Unit 1 · Algebra Foundations
Memory Key
SAME BASE → ADD exponents (multiply) · SUBTRACT exponents (divide)
Quick Example
x³ · x⁴ = x^(3+4) = x⁷
x⁶ ÷ x² = x^(6−2) = x⁴
x⁶ ÷ x² = x^(6−2) = x⁴
Simplify:
x⁵ · x⁻² ÷ x
What is the result?
✓ Correct! Well done.
✗ Not quite — see explanation.
📖 Explanation
Apply the exponent rules step by step:
x⁵ · x⁻² = x^(5+(−2)) = x³
x³ ÷ x¹ = x^(3−1) = x²
The answer is x². Remember: multiply → add, divide → subtract!
Memory Key
DIFFERENCE OF SQUARES: a² − b² = (a+b)(a−b) · Always check FIRST!
Quick Example
x² − 9 = x² − 3² = (x+3)(x−3)
Factor completely:
4x² − 25
✓ Correct! Well done.
✗ Not quite — see explanation.
📖 Explanation
Recognize the difference of squares pattern: a² − b²
4x² = (2x)², so a = 2x
25 = 5², so b = 5
4x² − 25 = (2x+5)(2x−5) ✓
Trick: (2x − 5)² would give 4x² − 20x + 25, which has a middle term — not what we have!
Memory Key
DISCRIMINANT b²−4ac: >0 two real roots · =0 one root · <0 no real roots
Quick Example
For x²−5x+6=0: Discriminant = 25−24=1 > 0 → two real solutions
How many real solutions does this equation have?
3x² − 4x + 5 = 0
✓ Correct! Well done.
✗ Not quite — see explanation.
📖 Explanation
Compute the discriminant: b² − 4ac
a = 3, b = −4, c = 5
Δ = (−4)² − 4(3)(5) = 16 − 60 = −44
Since Δ = −44 < 0, the quadratic has no real solutions. The parabola never crosses the x-axis!
Unit 2 · Functions & Their Graphs
Memory Key
SQUARE ROOT → inside ≥ 0 · FRACTION → denominator ≠ 0 · LOG → inside > 0
Quick Example
Domain of √(x−3): set x−3 ≥ 0, so x ≥ 3 → Domain: [3, ∞)
What is the domain of:
f(x) = √(2x − 6)
✓ Correct! Well done.
✗ Not quite — see explanation.
📖 Explanation
For a square root, the expression inside must be ≥ 0:
2x − 6 ≥ 0
2x ≥ 6
x ≥ 3
Domain: [3, ∞). Note: we use ≥ (not >) because √0 = 0 is defined!
Memory Key
f∘g means f(g(x)) → plug g INTO f · ORDER MATTERS: f∘g ≠ g∘f usually!
Quick Example
If f(x)=x+1 and g(x)=x², then f(g(x))=x²+1 but g(f(x))=(x+1)²
Given f(x) = 2x + 1 and g(x) = x², find:
f(g(3)) = ?
✓ Correct! Well done.
✗ Not quite — see explanation.
📖 Explanation
Work from the INSIDE out:
Step 1: g(3) = 3² = 9
Step 2: f(9) = 2(9) + 1 = 19
Common mistake: doing f first then g (that would give g(f(3)) = g(7) = 49 — option B!). Always read inside → outside.
Memory Key
INVERSE: swap x↔y then solve for y · Graph = reflection over y=x line
Quick Example
f(x)=3x+2 → swap: x=3y+2 → solve: f⁻¹(x)=(x−2)/3
Find the inverse of:
f(x) = 5x − 3
✓ Correct! Well done.
✗ Not quite — see explanation.
📖 Explanation
Swap x and y, then solve:
y = 5x − 3 → swap → x = 5y − 3
x + 3 = 5y
y = (x + 3) / 5 ✓
Option C has (x−3)/5 — that's what you get if you forget to add 3 to both sides!
Unit 3 · Polynomials & Rational Functions
Memory Key
LEADING TERM rules end behavior: even degree → same ends · odd degree → opposite ends
Quick Example
f(x)=−2x³: odd degree, negative lead → as x→+∞, f→−∞ and as x→−∞, f→+∞
Describe the end behavior of:
f(x) = −3x⁴ + 7x² − 1
✓ Correct! Well done.
✗ Not quite — see explanation.
📖 Explanation
Focus only on the leading term −3x⁴:
Degree = 4 (even) → both ends go the SAME direction
Coefficient = −3 (negative) → both ends go DOWN
As x→±∞, f(x)→−∞. The graph looks like an upside-down "U" shape on both sides.
Memory Key
VERTICAL: set denominator=0 · HORIZONTAL: compare degrees (top vs bottom)
Quick Example
f(x) = 2x/(x−1): VA at x=1 · HA: same degree → ratio of leading = 2/1 → y=2
Find the horizontal asymptote of:
f(x) = (3x² + 1) / (x² − 4)
✓ Correct! Well done.
✗ Not quite — see explanation.
📖 Explanation
Compare the degrees of numerator and denominator:
Numerator degree = 2 · Denominator degree = 2 → EQUAL degrees
When degrees are equal: HA = leading coefficient ratio = 3/1 = 3
Rule summary: top < bottom → y=0 · top = bottom → leading ratio · top > bottom → no HA (slant instead)
Unit 4 · Exponential & Logarithmic Functions
Memory Key
LOG RULES: log(AB)=logA+logB · log(A/B)=logA−logB · log(Aⁿ)=n·logA
Quick Example
log(x²y) = 2log(x) + log(y)
Expand completely:
log₂(8x³)
✓ Correct! Well done.
✗ Not quite — see explanation.
📖 Explanation
log₂(8x³) = log₂(8) + log₂(x³) ← Product rule
log₂(8) = log₂(2³) = 3 ← Since 2³=8
log₂(x³) = 3·log₂(x) ← Power rule
Result: 3 + 3log₂(x) ✓
Common mistake: option B forgets to apply the power rule to x³.
Memory Key
SAME BASE → set exponents equal · DIFFERENT BASE → take log of both sides
Quick Example
2^x = 32 → 2^x = 2^5 → x = 5
Solve for x:
3^(2x) = 27
✓ Correct! Well done.
✗ Not quite — see explanation.
📖 Explanation
Rewrite 27 with base 3:
27 = 3³
3^(2x) = 3³ → same base, set exponents equal
2x = 3 → x = 3/2 ✓
Trap: option A is x=3, which would give 3⁶=729, not 27!
Unit 5 · Trigonometry
Memory Key
ASTC: All Students Take Calculus → Q1: All positive · Q2: Sin+ · Q3: Tan+ · Q4: Cos+
Quick Example
sin(π/6) = 1/2 · cos(π/3) = 1/2 · tan(π/4) = 1
What is the exact value of:
cos(3π/4)
✓ Correct! Well done.
✗ Not quite — see explanation.
📖 Explanation
3π/4 is in Quadrant 2 (between π/2 and π).
Reference angle: π − 3π/4 = π/4
cos(π/4) = √2/2
In Q2, cosine is NEGATIVE → cos(3π/4) = −√2/2 ✓
Remember ASTC: in Quadrant 2, only Sine is positive!
Memory Key
PYTHAGOREAN: sin²θ + cos²θ = 1 · 1+tan²θ = sec²θ · 1+cot²θ = csc²θ
Quick Example
Simplify 1 − sin²x = cos²x (rearranging Pythagorean identity)
Simplify the expression:
sin²x + cos²x + tan²x
✓ Correct! Well done.
✗ Not quite — see explanation.
📖 Explanation
sin²x + cos²x = 1 (Pythagorean identity)
So the expression = 1 + tan²x
1 + tan²x = sec²x (second Pythagorean identity)
Note: Options A and D are both wrong. D = tan²x + 1 = sec²x, so D and C are actually the SAME — but the fully simplified form is sec²x.
Memory Key
y=A·sin(Bx+C)+D → Amplitude=|A| · Period=2π/|B| · Phase shift=−C/B
Quick Example
y = 3sin(2x): Amplitude = 3, Period = 2π/2 = π
What is the period of:
y = 4 cos(3x − π)
✓ Correct! Well done.
✗ Not quite — see explanation.
📖 Explanation
Identify B from y = 4cos(3x − π): here B = 3
Period = 2π / |B| = 2π / 3 ✓
The amplitude = 4 (not the period!). The phase shift = π/3 (not the period!). Don't mix these up.
Unit 6 · Sequences & Series
Memory Key
ARITHMETIC: aₙ = a₁ + (n−1)d · d = common difference (constant ADD or SUBTRACT)
Quick Example
Sequence 2, 5, 8, 11... → d=3 → a₁₀ = 2 + (10−1)·3 = 29
Find the 15th term of the arithmetic sequence:
4, 7, 10, 13, ...
✓ Correct! Well done.
✗ Not quite — see explanation.
📖 Explanation
a₁ = 4, d = 7 − 4 = 3, n = 15
a₁₅ = 4 + (15−1)·3 = 4 + 14·3 = 4 + 42 = 46 ✓
Classic error: using n instead of (n−1) → that gives 4+15·3=49. The formula uses (n−1) because you add d one fewer time than the term number!
Memory Key
GEOMETRIC: aₙ = a₁ · r^(n−1) · r = common ratio (constant MULTIPLY or DIVIDE)
Quick Example
Sequence 3, 6, 12, 24... → r=2 → a₅ = 3 · 2⁴ = 48
In a geometric sequence where a₁ = 5 and r = −2, find a₅.
a₅ = ?
✓ Correct! Well done.
✗ Not quite — see explanation.
📖 Explanation
a₅ = 5 · (−2)^(5−1) = 5 · (−2)⁴
(−2)⁴ = (−2)·(−2)·(−2)·(−2) = +16 (even power → positive!)
a₅ = 5 · 16 = 80 ✓
Trap: (−2)⁵ = −32 (odd power), but we only raise to the 4th power here. A negative ratio alternates signs: 5, −10, 20, −40, 80.
Unit 7 · Conic Sections & Analytic Geometry
Memory Key
VERTEX FORM: y=a(x−h)²+k → vertex=(h,k) · a>0 opens UP · a<0 opens DOWN
Quick Example
y = −2(x−3)² + 5: vertex = (3, 5), opens downward
Find the vertex of the parabola:
y = 2(x + 4)² − 7
✓ Correct! Well done.
✗ Not quite — see explanation.
📖 Explanation
Vertex form: y = a(x−h)² + k, vertex = (h, k)
Rewrite: y = 2(x−(−4))² + (−7)
So h = −4 and k = −7
Vertex = (−4, −7) ✓
The most common mistake: reading h=+4 instead of −4. The formula has (x−h), so (x+4) means h=−4!
Memory Key
CIRCLE: (x−h)² + (y−k)² = r² → center=(h,k) · radius=r (not r²!)
Quick Example
(x−2)² + (y+1)² = 25: center=(2,−1), radius=5
What is the center and radius of:
(x + 3)² + (y − 5)² = 36
✓ Correct! Well done.
✗ Not quite — see explanation.
📖 Explanation
Match to (x−h)² + (y−k)² = r²:
(x+3)² = (x−(−3))² → h = −3
(y−5)² → k = 5
r² = 36 → r = √36 = 6
Two traps: (1) reading h=+3 instead of −3, and (2) saying radius = 36 instead of 6!
Unit 8 · Introduction to Limits
Memory Key
GET 0/0? → FACTOR and CANCEL · Then substitute again. Don't give up at 0/0!
Quick Example
lim(x→2) (x²−4)/(x−2) = lim (x+2)(x−2)/(x−2) = lim (x+2) = 4
Evaluate:
lim(x→3) (x² − 9) / (x − 3)
✓ Correct! Well done.
✗ Not quite — see explanation.
📖 Explanation
Direct substitution gives 0/0 — factor first!
x² − 9 = (x+3)(x−3)
(x+3)(x−3) / (x−3) = x+3 (cancel x−3, valid since x≠3)
lim(x→3) (x+3) = 3+3 = 6 ✓
A limit asks "what value does the function approach?" — the function doesn't need to equal that value at x=3!
Memory Key
LIMIT at ∞: DIVIDE everything by highest power of x in denominator → small terms vanish
Quick Example
lim(x→∞) (5x+1)/(2x−3): divide by x → (5+1/x)/(2−3/x) → 5/2
Evaluate:
lim(x→∞) (4x³ − 2x) / (x³ + 5)
✓ Correct! Well done.
✗ Not quite — see explanation.
📖 Explanation
Divide numerator and denominator by x³ (highest power):
(4x³−2x)/x³ = 4 − 2/x²
(x³+5)/x³ = 1 + 5/x³
As x→∞: (4−0)/(1+0) = 4 ✓
Equal degrees → ratio of leading coefficients = 4/1 = 4. Same rule as horizontal asymptotes!
Memory Key
CONTINUITY 3-CHECK: (1) f(a) exists · (2) lim exists · (3) lim = f(a) · ALL THREE needed!
Quick Example
If f(2)=5 but lim(x→2)f(x)=3, the function is DISCONTINUOUS at x=2 (condition 3 fails)
A function f has: f(2) = 5, and lim(x→2) f(x) = 5.
Which statement is correct?
Which statement is correct?
✓ Correct! Well done.
✗ Not quite — see explanation.
📖 Explanation
Check all three continuity conditions at x=2:
✓ Condition 1: f(2) = 5 exists
✓ Condition 2: lim(x→2) f(x) = 5 exists
✓ Condition 3: lim(x→2) f(x) = f(2) → 5 = 5 ✓
All three conditions are satisfied, so f is continuous at x = 2. We don't need a formula — the given information is enough!
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