Core Concepts & Key Formulas

📐 Functions

Domain, range, composition, inverses, even/odd symmetry, transformations.

(f∘g)(x) = f(g(x))

🔢 Polynomials

Factor Theorem, Remainder Theorem, Rational Root Test, end behavior.

f(a)=0 ⟺ (x−a) is a factor

📊 Rational Functions

Vertical asymptotes: denom = 0. Horizontal: compare degrees. Holes: common factors.

HA: y = leading coeff ratio

🔬 Exponential & Logarithms

Change of base, log rules, exponential equations, natural log.

log_b(xy) = log_b x + log_b y

🌊 Trigonometry

Unit circle, identities, amplitude/period/phase shift, inverse trig.

sin²θ + cos²θ = 1

∑ Sequences & Series

Arithmetic: common difference d. Geometric: common ratio r. Sum formulas.

S_n = n/2 · (a₁ + aₙ)

⭕ Conic Sections

Parabola, ellipse, circle, hyperbola — standard forms and key features.

(x−h)²/a² + (y−k)²/b² = 1

→ Vectors & Intro Limits

Magnitude, dot product, unit vector; limit intuition, continuity.

|v| = √(vₓ² + vᵧ²)
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01
Functions
★★☆
Domain · Inverse

Concept: The inverse of a function swaps input and output. To find the domain of \(f^{-1}(x)\), note it equals the range of \(f(x)\).

If \(f(x) = \sqrt{x - 3} + 2\), which of the following is the correct expression for \(f^{-1}(x)\)?

A\((x-2)^2 + 3,\; x \ge 2\)
B\((x+2)^2 - 3,\; x \ge 0\)
C\(\sqrt{x+2} - 3,\; x \ge -2\)
D\((x-2)^2 - 3,\; x \ge 2\)
✓ Solution
Let \(y = \sqrt{x-3}+2\). Swap \(x\) and \(y\): \(x = \sqrt{y-3}+2\).
\(x - 2 = \sqrt{y-3}\) → \((x-2)^2 = y-3\) → \(y = (x-2)^2 + 3\).
Domain of \(f^{-1}\): range of \(f\) = \([2,\infty)\), so \(x \ge 2\).
Answer: A
02
Functions
★★☆
Composition

Concept: \((f \circ g)(x) = f(g(x))\). Substitute \(g(x)\) into \(f\).

If \(f(x) = 2x^2 - 1\) and \(g(x) = x + 3\), what is \((f \circ g)(x)\)?

A\(2x^2 + 12x + 17\)
B\(2x^2 + 6x + 8\)
C\(2x^2 + 5\)
D\(4x^2 + 12x + 8\)
✓ Solution
\(f(g(x)) = f(x+3) = 2(x+3)^2 - 1 = 2(x^2+6x+9)-1 = 2x^2+12x+18-1 = 2x^2+12x+17\).
Answer: A
03
Functions
★☆☆
Even / Odd

Concept: Even: \(f(-x)=f(x)\) (y-axis symmetry). Odd: \(f(-x)=-f(x)\) (origin symmetry).

Which function is odd?

A\(f(x) = x^4 - x^2\)
B\(f(x) = x^3 - x\)
C\(f(x) = x^2 + 1\)
D\(f(x) = |x| + 2\)
✓ Solution
Test B: \(f(-x) = (-x)^3 - (-x) = -x^3 + x = -(x^3-x) = -f(x)\). ✓ Odd.
A: \(f(-x)=x^4-x^2=f(x)\) → even. C and D: both even.
Answer: B
04
Polynomials
★★☆
Remainder Theorem

Concept: The Remainder Theorem states: when \(p(x)\) is divided by \((x-c)\), the remainder is \(p(c)\).

What is the remainder when \(p(x) = 2x^3 - x^2 + 3x - 1\) is divided by \((x - 2)\)?

A15
B17
C11
D9
✓ Solution
By the Remainder Theorem, the remainder = \(p(2)\).
\(p(2) = 2(8) - (4) + 3(2) - 1 = 16 - 4 + 6 - 1 = 17\).
Answer: B
05
Polynomials
★★★
Factor Theorem · Roots

Concept: Factor Theorem: \((x-a)\) is a factor iff \(f(a)=0\). Rational Root Test: possible rational roots = \(\pm\frac{p}{q}\) where \(p\mid\text{constant},\; q\mid\text{leading coeff}\).

Given \(f(x) = x^3 - 6x^2 + 11x - 6\), which set lists all real roots?

A\(\{1, 2, 3\}\)
B\(\{1, -2, 3\}\)
C\(\{-1, 2, -3\}\)
D\(\{1, 2, -3\}\)
✓ Solution
Try \(x=1\): \(1-6+11-6=0\) ✓. Factor: \((x-1)(x^2-5x+6)=(x-1)(x-2)(x-3)\).
Roots: \(x=1,2,3\).
Answer: A
06
Rational Functions
★★☆
Asymptotes

Concept: Vertical asymptotes occur where the denominator = 0 (and numerator ≠ 0). Horizontal asymptote: if deg(num) = deg(den), HA = ratio of leading coefficients.

For \(f(x) = \dfrac{3x^2 - 1}{2x^2 + 5}\), identify the horizontal asymptote.

A\(y = 0\)
B\(y = \tfrac{3}{2}\)
C\(y = \tfrac{2}{3}\)
DNo horizontal asymptote
✓ Solution
Degrees of numerator and denominator are both 2 (equal). HA = leading coefficient of numerator ÷ leading coefficient of denominator = \(\frac{3}{2}\).
Answer: B
07
Rational Functions
★★★
Holes · Vertical Asymptotes

Concept: A hole occurs when a factor cancels from both numerator and denominator. A vertical asymptote occurs at the remaining denominator zeros.

\(f(x) = \dfrac{x^2 - 4}{x^2 - x - 2}\). Does the graph have a hole, a vertical asymptote, or both?

AHole at \(x=2\), VA at \(x=-1\)
BVA at \(x=2\) and \(x=-1\)
CHole at \(x=-1\), VA at \(x=2\)
DHole at \(x=2\) only
✓ Solution
Factor: \(\frac{(x-2)(x+2)}{(x-2)(x+1)}\). The factor \((x-2)\) cancels → hole at \(x=2\).
Remaining denominator: \((x+1)=0\) → VA at \(x=-1\).
Answer: A
08
Exponential & Logarithms
★★☆
Logarithm Properties

Key rules: \(\log_b(MN)=\log_b M+\log_b N\), \(\log_b\!\left(\frac{M}{N}\right)=\log_b M - \log_b N\), \(\log_b(M^k)=k\log_b M\).

Simplify: \(\log_2 96 - \log_2 3\)

A4
B5
C6
D8
✓ Solution
\(\log_2 96 - \log_2 3 = \log_2\!\left(\frac{96}{3}\right) = \log_2 32 = \log_2 2^5 = 5\).
Answer: B
09
Exponential & Logarithms
★★★
Exponential Equations

Concept: To solve \(a^{f(x)}=a^{g(x)}\), set exponents equal. If bases differ, take logarithm of both sides.

Solve for \(x\): \(4^{x+1} = 8^{x-1}\)

A\(x = 3\)
B\(x = 5\)
C\(x = 7\)
D\(x = 9\)
✓ Solution
Write both as powers of 2: \(4^{x+1}=2^{2(x+1)}\), \(8^{x-1}=2^{3(x-1)}\).
Set exponents equal: \(2(x+1)=3(x-1)\) → \(2x+2=3x-3\) → \(x=5\).
Check: \(4^6=4096\), \(8^4=4096\). ✓
Answer: B
10
Exponential & Logarithms
★★☆
Change of Base

Formula: \(\log_b a = \dfrac{\ln a}{\ln b} = \dfrac{\log a}{\log b}\).

Using \(\log 2 \approx 0.301\) and \(\log 3 \approx 0.477\), evaluate \(\log_9 8\) to the nearest hundredth.

A0.95
B0.85
C0.94
D1.05
✓ Solution
\(\log_9 8 = \dfrac{\log 8}{\log 9} = \dfrac{3\log 2}{2\log 3} = \dfrac{3(0.301)}{2(0.477)} = \dfrac{0.903}{0.954} \approx 0.947 \approx 0.95\).
Answer: A
11
Trigonometry
★★☆
Unit Circle · Exact Values

Memorize: \(\sin 30°=\frac{1}{2}\), \(\cos 30°=\frac{\sqrt{3}}{2}\), \(\sin 45°=\frac{\sqrt{2}}{2}\), \(\sin 60°=\frac{\sqrt{3}}{2}\).

What is the exact value of \(\sin\!\left(\dfrac{7\pi}{6}\right)\)?

A\(-\dfrac{1}{2}\)
B\(\dfrac{1}{2}\)
C\(-\dfrac{\sqrt{3}}{2}\)
D\(\dfrac{\sqrt{2}}{2}\)
✓ Solution
\(\frac{7\pi}{6} = \pi + \frac{\pi}{6}\) → Quadrant III. Reference angle = \(\frac{\pi}{6}\) (30°).
In Q III, sin is negative: \(\sin\!\left(\frac{7\pi}{6}\right) = -\sin\!\left(\frac{\pi}{6}\right) = -\frac{1}{2}\).
Answer: A
12
Trigonometry
★★★
Amplitude · Period · Phase Shift

For \(y = A\sin(Bx - C) + D\): Amplitude = \(|A|\), Period = \(\frac{2\pi}{|B|}\), Phase Shift = \(\frac{C}{B}\), Vertical Shift = \(D\).

For \(y = 3\sin\!\left(2x - \dfrac{\pi}{3}\right) + 1\), what is the phase shift?

A\(\dfrac{\pi}{6}\) to the right
B\(\dfrac{\pi}{3}\) to the right
C\(\dfrac{\pi}{3}\) to the left
D\(\dfrac{\pi}{6}\) to the left
✓ Solution
Phase shift = \(\dfrac{C}{B} = \dfrac{\pi/3}{2} = \dfrac{\pi}{6}\).
Since \(C > 0\) in \(Bx - C\), the graph shifts to the right.
Answer: A
13
Trigonometry
★★★
Pythagorean Identities

Identities: \(\sin^2\theta+\cos^2\theta=1\), \(1+\tan^2\theta=\sec^2\theta\), \(1+\cot^2\theta=\csc^2\theta\).

If \(\cos\theta = -\dfrac{3}{5}\) and \(\theta\) is in Quadrant II, what is \(\tan\theta\)?

A\(\dfrac{4}{3}\)
B\(-\dfrac{4}{3}\)
C\(-\dfrac{3}{4}\)
D\(\dfrac{3}{4}\)
✓ Solution
\(\sin^2\theta = 1 - \cos^2\theta = 1 - \frac{9}{25} = \frac{16}{25}\), so \(\sin\theta = \frac{4}{5}\) (positive in Q II).
\(\tan\theta = \dfrac{\sin\theta}{\cos\theta} = \dfrac{4/5}{-3/5} = -\dfrac{4}{3}\).
Answer: B
14
Trigonometry
★★★
Double Angle Formula

Formula: \(\sin(2\theta) = 2\sin\theta\cos\theta\), \(\cos(2\theta) = \cos^2\theta - \sin^2\theta\).

If \(\sin\theta = \dfrac{5}{13}\) and \(\theta\) is in Quadrant I, find \(\sin(2\theta)\).

A\(\dfrac{60}{169}\)
B\(\dfrac{120}{169}\)
C\(\dfrac{119}{169}\)
D\(\dfrac{10}{13}\)
✓ Solution
\(\cos\theta = \sqrt{1 - 25/169} = \sqrt{144/169} = \frac{12}{13}\) (Q I, positive).
\(\sin(2\theta) = 2\cdot\frac{5}{13}\cdot\frac{12}{13} = \frac{120}{169}\).
Answer: B
15
Sequences & Series
★★☆
Arithmetic Series

Formula: \(S_n = \dfrac{n}{2}(a_1 + a_n)\) or \(S_n = \dfrac{n}{2}[2a_1 + (n-1)d]\).

Find the sum of the first 20 terms of the arithmetic sequence: \(3, 7, 11, 15, \ldots\)

A820
B780
C840
D800
✓ Solution
\(a_1=3,\; d=4,\; n=20\).
\(S_{20} = \frac{20}{2}[2(3)+(19)(4)] = 10[6+76] = 10(82) = 820\).
Answer: A
16
Sequences & Series
★★★
Infinite Geometric Series

Formula: \(S_\infty = \dfrac{a_1}{1-r}\), valid when \(|r| < 1\).

Find the sum of the infinite geometric series: \(16 + 8 + 4 + 2 + \cdots\)

A24
B32
C28
D30
✓ Solution
\(a_1=16,\; r=\frac{1}{2}\). Since \(|r|=\frac{1}{2}<1\), the series converges.
\(S = \dfrac{16}{1-\frac{1}{2}} = \dfrac{16}{\frac{1}{2}} = 32\).
Answer: B
17
Conic Sections
★★☆
Ellipse · Standard Form

Standard form of ellipse: \(\dfrac{(x-h)^2}{a^2}+\dfrac{(y-k)^2}{b^2}=1\). If \(a>b\), major axis is horizontal; if \(b>a\), vertical.

For \(\dfrac{(x-1)^2}{25}+\dfrac{(y+2)^2}{9}=1\), what are the coordinates of the center and the length of the major axis?

ACenter \((1,-2)\); major axis length \(10\)
BCenter \((-1,2)\); major axis length \(5\)
CCenter \((1,-2)\); major axis length \(5\)
DCenter \((1,2)\); major axis length \(10\)
✓ Solution
Center: \((h,k) = (1,-2)\). \(a^2=25 \Rightarrow a=5\); \(b^2=9 \Rightarrow b=3\).
Since \(a > b\), major axis is horizontal with length \(2a = 10\).
Answer: A
18
Conic Sections
★★★
Parabola · Focus & Directrix

Standard form: \((x-h)^2 = 4p(y-k)\). Focus at \((h, k+p)\). Directrix: \(y = k-p\).

For the parabola \((x-2)^2 = 8(y+1)\), what is the focus?

A\((2,\; 1)\)
B\((2,\; -3)\)
C\((2,\; 3)\)
D\((0,\; 1)\)
✓ Solution
\(4p = 8 \Rightarrow p = 2\). Vertex: \((h,k)=(2,-1)\).
Focus: \((h,\; k+p) = (2,\; -1+2) = (2,\; 1)\).
Answer: A
19
Vectors
★★☆
Dot Product · Angle

Formula: \(\vec{u}\cdot\vec{v} = |\vec{u}||\vec{v}|\cos\theta\), so \(\cos\theta = \dfrac{\vec{u}\cdot\vec{v}}{|\vec{u}||\vec{v}|}\).

Let \(\vec{u} = \langle 3, 4 \rangle\) and \(\vec{v} = \langle 0, 1 \rangle\). What is the angle between \(\vec{u}\) and \(\vec{v}\)?

A\(\arccos\!\left(\dfrac{4}{5}\right)\)
B\(\arccos\!\left(\dfrac{3}{5}\right)\)
C\(\arccos\!\left(\dfrac{1}{5}\right)\)
D\(90°\)
✓ Solution
\(\vec{u}\cdot\vec{v} = (3)(0)+(4)(1) = 4\).
\(|\vec{u}|=\sqrt{9+16}=5\), \(|\vec{v}|=1\).
\(\cos\theta = \dfrac{4}{5\cdot 1} = \dfrac{4}{5}\), so \(\theta = \arccos\!\left(\dfrac{4}{5}\right)\).
Answer: A
20
Intro to Limits
★★★
Limits · Factoring

Concept: When direct substitution gives \(\frac{0}{0}\), factor and cancel.

Evaluate: \(\displaystyle\lim_{x \to 3}\;\dfrac{x^2 - 9}{x - 3}\)

A3
B0
C6
Dundefined
✓ Solution
Factor: \(\dfrac{x^2-9}{x-3} = \dfrac{(x-3)(x+3)}{x-3} = x+3\) for \(x\neq 3\).
\(\lim_{x\to 3}(x+3) = 3+3 = 6\).
Answer: C
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