Unit 1 · Functions & Graphs
01
MEMORY KEY: Domain = "allowed x values" · Denominator ≠ 0, Inside √ ≥ 0
Quick Example
f(x) = 1/(x−2) → domain: all real numbers except x = 2
Which of the following is the domain of f(x) = √(x − 3) / (x − 7) ?
📖 Explanation
Two conditions must be met simultaneously:
① The expression inside the square root must be non-negative: x − 3 ≥ 0, so x ≥ 3.
② The denominator cannot be zero: x − 7 ≠ 0, so x ≠ 7.
Combining both: domain = [3, 7) ∪ (7, ∞), i.e., x ≥ 3 and x ≠ 7.
① The expression inside the square root must be non-negative: x − 3 ≥ 0, so x ≥ 3.
② The denominator cannot be zero: x − 7 ≠ 0, so x ≠ 7.
Combining both: domain = [3, 7) ∪ (7, ∞), i.e., x ≥ 3 and x ≠ 7.
02
SHIFT RULE: f(x − h) + k → shift RIGHT h, UP k · f(x + h) → shift LEFT (opposite sign!)
Quick Example
y = (x−2)² + 3 → vertex (2, 3) · shifted right 2, up 3 from y = x²
The graph of y = f(x) is shifted left 4 and down 2. What is the new equation?
📖 Explanation
Horizontal shift: moving left by 4 means replacing x with x + 4 (think: opposite direction!).
Vertical shift: moving down by 2 means subtracting 2 outside: −2.
Combined: y = f(x + 4) − 2. Many students accidentally use x − 4 for left shift — watch out!
Vertical shift: moving down by 2 means subtracting 2 outside: −2.
Combined: y = f(x + 4) − 2. Many students accidentally use x − 4 for left shift — watch out!
Unit 2 · Polynomial & Rational Functions
03
END BEHAVIOR: Degree EVEN → both ends same direction · Degree ODD → opposite ends · Leading coefficient + → right end goes UP
What is the end behavior of f(x) = −3x⁴ + 2x² − x + 5?
📖 Explanation
Only the leading term matters: −3x⁴.
• Degree 4 = EVEN → both ends behave the same direction.
• Leading coefficient −3 is negative → both ends go DOWN (→ −∞).
Answer: Both ends → −∞. This is choice D.
• Degree 4 = EVEN → both ends behave the same direction.
• Leading coefficient −3 is negative → both ends go DOWN (→ −∞).
Answer: Both ends → −∞. This is choice D.
04
ASYMPTOTES: Vertical → denominator = 0 (after simplifying) · Horizontal → compare degrees of top vs bottom · Hole → factor cancels
Quick Example
(x+2)/((x+2)(x−3)) → hole at x = −2, vertical asymptote at x = 3
Given f(x) = (x² − 4) / (x − 2), which statement is correct?
📖 Explanation
Factor the numerator: x² − 4 = (x−2)(x+2).
So f(x) = (x−2)(x+2)/(x−2). The (x−2) cancels — this creates a HOLE at x = 2 (not a vertical asymptote).
After cancelling: f(x) = x + 2, x ≠ 2. Answer: B.
So f(x) = (x−2)(x+2)/(x−2). The (x−2) cancels — this creates a HOLE at x = 2 (not a vertical asymptote).
After cancelling: f(x) = x + 2, x ≠ 2. Answer: B.
Unit 3 · Exponential & Logarithmic Functions
05
LOG LAWS: log(ab) = log a + log b · log(a/b) = log a − log b · log(aⁿ) = n·log a · log_b(b) = 1 · log_b(1) = 0
Simplify: log₂(32) − log₂(4)
📖 Explanation
Use the quotient rule: log₂(32) − log₂(4) = log₂(32/4) = log₂(8).
Now ask: "2 to what power = 8?" Answer: 2³ = 8, so log₂(8) = 3.
✓ Answer: B.
Now ask: "2 to what power = 8?" Answer: 2³ = 8, so log₂(8) = 3.
✓ Answer: B.
06
SOLVE EXP EQUATIONS: Get same base → set exponents equal · OR take log of both sides · eˣ equation → use ln
Solve for x: 2^(2x−1) = 16
📖 Explanation
Rewrite 16 as a power of 2: 16 = 2⁴.
Now set exponents equal: 2x − 1 = 4.
Solve: 2x = 5 → x = 5/2.
✓ Answer: B.
Now set exponents equal: 2x − 1 = 4.
Solve: 2x = 5 → x = 5/2.
✓ Answer: B.
Unit 4 · Trigonometry
07
UNIT CIRCLE FAST: sin = y-coord · cos = x-coord · "All Students Take Calculus" → Q1: All+, Q2: Sin+, Q3: Tan+, Q4: Cos+
If sin θ = −3/5 and θ is in Quadrant III, find cos θ.
📖 Explanation
Use the Pythagorean identity: sin²θ + cos²θ = 1.
(−3/5)² + cos²θ = 1 → 9/25 + cos²θ = 1 → cos²θ = 16/25 → cosθ = ±4/5.
In Quadrant III: both sin and cos are negative → cosθ = −4/5.
✓ Answer: B.
(−3/5)² + cos²θ = 1 → 9/25 + cos²θ = 1 → cos²θ = 16/25 → cosθ = ±4/5.
In Quadrant III: both sin and cos are negative → cosθ = −4/5.
✓ Answer: B.
08
KEY IDENTITIES: sin²x + cos²x = 1 · tan x = sin x/cos x · 1 + tan²x = sec²x · 1 + cot²x = csc²x
Which expression is equivalent to tan²x − sec²x?
📖 Explanation
From the Pythagorean identity: 1 + tan²x = sec²x.
Rearranging: tan²x − sec²x = −1.
This is a constant, regardless of x! ✓ Answer: B.
Rearranging: tan²x − sec²x = −1.
This is a constant, regardless of x! ✓ Answer: B.
09
SINUSOIDAL: y = A·sin(Bx + C) + D · Amplitude = |A| · Period = 2π/B · Phase shift = −C/B
For y = −3 sin(2x − π) + 1, what are the amplitude and period?
📖 Explanation
Here A = −3, B = 2, D = 1.
• Amplitude = |A| = |−3| = 3 (always positive — it's a distance!)
• Period = 2π/B = 2π/2 = π
The negative sign in front of 3 causes a vertical reflection, but does NOT change the amplitude. ✓ Answer: B.
• Amplitude = |A| = |−3| = 3 (always positive — it's a distance!)
• Period = 2π/B = 2π/2 = π
The negative sign in front of 3 causes a vertical reflection, but does NOT change the amplitude. ✓ Answer: B.
Unit 5 · Systems of Equations & Matrices
10
SYSTEMS: 1 solution → lines intersect · No solution → parallel lines (inconsistent) · Infinite solutions → same line (dependent) · Check: substitute back!
The system 2x + 3y = 12 and 4x + 6y = 18 has:
📖 Explanation
Multiply the first equation by 2: 4x + 6y = 24.
Compare with the second: 4x + 6y = 18.
Same left side, different right side → 24 ≠ 18 → contradiction → No solution (the lines are parallel). ✓ Answer: C.
Compare with the second: 4x + 6y = 18.
Same left side, different right side → 24 ≠ 18 → contradiction → No solution (the lines are parallel). ✓ Answer: C.
11
2×2 INVERSE: If A = [a b / c d], then A⁻¹ = (1/det)·[d −b / −c a] where det = ad − bc · det = 0 → NO inverse!
Find the determinant of the matrix:
⎡ 3 −1 ⎤
⎣ 5 2 ⎦
⎣ 5 2 ⎦
📖 Explanation
For a 2×2 matrix [a b / c d], det = ad − bc.
Here: a=3, b=−1, c=5, d=2.
det = (3)(2) − (−1)(5) = 6 − (−5) = 6 + 5 = 11.
Watch the double negative! ✓ Answer: B.
Here: a=3, b=−1, c=5, d=2.
det = (3)(2) − (−1)(5) = 6 − (−5) = 6 + 5 = 11.
Watch the double negative! ✓ Answer: B.
Unit 6 · Sequences, Series & Binomial Theorem
12
GEOMETRIC SUM: Sₙ = a(1 − rⁿ)/(1 − r) · Infinite sum (|r|<1): S∞ = a/(1−r) · r = common ratio = any term ÷ previous term
Find the sum of the infinite geometric series: 12 + 4 + 4/3 + 4/9 + ...
📖 Explanation
First term a = 12. Common ratio r = 4/12 = 1/3. Since |r| = 1/3 < 1, the sum converges.
S∞ = a/(1−r) = 12/(1 − 1/3) = 12/(2/3) = 12 × 3/2 = 18.
✓ Answer: A.
S∞ = a/(1−r) = 12/(1 − 1/3) = 12/(2/3) = 12 × 3/2 = 18.
✓ Answer: A.
13
BINOMIAL TERM: kth term of (a+b)ⁿ = C(n, k−1)·aⁿ⁻⁽ᵏ⁻¹⁾·bᵏ⁻¹ · C(n,r) = n!/(r!(n−r)!) · Powers of a DECREASE, b INCREASE
What is the coefficient of x² in the expansion of (x + 3)⁵?
📖 Explanation
The term with x² has the form: C(5,2)·x²·3³ (power of x = 2, power of 3 = 5−2 = 3).
C(5,2) = 10, 3³ = 27.
Coefficient = 10 × 27 = 270. ✓ Answer: B.
C(5,2) = 10, 3³ = 27.
Coefficient = 10 × 27 = 270. ✓ Answer: B.
Unit 7 · Conic Sections
14
CONICS ID: Circle: x²+y²=r² · Parabola: one variable squared · Ellipse: x²/a² + y²/b² = 1 · Hyperbola: minus sign between terms
What is the vertex of the parabola y = 2(x − 3)² + 5?
📖 Explanation
Vertex form: y = a(x − h)² + k → vertex is (h, k).
Here h = 3, k = 5. Beware: the formula says (x − h), and we see (x − 3), so h = +3 (not −3!).
✓ Vertex = (3, 5). Answer: B.
Here h = 3, k = 5. Beware: the formula says (x − h), and we see (x − 3), so h = +3 (not −3!).
✓ Vertex = (3, 5). Answer: B.
15
ELLIPSE: x²/a² + y²/b² = 1 · If a>b → major axis horizontal · If b>a → major axis vertical · Foci: c² = a² − b² (a>b)
For the ellipse x²/25 + y²/9 = 1, what are the foci?
📖 Explanation
a² = 25, b² = 9. Since a² > b², major axis is horizontal.
c² = a² − b² = 25 − 9 = 16 → c = 4.
Foci lie on horizontal axis: (±4, 0). ✓ Answer: C.
c² = a² − b² = 25 − 9 = 16 → c = 4.
Foci lie on horizontal axis: (±4, 0). ✓ Answer: C.
Unit 8 · Limits & Introduction to Calculus Concepts
16
LIMITS AT INFINITY: For rational functions: compare highest degrees · top degree > bottom → ∞ · top = bottom → ratio of coefficients · top < bottom → 0
Quick Example
lim(x→∞) of 3x²/(x²+1) = 3/1 = 3 (same degree → use leading coefficients)
Evaluate: lim(x→∞) of (5x³ − 2x) / (3x³ + x²)
📖 Explanation
Highest degree in numerator and denominator is both x³ (same degree!).
Divide everything by x³: (5 − 2/x²) / (3 + 1/x).
As x→∞: 2/x² → 0, 1/x → 0.
Limit = 5/3. ✓ Answer: B.
Divide everything by x³: (5 − 2/x²) / (3 + 1/x).
As x→∞: 2/x² → 0, 1/x → 0.
Limit = 5/3. ✓ Answer: B.
17
0/0 FORM: Factor and cancel · OR multiply by conjugate · Direct substitution gives 0/0 → it's a SIGN to factor, not the answer!
Evaluate: lim(x→2) of (x² − 4) / (x − 2)
📖 Explanation
Direct substitution → 0/0 (indeterminate form). Factor first!
(x² − 4)/(x − 2) = (x+2)(x−2)/(x−2) = x + 2 for x ≠ 2.
Now take the limit: lim(x→2) of (x+2) = 2 + 2 = 4.
✓ Answer: C.
(x² − 4)/(x − 2) = (x+2)(x−2)/(x−2) = x + 2 for x ≠ 2.
Now take the limit: lim(x→2) of (x+2) = 2 + 2 = 4.
✓ Answer: C.
18
INVERSE FUNCTION: Swap x and y, then solve for y · f(f⁻¹(x)) = x · Domain of f = Range of f⁻¹ and vice versa · Graph: reflect over y = x
If f(x) = 2x − 6, find f⁻¹(x).
📖 Explanation
Step 1: Replace f(x) with y: y = 2x − 6.
Step 2: Swap x and y: x = 2y − 6.
Step 3: Solve for y: x + 6 = 2y → y = (x + 6)/2.
So f⁻¹(x) = (x + 6)/2. ✓ Answer: A.
Step 2: Swap x and y: x = 2y − 6.
Step 3: Solve for y: x + 6 = 2y → y = (x + 6)/2.
So f⁻¹(x) = (x + 6)/2. ✓ Answer: A.
19
COMPOSITION: (f∘g)(x) = f(g(x)) → plug g(x) INTO f · Work INSIDE → OUT · (f∘g) ≠ (g∘f) in general!
Let f(x) = x² + 1 and g(x) = 2x − 3. Find (f ∘ g)(2).
📖 Explanation
(f∘g)(2) = f(g(2)).
Step 1: Find g(2) = 2(2) − 3 = 4 − 3 = 1.
Step 2: Find f(1) = 1² + 1 = 2.
Wait — let's check: f(1) = 1 + 1 = 2. ✓ Answer: B.
Step 1: Find g(2) = 2(2) − 3 = 4 − 3 = 1.
Step 2: Find f(1) = 1² + 1 = 2.
Wait — let's check: f(1) = 1 + 1 = 2. ✓ Answer: B.
20
COMPLEX DIVISION: Multiply top & bottom by the CONJUGATE of denominator · Conjugate of (a+bi) = (a−bi) · i² = −1 always!
Quick Example
1/(2+i) = (2−i)/((2+i)(2−i)) = (2−i)/(4+1) = (2−i)/5
Simplify: (3 + 2i) / (1 − i)
📖 Explanation
Multiply by conjugate of denominator (1+i)/(1+i):
Numerator: (3+2i)(1+i) = 3 + 3i + 2i + 2i² = 3 + 5i + 2(−1) = 1 + 5i
Denominator: (1−i)(1+i) = 1 − i² = 1−(−1) = 2
Result: (1 + 5i)/2. ✓ Answer: A.
Numerator: (3+2i)(1+i) = 3 + 3i + 2i + 2i² = 3 + 5i + 2(−1) = 1 + 5i
Denominator: (1−i)(1+i) = 1 − i² = 1−(−1) = 2
Result: (1 + 5i)/2. ✓ Answer: A.
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