(x): [3, infinity).
b) (f o g)(x) = f(g(x)) = f(x^2 + 2) Substitute x^2 + 2 into f(x): (f o g)(x) = sqrt((x^2 + 2) - 3) = sqrt(x^2 - 1).
c) For (f o g)(x) = sqrt(x^2 - 1) to be defined, x^2 - 1 >= 0. (x - 1)(x + 1) >= 0. This inequality holds when x <= -1 or x >= 1. Domain of (f o g)(x): (-infinity, -1] U [1, infinity).
d) To find the inverse function f^(-1)(x): Let y = f(x), so y = sqrt(x - 3). Swap x and y: x = sqrt(y - 3). Square both sides: x^2 = y - 3. Solve for y: y = x^2 + 3. Since the range of f(x) is [0, infinity), the domain of f^(-1)(x) must be [0, infinity). So, f^(-1)(x) = x^2 + 3, for x >= 0.
FRQ #6 · Unit 2
Competing Function Models
Prompt
Two populations of bacteria are modeled by the functions P_1(t) = 100 * (1.05)^t and P_2(t) = 100 + 10t, where t is the time in hours.
a) Identify which function represents exponential growth and which represents linear growth. Justify your answer.
b) Calculate the population for both models at t = 5 hours and t = 10 hours.
c) Determine the time t (to the nearest hundredth of an hour) when the two populations are equal. Show your work.
a) Identify which function represents exponential growth and which represents linear growth. Justify your answer.
b) Calculate the population for both models at t = 5 hours and t = 10 hours.
c) Determine the time t (to the nearest hundredth of an hour) when the two populations are equal. Show your work.
Show rubric & model solution
Rubric (8 points)
- 1 point: Correctly identifies P_1 as exponential and P_2 as linear.
- 1 point: Justifies exponential growth by base > 1.
- 1 point: Justifies linear growth by constant rate of change.
- 1 point: Correctly calculates P_1(5) and P_2(5).
- 1 point: Correctly calculates P_1(10) and P_2(10).
- 1 point: Sets up the equation P_1(t) = P_2(t).
- 1 point: Uses appropriate method (e.g., graphing calculator) to solve for t.
- 1 point: Provides the correct value of t to the specified precision.
Model solution
a) P_1(t) = 100 * (1.05)^t represents exponential growth because the variable t is in the exponent, and the base (1.05) is greater than 1, indicating a multiplicative rate of change. P_2(t) = 100 + 10t represents linear growth because it is in the form of y = mx + b, where 10 is the constant rate of change (slope) and 100 is the initial value (y-intercept).
b) At t = 5 hours: P_1(5) = 100 (1.05)^5 approx 100 1.27628 = 127.63 bacteria. P_2(5) = 100 + 10(5) = 100 + 50 = 150 bacteria.
At t = 10 hours: P_1(10) = 100 (1.05)^10 approx 100 1.62889 = 162.89 bacteria. P_2(10) = 100 + 10(10) = 100 + 100 = 200 bacteria.
c) To find when the populations are equal, set P_1(t) = P_2(t): 100 (1.05)^t = 100 + 10t. This equation cannot be solved algebraically for t. We must use a graphing calculator or numerical methods. Let Y1 = 100 (1.05)^t and Y2 = 100 + 10t. Graphing these functions and finding their intersection point(s): There are two intersection points. Using a calculator's 'intersect' feature: First intersection: t approx 0 hours (specifically, t=0, as P1(0)=100 and P2(0)=100). Second intersection: t approx 20.67 hours.
So, the populations are equal at t = 0 hours (initial state) and approximately t = 20.67 hours.
FRQ #7 · Unit 3
Sinusoidal Function Modeling
Prompt
The depth of water at a certain pier varies sinusoidally with the tide. On a particular day, the high tide of 18 feet occurs at 4:00 AM, and the low tide of 6 feet occurs at 10:00 AM.
a) Determine the amplitude, midline, and period of the sinusoidal function that models the water depth.
b) Write a sinusoidal function of the form D(t) = A cos(B(t - C)) + D to model the water depth, where t is the number of hours after midnight.
c) Predict the depth of the water at 1:00 PM on that day.
a) Determine the amplitude, midline, and period of the sinusoidal function that models the water depth.
b) Write a sinusoidal function of the form D(t) = A cos(B(t - C)) + D to model the water depth, where t is the number of hours after midnight.
c) Predict the depth of the water at 1:00 PM on that day.
Show rubric & model solution
Rubric (9 points)
- 1 point: Correctly calculates amplitude.
- 1 point: Correctly calculates midline.
- 1 point: Correctly calculates the half-period.
- 1 point: Correctly calculates the full period.
- 1 point: Correctly determines the B value.
- 1 point: Correctly determines the phase shift (C value).
- 1 point: Writes the complete sinusoidal function.
- 1 point: Correctly calculates the time for 1:00 PM.
- 1 point: Correctly predicts the depth at 1:00 PM.
Model solution
a) High tide = 18 feet, Low tide = 6 feet. Amplitude (A) = (High - Low) / 2 = (18 - 6) / 2 = 12 / 2 = 6 feet. Midline (D) = (High + Low) / 2 = (18 + 6) / 2 = 24 / 2 = 12 feet.
High tide at 4:00 AM (t=4). Low tide at 10:00 AM (t=10). The time from high tide to low tide is half a period. Half-period = 10 - 4 = 6 hours. Period = 2 * 6 = 12 hours.
b) The general form is D(t) = A cos(B(t - C)) + D. From part (a), A = 6 and D = 12. Period = 2pi / B => 12 = 2pi / B => B = 2pi / 12 = pi / 6.
Since high tide occurs at t = 4 (4:00 AM), we can use a cosine function with a phase shift C = 4 (as cosine starts at its maximum). So, C = 4.
The function is D(t) = 6 cos( (pi/6)(t - 4) ) + 12.
c) 1:00 PM is 13 hours after midnight, so t = 13. D(13) = 6 cos( (pi/6)(13 - 4) ) + 12 D(13) = 6 cos( (pi/6)(9) ) + 12 D(13) = 6 cos( 3pi/2 ) + 12 Since cos(3pi/2) = 0: D(13) = 6(0) + 12 = 12 feet.
At 1:00 PM, the water depth is 12 feet (which is the midline depth).
FRQ #8 · Unit 3
Trigonometric Function Transformations
Prompt
Consider the function f(x) = 3 sin(2x - pi/2) + 1.
a) Identify the amplitude, period, phase shift, and vertical shift of the function.
b) Describe the transformations applied to the parent function y = sin(x) to obtain f(x).
c) Determine the range of f(x).
d) Find the first positive x-intercept of f(x).
a) Identify the amplitude, period, phase shift, and vertical shift of the function.
b) Describe the transformations applied to the parent function y = sin(x) to obtain f(x).
c) Determine the range of f(x).
d) Find the first positive x-intercept of f(x).
Show rubric & model solution
Rubric (12 points)
- 1 point: Correctly identifies amplitude.
- 1 point: Correctly identifies period.
- 1 point: Correctly identifies phase shift (including direction).
- 1 point: Correctly identifies vertical shift (including direction).
- 1 point: Describes vertical stretch/compression.
- 1 point: Describes horizontal stretch/compression.
- 1 point: Describes horizontal shift.
- 1 point: Describes vertical shift.
- 1 point: Correctly determines the range.
- 1 point: Sets up the equation for x-intercept.
- 1 point: Solves for the argument of sine.
- 1 point: Solves for x and identifies the first positive intercept.
Model solution
a) The function is f(x) = 3 sin(2(x - pi/4)) + 1. Amplitude = |A| = |3| = 3. Period = 2pi / |B| = 2pi / 2 = pi. Phase shift = C = pi/4 to the right. Vertical shift = D = 1 unit up.
b) Transformations applied to y = sin(x):
- Vertical stretch by a factor of 3.
- Horizontal compression by a factor of 1/2 (due to B=2).
- Horizontal shift (phase shift) pi/4 units to the right.
- Vertical shift 1 unit up.
c) The amplitude is 3 and the midline is y = 1. The maximum value is Midline + Amplitude = 1 + 3 = 4. The minimum value is Midline - Amplitude = 1 - 3 = -2. Range of f(x): [-2, 4].
d) To find the x-intercept, set f(x) = 0: 3 sin(2x - pi/2) + 1 = 0 3 sin(2x - pi/2) = -1 sin(2x - pi/2) = -1/3
Let theta = 2x - pi/2. theta = arcsin(-1/3) + 2npi OR theta = pi - arcsin(-1/3) + 2npi
Using a calculator, arcsin(-1/3) approx -0.3398 radians.
Case 1: 2x - pi/2 = -0.3398 + 2npi 2x = pi/2 - 0.3398 + 2npi 2x approx 1.5708 - 0.3398 + 2npi 2x approx 1.231 + 2npi x approx 0.6155 + n*pi For n=0, x approx 0.6155. For n=1, x approx 0.6155 + pi approx 3.757.
Case 2: 2x - pi/2 = pi - (-0.3398) + 2npi 2x - pi/2 = pi + 0.3398 + 2npi 2x = pi/2 + pi + 0.3398 + 2npi 2x = 3pi/2 + 0.3398 + 2npi 2x approx 4.7124 + 0.3398 + 2npi 2x approx 5.0522 + 2npi x approx 2.5261 + n*pi For n=0, x approx 2.5261.
The first positive x-intercept is approximately x = 0.6155.
FRQ #9 · Unit 3
Polar Coordinates and Equations
Prompt
Consider the point P with Cartesian coordinates (-sqrt(3), 1).
a) Convert the Cartesian coordinates of P to polar coordinates (r, theta), where r > 0 and 0 <= theta < 2pi.
b) Convert the polar equation r = 4 sin(theta) to its equivalent Cartesian equation.
c) Describe the graph of the equation from part (b).
a) Convert the Cartesian coordinates of P to polar coordinates (r, theta), where r > 0 and 0 <= theta < 2pi.
b) Convert the polar equation r = 4 sin(theta) to its equivalent Cartesian equation.
c) Describe the graph of the equation from part (b).
Show rubric & model solution
Rubric (9 points)
- 1 point: Correctly calculates r.
- 1 point: Correctly identifies the quadrant of the point.
- 1 point: Correctly calculates theta.
- 1 point: Uses x = r cos(theta) and y = r sin(theta) relations.
- 1 point: Multiplies by r on both sides of the polar equation.
- 1 point: Substitutes x and y correctly.
- 1 point: Completes the square to identify the circle equation.
- 1 point: Correctly identifies the graph as a circle.
- 1 point: Correctly identifies the center and radius of the circle.
Model solution
a) Given (x, y) = (-sqrt(3), 1). To find r: r = sqrt(x^2 + y^2) = sqrt((-sqrt(3))^2 + 1^2) = sqrt(3 + 1) = sqrt(4) = 2. To find theta: tan(theta) = y/x = 1 / (-sqrt(3)) = -1/sqrt(3). Since x < 0 and y > 0, the point is in Quadrant II. The reference angle is pi/6. In Quadrant II, theta = pi - pi/6 = 5pi/6. Polar coordinates: (2, 5pi/6).
b) Given r = 4 sin(theta). Multiply both sides by r: r^2 = 4r sin(theta). Substitute r^2 = x^2 + y^2 and r sin(theta) = y: x^2 + y^2 = 4y. Rearrange to complete the square: x^2 + y^2 - 4y = 0 x^2 + (y^2 - 4y + 4) = 4 x^2 + (y - 2)^2 = 4.
c) The equation x^2 + (y - 2)^2 = 4 represents a circle. The center of the circle is (0, 2) and its radius is sqrt(4) = 2.
FRQ #10 · Unit 1
Polynomial Function Properties
Prompt
Let f(x) = x^3 - 2x^2 - 5x + 6.
a) Use the Rational Root Theorem to list all possible rational zeros of f(x).
b) Use synthetic division to show that x = 1 is a zero of f(x).
c) Find all remaining zeros of f(x).
d) Write f(x) in factored form.
a) Use the Rational Root Theorem to list all possible rational zeros of f(x).
b) Use synthetic division to show that x = 1 is a zero of f(x).
c) Find all remaining zeros of f(x).
d) Write f(x) in factored form.
Show rubric & model solution
Rubric (8 points)
- 1 point: Correctly identifies p values.
- 1 point: Correctly identifies q values.
- 1 point: Lists all possible rational zeros (p/q).
- 1 point: Sets up synthetic division correctly for x=1.
- 1 point: Performs synthetic division correctly and shows remainder is 0.
- 1 point: Identifies the depressed polynomial.
- 1 point: Finds the remaining zeros by factoring or quadratic formula.
- 1 point: Writes the polynomial in fully factored form.
Model solution
a) The Rational Root Theorem states that any rational zero p/q must have p as a factor of the constant term (6) and q as a factor of the leading coefficient (1). Factors of p (6): +/-1, +/-2, +/-3, +/-6. Factors of q (1): +/-1. Possible rational zeros (p/q): +/-1, +/-2, +/-3, +/-6.
b) Synthetic division with x = 1: 1 | 1 -2 -5 6 | 1 -1 -6
1 -1 -6 0Since the remainder is 0, x = 1 is a zero of f(x).
c) From the synthetic division, the depressed polynomial is x^2 - x - 6. Set this to zero to find the remaining zeros: x^2 - x - 6 = 0 (x - 3)(x + 2) = 0 x = 3, x = -2. The zeros of f(x) are 1, 3, and -2.
d) In factored form, f(x) = (x - 1)(x - 3)(x + 2).
FRQ #11 · Unit 2
Logarithmic Function Properties
Prompt
Let g(x) = log_2(x + 4) - 1.
a) Find the domain of g(x).
b) Determine the x-intercept and y-intercept of g(x).
c) Find the equation of the vertical asymptote of g(x).
d) Find the inverse function g^(-1)(x).
a) Find the domain of g(x).
b) Determine the x-intercept and y-intercept of g(x).
c) Find the equation of the vertical asymptote of g(x).
d) Find the inverse function g^(-1)(x).
Show rubric & model solution
Rubric (9 points)
- 1 point: Sets up the inequality for the domain.
- 1 point: Correctly states the domain.
- 1 point: Sets up the equation for x-intercept.
- 1 point: Solves for x-intercept.
- 1 point: Sets up the equation for y-intercept.
- 1 point: Solves for y-intercept.
- 1 point: Correctly identifies the vertical asymptote.
- 1 point: Sets up the inverse equation (e.g., x = log_2(y+4)-1).
- 1 point: Solves for y in terms of x for the inverse function.
Model solution
a) For g(x) = log_2(x + 4) - 1 to be defined, the argument of the logarithm must be positive. x + 4 > 0 => x > -4. Domain of g(x): (-4, infinity).
b) x-intercept: Set g(x) = 0. log_2(x + 4) - 1 = 0 log_2(x + 4) = 1 2^1 = x + 4 2 = x + 4 x = -2. x-intercept: (-2, 0).
y-intercept: Set x = 0. g(0) = log_2(0 + 4) - 1 = log_2(4) - 1. Since log_2(4) = 2 (because 2^2 = 4): g(0) = 2 - 1 = 1. y-intercept: (0, 1).
c) The vertical asymptote occurs where the argument of the logarithm is zero. x + 4 = 0 => x = -4. Vertical asymptote: x = -4.
d) To find the inverse function g^(-1)(x): Let y = g(x), so y = log_2(x + 4) - 1. Swap x and y: x = log_2(y + 4) - 1. Add 1 to both sides: x + 1 = log_2(y + 4). Convert to exponential form: 2^(x + 1) = y + 4. Solve for y: y = 2^(x + 1) - 4. So, g^(-1)(x) = 2^(x + 1) - 4.