Overview
AP Precalculus is a College Board course that deepens understanding of functions as dynamic objects whose properties model the real world. Unlike a traditional precalculus course, AP Precalculus emphasises rates of change, transformations, and function behaviour — the conceptual language of calculus.
Exam: Section I — 40 MCQ (Part A: 28 Qs, 80 min, no calculator; Part B: 12 Qs, 50 min, graphing calc). Section II — 4 FRQ (graphing calc permitted, 60 min).Unit Topics Exam weight MCQ type 1 — Polynomial & Rational Functions AROC, concavity, zeros, end behaviour, asymptotes, transformations 30–40% Part A & B 2 — Exponential & Logarithmic Multiplicative change, log properties, growth/decay models 27–40% Part A & B 3 — Trigonometric & Polar Unit circle, sinusoidal transformations, inverse trig, polar coords 25–35% Part A & B 4 — Functions & Parameters Composition, inverses, piecewise, parametric, implicit, conic sections Optional (AP Exam weighted toward Units 1–3) Part B
Exam format: Section I — 40 MCQ (Part A: 28 Qs, no calculator; Part B: 12 Qs, graphing calculator). Section II — 4 FRQ (graphing calculator permitted throughout).
Polynomial & Rational Functions
Unit 1 examines polynomial and rational functions through the lens of rates of change and function behaviour — the conceptual language of calculus. Emphasis is on concavity, zeros, asymptotes, and transformations.
The average rate of change (AROC) of \(f\) over \([a,b]\) is the slope of the secant line: Concavity: If the AROC is increasing as \(x\) increases, the function is concave up; if AROC is decreasing, it is concave down. An inflection point is where concavity changes.
| x | f(x) |
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Curve is \(f(x)=x^3/3-x\). Slide the interval right and watch the AROC itself start increasing — that's concave up, live.
End behaviour — determined by the leading term \(a_nx^n\): Multiplicity of zeros: If \((x-c)^m\) is a factor, zero at \(x=c\) has multiplicity \(m\). Odd multiplicity → graph crosses \(x\)-axis; Even multiplicity → graph touches (bounces).Leading term Left end Right end \(a_n>0,\;n\) even \(\to+\infty\) \(\to+\infty\) \(a_n<0,\;n\) even \(\to-\infty\) \(\to-\infty\) \(a_n>0,\;n\) odd \(\to-\infty\) \(\to+\infty\) \(a_n<0,\;n\) odd \(\to+\infty\) \(\to-\infty\)
When \(p(x)\) is divided by \((x-c)\), the remainder is \(p(c)\). Thus \((x-c)\) is a factor \(\Leftrightarrow p(c)=0.\) Rational Zero Theorem: If \(p(x)=a_nx^n+\cdots+a_0\) has integer coefficients, any rational zero has the form \(\pm\dfrac{\text{factor of }a_0}{\text{factor of }a_n}.\)
For \(f(x)=\dfrac{p(x)}{q(x)}\) fully reduced, \(\deg p=n,\;\deg q=d\): Vertical asymptote: \(x=a\) where \(q(a)=0\) after full reduction. Hole: common factor cancels; evaluate limit at that \(x\) for \(y\)-coordinate. Horizontal asymptote: \(n
Starting from \(y=f(x)\):Transformation Equation Vertical shift up \(k\) \(y=f(x)+k\) Horizontal shift right \(h\) \(y=f(x-h)\) Vertical stretch by \(a\) \(y=a\cdot f(x)\), \(|a|>1\) Reflection over \(x\)-axis \(y=-f(x)\) Reflection over \(y\)-axis \(y=f(-x)\)
Problem. Find all zeros and their multiplicities for \(p(x)=x^4-2x^3-3x^2\). Describe end behaviour.
Factor: \(p(x)=x^2(x^2-2x-3)=x^2(x-3)(x+1).\)
Zeros: \(x=0\) (multiplicity 2 — touches axis), \(x=3\) (mult 1 — crosses), \(x=-1\) (mult 1 — crosses).
End behaviour: \(n=4\) even, leading coefficient \(+1>0\) → both ends rise to \(+\infty.\)
Problem. Fully analyse \(f(x)=\dfrac{x^2-9}{x^2+x-6}\): find holes, VA, HA, and \(x\)-intercepts.
Factor: \(\dfrac{(x-3)(x+3)}{(x+3)(x-2)}.\) Cancel \((x+3)\): hole at \(x=-3\).
Simplified: \(\dfrac{x-3}{x-2}\) for \(x\neq-3.\) VA: \(x=2.\) HA: \(n=d=1\Rightarrow y=1.\) \(x\)-intercept: \(x=3.\)
Hole coordinates: \(y=\dfrac{-3-3}{-3-2}=\dfrac{-6}{-5}=\dfrac{6}{5}.\) So hole at \(\left(-3,\tfrac{6}{5}\right).\)
Problem. The AROC of \(f(x)=x^3\) on \([1,3]\) is 13 and on \([1,2]\) is 7. Is the function concave up or down on \([1,3]\)?
AROC increases from 7 (on \([1,2]\)) to 13 (on \([1,3]\)). Since AROC is increasing, \(f\) is concave up on \([1,3].\)
The function \(f(x)=x^3-3x^2-9x+2\) has a local maximum and local minimum. Determine the intervals on which \(f\) is concave up and concave down, and find the inflection point. AROC analysis: the inflection point is where concavity changes, i.e. where the AROC switches from decreasing to increasing (or vice versa). \(f''(x)=6x-6=0\Rightarrow x=1.\) Concave down on \((-\infty,1)\); concave up on \((1,\infty).\) Inflection point: \(f(1)=1-3-9+2=-9.\) Point: \((1,-9).\)Show solution
Exponential & Logarithmic Functions
Unit 2 treats exponential and logarithmic functions as models of multiplicative change — quantities that change by a constant factor over equal intervals, rather than a constant additive amount.
Domain: all reals. Range: \((0,\infty)\) when \(a>0.\) Horizontal asymptote: \(y=0.\) Key property: Over any equal-length interval \(\Delta x\), the output is multiplied by the same factor \(b^{\Delta x}.\) Growth: \(b>1.\) Decay: \(0
\[P(t)=P_0\,b^t,\quad\text{Continuous: }P(t)=P_0\,e^{rt},\]\[\text{Half-life: }A(t)=A_0\!\left(\tfrac{1}{2}\right)^{t/h},\quad\text{Doubling time: }P(t)=P_0\cdot 2^{t/d}.\] Compound interest: \(A=P\!\left(1+\dfrac{r}{n}\right)^{nt},\quad A=Pe^{rt}\text{ (continuously)}.\)
| t | P(t) |
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\(P(t)=P_0e^{rt}\) — the continuous-compounding form from Worked Example 2.3's car-depreciation model.
\[\log_b(MN)=\log_b M+\log_b N,\quad\log_b\!\left(\frac{M}{N}\right)=\log_b M-\log_b N,\]\[\log_b(M^p)=p\log_b M,\quad\log_b x=\frac{\ln x}{\ln b}\quad(\text{change of base}).\] Inverse relationship: \(b^{\log_b x}=x\) and \(\log_b(b^x)=x.\) Key values: \(\log_b 1=0,\;\log_b b=1.\)
Exponential: isolate the exponential, apply \(\ln\) or \(\log\) to both sides. Logarithmic: combine logs using properties, convert to exponential form. Extraneous solutions: always check that arguments of logs are positive in the original equation.
Problem. A bacteria culture starts with 400 and triples every 6 hours. Write \(P(t)\) and find when it reaches 10 800.
\(P(t)=400\cdot3^{t/6}.\) Set \(400\cdot3^{t/6}=10800:\quad 3^{t/6}=27=3^3.\)
\(\Rightarrow t/6=3\Rightarrow t=18\) hours.
Problem. Solve \(\log_2(x+4)+\log_2(x-2)=4.\)
Combine: \(\log_2[(x+4)(x-2)]=4\Rightarrow(x+4)(x-2)=2^4=16.\)
\(x^2+2x-8=16\Rightarrow x^2+2x-24=0\Rightarrow(x+6)(x-4)=0.\)
Check: \(x=-6\) makes \(x-2=-8<0\) — reject. Answer: \(x=4.\)
Problem. The value of a car is modelled by \(V(t)=25000\,e^{-0.12t}.\) (a) Find the initial value. (b) After how many years is the car worth $10,000? (c) What is the continuous rate of decay?
(a) \(V(0)=25000\,e^0=\$25{,}000.\)
(b) \(25000\,e^{-0.12t}=10000\Rightarrow e^{-0.12t}=0.4\Rightarrow -0.12t=\ln 0.4\Rightarrow t=\dfrac{\ln 0.4}{-0.12}\approx7.6\) years.
(c) The continuous decay rate is \(|-0.12|=12\%\) per year.
Carbon-14 has a half-life of 5730 years. A fossil contains 30% of its original \({}^{14}\text{C}.\) How old is the fossil? (Give answer to the nearest 100 years.) Model: \(A(t)=A_0\left(\frac{1}{2}\right)^{t/5730}.\) Set \(0.30=\left(\frac{1}{2}\right)^{t/5730}.\) \(\ln 0.30=\frac{t}{5730}\ln\frac{1}{2}\Rightarrow t=\frac{5730\ln 0.30}{\ln 0.5}=\frac{5730(-1.2040)}{-0.6931}\approx9950\) years. The fossil is approximately 9,950 years old.Show solution
Trigonometric & Polar Functions
Unit 3 treats trigonometric functions as circular functions defined for all real numbers — not just angles in triangles. The emphasis is on periodicity, transformations, inverses, and polar coordinates.
\[s=r\theta\quad(\text{arc length}),\quad\theta\text{ in radians.}\quad 360°=2\pi\text{ rad},\;1\text{ rad}=\frac{180°}{\pi}.\] On the unit circle: \((\cos\theta,\sin\theta)\) gives the coordinates of the terminal point. Period of \(\sin\) and \(\cos\): \(2\pi.\) Period of \(\tan\): \(\pi.\)
\[y=A\sin(B(x-h))+k\quad\text{or}\quad y=A\cos(B(x-h))+k.\]Parameter Effect Formula \(|A|\) Amplitude Half the range: \(\dfrac{\max-\min}{2}\) \(B\) Angular frequency Period \(=\dfrac{2\pi}{|B|}\) \(h\) Horizontal (phase) shift Right if \(h>0\) \(k\) Vertical shift (midline) Midline: \(y=k\)
\[\sin^{-1}x:\text{ domain }[-1,1],\text{ range }\left[-\tfrac{\pi}{2},\tfrac{\pi}{2}\right].\]\[\cos^{-1}x:\text{ domain }[-1,1],\text{ range }[0,\pi].\]\[\tan^{-1}x:\text{ domain }(-\infty,\infty),\text{ range }\left(-\tfrac{\pi}{2},\tfrac{\pi}{2}\right).\] These are restricted domains to ensure the inverse is a function.
\[\sin^2\theta+\cos^2\theta=1,\quad\tan\theta=\frac{\sin\theta}{\cos\theta},\]\[\sin(2\theta)=2\sin\theta\cos\theta,\quad\cos(2\theta)=1-2\sin^2\theta=2\cos^2\theta-1.\]
Point \((r,\theta)\): distance \(r\) from origin, angle \(\theta\) from positive \(x\)-axis. Common polar curves: \(r=a\) (circle), \(r=a(1\pm\cos\theta)\) (cardioid), \(r=a\cos(n\theta)\) (rose with \(n\) or \(2n\) petals).
| θ (rad) | r = A cos(nθ) | (x, y) |
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a=3 fixed. Odd n → n petals; even n → 2n petals — exactly the rule stated above.
Problem. A function has amplitude 3, period \(\pi\), phase shift right \(\pi/4\), midline \(y=1\). Write its equation as a cosine.
\(A=3,\;B=2\pi/\pi=2,\;h=\pi/4,\;k=1.\)
\(y=3\cos\!\left(2\!\left(x-\dfrac{\pi}{4}\right)\right)+1.\)
Problem. Find all solutions in \([0,2\pi)\): \(2\sin^2\theta-\sin\theta-1=0.\)
Factor: \((2\sin\theta+1)(\sin\theta-1)=0.\)
\(\sin\theta=1\Rightarrow\theta=\dfrac{\pi}{2}.\)
\(\sin\theta=-\dfrac{1}{2}\Rightarrow\theta=\dfrac{7\pi}{6},\;\dfrac{11\pi}{6}.\)
Solutions: \(\left\{\dfrac{\pi}{2},\;\dfrac{7\pi}{6},\;\dfrac{11\pi}{6}\right\}.\)
Problem. Convert \(r=\dfrac{4}{1-2\cos\theta}\) to Cartesian form and identify the conic.
\(r(1-2\cos\theta)=4\Rightarrow r-2r\cos\theta=4\Rightarrow r-2x=4\Rightarrow r=2x+4.\)
Square: \(x^2+y^2=(2x+4)^2=4x^2+16x+16.\)
\(y^2-3x^2-16x-16=0.\) Since the \(x^2\) and \(y^2\) terms have opposite signs, this is a hyperbola. (Eccentricity \(e=2>1.\))
The height of a Ferris wheel seat above the ground is modelled by \(h(t)=20\cos\!\left(\frac{\pi}{15}(t-15)\right)+22\), where \(t\) is seconds. Find (a) the amplitude, (b) the period, (c) the minimum height, and (d) the first time \(t>0\) the seat is at maximum height. (a) Amplitude \(=|A|=20\) m. (b) Period \(=\frac{2\pi}{\pi/15}=30\) seconds. (c) Minimum height \(=22-20=2\) m. (d) Maximum when \(\cos(\cdots)=1\): \(\frac{\pi}{15}(t-15)=0\Rightarrow t=15\) s.Show solution
Functions & Their Parameters
Unit 4 is a capstone: functions as dynamic models where parameters control behaviour. This includes inverse functions, composite functions, implicit functions, and parametric curves — all viewed as tools for modelling the real world.
Composition: \((f\circ g)(x)=f(g(x)).\) Domain of \(f\circ g\) = domain of \(g\) restricted so that \(g(x)\) is in the domain of \(f.\) Inverse: \(f^{-1}\) exists iff \(f\) is one-to-one (passes horizontal line test). Graph of \(f^{-1}\) is the reflection of \(f\) over \(y=x.\)
A piecewise function uses different rules for different intervals of the domain. Check continuity at breakpoints: left- and right-hand values must agree. Implicit function: defined by a relation \(F(x,y)=0\) rather than explicitly as \(y=f(x)\). A circle \(x^2+y^2=r^2\) is implicit.
A parametric curve is given by \(x=f(t),\;y=g(t)\) for parameter \(t.\) Eliminating the parameter: solve one equation for \(t\), substitute into the other. Or use a Pythagorean identity when \(x=a\cos t,\;y=b\sin t\).
Model Equation Growth type Linear \(f(x)=mx+b\) Constant AROC (additive) Quadratic \(f(x)=ax^2+bx+c\) AROC increases/decreases linearly Exponential \(f(x)=ab^x\) Constant percentage change (multiplicative) Logarithmic \(f(x)=a\log_b x+c\) Inverse of exponential; slow growth Power \(f(x)=ax^n\) Scales with \(x^n\) Sinusoidal \(f(x)=A\sin(Bx+C)+D\) Periodic / cyclic
Problem. Find \(f^{-1}(x)\) for \(f(x)=\dfrac{3x-1}{x+2}\), then verify.
Set \(y=\dfrac{3x-1}{x+2}\). Solve for \(x\): \(y(x+2)=3x-1\Rightarrow yx+2y=3x-1\Rightarrow x(y-3)=-1-2y.\)
\(x=\dfrac{-1-2y}{y-3}=\dfrac{2y+1}{3-y}.\) So \(f^{-1}(x)=\dfrac{2x+1}{3-x}.\)
Verify: \(f(f^{-1}(x))=\dfrac{3\cdot\frac{2x+1}{3-x}-1}{\frac{2x+1}{3-x}+2}=\dfrac{\frac{6x+3-3+x}{3-x}}{\frac{2x+1+6-2x}{3-x}}=\dfrac{7x}{7}=x.\;\checkmark\)
Problem. A projectile: \(x=30t,\;y=40t-5t^2\). (a) Eliminate \(t\) to find the Cartesian equation. (b) Find the maximum height.
(a) \(t=x/30\). Then \(y=40\!\left(\dfrac{x}{30}\right)-5\!\left(\dfrac{x}{30}\right)^2=\dfrac{4x}{3}-\dfrac{x^2}{180}.\)
(b) Complete the square or use vertex formula: \(t_{\max}=40/(2\cdot5)=4.\) \(y_{\max}=40(4)-5(16)=160-80=80\) m.
Problem. Data: \(x\): 0, 1, 2, 3; \(y\): 3, 6, 12, 24. Identify the model and write its equation.
Check ratios: \(6/3=2,\;12/6=2,\;24/12=2\) — constant ratio → exponential model.
\(y=3\cdot2^x.\) Verify: \(x=0: 3\checkmark;\;x=3: 24\checkmark.\)
A parametric curve is defined by \(x(t)=2\cos t\) and \(y(t)=3\sin t.\) (a) Eliminate the parameter to find the Cartesian equation. (b) Identify the curve and state the domain of \(x.\) (a) \(\cos t=\frac{x}{2},\;\sin t=\frac{y}{3}.\) Use \(\cos^2 t+\sin^2 t=1:\) \(\frac{x^2}{4}+\frac{y^2}{9}=1.\) (b) This is an ellipse with semi-major axis \(3\) (along \(y\)) and semi-minor axis \(2\) (along \(x\)). Domain: \(-2\le x\le2.\)Show solution
Practice Set
State the amplitude, period, phase shift, and midline of \(f(x)=3\sin(2x-\pi)+1\).
Amplitude 3, period \(\pi\), phase shift \(\pi/2\) right, midline \(y=1.\)
A population doubles every 12 years, starting from \(P_0=500\). When does it reach 4000?
\(P(t)=500\cdot2^{t/12}=4000\Rightarrow 2^{t/12}=8=2^3\Rightarrow t=36\) years.
Find all zeros of \(p(x)=x^3-2x^2-x+2.\)
\(p(1)=0\Rightarrow(x-1)(x-2)(x+1).\) Zeros: \(x=1,2,-1.\)
Solve \(\ln(2x+1)=3\).
\(2x+1=e^3\Rightarrow x=\dfrac{e^3-1}{2}\approx9.04.\)
The half-life of Carbon-14 is 5730 years. (a) Write \(A(t)\). (b) How much of 200 g remains after 11 460 years? (c) When is only 10% left?
(a) \(A(t)=200\cdot\left(\tfrac{1}{2}\right)^{t/5730}.\)
(b) \(t=2\times5730\): \(A=200\cdot(1/4)=50\) g.
(c) \((1/2)^{t/5730}=0.1\Rightarrow t=5730\cdot\dfrac{\ln0.1}{\ln0.5}\approx19{,}035\) years.
A particle moves along \(x(t)=2t+1,\;y(t)=t^2-3.\) Find the Cartesian equation of the path.
\(t=\dfrac{x-1}{2}.\quad y=\left(\dfrac{x-1}{2}\right)^2-3=\dfrac{(x-1)^2}{4}-3.\) A parabola.
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