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ACT 2 Math Level 1

Provider · ACT Inc. Format · Subject Test · Advanced Chapters · 8 Level · Pre-calculus & beyond

Test format & strategy

🗻 ACT Math Advanced — Score Range 28–36 Topics

Reporting Category	% of Test	Key Topics
Advanced Number Theory	~8%	Sequences, integer properties, modular arithmetic, complex numbers
Advanced Algebra	~15%	Polynomial systems, rational equations, logarithms, exponentials
Advanced Functions	~15%	Piecewise, recursive, parametric, function composition chains
Advanced Trigonometry	~10%	Radians, unit circle, identities, inverse trig, law of sines/cosines
Pre-Calculus Concepts	~12%	Limits (informal), sequences/series, conic sections, polar basics
Advanced Coordinate Geometry	~10%	Conic equations, transformations, distance in 3D
Advanced Plane Geometry	~10%	Circle theorems, similarity, 3D solids, geometric proof
Statistics, Counting & Probability	~10%	Permutations, combinations, expected value, normal dist., regression

ACT Math 2 is an advanced subject-level mathematics course for students who have completed standard ACT Math preparation and wish to extend their skills into pre-calculus territory — the topics that appear in the hardest ACT questions and on ACT subject tests.

What to expect

This course covers the top-difficulty ACT Math topics: advanced algebra, pre-calculus functions, matrices, advanced trigonometry, and non-routine geometry. It is also ideal preparation for the ACT Subject Test in Mathematics, which tests material through pre-calculus.

Prerequisite

ACT Math (standard course)

Level

Pre-calculus and beyond

Ideal for

Score 28+ target students

Calculator

permitted throughout

⏱ ACT Subject Math 1 Pacing & Time Management Strategy

Pacing Baseline: 50 questions in 60 minutes = 72 seconds per question. There is no guessing penalty — every unanswered question is a lost opportunity.
Three-Pass Strategy: Pass 1 (Questions 1–25, ~20 min): bank easy points rapidly. Pass 2 (Questions 26–45, ~25 min): methodical solving of mid-level algebra, trig, and coordinate geometry. Pass 3 (Questions 46–50 & review, ~15 min): tackle non-routine multi-step items and ensure all bubble answers are filled.
Approved Calculator Fluency: Graphing calculators are permitted throughout. Store formulas, use matrix inverse/determinant functions, and verify complex number arithmetic $a + bi$ directly.

▲ The 4 Core Commandments for Scoring 30+ on ACT Advanced Math

1. Never Leave Blanks: Unlike old SAT Subject tests, there is zero penalty for incorrect answers. If time is running short with 2 minutes remaining, pick a favorite letter (e.g., C or D) and fill every remaining bubble.
2. Identify Conic Signatures at a Glance: In $Ax^2 + Bxy + Cy^2 + Dx + Ey + F = 0$ , same signs on $x^2$ and $y^2$ with $A = C$ indicates a circle; same signs with $A \neq C$ indicates an ellipse; opposite signs indicates a hyperbola; only one squared term indicates a parabola.
3. Exploit Backsolving & Number Substitution: When an algebraic equation has complex fractions, logs, or radicals, testing answer choice C (the median value) first often eliminates 2–3 options in under 30 seconds.
4. Memorize the Critical Edge-Case Formulas: Circle inscribed angle $=\frac{1}{2}\text{central}$ , Heron's formula $A = \sqrt{s(s-a)(s-b)(s-c)}$ , sum of infinite geometric series $S_\infty = \frac{a}{1-r}$ , and vector dot product $\vec{u}\cdot\vec{v} = u_1 v_1 + u_2 v_2$ .

ACT Math 2 (Advanced Math) topic distribution across primary subject domains. Advanced Algebra, Functions, and Pre-Calculus comprise over 40% of the tested curriculum, supported by Advanced Trigonometry, Coordinate Geometry, and Plane Geometry.

The ACT Subject Test in Mathematics consists of 50 questions in 60 minutes. If a student spends an average of 55 seconds on each of the first 30 questions, how many seconds per question are available for the remaining 20 questions?

\text{Time spent on first 30 questions} = 30 \times 55 = 1650\text{ seconds}

(27.5 minutes).

\text{Total time available} = 60 \times 60 = 3600\text{ seconds}

\text{Remaining time} = 3600 - 1650 = 1950\text{ seconds}

. For the final 20 questions:

\frac{1950}{20} = 97.5\text{ seconds per question}

📝 Chapter Quiz

Questions Level

Advanced Number Theory

🗻 Advanced Number Theory ~8% · Integer properties · Sequences & series · Modular arithmetic · Complex number operations

Number theory questions on hard ACT items test divisibility rules, prime factorisation, GCD and LCM, modular arithmetic, and properties of integers.

GCD

greatest common divisor

LCM

\text{LCM}(a,b)=\dfrac{ab}{\gcd(a,b)}

Modular arithmetic

a \equiv r \pmod{n}

Prime factorisation

unique decomposition theorem

Terms of an arithmetic progression on the number line with initial term

a_1 = 5

and common difference

d = 5

. The sum of the first 5 terms is

S_5 = \frac{5}{2}(5 + 25) = 75

Worked example 1.A

Find LCM(12, 18). $12=2^2\cdot 3,\quad 18=2\cdot 3^2.\quad \text{LCM}=2^2\cdot 3^2=36.$

Complex numbers revisited

On advanced ACT items, complex numbers appear in equation-solving contexts:

x^2+4=0 \Rightarrow x=\pm 2i

. The modulus of

a+bi

\sqrt{a^2+b^2}

● Sequences & Modular Arithmetic

An arithmetic sequence: $a_n = a_1 + (n-1)d$ . A geometric sequence: $a_n = a_1 r^{n-1}$ . Modular arithmetic: $a \equiv b \pmod{m}$ means $m \mid (a-b)$ . Useful for finding last digits (mod 10) or remainders.

Worked example — Last digit of a power

What is the units digit of $7^{2026}$ ?

Powers of 7 cycle: $7^1=7, 7^2=49, 7^3=343, 7^4=2401, 7^5=\ldots7$ . Cycle length 4. $2026 \div 4$ remainder $2$ . So units digit = same as $7^2$ = 9.

▲ ACT Math 2 Traps & Pitfalls — Number Theory & Complex Numbers

Powers of $i$ Cycle Modulo 4: $i^1 = i$ , $i^2 = -1$ , $i^3 = -i$ , and $i^4 = 1$ . To evaluate $i^k$ , find $k \pmod 4$ : remainder 1 gives $i$ , remainder 2 gives $-1$ , remainder 3 gives $-i$ , and remainder 0 gives $1$ .
Complex Fraction Rationalization: To simplify $\dfrac{a+bi}{c+di}$ , multiply both numerator and denominator by the complex conjugate $c - di$ . The denominator becomes the real number $c^2 + d^2$ .
Infinite Geometric Series Convergence: The sum formula $S_\infty = \dfrac{a_1}{1 - r}$ is strictly valid ONLY when $|r| < 1$ . If $|r| \ge 1$ , the series diverges and has no finite sum.
GCD vs. LCM Relationship: Remember the product identity $\text{GCD}(a,b) \times \text{LCM}(a,b) = a \times b$ . If a question asks for the LCM given prime factorizations, take the highest power of each distinct prime; for GCD, take the lowest common power.

The sum of an arithmetic sequence is 150, with first term 5 and last term 25. How many terms are there?

S = \dfrac{n}{2}(a_1+a_n) = \dfrac{n}{2}(30) = 150 \Rightarrow n = 10

Express the complex number

z = \dfrac{5 + 2i}{3 - 4i}

in standard form

a + bi

, and determine its modulus

|z|

(a) Standard form $a + bi$ : Multiply numerator and denominator by the complex conjugate $3 + 4i$ : $z = \dfrac{(5 + 2i)(3 + 4i)}{(3 - 4i)(3 + 4i)} = \dfrac{15 + 20i + 6i + 8i^2}{3^2 - (4i)^2} = \dfrac{15 + 26i - 8}{9 - (-16)} = \dfrac{7 + 26i}{25} = \dfrac{7}{25} + \dfrac{26}{25}i.$

(b) Modulus $|z|$ : Using the modulus quotient property $|z| = \dfrac{|5+2i|}{|3-4i|}$ : $|5+2i| = \sqrt{5^2 + 2^2} = \sqrt{29}$ , and $|3-4i| = \sqrt{3^2 + (-4)^2} = \sqrt{25} = 5$ . Thus $|z| = \dfrac{\sqrt{29}}{5}$ (or $\approx 1.077$ ).

📝 Chapter Quiz

Questions Level

Advanced Algebra & Systems

🗻 Advanced Algebra ~15% · Polynomial systems · Rational equations · Logarithms & exponentials · Absolute value in systems

Advanced ACT algebra includes matrices, non-linear systems, and applications of polynomial and rational functions.

Matrix addition

add corresponding entries

Matrix product

AB

row × column dot products

2×2 determinant

ad-bc

System via matrices

A\mathbf{x}=\mathbf{b}

Worked example 2.A

Solve the non-linear system: $y=x^2$ and $y=x+2$ . $x^2=x+2 \implies x^2-x-2=0 \implies (x-2)(x+1)=0 \implies x=2 \text{ or } x=-1.$

Graphical solution of the non-linear system

y = x^2

and

y = x + 2

. The line intersects the upward parabola at precisely two coordinate points:

(-1, 1)

and

(2, 4)

● Logarithms & Exponential Equations

Key log laws: $\log_b(xy) = \log_b x + \log_b y$ ; $\log_b(x/y) = \log_b x - \log_b y$ ; $\log_b(x^n) = n\log_b x$ . To solve $a^{f(x)} = b$ : take $\log$ of both sides. Change of base: $\log_b x = \dfrac{\ln x}{\ln b}$ .

Worked example — Solving a logarithmic equation

Solve $\log_2(x+3) + \log_2(x-1) = 5$ .

$\log_2[(x+3)(x-1)] = 5 \Rightarrow (x+3)(x-1) = 32 \Rightarrow x^2+2x-35 = 0 \Rightarrow (x+7)(x-5) = 0$ .

$x = 5$ (reject $x=-7$ since $x-1 > 0$ required). Answer: $x=5$ .

▲ ACT Math 2 Traps & Pitfalls — Matrices & Advanced Systems

Matrix Multiplication Order & Dimensions: To compute $AB$ , the number of columns in $A$ (inner dimension) must equal the number of rows in $B$ : $(m \times k) \times (k \times n) = (m \times n)$ . Furthermore, $AB \neq BA$ in general — matrix multiplication is non-commutative!
Determinant & Singularity: For $A = \begin{pmatrix} a & b \\ c & d \end{pmatrix}$ , $\det(A) = ad - bc$ . The matrix inverse $A^{-1}$ exists if and only if $\det(A) \neq 0$ . If $\det(A) = 0$ , the corresponding linear system has either no unique solution (inconsistent or dependent).
Extraneous Solutions in Rational Equations: When clearing denominators by multiplying by a common multiple, values that make any original denominator equal to zero are extraneous and must be discarded.
Logarithm Bases & Power Properties: Remember $\log_b(A + B) \neq \log_b A + \log_b B$ . The addition rule applies only inside the logarithm: $\log_b(AB) = \log_b A + \log_b B$ .

Solve for

x

3^{x+1} = 5^x

. Give your answer in terms of logarithms.

Take

\ln

(x+1)\ln 3 = x\ln 5 \Rightarrow x\ln 3 + \ln 3 = x\ln 5 \Rightarrow x = \dfrac{\ln 3}{\ln 5 - \ln 3} = \dfrac{\ln 3}{\ln(5/3)}

Let matrices

A = \begin{pmatrix} 3 & -1 \\ 2 & 4 \end{pmatrix}

and

B = \begin{pmatrix} 1 & 5 \\ 0 & -2 \end{pmatrix}

. (a) Compute the matrix product

AB

. (b) Calculate

\det(A)

\det(B)

, and verify that

\det(AB) = \det(A)\det(B)

(a) Matrix product $AB$ : $AB = \begin{pmatrix} (3)(1) + (-1)(0) & (3)(5) + (-1)(-2) \\ (2)(1) + (4)(0) & (2)(5) + (4)(-2) \end{pmatrix} = \begin{pmatrix} 3 & 15 + 2 \\ 2 & 10 - 8 \end{pmatrix} = \begin{pmatrix} 3 & 17 \\ 2 & 2 \end{pmatrix}.$

(b) Determinant calculation: $\det(A) = (3)(4) - (-1)(2) = 12 + 2 = 14$ . $\det(B) = (1)(-2) - (5)(0) = -2$ . Then $\det(A)\det(B) = 14(-2) = -28$ . From the product: $\det(AB) = (3)(2) - (17)(2) = 6 - 34 = -28$ . Thus $\det(AB) = \det(A)\det(B) = -28$ . $\checkmark$

📝 Chapter Quiz

Questions Level

Advanced Functions

🗻 Advanced Functions ~15% · Piecewise & recursive · Parametric · Composition chains · Inverse functions

Pre-calculus functions — rational, piecewise, exponential, and logarithmic — with composition, inverses, and detailed graph analysis.

Rational function

\dfrac{P(x)}{Q(x)}

— vertical asymptotes where

Q=0

Horizontal asymptote

compare degrees of

P

and

Q

Piecewise

different rules on different intervals

Even / odd

f(-x)=f(x)

f(-x)=-f(x)

Worked example 3.A

Find the horizontal asymptote of $f(x)=\dfrac{3x^2+1}{x^2-4}$ .

Both numerator and denominator have degree 2; ratio of leading coefficients: $y=\dfrac{3}{1}=3$ .

⊙ Explorer · Rational function asymptotesinteractive

vertical asymptote p: 1

horizontal asymptote q: 0

vertical asymptotex = 1

horizontal asymptotey = 0

$f(x)=\dfrac{2}{x-p}+q$ — the denominator vanishing at $x=p$ makes the vertical asymptote; as $x\to\pm\infty$ , $f(x)\to q$ .

● Piecewise & Recursive Functions

A piecewise function uses different rules for different intervals. A recursive function is defined by its previous terms: e.g. $a_1 = 2$ , $a_n = 3a_{n-1} - 1$ . Parametric equations express $x$ and $y$ as separate functions of a parameter $t$ .

Worked example — Parametric to Cartesian

Eliminate $t$ from $x = t+2$ , $y = t^2 - 1$ .

From $x = t+2$ : $t = x-2$ . Substitute: $y = (x-2)^2 - 1 = x^2 - 4x + 3$ . Parabola.

Graph of the rational function

f(x) = \dfrac{2x - 1}{x - 2}

. The denominator zero produces the vertical asymptote

x = 2

, while the ratio of equal leading coefficients produces the horizontal asymptote

y = 2

▲ ACT Math 2 Traps & Pitfalls — Rational, Piecewise & Inverse Functions

Holes vs. Vertical Asymptotes: If a factor cancels from both numerator and denominator, it creates a removable hole $(c, y_0)$ , NOT a vertical asymptote. A vertical asymptote occurs ONLY where the reduced denominator equals zero.
Piecewise Boundary Conditions: When evaluating piecewise functions at domain boundaries (e.g. $x = 0$ ), check the inequality symbols carefully ( $<$ vs. $\le$ ) to apply the correct rule.
Inverse Function Domain & Range Swap: The domain of $f^{-1}$ is the range of $f$ , and the range of $f^{-1}$ is the domain of $f$ . A function only has an inverse function if it is strictly one-to-one (passes the horizontal line test).
Composite Function Order: $(f \circ g)(x) = f(g(x))$ means evaluate $g(x)$ first, then substitute that output into $f$ . In general, $f(g(x)) \neq g(f(x))$ .

A function is defined piecewise:

f(x) = x^2

for

x < 0

, and

f(x) = 2x+1

for

x \geq 0

. Find

f(-3) + f(2)

f(-3) = (-3)^2 = 9

f(2) = 2(2)+1 = 5

. Sum

= 14

Let

f(x) = \dfrac{3x + 4}{x - 2}

for

x \neq 2

. (a) Determine the equations of all vertical and horizontal asymptotes of

f(x)

. (b) Derive the inverse function formula

f^{-1}(x)

(a) Asymptotes: Vertical asymptote: denominator zero $x - 2 = 0 \Rightarrow x = 2$ . Horizontal asymptote: degrees of numerator and denominator are both 1, so $y = \dfrac{3}{1} = 3$ .

(b) Inverse function: Set $y = \dfrac{3x + 4}{x - 2}$ . Solve for $x$ : $y(x - 2) = 3x + 4 \implies xy - 2y = 3x + 4 \implies xy - 3x = 2y + 4 \implies x(y - 3) = 2y + 4.$ Thus $x = \dfrac{2y + 4}{y - 3}$ . Swapping variables: $f^{-1}(x) = \dfrac{2x + 4}{x - 3}$ for $x \neq 3$ .

📝 Chapter Quiz

Questions Level

Advanced Trigonometry

🗻 Advanced Trigonometry ~10% · Radians & unit circle · Trig identities · Law of sines/cosines · Inverse trig

Beyond SOHCAHTOA — the unit circle, inverse trig functions, sum/difference identities, and solving trig equations over a given interval.

Sum formula sin

\sin(A+B)=\sin A\cos B+\cos A\sin B

Sum formula cos

\cos(A+B)=\cos A\cos B-\sin A\sin B

Law of sines

\dfrac{a}{\sin A}=\dfrac{b}{\sin B}

Inverse trig

\sin^{-1},\cos^{-1},\tan^{-1}

Worked example 4.A

Find all solutions to $2\sin\theta-1=0$ on $[0,2\pi)$ . $\sin\theta=\tfrac{1}{2} \implies \theta=\tfrac{\pi}{6} \text{ or } \theta=\tfrac{5\pi}{6}.$

⊙ Explorer · Law of Cosinesinteractive

side a: 7

side b: 10

included angle C: 60

c² = a² + b² − 2ab cos C8.89

● Law of Sines, Cosines & Identities

Law of Sines: $\dfrac{a}{\sin A} = \dfrac{b}{\sin B} = \dfrac{c}{\sin C}$ . Law of Cosines: $c^2 = a^2+b^2-2ab\cos C$ . Key identities: $\sin^2\theta+\cos^2\theta=1$ ; $\sin 2\theta = 2\sin\theta\cos\theta$ ; $\cos 2\theta = \cos^2\theta - \sin^2\theta$ .

Worked example — Law of Cosines

A triangle has sides $a=7, b=10$ and included angle $C=60°$ . Find side $c$ .

$c^2 = 49 + 100 - 2(7)(10)\cos 60° = 149 - 70 = 79$ . $c = \sqrt{79} \approx 8.89$ .

Unit circle coordinates for

\theta = 120^\circ = \frac{2\pi}{3}\text{ rad}

. The terminal point has coordinates

(\cos 120^\circ, \sin 120^\circ) = \left(-\frac{1}{2}, \frac{\sqrt{3}}{2}\right)

with reference angle

60^\circ

in Quadrant II.

▲ ACT Math 2 Traps & Pitfalls — Trigonometry & Angle Measures

Radian vs. Degree Calculator Mode: ACT questions mix degrees (e.g. $120°$ ) and radians (e.g. $2\pi/3$ ). Check your calculator mode before every trig calculation! If you get unexpected decimals for standard values, check the mode first.
Signs by Quadrant (ASTC): In Quadrant II, sine is positive but cosine and tangent are negative. In Quadrant III, tangent is positive but sine and cosine are negative. In Quadrant IV, cosine is positive but sine and tangent are negative.
The Ambiguous Case (SSA): When using the Law of Sines with two sides and an opposite angle (SSA), always check whether the supplementary angle $180° - B$ can form a valid second triangle (sum of angles < 180°).
Double-Angle Cosine Forms: Remember the three equivalent forms: $\cos(2\theta) = \cos^2\theta - \sin^2\theta = 2\cos^2\theta - 1 = 1 - 2\sin^2\theta$ . Choosing the form matching the other terms simplifies quadratic trig equations immediately.

Prove that

\dfrac{1-\cos 2\theta}{\sin 2\theta} = \tan\theta

\dfrac{1-(1-2\sin^2\theta)}{2\sin\theta\cos\theta} = \dfrac{2\sin^2\theta}{2\sin\theta\cos\theta} = \dfrac{\sin\theta}{\cos\theta} = \tan\theta

. ✓

Solve the quadratic trigonometric equation

2\sin^2\theta - 3\sin\theta + 1 = 0

for all solutions

\theta \in [0, 2\pi)

Factoring: Factor as a quadratic in $\sin\theta$ : $(2\sin\theta - 1)(\sin\theta - 1) = 0.$

Case 1: $2\sin\theta - 1 = 0 \implies \sin\theta = \dfrac{1}{2}$ . In $[0, 2\pi)$ , sine is positive in Quadrants I and II with reference angle $\pi/6$ : $\theta = \dfrac{\pi}{6},\quad \theta = \pi - \dfrac{\pi}{6} = \dfrac{5\pi}{6}.$

Case 2: $\sin\theta - 1 = 0 \implies \sin\theta = 1$ . In $[0, 2\pi)$ , sine equals 1 at $\theta = \dfrac{\pi}{2}$ .

Complete solution set: $\left\{ \dfrac{\pi}{6},\, \dfrac{\pi}{2},\, \dfrac{5\pi}{6} \right\}$ .

📝 Chapter Quiz

Questions Level

Pre-Calculus Concepts

🗻 Pre-Calculus Concepts ~12% · Limits (informal) · Sequences & series · Conic sections · Polar coordinates

Limits, sequences, series, and polynomial behaviour give students the conceptual scaffolding for calculus — and appear on the hardest ACT questions.

Limit intuition

\lim_{x\to a}f(x)

— value

f

approaches

Geometric series

\dfrac{a_1}{1-r},\ |r|<1

Binomial theorem

(a+b)^n=\sum\binom{n}{k}a^{n-k}b^k

Parametric

x=f(t),\ y=g(t)

Worked example 5.A

Evaluate $\displaystyle\sum_{n=1}^{\infty}3\left(\tfrac{1}{2}\right)^{n-1}$ . $S=\frac{3}{1-\tfrac{1}{2}}=\frac{3}{\tfrac{1}{2}}=6.$

● Limits & Conic Sections

Informal limit: $\lim_{x \to a}f(x) = L$ means $f(x)$ approaches $L$ as $x$ approaches $a$ . Conic sections — standard forms: circle $(x-h)^2+(y-k)^2=r^2$ ; ellipse $\dfrac{x^2}{a^2}+\dfrac{y^2}{b^2}=1$ ; hyperbola $\dfrac{x^2}{a^2}-\dfrac{y^2}{b^2}=1$ ; parabola $y=ax^2+bx+c$ .

Worked example — Identify and graph a conic

Identify the conic: $4x^2 + 9y^2 = 36$ .

Divide by 36: $\dfrac{x^2}{9}+\dfrac{y^2}{4}=1$ . This is an ellipse with $a=3$ along $x$ -axis and $b=2$ along $y$ -axis, centre $(0,0)$ .

Standard ellipse

\dfrac{x^2}{9} + \dfrac{y^2}{4} = 1

centered at

(0,0)

. The major horizontal axis has length

2a = 6

(vertices

(\pm 3, 0)

) and the minor vertical axis has length

2b = 4

(co-vertices

(0, \pm 2)

▲ ACT Math 2 Traps & Pitfalls — Limits, Conics & Binomial Expansion

Indeterminate Form $0/0$ vs. Undefined: When direct substitution into $\lim_{x \to a} \dfrac{f(x)}{g(x)}$ yields $0/0$ , the limit is indeterminate — factor and cancel the common factor $(x - a)$ before re-evaluating. Do NOT write 'undefined' or 'does not exist'!
Focal Distance in Ellipses vs. Hyperbolas: For an ellipse, the focal distance satisfies $c^2 = a^2 - b^2$ (where $a > b$ ). For a hyperbola, the focal distance satisfies $c^2 = a^2 + b^2$ . Confusing these two relation signs is one of the most common mistakes on ACT Math 2.
The $k$ -th Term Index in Binomial Expansion: In the expansion of $(a+b)^n$ , the general term is $T_{k+1} = \binom{n}{k} a^{n-k} b^k$ . The 4th term corresponds to index $k = 3$ , not $k = 4$ ! Watch negative signs when expanding expressions like $(2x - 3)^n$ .
Infinite Series Ratio Magnitude: For $\sum a_1 r^{n-1}$ , if $|r| \ge 1$ , the series does not converge, so applying $S = \dfrac{a_1}{1-r}$ produces a false number.

Find

\lim_{x \to 3}\dfrac{x^2-9}{x-3}

Factor:

\dfrac{(x-3)(x+3)}{x-3} = x+3

. As

x \to 3

: limit

= 6

Find the 4th term in the binomial expansion of

(2x - 3)^6

Using the binomial theorem general term formula $T_{k+1} = \binom{n}{k} a^{n-k} b^k$ with $n = 6$ , $a = 2x$ , $b = -3$ , and $k = 3$ (for the 4th term): $T_4 = \binom{6}{3} (2x)^{6-3} (-3)^3.$ Compute the binomial coefficient: $\binom{6}{3} = \dfrac{6 \times 5 \times 4}{3 \times 2 \times 1} = 20$ . $T_4 = 20 \cdot (2x)^3 \cdot (-27) = 20 \cdot (8x^3) \cdot (-27) = -4320x^3.$

📝 Chapter Quiz

Questions Level

Advanced Coordinate Geometry

🗻 Advanced Coordinate Geometry ~10% · Conic equations · Rotations & transformations · Distance in 3D · Locus of points

Conics in standard and general form, transformations, and polar coordinates are tested on advanced ACT and ACT subject items.

Parabola (vertical)

(x-h)^2=4p(y-k)

Ellipse

\dfrac{x^2}{a^2}+\dfrac{y^2}{b^2}=1

Hyperbola

\dfrac{x^2}{a^2}-\dfrac{y^2}{b^2}=1

Polar

x=r\cos\theta,\ y=r\sin\theta

Worked example 6.A

Convert the polar point $\left(4,\dfrac{\pi}{3}\right)$ to Cartesian. $x=4\cos\tfrac{\pi}{3}=4\cdot\tfrac{1}{2}=2,\quad y=4\sin\tfrac{\pi}{3}=4\cdot\tfrac{\sqrt3}{2}=2\sqrt3.$

● Conics in Standard Form & 3D Distance

Completing the square converts a general conic to standard form. 3D distance: $d = \sqrt{(x_2-x_1)^2+(y_2-y_1)^2+(z_2-z_1)^2}$ . A rotation of axes by $\theta$ transforms $x = X\cos\theta - Y\sin\theta$ , $y = X\sin\theta + Y\cos\theta$ .

Worked example — Complete the square to find a circle

Write $x^2+y^2-6x+4y-3=0$ in standard form.

$(x^2-6x+9)+(y^2+4y+4) = 3+9+4 = 16$ . Centre $(3,-2)$ , radius $4$ .

The circle

(x - 3)^2 + (y + 2)^2 = 16

plotted on the Cartesian plane. The center is located at

(h, k) = (3, -2)

and the radius is

r = \sqrt{16} = 4

▲ ACT Math 2 Traps & Pitfalls — Coordinate Geometry & 3D Space

Center Signs & Radius in Circle Equations: In $(x - h)^2 + (y - k)^2 = r^2$ , the center is $(h, k)$ . Watch the signs carefully: $(x - 3)^2 + (y + 2)^2 = 16$ has center $(3, -2)$ , NOT $(-3, 2)$ , and its radius is $\sqrt{16} = 4$ , NOT 16.
Perpendicular Slopes: Two lines are perpendicular if and only if their slopes are negative reciprocals: $m_\perp = -1/m$ . A common pitfall is forgetting the negative sign and taking merely the reciprocal.
Standard 2D Rotations About the Origin: A counterclockwise rotation of $90°$ maps $(x, y) \to (-y, x)$ ; $180°$ maps $(x, y) \to (-x, -y)$ ; and $270°$ (or $90°$ clockwise) maps $(x, y) \to (y, -x)$ .
3D Distance Extension: Do not collapse 3D distance into 2D: $d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2 + (z_2 - z_1)^2}$ . All three coordinate differences must be squared.

Find the distance between points

A(1,2,3)

and

B(4,6,3)

d = \sqrt{(4-1)^2+(6-2)^2+(3-3)^2} = \sqrt{9+16+0} = \sqrt{25} = 5

Find the standard equation

Ax + By = C

of the line perpendicular to

3x - 4y = 12

that passes through the midpoint of the line segment joining

P(-2, 7)

and

Q(6, 1)

(a) Midpoint $M$ : $M = \left( \dfrac{-2 + 6}{2}, \dfrac{7 + 1}{2} \right) = \left( \dfrac{4}{2}, \dfrac{8}{2} \right) = (2, 4).$

(b) Perpendicular slope: Rewrite the given line in slope-intercept form: $4y = 3x - 12 \Rightarrow y = \dfrac{3}{4}x - 3$ . The slope is $m = \dfrac{3}{4}$ . The perpendicular slope is the negative reciprocal: $m_\perp = -\dfrac{4}{3}$ .

(c) Line equation: Using point-slope form with point $(2, 4)$ : $y - 4 = -\dfrac{4}{3}(x - 2) \implies 3(y - 4) = -4(x - 2) \implies 3y - 12 = -4x + 8 \implies 4x + 3y = 20.$

📝 Chapter Quiz

Questions Level

Advanced Plane Geometry

🗻 Advanced Plane Geometry ~10% · Circle theorems · Similarity & congruence proofs · 3D solids · Geometric inequalities

Non-routine geometry questions involve multiple theorems applied together — circle inscribed in a triangle, overlapping regions, and proof-based reasoning.

Inscribed angle

\tfrac{1}{2}\times

central angle

Power of a point

PA\cdot PB=PC\cdot PD

Heron's formula

\sqrt{s(s-a)(s-b)(s-c)}

Regular polygon area

\tfrac{1}{4}ns^2\cot(\pi/n)

Worked example 7.A

Find the area of a triangle with sides 7, 8, 9. $s=12,\quad A=\sqrt{12\cdot5\cdot4\cdot3}=\sqrt{720}=12\sqrt5.$

⊙ Explorer · Inscribed angle theoreminteractive

central angle ∠AOB: 140

central angle ∠AOB140.0°

inscribed angle ∠ACB70.0°

C sits anywhere on the major arc — drag the slider and the inscribed angle stays exactly half the central angle, every time.

● Circle Theorems & 3D Solids

Key circle theorems: (1) Angle at centre = 2 × angle at circumference. (2) Angles in same segment are equal. (3) Opposite angles in cyclic quadrilateral sum to 180°. (4) Tangent ⊥ radius. Surface area of sphere: $4\pi r^2$ ; volume: $\dfrac{4}{3}\pi r^3$ . Cone volume: $\dfrac{1}{3}\pi r^2 h$ .

Worked example — Circle theorem

O is the centre of a circle. Angle $AOB = 140°$ . Find the angle $ACB$ where $C$ is on the major arc.

Angle at circumference on minor arc $= 140°/2 = 70°$ . $C$ is on major arc, so $\angle ACB = 180° - 70° = 110°$ (opposite angles in cyclic quad sum to $180°$ ).

Scalene triangle with side lengths

a = 7

b = 8

, and

c = 9

. Using Heron's formula with semi-perimeter

s = 12

, the area is

A = \sqrt{12(5)(4)(3)} = 12\sqrt{5} \approx 26.83

▲ ACT Math 2 Traps & Pitfalls — Circle Theorems & 3D Geometry

Inscribed Angle vs. Central Angle: The inscribed angle $\angle ACB$ subtended by an arc is exactly half the central angle $\angle AOB$ : $\angle ACB = \frac{1}{2} \angle AOB$ . Inscribed angles subtending the same chord from opposite sides of the circle are supplementary, not equal!
Cyclic Quadrilateral Angles: A quadrilateral is cyclic (inscribed in a circle) if and only if opposite angles sum to $180°$ : $\angle A + \angle C = 180°$ and $\angle B + \angle D = 180°$ .
Vertical Height vs. Slant Height in 3D Solids: In cones and pyramids, the volume formula $V = \frac{1}{3}Bh$ uses the perpendicular vertical height $h$ , NOT the slant height $l$ . Use the Pythagorean theorem $l^2 = r^2 + h^2$ to solve for $h$ first!
Tangent-Secant Power of a Point: If a tangent segment $PT$ and a secant line $PAB$ are drawn from an external point $P$ , the power of a point theorem states $PT^2 = PA \cdot PB$ , where $PB$ is the entire external secant length.

A cone has base radius 6 cm and slant height 10 cm. Find its total surface area.

Height:

h = \sqrt{100-36} = 8

cm. TSA

= \pi r^2 + \pi r l = 36\pi + 60\pi = 96\pi \approx 301.6

^2

A quadrilateral

ABCD

is inscribed in a circle. If

\angle DAB = (2x + 15)^\circ

and the opposite angle

\angle DCB = (3x - 10)^\circ

, determine the value of

x

and calculate the measure of

\angle DAB

(a) Supplementary property: Opposite angles of any cyclic quadrilateral sum to $180^\circ$ : $(2x + 15) + (3x - 10) = 180 \implies 5x + 5 = 180 \implies 5x = 175 \implies x = 35.$

(b) Angle measure: Substitute $x = 35$ into $\angle DAB$ : $\angle DAB = 2(35) + 15 = 70 + 15 = 85^\circ.$ (Check: $\angle DCB = 3(35) - 10 = 105 - 10 = 95^\circ$ , and $85^\circ + 95^\circ = 180^\circ$ ).

📝 Chapter Quiz

Questions Level

Statistics, Counting & Probability

🗻 Statistics, Counting & Probability ~10% · Permutations & combinations · Expected value · Normal distribution · Regression & residuals

Advanced ACT Math includes combinations, permutations, compound probability, and normal distribution concepts.

Permutations

_nP_r=\dfrac{n!}{(n-r)!}

Combinations

\binom{n}{r}=\dfrac{n!}{r!(n-r)!}

Bayes (simple)

P(A|B)=\dfrac{P(B|A)P(A)}{P(B)}

Expected value

E=\sum x_i P(x_i)

Worked example 8.A

How many 3-letter arrangements (order matters) can be made from the word MATH? $_4P_3=\frac{4!}{1!}=24.$

● Counting, Expected Value & Normal Distribution

Fundamental counting principle: $m \times n$ ways for two independent events. Expected value: $E(X) = \sum x_i P(x_i)$ . Standard normal: $z = \dfrac{x-\mu}{\sigma}$ ; about 68% of data within $1\sigma$ , 95% within $2\sigma$ , 99.7% within $3\sigma$ .

Worked example — Expected value

A game pays $8 if you roll a 6 on a fair die, and you lose $2 otherwise. What is the expected gain per roll?

$E = 8 \cdot \dfrac{1}{6} + (-2) \cdot \dfrac{5}{6} = \dfrac{8-10}{6} = -\dfrac{2}{6} \approx -\$0.33$ .

The game is unfair — you lose on average 33 cents per roll.

Normal distribution model with mean

\mu = 100

and standard deviation

\sigma = 15

. The shaded region represents the interval

[\mu - \sigma, \mu + \sigma] = [85, 115]

, capturing approximately

68.2\%

of the distribution.

▲ ACT Math 2 Traps & Pitfalls — Probability, Counting & Normal Curves

Permutations vs. Combinations: Ask: 'Does order matter?' If order matters (ranking, electing distinct officers, line arrangements), use permutations $_nP_r$ . If group membership is all that counts (committees, picking toppings, drawing cards), use combinations $\binom{n}{r}$ .
The Complement Shortcut for 'At Least One': Questions asking for the probability of 'at least one' of an event occurring are almost always solved fastest using the complement: $P(\text{at least one}) = 1 - P(\text{none})$ .
Signs in Expected Value: Losses, entry fees, or payouts must include their respective positive or negative signs in $E = \sum x_i P(x_i)$ . If a ticket costs $\$5$ with a prize of $\$20$ , the net gain when winning is $+\$15$ , not $\$20$ .
Empirical Rule Sub-Intervals: Between the mean and $+1\sigma$ lies $34\%$ ; between $+1\sigma$ and $+2\sigma$ lies $13.5\%$ ; between $+2\sigma$ and $+3\sigma$ lies $2.35\%$ ; and above $+3\sigma$ lies approximately $0.15\%$ .

In a class of 30, 18 play football (F), 14 play basketball (B), and 6 play both. How many play neither sport?

|F \cup B| = 18+14-6 = 26

. Neither

= 30-26 = 4

A school committee of 4 members is to be selected from a club consisting of 6 juniors and 5 seniors. (a) In how many total ways can the 4 members be selected? (b) In how many ways can the committee be formed if it must contain exactly 2 juniors and 2 seniors?

(a) Unrestricted selection: Total students $= 6 + 5 = 11$ . Choosing 4 where order does not matter: $\binom{11}{4} = \dfrac{11 \times 10 \times 9 \times 8}{4 \times 3 \times 2 \times 1} = 11 \times 10 \times 3 = 330 \text{ ways}.$

(b) Exactly 2 juniors and 2 seniors: Select 2 juniors from 6, AND 2 seniors from 5: $\binom{6}{2} \times \binom{5}{2} = \left( \dfrac{6 \times 5}{2} \right) \times \left( \dfrac{5 \times 4}{2} \right) = 15 \times 10 = 150 \text{ ways}.$

📝 Chapter Quiz

Questions Level

Practice set

Fifty advanced ACT Math problems — pre-algebra and number theory, advanced algebra, functions, trigonometry, coordinate geometry, plane geometry, and statistics — all graded by difficulty.

Simplify $\dfrac{4!}{2!\cdot2!}$ .

Basic

\dfrac{24}{2\times2}=6.

Find $\gcd(48,\,36)$ .

Basic

\gcd(48,36)=12.

Express $0.\overline{27}$ as a fraction.

Intermediate

Let $x=0.\overline{27}$ . Then $100x=27.\overline{27}$ , so $99x=27$ .

x=\dfrac{27}{99}=\dfrac{3}{11}.

A $12\%$ annual interest rate is applied to 5000. Find the amount after 2 years with compound interest.

Intermediate

A=5000(1.12)^2=5000\times1.2544=6272.

Find the LCM of 12 and 18.

Basic

\text{LCM}(12,18)=36.

Solve $3x^2 - 5x - 2 = 0$ .

Intermediate

(3x+1)(x-2)=0 \Rightarrow x=2\text{ or }x=-\tfrac{1}{3}.

For what values of $k$ does $x^2-kx+9=0$ have equal roots?

Advanced

Discriminant $=k^2-36=0$ .

k=\pm6.

Simplify $\dfrac{x^2-x-6}{x^2-4}$ .

Intermediate

\dfrac{(x-3)(x+2)}{(x-2)(x+2)}=\dfrac{x-3}{x-2}.

Solve $\sqrt{x+6}-\sqrt{x}=1$ .

Advanced

Square both sides after isolating: $x+6=x+2\sqrt{x}+1 \Rightarrow 5=2\sqrt{x} \Rightarrow x=\tfrac{25}{4}$ .

Solve the system: $x^2-y=3$ and $x+y=1$ .

Advanced

$y=1-x$ . Then $x^2-1+x=3\Rightarrow x^2+x-4=0 \Rightarrow x=\dfrac{-1\pm\sqrt{17}}{2}$ .

If $f(x)=|2x-3|$ , find $f(-1)$ .

Basic

|2(-1)-3|=|-5|=5.

Find the vertical asymptote of $f(x)=\dfrac{3x}{x^2-4}$ .

Intermediate

x=2\text{ and }x=-2.

What is the horizontal asymptote of $f(x)=\dfrac{3x^2}{x^2+1}$ ?

Intermediate

Degrees equal; ratio of leading coefficients.

y=3.

Is $f(x)=x^4-2x^2$ even, odd, or neither?

Intermediate

$f(-x)=x^4-2x^2=f(x)$ .

\text{Even.}

Find the inverse of $f(x)=\dfrac{2x-1}{x+3}$ .

Advanced

Swap $x,y$ and solve: $x(y+3)=2y-1 \Rightarrow y(x-2)=-1-3x$ .

f^{-1}(x)=\dfrac{-3x-1}{x-2}.

Find the exact value of $\sin 135^\circ$ .

Intermediate

\sin(180^\circ-45^\circ)=\sin45^\circ=\dfrac{\sqrt{2}}{2}.

Solve $2\sin\theta-1=0$ for $0^\circ\le\theta<360^\circ$ .

Intermediate

\sin\theta=\tfrac{1}{2} \Rightarrow \theta=30^\circ\text{ or }150^\circ.

Find the period of $y=3\sin(2x)$ .

Intermediate

T=\dfrac{2\pi}{2}=\pi.

Simplify $\dfrac{\tan\theta}{\sin\theta}$ .

Advanced

\dfrac{\sin\theta/\cos\theta}{\sin\theta}=\dfrac{1}{\cos\theta}=\sec\theta.

In triangle $ABC$ , $a=8$ , $A=30^\circ$ , $B=45^\circ$ . Find $b$ .

Advanced

\dfrac{b}{\sin45^\circ}=\dfrac{8}{\sin30^\circ} \Rightarrow b=8\sqrt{2}.

Find the amplitude and period of $y=-4\cos\!\left(\dfrac{\pi x}{3}\right)$ .

Intermediate

\text{Amplitude}=4;\quad T=\dfrac{2\pi}{\pi/3}=6.

Convert $210^\circ$ to radians.

Intermediate

210\times\dfrac{\pi}{180}=\dfrac{7\pi}{6}.

Find the distance from $(1,\,-2)$ to $(5,\,1)$ .

Basic

\sqrt{16+9}=5.

Find the midpoint of the segment from $(-3,\,4)$ to $(7,\,-2)$ .

Basic

\left(\dfrac{-3+7}{2},\,\dfrac{4-2}{2}\right)=(2,\,1).

Find the equation of the ellipse with semi-axes $a=5$ and $b=3$ .

Intermediate

\dfrac{x^2}{25}+\dfrac{y^2}{9}=1.

Find the focus of the parabola $y^2=12x$ .

Advanced

$4p=12 \Rightarrow p=3$ .

\text{Focus: }(3,\,0).

Find the asymptotes of $\dfrac{x^2}{4}-\dfrac{y^2}{9}=1$ .

Advanced

y=\pm\dfrac{3}{2}x.

A line passes through $(2,3)$ and is parallel to $2x-y=5$ . Find its equation.

Intermediate

Slope $=2$ ; point-slope: $y-3=2(x-2)$ .

y=2x-1.

Find the area enclosed by $y=x^2$ and $y=x+2$ . (Use integration concept only.)

Advanced

Intersect at $x=-1$ and $x=2$ .

A=\int_{-1}^{2}(x+2-x^2)\,dx=\left[\tfrac{x^2}{2}+2x-\tfrac{x^3}{3}\right]_{-1}^{2}=\dfrac{9}{2}.

The vertices of a triangle are $(0,0)$ , $(4,0)$ , $(0,3)$ . Find the area.

Basic

A=\tfrac{1}{2}\times4\times3=6.

Find the exterior angle of a regular hexagon.

Intermediate

\dfrac{360^\circ}{6}=60^\circ.

Two parallel lines are cut by a transversal. Co-interior angles sum to what?

Basic

180^\circ.

Find the area of a regular hexagon with side 4.

Advanced

A=\dfrac{3\sqrt{3}}{2}s^2=\dfrac{3\sqrt{3}}{2}\times16=24\sqrt{3}.

Two similar triangles have sides in ratio $3:5$ . Find the ratio of their areas.

Intermediate

\left(\dfrac{3}{5}\right)^2=\dfrac{9}{25}.

A tangent from an external point touches the circle at $T$ . The distance from the external point to the centre is 13 and radius is 5. Find the tangent length.

Intermediate

PT=\sqrt{13^2-5^2}=\sqrt{144}=12.

Find the volume of a cone with base radius 6 and height 8.

Intermediate

V=\tfrac{1}{3}\pi r^2h=\tfrac{1}{3}\pi(36)(8)=96\pi.

Find the sum of interior angles of a polygon with 9 sides.

Intermediate

(9-2)\times180^\circ=1260^\circ.

The mean of 10 numbers is 15. One number is removed and the new mean is 14. Find the removed number.

Intermediate

Total was 150; new total $=9\times14=126$ .

150-126=24.

A data set has standard deviation 3 and mean 20. All values increase by 5. Find the new mean and new standard deviation.

Advanced

\text{New mean}=25;\quad\text{Std dev unchanged}=3.

A box has 3 red, 4 white, 5 blue cards. Find $P(\text{not white})$ .

Intermediate

P(\text{not white})=\dfrac{8}{12}=\dfrac{2}{3}.

Find $\lim_{x\to\infty}\dfrac{3x^2+1}{x^2-2}$ .

Advanced

\lim_{x\to\infty}\dfrac{3+1/x^2}{1-2/x^2}=3.

How many arrangements of the letters CALCULUS exist?

Advanced

8 letters: C×2, A×1, L×3, U×2.

\dfrac{8!}{2!\cdot1!\cdot3!\cdot2!}=1680.

Find the parametric point on $x=2t-1$ , $y=t^2$ at $t=3$ .

Intermediate

(2(3)-1,\;3^2)=(5,\,9).

Convert the polar point $\left(4,\,\dfrac{\pi}{3}\right)$ to Cartesian coordinates.

Intermediate

x=4\cos\dfrac{\pi}{3}=2;\quad y=4\sin\dfrac{\pi}{3}=2\sqrt{3}.

Expand $(a+b)^5$ and find the coefficient of $a^2b^3$ .

Advanced

\binom{5}{3}=10.

A geometric series has $a_1=5$ and $r=0.6$ . Find $S_\infty$ .

Intermediate

S_\infty=\dfrac{5}{1-0.6}=12.5.

Evaluate $\sum_{k=1}^{4}(2k-1)$ .

Intermediate

1+3+5+7=16.

A box has 5 red and 3 blue balls. Find the probability that 3 drawn (without replacement) are all red.

Advanced

\dfrac{5}{8}\times\dfrac{4}{7}\times\dfrac{3}{6}=\dfrac{60}{336}=\dfrac{5}{28}.

If $P(A)=0.6$ and $P(B|A)=0.4$ , find $P(A\cap B)$ .

Advanced

P(A\cap B)=P(B|A)\cdot P(A)=0.4\times0.6=0.24.

Using Bayes' theorem: $P(A)=0.3$ , $P(B|A)=0.8$ , $P(B|\overline{A})=0.2$ . Find $P(A|B)$ .

Advanced

$P(B)=0.3\times0.8+0.7\times0.2=0.38$ .

P(A|B)=\dfrac{0.3\times0.8}{0.38}=\dfrac{0.24}{0.38}\approx0.632.

✎

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ACT 2 Math Level 1

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Advanced Number Theory

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