Crash Course — Full Course Cram Sheet
Crash Course — Full Course Cram Sheet
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.
(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).82 high-yield questions
Distributed across all units. Tap a choice to lock in your answer.
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
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
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
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) 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
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
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
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
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
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
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 0
Since 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
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.