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Based on this guide's real question bank — 190 practice questions across 9 units. Slide to match your situation.
Unit 1: Functions & Their Graphs
▾Function notation & evaluation
- f(x) is read 'f of x' — NOT f times x
- To evaluate f(3), substitute 3 for every x in the rule: $f(x)=2x²−1$ → $f(3)=2(9)−1=17$
- Vertical Line Test: a graph represents a function iff every vertical line crosses it at most once
- Composite functions: $(f∘g)(x)=f(g(x))$ — evaluate the INSIDE function first
- Example: $f(x)=x+2, g(x)=x². (f∘g)(3)=f(9)=11. (g∘f)(3)=g(5)=25$ — order matters!
- Difference quotient: [f(x+h)−f(x)]/h, the building block for average rate of change and derivatives
Domain and range
- Rational functions: exclude x-values that make the denominator 0 (e.g., $f(x)=1/(x−3)$: domain is x≠3)
- Even-root functions: the radicand must be ≥0 (e.g., $f(x)=√(x−4)$: domain is x≥4)
- Polynomial functions: domain is always all real numbers (−∞,∞)
- Range is found from the graph's minimum/maximum behavior (e.g., $y=x²$ has range y≥0)
- Interval notation: use parentheses for excluded endpoints, brackets for included endpoints
- Domain of a sum/product/quotient of functions $=$ intersection of each piece's domain (and denominator ≠0 for quotients)
Transformations of functions
- f(x)+k: shifts UP k units (k>0) | f(x)−k: shifts DOWN k units
- f(x−h): shifts RIGHT h units | f(x+h): shifts LEFT h units (opposite of the sign!)
- a·f(x): vertical stretch if |a|>1, vertical compression if 0<|a|<1
- f(bx): horizontal compression if |b|>1, horizontal stretch if 0<|b|<1
- −f(x): reflects over the x-axis | f(−x): reflects over the y-axis
- Order of operations for combined transformations: horizontal shift/stretch first (inside), then vertical stretch, then vertical shift (outside)
Even and odd functions
- Even function: $f(−x)=f(x)$ for all x — graph is symmetric about the y-axis (e.g., $f(x)=x², f(x)=cos$ x)
- Odd function: $f(−x)=−f(x)$ for all x — graph has 180° rotational symmetry about the origin (e.g., $f(x)=x³, f(x)=sin$ x)
- To test algebraically: plug −x into f(x), simplify, and compare to f(x) and −f(x)
- A function can be neither even nor odd (most functions), but cannot be both unless $f(x)=0$ for all x
- Sum of two even functions is even; sum of two odd functions is odd; product of two odd functions is even
Piecewise functions
- Notation: f(x)={rule₁ if x<a; rule₂ if x≥a} — check the given x against each interval's boundary
- Example: f(x)={x+3 if x<2; x² if x≥2}. $f(2)=2²=4$ (use second rule since 2≥2); $f(−1)=−1+3=2$
- Graph each piece only over its stated domain; use open circles for excluded endpoints, closed circles for included
- Step functions (e.g., the greatest integer/floor function ⌊x⌋) are a special piecewise case with constant horizontal segments
- Continuity at a breakpoint requires the two pieces to meet at the same y-value there
Difference quotient: [f(x+h)−f(x)]/h
$f(−x)=f(x)$ even; $f(−x)=−f(x)$ odd
Unit 2: Polynomial & Rational Functions
▾End behavior of polynomials
- Odd degree, positive leading coefficient: falls left, rises right (like $y=x³$)
- Odd degree, negative leading coefficient: rises left, falls right (like $y=−x³$)
- Even degree, positive leading coefficient: rises left and right (like $y=x²$)
- Even degree, negative leading coefficient: falls left and right (like $y=−x²$)
- Degree $=$ maximum number of real zeros (counting multiplicity) and one less than the max number of turning points
- Example: $f(x)=−2x⁵+3x²−1$ has odd degree (5) and negative leading coefficient → rises left, falls right
Factoring & finding zeros
- Factor completely, set each factor to 0, solve for x (zero product property)
- Multiplicity 1 (odd, simple): graph CROSSES the x-axis at that zero
- Multiplicity 2 (even): graph TOUCHES the x-axis and turns back (bounces)
- Multiplicity 3 (odd, ≥3): graph crosses but flattens (inflects) at that zero
- Example: $f(x)=x³−4x =$ x(x−2)(x+2) → zeros at $x=0,2,−2$, each multiplicity 1, all crossings
- Difference of squares: $a²−b²=(a−b)(a+b)$; sum/difference of cubes: $a³±b³=(a±b)(a²∓ab+b²)$
Rational Root Theorem & synthetic division
- Possible rational roots $=$ ±(factors of constant term)/(factors of leading coefficient)
- Example: $2x³−3x²−11x+6=0$ → $constant=6$ (factors 1,2,3,6), $leading=2$ (factors 1,2) → candidates ±1,2,3,6,1/2,3/2
- Test candidates using synthetic division; a remainder of 0 confirms a root and gives the depressed (reduced) polynomial
- Synthetic division steps: bring down leading coefficient, multiply by the root, add to next coefficient, repeat
- Example: dividing 2x³−3x²−11x+6 by (x−3) using synthetic division with root 3 gives 2x²+3x−2 remainder 0, confirming $x=3$ is a root
- Once reduced to a quadratic, use factoring or the quadratic formula to find remaining roots
Rational functions & asymptotes
- Vertical asymptote: at x-values where the denominator $=$ 0 AND the numerator ≠ 0 there (a hole occurs if both are 0 after cancelling common factors)
- Horizontal asymptote (compare degrees of numerator n and denominator m): if n<m, $y=0$; if $n=m, y=$(ratio of leading coefficients); if n>m, no horizontal asymptote
- Slant (oblique) asymptote exists when $n=m+1$ exactly one degree higher — found via polynomial long division, ignoring the remainder
- Example: $f(x)=(x²−1)/(x−1)$ simplifies to x+1 with a HOLE at $x=1$ (not a vertical asymptote) since (x−1) cancels
- Example: $f(x)=(3x²+1)/(x²−4)$ has horizontal asymptote $y=3$ (equal degrees, ratio of leading coefficients 3/1) and vertical asymptotes at $x=±2$
Solving rational equations & inequalities
- Multiply every term by the LCD to eliminate fractions, then solve the resulting polynomial equation
- Always check solutions back in the ORIGINAL equation — any value making a denominator 0 must be rejected (extraneous)
- For rational inequalities: move everything to one side, find critical values (zeros and undefined points), test each interval on a sign chart
- Example: solve (x−1)/(x+2)≥0 → critical values $x=1, x=−2$ (excluded, open circle); test intervals → solution x<−2 or x≥1
HA rules: $n<m→y=0; n=m→y=leading$ ratio; n>m→none
$a³±b³=(a±b)(a²∓ab+b²)$
Unit 3: Exponential & Logarithmic Functions
▾Exponential growth and decay
- A₀ $=$ initial amount, r $=$ rate as a decimal, t $=$ number of time periods, $b=(1+r)$ or (1−r) is the growth/decay factor
- If b>1 the function models growth; if 0<b<1 it models decay
- Example: population growing 5%/year from 2000: $A(t)=2000(1.05)^t$ → after 10 years $A(10)=2000(1.05)¹⁰≈3257.8$
- Continuous growth/decay uses $A(t)=A₀e^(rt)$ where e≈2.71828 (r>0 growth, r<0 decay)
- Half-life problems: $A(t)=A₀(1/2)^(t/h)$ where h is the half-life
- Compound interest: $A=P(1+r/n)^(nt)$; continuously compounded: $A=Pe^(rt)$
Laws of exponents
- Product rule: $bᵐ·bⁿ=bᵐ⁺ⁿ$ | Quotient rule: $bᵐ/bⁿ=bᵐ⁻ⁿ$
- Power rule: $(bᵐ)ⁿ=bᵐⁿ$ | Power of a product: $(ab)ⁿ=aⁿbⁿ$
- Zero exponent: $b⁰=1$ (b≠0) | Negative exponent: $b⁻ⁿ=1/bⁿ$
- Rational (fractional) exponent: $b^(m/n)=ⁿ√(bᵐ)=(ⁿ√b)ᵐ$
- Example: $8^(2/3)=(³√8)²=2²=4$
- To solve exponential equations with matching bases: set exponents equal once bases match (e.g., $2^(x+1)=8$ → $2^(x+1)=2³$ → $x+1=3$ → $x=2$)
Logarithm definition & properties
- $log_b(x)=y$ ⟺ $bʸ=x$ (b>0, b≠1, x>0)
- Product rule: $log_b(MN)=log_b(M)+log_b(N)$
- Quotient rule: $log_b(M/N)=log_b(M)−log_b(N)$
- Power rule: $log_b(Mᵏ)=k·log_b(M)$
- $log_b(1)=0$ always (since $b⁰=1$); $log_b(b)=1$ always (since $b¹=b$)
- Change of base formula: $log_b(x)=ln(x)/ln(b)=log(x)/log(b)$ — needed for bases other than 10 or e on a calculator
Solving exponential & logarithmic equations
- To solve $bˣ=k$ (unlike bases): take ln or log of both sides → $x=ln(k)/ln(b)$
- Example: $5^x=40$ → $x=ln(40)/ln(5)≈2.292$
- To solve $log_b(x)=k$: rewrite in exponential form $x=bᵏ$
- Example: $log₂(x)=5$ → $x=2⁵=32$
- When both sides are logs with the same base: $log_b(A)=log_b(B)$ ⟹ $A=B$ (then check domain — argument must stay positive)
- Extraneous solutions: after solving a log equation, reject any solution that makes an argument of a log ≤0 in the ORIGINAL equation
Natural log and e
- ln(x) and eˣ are inverse functions: $ln(eˣ)=x$ and $e^(ln x)=x$
- $ln(1)=0, ln(e)=1$
- Solve $eˣ=k$ by taking ln of both sides: $x=ln(k)$
- Solve $ln(x)=k$ by exponentiating: $x=eᵏ$
- Continuous compounding/growth: $A=Pe^(rt)$ — appears in population models, radioactive decay, and continuously compounded interest
- Example: solve $e^(2x)=15$ → $2x=ln(15)$ → $x=ln(15)/2≈1.354$
Change of base: $log_b(x)=ln(x)/ln(b)$
$A=Pe^(rt); A=P(1+r/n)^(nt)$
Unit 4: Trigonometric Functions
▾The unit circle
- cos $θ =$ x-coordinate, sin $θ =$ y-coordinate of the point where the terminal ray meets the circle
- tan $θ =$ sin $θ$ / cos $θ =$ y/x (undefined when cos $θ=0$, i.e., $θ=90°,270°$)
- Quadrant signs (ASTC, 'All Students Take Calculus'): QI all positive, QII sine positive, QIII tangent positive, QIV cosine positive
- Key points to memorize: (1,0) at 0°, (0,1) at 90°, (−1,0) at 180°, (0,−1) at 270°
- The unit circle repeats every 360° ($2π$ radians) — this is why trig functions are periodic
Radians vs degrees
- Conversion: radians $=$ degrees × ($π/180$) | degrees $=$ radians × ($180/π$)
- $180°=π$ rad, $90°=π/2$ rad, $60°=π/3$ rad, $45°=π/4$ rad, $30°=π/6$ rad
- Example: convert 150° to radians: $150×(π/180)=5π/6$
- Example: convert $7π/4$ rad to degrees: $(7π/4)×(180/π)=315°$
- Arc length formula: $s=rθ$ (θ MUST be in radians)
- Angular/linear speed problems use $s=rθ$ and $v=s/t$ or $ω=θ/t$ (ω in rad/time)
Graphs of sine, cosine, and tangent
- Amplitude $=$ |A| (half the distance between max and min); period $= 2π/|B|$
- Phase shift $=$ C/B (shift right if positive); vertical shift $=$ D (midline $y=D$)
- $y=sin$ x: starts at midline going up, zeros at $0,π,2π$, max at $π/2$, min at $3π/2$
- $y=cos$ x: starts at max, zeros at $π/2,3π/2$, min at $π$
- $y=tan$ x has period $π$ (not $2π$) and vertical asymptotes at $x=π/2+kπ$; no amplitude (unbounded range)
- Example: $y=3sin(2x)+1$ has amplitude 3, period $2π/2=π$, midline $y=1$, no phase shift
Reference angles
- QI: reference angle $= θ$ itself
- QII: reference angle $= 180°−θ$ (or $π−θ$)
- QIII: reference angle $= θ−180°$ (or $θ−π$)
- QIV: reference angle $= 360°−θ$ (or $2π−θ$)
- Find the trig value of the reference angle, then apply the correct sign for the original quadrant (ASTC)
- Example: sin(210°): QIII, reference $angle=210−180=30°, \sin(30°)=1/2$, QIII sine is negative → $\sin(210°)=−1/2$
Special angle trig values
- 0°(0): $sin=0, cos=1, tan=0$
- $30°(π/6): sin=1/2, cos=√3/2, tan=√3/3 (=1/√3)$
- $45°(π/4): sin=√2/2, cos=√2/2, tan=1$
- $60°(π/3): sin=√3/2, cos=1/2, tan=√3$
- $90°(π/2): sin=1, cos=0, tan=undefined$
- Memory trick: sin values at 0°,30°,45°,60°,90° follow √0/2,√1/2,√2/2,√3/2,√4/2 (simplify each); cos is the reverse order
Period: sin/cos $= 2π/|B|$; tan $= π/|B|$
tan $θ =$ sin $θ/cos θ$
Unit 5: Trigonometric Identities & Equations
▾Pythagorean identities
- $sin²θ+cos²θ=1$ (the core identity — memorize this one first)
- Divide by $cos²θ: 1+tan²θ=sec²θ$
- Divide by $sin²θ: 1+cot²θ=csc²θ$
- Rearranged forms are useful for simplifying: $sin²θ=1−cos²θ, cos²θ=1−sin²θ$
- Example: simplify $sin²θ/(1−cos θ)$ using $1−cos²θ=(1−\cosθ)(1+\cosθ): sin²θ/(1−\cosθ)=(1−\cosθ)(1+\cosθ)/(1−\cosθ)=1+\cosθ$
Sum and difference formulas
- $\sin(A±B)=sin$ A cos B ± cos A sin B
- $\cos(A±B)=cos$ A cos B ∓ sin A sin B (note the sign FLIPS for cosine)
- $\tan(A±B)=(tan A ± tan B)$/(1 ∓ tan A tan B)
- Example: find $\cos(75°)=\cos(45°+30°)=cos45°cos30°−sin45°sin30°=(√2/2)(√3/2)−(√2/2)(1/2)=(√6−√2)/4$
- Cofunction identities follow from these: $\sin(90°−θ)=cos θ, \cos(90°−θ)=sin θ$
Double-angle formulas
- $\sin(2θ)=2$ sin $θ$ cos $θ$
- $\cos(2θ)=cos²θ−sin²θ = 2cos²θ−1 = 1−2sin²θ$ (three equivalent forms — pick based on what's given)
- $\tan(2θ)=2tanθ/(1−tan²θ)$
- Example: if $\sinθ=3/5$ and $θ$ in QI ($\cosθ=4/5$), then $\sin(2θ)=2(3/5)(4/5)=24/25$
- Half-angle formulas (derived from double-angle, less common but related): $\sin(θ/2)=±√[(1−\cosθ)/2], \cos(θ/2)=±√[(1+\cosθ)/2]$
Solving trigonometric equations
- Isolate the trig function first, then find the reference angle from its exact value
- Find every solution in [$0,2π$) or [0°,360°) by applying the correct quadrant signs (ASTC), then add multiples of the period for a general solution ($+2πk$ or +360°k)
- Example: solve $2sinθ−1=0$ on $[0,2π)$: $\sinθ=1/2$ → reference angle $π/6$ → sine positive in QI,QII → $θ=π/6, 5π/6$
- For quadratics in a trig function (e.g., $2sin²θ−\sinθ−1=0$), factor like a normal quadratic: $(2sinθ+1)(\sinθ−1)=0$ → $\sinθ=−1/2$ or $\sinθ=1$
- When an equation has multiple trig functions, use identities (Pythagorean, double-angle) to rewrite everything in terms of ONE function before solving
$\sin(2θ)=2sinθcosθ; \cos(2θ)=cos²θ−sin²θ$
$\sin(A±B)=sinAcosB±cosAsinB$
Unit 6: Analytic Trigonometry & Applications
▾Law of Sines
- Use when given: AAS, ASA, or SSA (the 'ambiguous case')
- Set up as a proportion and cross-multiply to solve for the unknown side or angle
- Example: $A=40°, a=15, B=60°$. Find b: $15/sin40°=b/sin60°$ → $b=15·sin60°/sin40°≈20.2$
- Ambiguous case (SSA): given two sides and a non-included angle, there may be 0, 1, or 2 valid triangles — check by comparing the given side to $h=b·sinA$
- Sum of angles in any triangle $=$ 180°, so the third angle is always found by subtraction once two are known
Law of Cosines
- Use when given: SAS (two sides and the included angle) or SSS (three sides, no angles)
- Solving for a side (SAS): $c²=a²+b²−2ab·cos$ C, then take the square root
- Solving for an angle (SSS): rearrange to cos $C=(a²+b²−c²)/(2ab)$, then take cos⁻¹
- Example (SAS): $a=8, b=10, C=50°$ → $c²=64+100−2(8)(10)cos50°≈61.13$ → c≈7.82
- Example (SSS): $a=5,b=7,c=10$ → $cosC=(25+49−100)/(2·5·7)=−26/70≈−0.371$ → C≈111.8°
- If cos C comes out negative, the angle C is obtuse (>90°) — this is a normal, valid result
Solving triangles & area
- SSS or SAS → start with Law of Cosines; AAS/ASA/SSA → start with Law of Sines
- Once one angle is found via Law of Cosines, use Law of Sines for the rest (often easier algebra)
- Area of any triangle given SAS: $Area=(1/2)ab·sinC$ (works without needing the height)
- Example: $a=6,b=9,C=30°$ → $Area=(1/2)(6)(9)sin30°=27(0.5)=13.5$
- Heron's formula (SSS, no angles needed): $s=(a+b+c)/2, Area=√[s(s−a)(s−b)(s−c)]$
Inverse trigonometric functions
- sin⁻¹x (arcsin): domain [−1,1], range $[−π/2,π/2]$ (QI and QIV)
- cos⁻¹x (arccos): domain [−1,1], range $[0,π]$ (QI and QII)
- tan⁻¹x (arctan): domain all reals, range ($−π/2,π/2$)
- $sin⁻¹(\sinθ)=θ$ only if $θ$ is already in $[−π/2,π/2]$; otherwise find the equivalent angle in range
- Example: $sin⁻¹(1/2)=π/6$ (30°) — the answer must be in $[−π/2,π/2]$, so 150° is NOT the output even though $sin150°=1/2$ too
- Used to solve for an unknown ANGLE once a trig ratio is known (e.g., in right-triangle applications: angle of elevation/depression)
$c²=a²+b²−2ab·cosC$
$Area=(1/2)ab·sinC$
Unit 7: Systems & Matrices
▾Solving systems by substitution & elimination
- Substitution: solve one equation for one variable, substitute into the other equation(s)
- Elimination: multiply equations by constants so one variable's coefficients are opposites, then add the equations
- Example (elimination): $2x+3y=7$ and $4x−3y=11$ → add directly: $6x=18$ → $x=3$, then $y=1/3(7−6)=1/3$
- A system with NO solution: equations simplify to a false statement (e.g., $0=5$) — lines are parallel
- A system with INFINITE solutions: equations simplify to a true statement (e.g., $0=0$) — lines coincide
- For 3-variable systems, eliminate one variable at a time to reduce to a 2-variable system
Matrices — basics
- A matrix with m rows and n columns is an m×n matrix
- Two matrices are equal only if they have the same dimensions AND every corresponding entry is equal
- A matrix can represent a system of equations as an augmented matrix (coefficients | constants)
- Row operations (swap rows, multiply a row by a nonzero constant, add a multiple of one row to another) don't change the system's solution — basis of row-reduction/Gaussian elimination
- The identity matrix I (1's on the diagonal, 0's elsewhere) acts like the number 1 in matrix multiplication: $AI=IA=A$
Matrix operations
- Add/subtract matrices by adding/subtracting corresponding entries — matrices MUST be the same dimensions
- Scalar multiplication: multiply every entry by the scalar
- Matrix multiplication AB is defined only if the number of COLUMNS of A equals the number of ROWS of B
- Result dimensions: if A is m×n and B is n×p, then AB is m×p
- Entry (row i, col j) of AB $=$ dot product of row i of A with column j of B
- Matrix multiplication is NOT commutative in general: AB ≠ BA
Determinants
- 2×2 matrix [[a,b],[c,d]]: determinant $=$ ad−bc
- A matrix has an inverse iff its determinant ≠ 0 (a matrix with $det=0$ is called 'singular')
- 3×3 determinant via cofactor expansion along the top row: expand and use the checkerboard sign pattern (+,−,+)
- Example: $det([[2,3],[1,4]])=2(4)−3(1)=8−3=5$
- The inverse of a 2×2 matrix [[a,b],[c,d]] is (1/det)·[[d,−b],[−c,a]]
Cramer's Rule
- For $ax+by=e, cx+dy=f: x=Dx/D, y=Dy/D$, where D is the coefficient matrix's determinant
- $D=det([[a,b],[c,d]]); Dx=det([[e,b],[f,d]])$ (replace the x-column with constants); $Dy=det([[a,e],[c,f]])$ (replace the y-column)
- Example: $2x+y=5, x−3y=−8$ → $D=2(−3)−1(1)=−7; Dx=5(−3)−1(−8)=−7; Dy=2(−8)−5(1)=−21$ → $x=1, y=3$
- If $D=0$, Cramer's Rule fails — the system either has no solution or infinitely many (cannot be determined by Cramer's Rule alone)
Cramer's Rule: $x=Dx/D, y=Dy/D$
Inverse of [[a,b],[c,d]] $=$ (1/det)[[d,−b],[−c,a]]
Unit 8: Sequences, Series & Conics
▾Arithmetic sequences
- Explicit (nth term) formula: $aₙ=a₁+(n−1)d$
- d $=$ common difference $=$ aₙ − aₙ₋₁ (any term minus the one before it)
- Example: 5,9,13,17,... has $a₁=5, d=4$ → $a₁₀=5+9(4)=41$
- Recursive form: $aₙ=aₙ₋₁+d$, with a₁ given
- To find d given two non-consecutive terms: $d=(aₙ−aₘ)/(n−m)$
Geometric sequences
- Explicit (nth term) formula: $aₙ=a₁·r^(n−1)$
- r $=$ common ratio $=$ aₙ/aₙ₋₁ (any term divided by the one before it)
- Example: 3,6,12,24,... has $a₁=3, r=2$ → $a₈=3·2⁷=384$
- If |r|<1, terms shrink toward 0 (e.g., 8,4,2,1,0.5,...); if |r|>1, terms grow without bound
- Recursive form: $aₙ=r·aₙ₋₁$, with a₁ given
Series and sigma notation
- Sigma notation: $Σ(from k=1 to n)$ f(k) means add f(1)+f(2)+...+f(n)
- Arithmetic series sum: $Sₙ=n(a₁+aₙ)/2 =$ n[2a₁+(n−1)d]/2
- Finite geometric series sum: $Sₙ=a₁(1−rⁿ)/(1−r)$, for r≠1
- Infinite geometric series sum (converges only if |r|<1): $S=a₁/(1−r)$
- Example: find S for 1+1/2+1/4+1/8+...: $a₁=1, r=1/2$ → $S=1/(1−1/2)=2$
- Example: sum of first 20 terms of 4,7,10,...: $a₁=4, d=3, a₂₀=4+19(3)=61$ → $S₂₀=20(4+61)/2=650$
Parabolas
- Vertex form (vertex at origin), opens up/down: $x²=4py$ (p=distance from vertex to focus, and to directrix)
- Opens left/right: $y²=4px$
- If p>0: opens up (or right); if p<0: opens down (or left)
- Focus is INSIDE the curve at distance |p| from vertex; directrix is a line OUTSIDE at distance |p| on the opposite side
- Example: $x²=8y$ → $4p=8$ → $p=2$ → focus at (0,2), directrix $y=−2$
- General vertex-form parabola (vertex (h,k)): $(x−h)²=4p(y−k)$ or $(y−k)²=4p(x−h)$
Ellipses and hyperbolas (basics)
- Ellipse standard form centered at origin: $x²/a²+y²/b²=1$, where a is the semi-major axis, b the semi-minor axis (a>b)
- Ellipse foci lie on the major axis at distance c from center, where $c²=a²−b²$
- Hyperbola standard form (opens left/right): $x²/a²−y²/b²=1$; (opens up/down): $y²/a²−x²/b²=1$
- Hyperbola foci satisfy $c²=a²+b²$ (note the PLUS sign, unlike the ellipse)
- Hyperbolas have asymptotes $y=±(b/a)x$ (for the $x²/a²−y²/b²=1$ form) that the branches approach but never touch
- Example: $x²/25+y²/9=1$ is an ellipse with $a=5,b=3$ → $c²=25−9=16$ → $c=4$, foci at (±4,0)
$Sₙ=a₁(1−rⁿ)/(1−r)$ ; $S∞=a₁/(1−r)$, |r|<1
Ellipse: $c²=a²−b²$ ; Hyperbola: $c²=a²+b²$
Unit 9: Limits & Intro to Calculus Concepts
▾The intuitive idea of a limit
- A limit describes the function's BEHAVIOR NEAR a point, not necessarily the value AT that point
- Two-sided limit exists only if the left-hand limit and right-hand limit are EQUAL: $lim(x→a⁻)f(x)=lim(x→a⁺)f(x)$
- Notation: lim(x→a⁻) means approaching from the left (values less than a); lim(x→a⁺) means approaching from the right
- A limit can exist at a point even if the function is undefined there (e.g., a hole in the graph)
- A limit fails to exist if the function jumps (left/right limits differ), oscillates infinitely, or increases/decreases without bound near a
Evaluating limits graphically and numerically
- Graphically: trace the curve from both sides toward $x=a$ and see what y-value it approaches (ignore any open/closed dot exactly at $x=a$)
- Numerically: build a table of x-values very close to a (e.g., a−0.1, a−0.01, a+0.01, a+0.1) and see what f(x) approaches
- Example: lim(x→2) (x²−4)/(x−2) — direct substitution gives 0/0 (indeterminate), but the table shows values approaching 4
- Algebraically confirm: $(x²−4)/(x−2)=(x−2)(x+2)/(x−2)=x+2$ (for x≠2) → limit as x→2 is $2+2=4$
- Direct substitution works to evaluate a limit whenever the function is continuous at that point (most polynomials, and rational functions where the denominator isn't 0)
Average rate of change
- Average rate of change $=$ [f(b)−f(a)] / (b−a)
- This is exactly a SLOPE calculation — same formula as (y₂−y₁)/(x₂−x₁)
- Example: $f(x)=x²$, find average rate of change on [1,4]: $[f(4)−f(1)]/(4−1)=(16−1)/3=5$
- As the interval [a,b] shrinks (b→a), the average rate of change approaches the INSTANTANEOUS rate of change at $x=a$ — this is the core idea that leads to the derivative
- Units matter in applications: e.g., average velocity $=$ change in position / change in time
Introduction to derivatives as slope of the tangent line
- Definition: $f'(a)=lim(h→0)$ [f(a+h)−f(a)]/h — the limit of the difference quotient as h→0
- This is the limiting slope of secant lines through (a,f(a)) and (a+h,f(a+h)) as h shrinks toward 0
- Geometrically: the tangent line touches the curve at exactly one point (locally) and matches its instantaneous direction there
- Example: for $f(x)=x², f'(a)=lim(h→0)$ [(a+h)²−a²]/h $=$ lim(h→0) [2ah+h²]/h $= lim(h→0)(2a+h)=2a$ — so the slope of the tangent to $y=x²$ at any point a is 2a
- This 'slope of the tangent line' interpretation is the geometric heart of every derivative computed in calculus
$f'(a)=lim(h→0)$ [f(a+h)−f(a)]/h
Limit exists iff $lim(x→a⁻)f(x)=lim(x→a⁺)f(x)$
Practice Question Bank — 190 questions
Unit 1: Functions & Their Graphs (20)
If $f(x)=2x²−3x+1$, find f(3).
- 10
- 16
- 22
- 4
$f(3)=2(9)−3(3)+1=18−9+1=10$.
If $f(x)=x+2$ and $g(x)=x²$, find (f∘g)(3).
- 11
- 25
- 20
- 7
$(f∘g)(3)=f(g(3))=f(9)=9+2=11$.
If $f(x)=x+2$ and $g(x)=x²$, find (g∘f)(3).
- 11
- 25
- 20
- 7
$(g∘f)(3)=g(f(3))=g(5)=25$. Composition order matters — this differs from (f∘g)(3).
What is the domain of $f(x)=1/(x−5)?$
- All real numbers
- All real numbers except $x=5$
- x≥5
- x≤5
The denominator cannot equal 0, so x≠5 is excluded.
What is the domain of $f(x)=√(x−4)?$
- x≥4
- x≤4
- x>0
- All real numbers
The radicand must be ≥0 for a real, even-root output: x−4≥0 → x≥4.
Which test determines whether a graph represents a function?
- Horizontal Line Test
- Vertical Line Test
- Origin symmetry test
- Intercept test
The Vertical Line Test: if any vertical line crosses the graph more than once, it is not a function.
The graph of $y=(x−3)²+5$ is the graph of $y=x²$ shifted:
- Right 3, up 5
- Left 3, up 5
- Right 3, down 5
- Left 3, down 5
(x−h) shifts right h units ($h=3$); +k shifts up k units ($k=5$).
The graph of $y=f(x+4)$ is obtained from $y=f(x)$ by shifting:
- Right 4 units
- Left 4 units
- Up 4 units
- Down 4 units
f(x+h) shifts LEFT h units — opposite of the sign inside the parentheses.
Which best describes the transformation from $y=x²$ to $y=3x²?$
- Vertical stretch by factor 3
- Vertical compression by factor 3
- Horizontal stretch by factor 3
- Shift up 3 units
Multiplying the whole function by a>1 (outside) vertically stretches the graph.
Which function is even?
- $f(x)=x³$
- $f(x)=x²+3$
- $f(x)=x+1$
- $f(x)=sin$ x
$f(−x)=(−x)²+3=x²+3=f(x)$, so it's even. It's symmetric about the y-axis.
Which function is odd?
- $f(x)=x²$
- $f(x)=cos$ x
- $f(x)=x³−x$
- $f(x)=|x|$
$f(−x)=(−x)³−(−x)=−x³+x=−(x³−x)=−f(x)$, so it's odd.
Given f(x)={2x+1 if x<3; x²−2 if x≥3}, find f(3).
- 7
- 6
- 3
- 9
Since 3≥3, use the second rule: $f(3)=3²−2=7$.
Given f(x)={2x+1 if x<3; x²−2 if x≥3}, find f(1).
- 3
- −1
- 1
- 2
Since 1<3, use the first rule: $f(1)=2(1)+1=3$.
What is the range of $f(x)=−x²+4$ (a downward parabola with vertex (0,4))?
- y≤4
- y≥4
- All real numbers
- y≤0
The parabola opens downward with maximum value 4 at the vertex, so the range is y≤4.
The difference quotient [f(x+h)−f(x)]/h for $f(x)=x²$ simplifies to:
- 2x+h
- 2x
- x+h
- 2x−h
$[(x+h)²−x²]/h=[2xh+h²]/h=2x+h$.
Which transformation reflects $y=f(x)$ over the x-axis?
- $y=f(−x)$
- $y=−f(x)$
- $y=f(x)−1$
- $y=1/f(x)$
−f(x) negates every output (y-value), flipping the graph over the x-axis. f(−x) reflects over the y-axis instead.
If f(x) = 2x + 3 and g(x) = x² − 1, then f(g(x)) =
- 4x² + 12x + 8
- 2x² − 1
- 2x² + 1
- 2x² + 2
Substitute g into f: 2(x² − 1) + 3 = 2x² + 1.
The inverse of f(x) = 3x − 6 is:
- f⁻¹(x) = (x + 6)/3
- f⁻¹(x) = 3x + 6
- f⁻¹(x) = x/3 − 6
- f⁻¹(x) = (x − 6)/3
Swap x and y and solve: x = 3y − 6 gives y = (x + 6)/3.
The domain of f(x) = √(x − 4) is:
- x > 4
- x ≤ 4
- all real numbers
- x ≥ 4
The expression under the square root must be at least 0.
The graph of y = f(x − 3) + 2 is the graph of f shifted:
- left 3 and up 2
- right 3 and up 2
- right 3 and down 2
- left 3 and down 2
Replacing x with x − 3 moves the graph right; adding 2 moves it up.
Unit 2: Polynomial & Rational Functions (21)
What is the end behavior of $f(x)=−3x⁴+2x−1?$
- Rises left, rises right
- Falls left, falls right
- Rises left, falls right
- Falls left, rises right
Even degree (4) with negative leading coefficient → falls on both ends.
What is the end behavior of $f(x)=2x⁵−x²+3?$
- Falls left, rises right
- Rises left, falls right
- Rises left, rises right
- Falls left, falls right
Odd degree (5) with positive leading coefficient → falls left, rises right.
Find the zeros of $f(x)=x²−4x−5$.
- $x=5$, −1
- $x=−5$, 1
- $x=5$, 1
- $x=−5$, −1
Factor: $(x−5)(x+1)=0$ → $x=5$ or $x=−1$.
For $f(x)=(x−2)²(x+3)$, what is the behavior at $x=2?$
- Crosses the x-axis
- Bounces off (touches) the x-axis
- Passes through a vertical asymptote
- Passes through a hole
The factor (x−2) has multiplicity 2 (even) → the graph touches and turns back at $x=2$.
Which are the possible rational roots of $2x³+x²−13x+6=0$ given by the Rational Root Theorem, considering $constant=6$ and leading $coefficient=2?$
- ±1,2,3,6,1/2,3/2
- ±1,2,3,6 only
- ±1,3,6 only
- ±2,3,6,1/6
Candidates $=$ ±(factors of 6)/(factors of 2) $=$ ±1,2,3,6,1/2,3/2.
Using synthetic division, divide x³−6x²+11x−6 by (x−1). What is the remainder?
- 0
- 6
- −6
- 1
Synthetic division with root 1: 1,−6,11,−6 → bring down 1, $1(1)−6=−5, −5(1)+11=6, 6(1)−6=0$. Remainder is 0, so $x=1$ is a root.
What is the vertical asymptote of $f(x)=(x+1)/(x−4)?$
- $x=4$
- $x=−1$
- $x=−4$
- $y=1$
Set the denominator to 0: $x−4=0$ → $x=4$. The numerator isn't zero there, so it's a true vertical asymptote.
What is the horizontal asymptote of $f(x)=(2x²+5)/(x²−9)?$
- $y=2$
- $y=0$
- No horizontal asymptote
- $y=5/9$
Degrees are equal (both 2), so HA $=$ ratio of leading coefficients $=$ 2/1 $=$ 2.
What is the horizontal asymptote of $f(x)=(3x+1)/(x²−4)?$
- $y=3$
- $y=0$
- No horizontal asymptote
- $y=1/4$
Numerator degree (1) is less than denominator degree (2), so HA is $y=0$.
For $f(x)=(x²+3x)/(x²−9)$, where is there a HOLE (not an asymptote)?
- $x=3$
- $x=−3$
- $x=0$
- $x=9$
$f(x)=x(x+3)/[(x−3)(x+3)]$. The (x+3) factor cancels, creating a hole at $x=−3. x=3$ remains a vertical asymptote.
Does $f(x)=(x²+1)/(x−2)$ have a horizontal or slant asymptote?
- Horizontal asymptote $y=0$ (denominator degree higher)
- Slant asymptote (numerator degree exactly 1 more)
- Horizontal asymptote $y=1$ (degrees are equal)
- No asymptote of either kind (degrees differ by one)
Numerator degree (2) is exactly one more than denominator degree (1) → a slant asymptote exists, found by long division.
Solve $(x−3)/(x+2)=0$.
- $x=3$
- $x=−2$
- $x=0$
- No solution
A fraction equals 0 only when its numerator is 0 (and denominator isn't): $x−3=0$ → $x=3$.
Solve $3/(x−1)=6/(x+2)$ for x.
- $x=4$
- $x=−4$
- $x=1$
- $x=2$
Cross-multiply: $3(x+2)=6(x−1)$ → $3x+6=6x−6$ → $12=3x$ → $x=4$. Check: neither denominator is 0.
Which factorization is correct for x³−8?
- (x−2)(x²+2x+4)
- (x−2)(x²−2x+4)
- (x+2)(x²−2x+4)
- (x−2)³
Difference of cubes: $a³−b³=(a−b)(a²+ab+b²)$, with $a=x,b=2$: (x−2)(x²+2x+4).
Solve the rational inequality (x−1)/(x+2)≥0.
- x<−2 or x≥1
- −2<x≤1
- x≤−2 or x≥1
- −2≤x≤1
Critical values $x=−2$ (excluded, undefined) and $x=1$ (included, zero). Testing intervals shows the expression is ≥0 for x<−2 or x≥1.
A polynomial of degree 4 can have at most how many real zeros (counting multiplicity)?
- 3
- 4
- 5
- 2
A degree-n polynomial has at most n real zeros, counting multiplicity — degree 4 → at most 4.
The remainder when x³ − 2x + 5 is divided by x − 2 is:
- 5
- 1
- 9
- 0
By the Remainder Theorem the remainder is f(2) = 8 − 4 + 5 = 9.
The zeros of f(x) = x³ − 7x + 6 are:
- 1, 2 and −3
- −1, −2 and 3
- 1, −2 and 3
- 6, 1 and −1
f(x) = (x − 1)(x − 2)(x + 3).
The horizontal asymptote of f(x) = (2x² + 1)/(x² − 4) is:
- y = 0
- y = 1/2
- There is none
- y = 2
Equal degrees: the asymptote is the ratio of the leading coefficients.
f(x) = (2x² + 1)/(x² − 4) has vertical asymptotes at:
- x = 4 only
- x = 2 and x = −2
- x = 0
- x = 2 only
The denominator is 0 at x = ±2, and the numerator is not 0 there.
A polynomial of degree 5 with a positive leading coefficient has which end behavior?
- falls to the left and rises to the right
- rises to the left and falls to the right
- falls to the left and falls to the right
- rises to the left and rises to the right
Odd degree with a positive leading coefficient: down on the left, up on the right.
Unit 3: Exponential & Logarithmic Functions (21)
A population of 2000 grows 5% per year. What is the population after 3 years, $A(t)=2000(1.05)^t?$
- ≈2315.3
- 2300
- ≈2310
- ≈2321.9
$A(3)=2000(1.05)³=2000(1.157625)≈2315.3$.
Simplify: $8^(2/3)$.
- 4
- 16
- 2
- 64
$8^(2/3)=(³√8)²=2²=4$.
Simplify: 2⁴·2³.
- 2⁷
- 2¹²
- 4⁷
- 2¹
Product rule: $bᵐ·bⁿ=bᵐ⁺ⁿ$ → $2⁴⁺³=2⁷$.
Simplify: (3²)³.
- 3⁵
- 3⁶
- 9⁶
- 3⁹
Power rule: $(bᵐ)ⁿ=bᵐⁿ$ → $3^(2·3)=3⁶$.
Solve $2^(x+1)=8$.
- $x=2$
- $x=3$
- $x=1$
- $x=4$
$8=2³$, so $2^(x+1)=2³$ → $x+1=3$ → $x=2$.
Rewrite $log₂(32)=5$ in exponential form.
- $2⁵=32$
- $5²=32$
- $2³²=5$
- $32⁵=2$
$log_b(x)=y$ means $bʸ=x$, so $log₂(32)=5$ means $2⁵=32$.
Simplify $log_b(x³y)$ using log properties.
- $3log_b(x)+log_b(y)$
- $log_b(3x)+log_b(y)$
- $3log_b(x)·log_b(y)$
- $log_b(x)+3log_b(y)$
Product rule then power rule: $log_b(x³y)=log_b(x³)+log_b(y)=3log_b(x)+log_b(y)$.
Simplify log₅(125)−log₅(5).
- 2
- 3
- 4
- 1
$log₅(125)=3$ (since $5³=125$); $log₅(5)=1$. $3−1=2$. Also equals $log₅(125/5)=log₅(25)=2$.
Solve $log₂(x)=5$.
- $x=32$
- $x=10$
- $x=25$
- $x=2.5$
Rewrite in exponential form: $x=2⁵=32$.
Solve $5^x=40$ (round to 3 decimal places).
- ≈2.292
- ≈1.723
- ≈8.000
- ≈3.401
$x=ln(40)/ln(5)≈3.689/1.609≈2.292$.
Solve $log(x−1)=2$.
- $x=101$
- $x=100$
- $x=21$
- $x=3$
$log(x−1)=2$ (base 10) means $x−1=10²=100$ → $x=101$.
Which value of x makes log₃(x−4) undefined?
- $x=4$
- $x=5$
- $x=7$
- $x=13$
The argument (x−4) must be >0, so $x=4$ makes it 0, which is not allowed — the expression is undefined there.
What is ln(e⁵)?
- 5
- e⁵
- 5e
- 1
ln and eˣ are inverse functions: $ln(eˣ)=x$, so $ln(e⁵)=5$.
Solve $e^(2x)=15$ (round to 3 decimal places).
- ≈1.354
- ≈2.708
- ≈0.677
- ≈3.750
Take ln of both sides: $2x=ln(15)≈2.708$ → x≈1.354.
$1000 is invested at 4% annual interest compounded continuously. Which formula gives the balance after t years?
- $A=1000e^(0.04t)$
- $A=1000(1.04)^t$
- $A=1000+0.04t$
- $A=1000(0.04)^t$
Continuous compounding uses $A=Pe^(rt)$, with $P=1000, r=0.04$.
A radioactive isotope has a half-life of 10 years. Using $A(t)=A₀(1/2)^(t/10)$, what fraction remains after 30 years?
- 1/8
- 1/2
- 1/4
- 1/16
$A(30)/A₀=(1/2)^(30/10)=(1/2)³=1/8$.
log₂ 8 =
- 3
- 2
- 4
- 8
2³ = 8.
Solve 2ˣ = 32.
- x = 4
- x = 6
- x = 16
- x = 5
32 = 2⁵.
log(ab) equals:
- log a · log b
- log a + log b
- log a − log b
- b log a
The product rule for logarithms.
Solve ln x = 2.
- x = 2e
- x = ln 2
- x = e²
- x = 100
Exponentiate both sides with base e.
A population of 1000 doubles every 4 years. After 12 years it is:
- 8000
- 3000
- 6000
- 4000
Three doublings: 1000 · 2³ = 8000.
Unit 4: Trigonometric Functions (22)
Convert 150° to radians.
- $5π/6$
- $2π/3$
- $π/6$
- $7π/6$
$150×(π/180)=150π/180=5π/6$.
Convert $7π/4$ radians to degrees.
- 315°
- 280°
- 210°
- 350°
$(7π/4)×(180/π)=7(180)/4=315°$.
What is sin(30°)?
- 1/2
- √2/2
- √3/2
- 1
$\sin(30°)=1/2$, one of the standard special-angle values.
What is cos(60°)?
- 1/2
- √3/2
- √2/2
- 0
$\cos(60°)=1/2$, a standard special-angle value.
What is tan(45°)?
- 1
- 0
- √2
- undefined
$\tan(45°)=\sin(45°)/\cos(45°)=(√2/2)/(√2/2)=1$.
In which quadrant is tan $θ$ positive but sin $θ$ negative?
- QIII
- QI
- QII
- QIV
By ASTC, QIII has tangent positive (both sine and cosine negative, so their ratio is positive) while sine itself is negative.
Find the reference angle for 210°.
- 30°
- 210°
- 150°
- 60°
QIII: reference angle $= θ−180° = 210−180=30°$.
Find sin(210°) using the reference angle.
- −1/2
- 1/2
- −√3/2
- √3/2
Reference angle is 30° ($sin=1/2$); QIII has sine negative → $\sin(210°)=−1/2$.
Find cos(315°).
- √2/2
- −√2/2
- √3/2
- −1/2
Reference angle for 315° (QIV) is $360−315=45°; \cos(45°)=√2/2$; QIV has cosine positive → $\cos(315°)=√2/2$.
What is the period of $y=2sin(3x)+1?$
- $2π/3$
- $π/3$
- $2π$
- $6π$
$Period=2π/|B|=2π/3$.
What is the amplitude of $y=−4cos(x)+2?$
- 4
- −4
- 2
- 6
$Amplitude=|A|=|−4|=4$.
What is the midline (vertical shift) of $y=3sin(2x)−5?$
- $y=−5$
- $y=3$
- $y=2$
- $y=5$
$D=−5$ gives the midline $y=−5$, since the function is written as A sin(Bx)+D.
What is the period of $y=\tan(x/2)?$
- $2π$
- $π$
- $π/2$
- $4π$
Period of tangent $= π/|B| = π/(1/2) = 2π$.
An angle sweeps through an arc of length 12 cm on a circle of radius 4 cm. What is $θ$ in radians?
- 3
- 0.33
- 48
- 12
$s=rθ$ → $θ=s/r=12/4=3$ radians.
At which angle(s) in [0°,360°) is tan $θ$ undefined?
- 90° and 270°
- 0° and 180°
- 45° and 135°
- 60° and 240°
tan $θ=\sinθ/\cosθ$ is undefined where cos $θ=0$, which occurs at 90° and 270°.
Which point lies on the unit circle at $θ=180°?$
- (−1,0)
- (1,0)
- (0,−1)
- (0,1)
At 180°, the terminal point is $(cos180°,sin180°)=(−1,0)$.
sin(π/6) =
- √3/2
- √2/2
- 1
- 1/2
On the unit circle, π/6 (30°) has y-coordinate 1/2.
cos π =
- 0
- −1
- 1
- 1/2
At angle π the point on the unit circle is (−1, 0).
150° in radians is:
- 2π/3
- 3π/4
- 5π/6
- 5π/3
Multiply by π/180: 150π/180 = 5π/6.
The period of y = sin(2x) is:
- π
- 2π
- 4π
- π/2
Period = 2π/|b| = 2π/2.
The amplitude of y = −3 cos x + 1 is:
- −3
- 1
- 4
- 3
Amplitude is the absolute value of the coefficient of the cosine.
tan(π/4) =
- 0
- 1
- √3
- √3/3
sin and cos are equal at π/4, so their ratio is 1.
Unit 5: Trigonometric Identities & Equations (21)
If $\sinθ=3/5$ and $θ$ is in QI, what is $\cosθ?$
- 4/5
- 3/4
- 5/4
- −4/5
$sin²θ+cos²θ=1$ → $cos²θ=1−9/25=16/25$ → $\cosθ=4/5$ (positive in QI).
Simplify $sin²θ$ + $cos²θ$ + $tan²θ$.
- $1+tan²θ$
- $sec²θ$ (equivalently $1+tan²θ$)
- 2
- $csc²θ$
$sin²θ+cos²θ=1$, so the expression is $1+tan²θ$, which equals $sec²θ$ by the Pythagorean identity.
Find the exact value of sin(75°) using sin(45°+30°).
- (√6+√2)/4
- (√6−√2)/4
- √2/2
- (√3+1)/2
$\sin(45+30)=sin45cos30+cos45sin30=(√2/2)(√3/2)+(√2/2)(1/2)=(√6+√2)/4$.
Find the exact value of cos(75°) using cos(45°+30°).
- (√6+√2)/4
- (√6−√2)/4
- (√2−√3)/4
- 1/2
$\cos(45+30)=cos45cos30−sin45sin30=(√2/2)(√3/2)−(√2/2)(1/2)=(√6−√2)/4$.
If $\sinθ=3/5$ and $\cosθ=4/5$ (θ in QI), find $\sin(2θ)$.
- 24/25
- 7/25
- 12/25
- 6/5
$\sin(2θ)=2sinθcosθ=2(3/5)(4/5)=24/25$.
If $\cosθ=4/5$ and $θ$ in QI, find $\cos(2θ)$ using $\cos(2θ)=2cos²θ−1$.
- 7/25
- 24/25
- −7/25
- 16/25
$\cos(2θ)=2(4/5)²−1=2(16/25)−1=32/25−25/25=7/25$.
Solve $2sinθ−1=0$ for $θ$ in [$0,2π$).
- $π/6, 5π/6$
- $π/6, 7π/6$
- $π/3, 2π/3$
- $π/6$ only
$\sinθ=1/2$ → reference angle $π/6$; sine positive in QI and QII → $θ=π/6$ and $5π/6$.
Solve $2cosθ+√3=0$ for $θ$ in [$0,2π$).
- $5π/6, 7π/6$
- $π/6, 11π/6$
- $5π/6, 11π/6$
- $π/3, 2π/3$
$\cosθ=−√3/2$ → reference angle $π/6$; cosine negative in QII and QIII → $θ=5π/6$ and $7π/6$.
Solve $2sin²θ−\sinθ−1=0$ for $θ$ in [$0,2π$). (Factor: $(2sinθ+1)(\sinθ−1)=0$)
- $π/2, 7π/6, 11π/6$
- $π/2$ only
- $7π/6, 11π/6$
- $π/6, 5π/6, π/2$
$\sinθ=1$ → $θ=π/2. \sinθ=−1/2$ → $θ=7π/6, 11π/6$. All three solutions satisfy the interval.
Simplify $(1−cos²θ)/\sinθ$.
- $\sinθ$
- $\cosθ$
- $\tanθ$
- 1
$1−cos²θ=sin²θ$ (Pythagorean identity), so $sin²θ/\sinθ=\sinθ$.
Which is the correct sum formula for tan(A+B)?
- (tanA+tanB)/(1−tanAtanB)
- tanA+tanB
- (tanA−tanB)/(1+tanAtanB)
- (tanA+tanB)/(1+tanAtanB)
$\tan(A+B)=(tanA+tanB)/(1−tanAtanB)$.
Simplify $\sinθ·cotθ$.
- $\cosθ$
- $\sinθ$
- $\tanθ$
- 1
$cotθ=\cosθ/\sinθ$, so $\sinθ·(\cosθ/\sinθ)=\cosθ$.
Which identity is used to derive $1+cot²θ=csc²θ?$
- Dividing $sin²θ+cos²θ=1$ by $sin²θ$
- Dividing $sin²θ+cos²θ=1$ by $cos²θ$
- Multiplying $sin²θ+cos²θ=1$ by $cos²θ$
- Subtracting $cos²θ$ from both sides
Dividing every term of $sin²θ+cos²θ=1$ by $sin²θ$ gives $1+cot²θ=csc²θ$.
Solve $\cos(2θ)=\cosθ$ for $θ$ in [$0,2π$) using $\cos(2θ)=2cos²θ−1: 2cos²θ−\cosθ−1=0$.
- 0, $2π/3, 4π/3$
- 0, $π/3$
- $2π/3, 4π/3$
- $π/2, 3π/2$
Factor $(2cosθ+1)(\cosθ−1)=0$ → $\cosθ=1 (θ=0)$ or $\cosθ=−1/2 (θ=2π/3, 4π/3)$.
What is $sec²θ$ − $tan²θ$ equal to for all $θ?$
- 1
- 0
- $2tan²θ$
- $sin²θ$
From $1+tan²θ=sec²θ$, rearranged: $sec²θ−tan²θ=1$ for all $θ$.
Find cos(15°) using cos(45°−30°).
- (√6+√2)/4
- (√6−√2)/4
- √3/2−1/2
- 1/2
$\cos(45−30)=cos45cos30+sin45sin30=(√2/2)(√3/2)+(√2/2)(1/2)=(√6+√2)/4$.
sin²θ + cos²θ =
- 0
- tan²θ
- 1
- 2
The Pythagorean identity.
sin(2θ) =
- 2 sin θ cos θ
- 2 sin θ
- sin²θ − cos²θ
- cos²θ − sin²θ
The double-angle identity for sine.
1 + tan²θ =
- csc²θ
- cot²θ
- cos²θ
- sec²θ
Divide sin²θ + cos²θ = 1 by cos²θ.
The solutions of sin x = 1/2 on [0, 2π) are:
- π/3 and 2π/3
- π/6 and 5π/6
- π/6 and 7π/6
- π/6 only
Sine is positive in quadrants I and II, with reference angle π/6.
cos(2θ) can be written as:
- 2 sin θ cos θ
- 1 + 2 sin²θ
- 1 − 2 sin²θ
- 2 cos θ
cos(2θ) = cos²θ − sin²θ = 1 − 2 sin²θ.
Unit 6: Analytic Trigonometry & Applications (21)
In a triangle, $A=40°, a=15, B=60°$. Find b using the Law of Sines.
- ≈20.2
- ≈17.5
- ≈23.4
- ≈12.9
$b=a·sinB/sinA=15·sin60°/sin40°=15(0.866)/(0.643)≈20.2$.
In a triangle, $a=8, b=10, C=50°$. Find c using the Law of Cosines.
- ≈7.82
- ≈9.15
- ≈6.40
- ≈8.90
$c²=8²+10²−2(8)(10)cos50°=164−160(0.643)≈61.1$ → c≈7.82.
In a triangle, $a=5, b=7, c=10$. Find angle C using the Law of Cosines.
- ≈111.8°
- ≈68.2°
- ≈90°
- ≈45.6°
$cosC=(25+49−100)/(2·5·7)=−26/70≈−0.371$ → $C=cos⁻¹(−0.371)≈111.8°$.
Find the area of a triangle with $a=6, b=9$, and included angle $C=30°$.
- 13.5
- 27
- 54
- 6.75
$Area=(1/2)ab·sinC=(1/2)(6)(9)(0.5)=13.5$.
Use Heron's formula to find the area of a triangle with sides $a=5, b=6, c=7 (s=9)$.
- ≈14.70
- ≈17.5
- ≈21
- ≈12.25
$Area=√[9(9−5)(9−6)(9−7)]=√[9·4·3·2]=√216≈14.70$.
Which case requires starting with the Law of Cosines rather than the Law of Sines?
- SSS (three sides, no angles)
- AAS
- ASA
- SSA
With SSS, no angle is known, so Law of Sines (which needs an angle-side pair) can't start the problem — Law of Cosines is needed first.
What is sin⁻¹(1/2)?
- 30° ($π/6$)
- 150° ($5π/6$)
- Both 30° and 150°
- 60° ($π/3$)
sin⁻¹ returns only values in [−90°,90°]. Even though $sin150°=1/2$ too, only 30° is in the restricted range.
What is cos⁻¹(−1/2)?
- 120°
- 60°
- −60°
- 240°
cos⁻¹ has range [0°,180°]. $\cos(120°)=−1/2$ and 120° is within that range.
What is the range of $y=tan⁻¹(x)?$
- (−90°,90°)
- [0°,180°]
- [−90°,90°]
- (0°,180°)
arctan's range is the open interval ($−π/2,π/2$), i.e., (−90°,90°), since tangent has vertical asymptotes there.
In triangle solving with SSA, given $a=7, b=10, A=40°$, compute $h=b·sinA$ to test the ambiguous case.
- ≈6.43
- ≈5.36
- ≈7.66
- ≈9.19
$h=b·sinA=10·sin40°≈10(0.643)≈6.43$. Since $a=7>h$, and a<b, there are two possible triangles.
A triangle has $a=8, b=8, c=8$ (equilateral). What is angle A?
- 60°
- 90°
- 45°
- 120°
An equilateral triangle has all angles equal, and they must sum to 180°, so each is 60°.
Find angle B in a triangle where $A=35°, C=75°$.
- 70°
- 65°
- 110°
- 55°
Angles sum to 180°: $B=180−35−75=70°$.
Which formula finds a triangle's area when only the three sides (SSS) are known?
- Heron's formula
- (1/2)ab sinC
- (1/2)base×height only
- Law of Sines
Heron's formula computes area directly from the three side lengths without needing any angle.
If cosC turns out to be negative when applying the Law of Cosines, what does this mean?
- Angle C is acute (between 0° and 90°)
- Angle C is exactly 90° (a right angle)
- The triangle is impossible (no such triangle)
- Angle C is obtuse (between 90° and 180°)
A negative cosine value corresponds to an angle between 90° and 180° — this is a valid, expected result.
Evaluate tan⁻¹(1).
- 45°
- −45°
- 135°
- 90°
$\tan(45°)=1$, and 45° lies within arctan's range of (−90°,90°).
A surveyor measures two sides of a triangular lot as 120 ft and 150 ft with an included angle of 65°. Find the area.
- ≈8157 ft²
- ≈9000 ft²
- ≈16314 ft²
- ≈4078 ft²
$Area=(1/2)(120)(150)sin65°=9000(0.9063)≈8157$ ft².
In a triangle with sides 5 and 7 and included angle 60°, the third side is:
- √39
- √74
- √109
- 12
Law of Cosines: c² = 25 + 49 − 2(5)(7)(1/2) = 39.
The area of a triangle with sides 6 and 8 and included angle 30° is:
- 24
- 48
- 6
- 12
Area = (1/2)ab sin C = (1/2)(6)(8)(1/2) = 12.
The Law of Sines states that:
- a² = b² + c²
- a/sin A = b/sin B = c/sin C
- a sin A = b sin B
- a/cos A = b/cos B
Each side divided by the sine of its opposite angle gives the same value.
The magnitude of the vector ⟨3, −4⟩ is:
- 7
- 1
- 5
- 25
√(3² + (−4)²) = 5.
The dot product ⟨3, 4⟩ · ⟨1, 2⟩ is:
- 11
- 10
- 24
- 7
3·1 + 4·2 = 11.
Unit 7: Systems & Matrices (21)
Solve the system: $2x+3y=7$ and $4x−3y=11$.
- $x=3, y=1/3$
- $x=3, y=−1/3$
- $x=−3, y=13/3$
- $x=2, y=1$
Add the equations: $6x=18$ → $x=3$. Substitute: $2(3)+3y=7$ → $3y=1$ → $y=1/3$.
Solve the system: $x+y=5$ and $x−y=1$.
- $x=3, y=2$
- $x=2, y=3$
- $x=4, y=1$
- $x=1, y=4$
Add: $2x=6$ → $x=3$. Then $y=5−3=2$.
Which describes a system with no solution?
- Elimination leads to a false statement like $0=5$
- Elimination leads to a true statement like $0=0$
- Elimination gives a unique x and y
- The lines have different slopes
A false numerical statement means the equations are contradictory — parallel lines that never intersect.
Which describes a system with infinitely many solutions?
- Elimination leads to a true statement like $0=0$
- Elimination leads to a false statement like $0=5$
- The system has a unique solution
- The lines are perpendicular
A true statement with no variables means the two equations represent the same line.
A matrix has 3 rows and 5 columns. What are its dimensions?
- 3×5
- 5×3
- 8
- 15
Matrix dimensions are given as rows × columns: 3×5.
For matrix multiplication AB to be defined, what must be true?
- Columns of A $=$ Rows of B
- Rows of A $=$ Rows of B
- Columns of A $=$ Columns of B
- A and B must have the same dimensions
Matrix multiplication AB requires the number of columns in A to equal the number of rows in B.
If A is 2×3 and B is 3×4, what are the dimensions of AB?
- 2×4
- 3×3
- 2×3
- 4×2
The product takes the outer dimensions: rows of A (2) × columns of B (4) $=$ 2×4.
Find the determinant of [[2,3],[1,4]].
- 5
- 11
- −5
- 10
$det=ad−bc=2(4)−3(1)=8−3=5$.
Find the determinant of [[5,2],[3,1]].
- −1
- 1
- 11
- −11
$det=5(1)−2(3)=5−6=−1$.
A matrix has determinant 0. What does this mean?
- The matrix is singular (has no inverse)
- The matrix equals the identity matrix
- The matrix has an inverse
- The matrix is a zero matrix
A determinant of 0 means the matrix is 'singular' and has no inverse.
Use Cramer's Rule to solve $2x+y=5, x−3y=−8$. What is D (the coefficient determinant)?
- −7
- −2
- 2
- 7
$D=det([[2,1],[1,−3]])=2(−3)−1(1)=−6−1=−7$.
Using the system $2x+y=5, x−3y=−8 (D=−7)$, find x using Cramer's Rule, where $Dx=det([[5,1],[−8,−3]])$.
- 1
- −1
- 3
- 7
$Dx=5(−3)−1(−8)=−15+8=−7. x=Dx/D=−7/−7=1$.
Find the inverse of [[2,1],[1,1]].
- [[1,−1],[−1,2]]
- [[1,1],[1,2]]
- [[2,−1],[−1,1]]
- [[1,−1],[−1,1]]
$det=2(1)−1(1)=1. Inverse=(1/1)[[1,−1],[−1,2]]=[[1,−1],[−1,2]]$.
If Cramer's Rule gives $D=0$ for a system, what can be concluded?
- The system has exactly one solution, at the point $(0,0)$ and nowhere else
- The system has no solution at all, so it is always an inconsistent system
- The system has exactly two solutions, one for each variable in the system
- The system has no unique solution — either none or infinitely many
$D=0$ means Cramer's Rule cannot determine a unique solution — further analysis (or another method) is needed to tell no-solution from infinite-solutions.
Is matrix multiplication commutative ($AB=BA$) in general?
- No, order matters
- Yes, always
- Only for square matrices
- Only for identity matrices
Matrix multiplication is generally NOT commutative — AB and BA can even have different dimensions or different values.
What is the identity matrix's role in matrix multiplication?
- $AI=IA=A$, like multiplying by 1
- It always produces the zero matrix
- It has determinant 0
- It reverses the sign of every entry
The identity matrix I acts as the multiplicative identity: multiplying any compatible matrix A by I returns A unchanged.
Solve the system 2x + y = 7 and x − y = 2.
- (2, 3)
- (1, 5)
- (4, −1)
- (3, 1)
Add the equations: 3x = 9, so x = 3 and y = 1.
The determinant of the matrix [[2, 3], [1, 4]] is:
- 11
- 5
- 8
- −5
ad − bc = (2)(4) − (3)(1) = 5.
A 2 × 3 matrix can be multiplied by a matrix with how many rows?
- 2
- 6
- 3
- any number
The number of columns of the first must equal the number of rows of the second.
A square matrix has no inverse when its determinant is:
- 0
- 1
- negative
- positive
A matrix is invertible only if its determinant is not zero.
[[1, 2], [3, 4]] + [[5, 6], [7, 8]] =
- [[5, 12], [21, 32]]
- [[6, 8], [10, 11]]
- [[4, 4], [4, 4]]
- [[6, 8], [10, 12]]
Add matching entries.
Unit 8: Sequences, Series & Conics (22)
Find the 10th term of the arithmetic sequence 5,9,13,17,...
- 41
- 45
- 37
- 44
$a₁=5, d=4. a₁₀=5+9(4)=5+36=41$.
Find the 8th term of the geometric sequence 3,6,12,24,...
- 384
- 192
- 768
- 96
$a₁=3, r=2. a₈=3·2⁷=3(128)=384$.
Find the sum of the first 20 terms of the arithmetic sequence 4,7,10,...
- 650
- 610
- 690
- 570
$a₂₀=4+19(3)=61. S₂₀=20(4+61)/2=20(65)/2=650$.
Find the sum of the infinite geometric series 1+1/2+1/4+1/8+...
- 2
- 1
- 4
- 1/2
$a₁=1, r=1/2$ (|r|<1, converges). $S=1/(1−1/2)=1/(1/2)=2$.
Find the sum of the first 6 terms of the geometric series 2+6+18+54+...
- 728
- 486
- 1458
- 242
$a₁=2, r=3. S₆=2(1−3⁶)/(1−3)=2(1−729)/(−2)=2(−728)/(−2)=728$.
Does the infinite series 3+6+12+24+... have a finite sum?
- No, since $r=2$ and |r|≥1
- Yes, it equals 6
- Yes, it equals −6
- Yes, it equals 3
$r=2$, and infinite geometric series only converge when |r|<1. Since |2|≥1, this series diverges (no finite sum).
Find the common difference of the arithmetic sequence where $a₅=17$ and $a₁₂=45$.
- 4
- 3
- 5
- $28/7=4$
$d=(a₁₂−a₅)/(12−5)=(45−17)/7=28/7=4$.
For the parabola $x²=8y$, find the focus.
- (0,2)
- (2,0)
- (0,4)
- (0,8)
$x²=4py$ → $4p=8$ → $p=2$. Vertex at origin, opens up, focus at $(0,p)=(0,2)$.
For the parabola $x²=8y$, find the directrix.
- $y=−2$
- $y=2$
- $x=−2$
- $y=−4$
With $p=2$, the directrix is the horizontal line $y=−p=−2$ (opposite side of vertex from the focus).
For the ellipse $x²/25+y²/9=1$, find the foci.
- (±4,0)
- (0,±4)
- (±3,0)
- (±5,0)
$a²=25,b²=9$ → $c²=25−9=16$ → $c=4$. Since the larger denominator is under x², foci are on the x-axis: (±4,0).
For the hyperbola $x²/16−y²/9=1$, find c (distance from center to each focus).
- 5
- 7
- √7
- 25
For a hyperbola, $c²=a²+b²=16+9=25$ → $c=5$.
What are the asymptotes of the hyperbola $x²/16−y²/9=1?$
- $y=±(3/4)x$
- $y=±(4/3)x$
- $y=±(9/16)x$
- $y=±(16/9)x$
For $x²/a²−y²/b²=1$, asymptotes are $y=±(b/a)x =$ ±(3/4)x ($a²=16→a=4, b²=9→b=3$).
Evaluate $Σ(k=1 to 5)$ (2k+1).
- 35
- 30
- 25
- 40
Terms: $3+5+7+9+11=35$ (an arithmetic series with $a₁=3, d=2, n=5: S=5(3+11)/2=35$).
A geometric sequence has $a₁=100$ and $r=0.5$. What is a₅?
- 6.25
- 12.5
- 3.125
- 25
$a₅=100(0.5)⁴=100(0.0625)=6.25$.
Which conic is $x²/9+y²/25=1?$
- An ellipse with major axis along the x-axis
- A hyperbola with vertices along the y-axis
- A circle of radius 5 centered at the origin
- An ellipse with major axis along the y-axis
Both terms are added (ellipse) and the larger denominator (25) is under y², so the major axis is vertical.
A ball is dropped from 10 ft and bounces back to 60% of its previous height each time. What is the total distance the ball has fallen for the infinite bounces (sum of all drop heights after the first, i.e., 10(0.6)+10(0.6)²+...)?
- 15 ft
- 20 ft
- 10 ft
- 6 ft
This is an infinite geometric series with $a₁=10(0.6)=6, r=0.6: S=6/(1−0.6)=6/0.4=15$ ft.
The 10th term of the arithmetic sequence 5, 8, 11, … is:
- 35
- 32
- 29
- 30
aₙ = 5 + (n − 1)(3), so a₁₀ = 5 + 27 = 32.
The sum 1 + 2 + 3 + … + 20 is:
- 200
- 190
- 210
- 420
n(n + 1)/2 = 20(21)/2 = 210.
The 5th term of the geometric sequence 2, 6, 18, … is:
- 162
- 54
- 486
- 48
aₙ = 2 · 3ⁿ⁻¹, so a₅ = 2 · 81 = 162.
The sum of the infinite geometric series 8 + 4 + 2 + … is:
- 14
- 15
- It diverges
- 16
a/(1 − r) = 8/(1 − 1/2) = 16.
The equation x²/9 + y²/4 = 1 describes:
- a hyperbola
- an ellipse
- a parabola
- a circle
A sum of squared terms with different positive denominators equal to 1 is an ellipse.
The center and radius of (x − 2)² + (y + 3)² = 25 are:
- center (−2, 3), radius 5
- center (2, −3), radius 25
- center (2, −3), radius 5
- center (2, 3), radius 5
Compare with (x − h)² + (y − k)² = r².
Unit 9: Limits & Intro to Calculus Concepts (21)
Evaluate lim(x→3) (x²−9)/(x−3).
- 6
- 0
- Undefined
- 9
Factor: $(x−3)(x+3)/(x−3)=x+3$ for x≠3. As x→3, this approaches $3+3=6$.
Evaluate lim(x→2) (x²+3x−1).
- 9
- 5
- 4
- 13
This is a polynomial (continuous everywhere) so use direct substitution: $4+6−1=9$.
If lim(x→2⁻) $f(x)=5$ and lim(x→2⁺) $f(x)=5$, what can be concluded about lim(x→2) f(x)?
- It equals 5
- It does not exist
- It equals f(2)
- It equals 0
Since the left-hand and right-hand limits agree (both 5), the two-sided limit exists and equals 5, regardless of f(2).
If lim(x→1⁻) $f(x)=3$ and lim(x→1⁺) $f(x)=7$, what can be concluded about lim(x→1) f(x)?
- It equals 5 (average of one-sided limits)
- It equals 3 (the left-hand limit decides)
- It equals 7 (the right-hand limit decides)
- It does not exist (jump discontinuity)
The one-sided limits disagree, so the overall (two-sided) limit does not exist there.
Find the average rate of change of $f(x)=x²$ over [1,4].
- 5
- 3
- 15
- 7
$[f(4)−f(1)]/(4−1)=(16−1)/3=15/3=5$.
Find the average rate of change of $f(x)=2x²+1$ over [0,3].
- 6
- 9
- 3
- 18
$[f(3)−f(0)]/(3−0)=[(19)−(1)]/3=18/3=6$.
What does the average rate of change of a function over [a,b] represent geometrically?
- The slope of the secant line through (a,f(a)) and (b,f(b))
- The area under the curve between $x=a$ and $x=b$ on the graph
- The slope of the tangent line at the single point $(a,f(a))$ only
- The vertical distance between $f(a)$ and $f(b)$ on the y-axis
Average rate of change is exactly the slope formula (y₂−y₁)/(x₂−x₁), which is the secant line's slope.
Using $f(x)=x²$, find f'(a) via the definition $f'(a)=lim(h→0)[f(a+h)−f(a)]/h$.
- 2a
- a²
- 2h
- a
$[(a+h)²−a²]/h=[2ah+h²]/h=2a+h$ → as h→0, this approaches 2a.
What does f'(a) represent geometrically?
- The slope of the tangent line to f at $x=a$
- The slope of the secant line over an interval
- The average value of f near a
- The y-intercept of f
The derivative f'(a) is defined as the limiting slope of secant lines as h→0, which is exactly the slope of the tangent line at $x=a$.
For $f(x)=x²+2x$, find f'(a) using the limit definition.
- 2a+2
- 2a
- a+2
- 2a+h
$[(a+h)²+2(a+h)−(a²+2a)]/h=[2ah+h²+2h]/h=2a+h+2$ → as h→0, this approaches 2a+2.
A function f is continuous at $x=a$. How can lim(x→a) f(x) be evaluated?
- By direct substitution: f(a)
- It cannot be found without a table
- Only using one-sided limits
- It always equals 0
If f is continuous at a, then $lim(x→a)f(x)=f(a)$, so plugging in a directly gives the limit.
Evaluate lim(x→0) (sin x)/x using a table of values approaching 0 (a well-known result).
- 1
- 0
- Undefined
- ∞
This classic limit approaches 1 as x→0 from both sides, even though direct substitution gives the indeterminate form 0/0.
A car travels so that its position is $s(t)=t²+3t$ (meters, seconds). Find the average velocity over [2,5].
- 10 m/s
- 8 m/s
- 12 m/s
- 6 m/s
$[s(5)−s(2)]/(5−2)=[(40)−(10)]/3=30/3=10$ m/s.
Which situation describes a limit that does NOT exist as x→a?
- The left and right limits approach the same value
- The left and right limits approach different values
- The function is undefined at $x=a$ but the limits agree
- The function has a hole at $x=a$ with matching sides
When the left-hand and right-hand limits disagree, no single limiting value exists — this is the defining condition for the limit failing to exist.
Given $f(x)=(x²−1)/(x−1)$ for x≠1, what is lim(x→1) f(x)?
- 2
- 0
- 1
- Undefined
Factor: $(x−1)(x+1)/(x−1)=x+1$ for x≠1. As x→1, this approaches $1+1=2$ (even though f(1) itself is undefined — a hole).
What is the relationship between average rate of change and the derivative as the interval [a,b] shrinks (b→a)?
- Average rate of change grows without bound and the derivative is undefined at $a$
- Average rate of change approaches the total change in $f$ over the interval, not the derivative
- Average rate of change approaches the instantaneous rate of change (the derivative) at a
- Average rate of change always equals the derivative at $a$ no matter how large the interval
As the interval shrinks to a single point, the secant slope (average rate) approaches the tangent slope (derivative) — this limiting process is the foundation of differential calculus.
lim_{x→2} (x² − 4)/(x − 2) =
- 4
- 0
- 2
- Does not exist
Factor and cancel: x + 2, which is 4 at x = 2.
lim_{x→∞} (3x + 1)/(x − 5) =
- 0
- ∞
- −1/5
- 3
Equal degrees, so the limit is the ratio of leading coefficients.
The average rate of change of f(x) = x² from x = 1 to x = 3 is:
- 2
- 4
- 8
- 3
(9 − 1)/(3 − 1) = 4.
The slope of the tangent line to a curve at a point is the:
- average of all nearby tangent slopes
- limit of the slopes of secant lines
- limit of areas beneath the secant lines
- value of the function at that point
As the second point approaches the first, secant slopes approach the tangent slope: the derivative.
Using the limit definition, the derivative of f(x) = x² at x = 3 is:
- 6
- 9
- 3
- 2
lim [(3 + h)² − 9]/h = lim (6 + h) = 6.
Hard Mode Questions — 14 questions
Unit 1: Functions & Their Graphs (1)
If f(x) = x^2 and g(x) = x + 1, then f(g(x)) equals:
- x^2 + 1
- (x + 1)^2
- x^2 + x
- x + 1
Substitute g(x) into f.
Unit 2: Polynomial & Rational Functions (2)
How many real zeros does x^3 - x have?
- 1
- 3
- 2
- 0
It factors as x(x - 1)(x + 1).
The vertical asymptote of f(x) = 1/(x - 2) is:
- x = 2
- y = 2
- x = 0
- y = 0
The denominator is zero at x = 2.
Unit 3: Exponential & Logarithmic Functions (2)
log base 2 of 8 equals:
- 3
- 4
- 2
- 8
2^3 = 8.
Solve 2^x = 16:
- x = 8
- x = 2
- x = 16
- x = 4
16 = 2^4.
Unit 4: Trigonometric Functions (1)
sin(pi/6) equals:
- sqrt(3)/2
- 1
- sqrt(2)/2
- 1/2
It is a standard unit circle value.
Unit 5: Trigonometric Identities & Equations (2)
Solve 2 sin x = 1 on [0, 2 pi):
- pi/3 and 2 pi/3
- pi/6 only
- pi/6 and 5 pi/6
- pi/4 and 3 pi/4
sin x = 1/2 has solutions in quadrants I and II.
The Pythagorean identity states that:
- sin x + cos x = 1
- sin^2 x + cos^2 x = 1
- tan x = sin x + cos x
- sin^2 x - cos^2 x = 1
It follows from the unit circle.
Unit 6: Analytic Trigonometry & Applications (2)
The Law of Cosines is most useful when you are given:
- Two angles and the side between them (ASA), or one side
- Two sides and a non-included angle, or all three angles
- Two sides and the included angle, or all three sides
- Only one side and the right angle of a right triangle
These cases cannot be solved directly with the Law of Sines.
In a right triangle with sides 3, 4, 5, the sine of the larger acute angle is:
- 4/5
- 3/5
- 3/4
- 5/4
The larger acute angle is opposite the side of length 4.
Unit 7: Systems & Matrices (1)
The solution of x + y = 3 and x - y = 1 is:
- (1, 2)
- (3, 0)
- (0, 3)
- (2, 1)
Add the equations to get 2x = 4.
Unit 8: Sequences, Series & Conics (2)
The sum of the first 10 positive integers is:
- 50
- 45
- 100
- 55
n(n + 1)/2 = 55.
A hyperbola has an eccentricity that is:
- Equal to 0
- Between 0 and 1
- Greater than 1
- Equal to 1
An ellipse has eccentricity less than 1 and a parabola exactly 1.
Unit 9: Limits & Intro to Calculus Concepts (1)
The limit as x approaches 2 of (x^2 - 4)/(x - 2) is:
- 0
- 2
- 4
- Does not exist
Factor and cancel: x + 2 approaches 4.
Browse all 81 flashcards as a list
Unit 1: Functions & Their Graphs
- Function
- A relation where every input (x) has EXACTLY ONE output. Tested visually with the Vertical Line Test.
- Domain
- The complete set of valid x-values (inputs) for a function.
- Range
- The complete set of possible y-values (outputs) a function can produce.
- Composite function (f∘g)(x)
- f(g(x)) — evaluate the inside function g first, then plug that result into f.
- Vertical shift f(x)+k
- Moves the graph UP k units if k>0, DOWN if k<0. Outside the parentheses = intuitive direction.
- Horizontal shift f(x−h)
- Moves the graph RIGHT h units if h>0 — opposite of the sign inside the parentheses.
- Even function
- f(−x)=f(x) for all x. Graph is symmetric about the y-axis (e.g., x², cos x).
- Odd function
- f(−x)=−f(x) for all x. Graph has 180° rotational symmetry about the origin (e.g., x³, sin x).
- Piecewise function
- A function defined by different formulas over different intervals of its domain.
Unit 2: Polynomial & Rational Functions
- End behavior (odd degree, +leading coeff)
- Falls to the left, rises to the right — like y=x³.
- Multiplicity of a zero
- How many times a factor repeats. Odd multiplicity → graph crosses x-axis; even → graph bounces (touches and turns).
- Rational Root Theorem
- Possible rational zeros = ±(factors of constant term)/(factors of leading coefficient).
- Synthetic division
- A shortcut method for dividing a polynomial by (x−c); a zero remainder confirms c is a root.
- Vertical asymptote (rational function)
- Occurs where the denominator = 0 and that factor does NOT cancel with the numerator.
- Hole (removable discontinuity)
- Occurs where a factor cancels from both numerator and denominator of a rational function.
- Horizontal asymptote rule (n=m)
- When numerator and denominator have equal degree, HA = ratio of leading coefficients.
- Slant (oblique) asymptote
- Occurs when the numerator's degree is exactly ONE more than the denominator's; found via long division.
- Extraneous solution
- A solution produced algebraically that fails in the ORIGINAL equation (often makes a denominator zero).
Unit 3: Exponential & Logarithmic Functions
- Exponential growth model
- A(t)=A₀(1+r)^t — quantity increases by a constant percent rate r per time period.
- Exponential decay model
- A(t)=A₀(1−r)^t — quantity decreases by a constant percent rate r per time period.
- log_b(x)=y definition
- Means bʸ=x. A logarithm is the exponent you raise the base b to, to get x.
- Product rule for logs
- log_b(MN)=log_b(M)+log_b(N).
- Power rule for logs
- log_b(Mᵏ)=k·log_b(M) — lets you bring an exponent down as a coefficient.
- Change of base formula
- log_b(x)=ln(x)/ln(b) — used to evaluate logs of any base on a calculator.
- Natural log ln(x)
- log_e(x), where e≈2.71828. Inverse of eˣ: ln(eˣ)=x and e^(ln x)=x.
- Continuous growth/decay
- A(t)=A₀e^(rt) — models growth (r>0) or decay (r<0) that compounds continuously.
- Solving bˣ=k
- Take ln (or log) of both sides: x=ln(k)/ln(b).
Unit 4: Trigonometric Functions
- Unit circle point at angle θ
- (cos θ, sin θ) — the x-coordinate is cosine, the y-coordinate is sine.
- Radian-degree conversion
- radians = degrees × π/180; degrees = radians × 180/π.
- Arc length formula
- s=rθ, where θ MUST be measured in radians.
- Amplitude
- |A| in y=A sin(Bx−C)+D — half the vertical distance between the max and min of the graph.
- Period of sine/cosine
- 2π/|B| in y=A sin(Bx−C)+D or y=A cos(Bx−C)+D.
- Period of tangent
- π/|B| — half the period of sine/cosine, and tan x has no amplitude (unbounded range).
- Reference angle
- The acute angle between the terminal side and the x-axis; used to find exact trig values with the correct sign.
- ASTC rule
- 'All Students Take Calculus' — signs of trig functions positive in QI (All), QII (Sine), QIII (Tangent), QIV (Cosine).
- sin(45°)=cos(45°)
- Both equal √2/2 ≈0.707, since 45° is symmetric between the axes on the unit circle.
Unit 5: Trigonometric Identities & Equations
- Pythagorean identity
- sin²θ+cos²θ=1 for all θ — derived from x²+y²=1 on the unit circle.
- 1+tan²θ=sec²θ
- Derived by dividing the Pythagorean identity sin²θ+cos²θ=1 by cos²θ.
- sin(A+B)
- sin A cos B + cos A sin B — the sine sum formula.
- cos(A+B)
- cos A cos B − sin A sin B — note the MINUS sign, unlike the sine formula.
- Double-angle for sine
- sin(2θ)=2 sin θ cos θ.
- Double-angle for cosine (3 forms)
- cos(2θ)=cos²θ−sin²θ = 2cos²θ−1 = 1−2sin²θ.
- General solution to a trig equation
- Add integer multiples of the period (+2πk or +360°k) to every solution found in one cycle.
- Solving a trig quadratic
- Treat sinθ, cosθ, or tanθ like a variable and factor normally, e.g., (2sinθ+1)(sinθ−1)=0.
- cot θ
- 1/tan θ = cos θ/sin θ — the reciprocal of tangent.
Unit 6: Analytic Trigonometry & Applications
- Law of Sines
- a/sinA = b/sinB = c/sinC. Use with AAS, ASA, or the ambiguous SSA case.
- Law of Cosines
- c²=a²+b²−2ab·cosC. Use with SAS or SSS (no Law of Sines shortcut applies yet).
- Ambiguous case (SSA)
- Given two sides and a non-included angle, there may be 0, 1, or 2 valid triangles — must be checked.
- Area of a triangle (SAS)
- Area=(1/2)ab·sinC, where C is the angle INCLUDED between sides a and b.
- Heron's Formula
- Area=√[s(s−a)(s−b)(s−c)], where s=(a+b+c)/2 (semi-perimeter). Works for SSS, no angle needed.
- arcsin (sin⁻¹) range
- [−π/2, π/2] — the restricted range so arcsin is a true function.
- arccos (cos⁻¹) range
- [0, π] — the restricted range so arccos is a true function.
- arctan (tan⁻¹) range
- (−π/2, π/2) — domain is all real numbers.
- Obtuse angle from Law of Cosines
- If cos C comes out negative when solving for an angle, C is obtuse (between 90° and 180°) — a valid result.
Unit 7: Systems & Matrices
- Solving by elimination
- Multiply equations so one variable's coefficients are opposites, then add the equations to eliminate it.
- No solution (linear system)
- Occurs when elimination/substitution leads to a FALSE statement, like 0=5 — the lines are parallel.
- Infinite solutions (linear system)
- Occurs when elimination/substitution leads to a TRUE statement, like 0=0 — the lines coincide.
- Matrix dimensions
- Given as rows × columns; an m×n matrix has m rows and n columns.
- Matrix multiplication requirement
- AB is defined only if the number of columns of A equals the number of rows of B.
- Determinant of a 2×2 matrix
- For [[a,b],[c,d]], det=ad−bc.
- Singular matrix
- A square matrix whose determinant equals 0 — it has NO inverse.
- Inverse of a 2×2 matrix
- (1/det)·[[d,−b],[−c,a]], valid only when det≠0.
- Cramer's Rule
- Solves a linear system using determinants: x=Dx/D, y=Dy/D. Fails (no unique solution) if D=0.
Unit 8: Sequences, Series & Conics
- Arithmetic sequence
- A sequence with a CONSTANT difference d between consecutive terms: aₙ=a₁+(n−1)d.
- Geometric sequence
- A sequence with a CONSTANT ratio r between consecutive terms: aₙ=a₁·r^(n−1).
- Sigma notation
- Σ(k=1 to n) f(k) — compact notation representing the sum f(1)+f(2)+...+f(n).
- Finite geometric series sum
- Sₙ=a₁(1−rⁿ)/(1−r), for r≠1.
- Infinite geometric series sum
- S=a₁/(1−r), but ONLY converges (has a finite sum) when |r|<1.
- Parabola x²=4py
- Vertex at origin; focus at (0,p); directrix y=−p. Opens up if p>0, down if p<0.
- Ellipse foci distance
- c²=a²−b² (a=semi-major axis, b=semi-minor axis), where c is the distance from center to each focus.
- Hyperbola foci distance
- c²=a²+b² — note the PLUS sign, unlike the ellipse's minus sign.
- Hyperbola asymptotes
- y=±(b/a)x for x²/a²−y²/b²=1 — lines the branches approach but never touch.
Unit 9: Limits & Intro to Calculus Concepts
- Limit lim(x→a) f(x)=L
- f(x) gets arbitrarily close to L as x approaches a from both sides, regardless of f(a).
- Two-sided limit existence
- Exists only if the left-hand limit equals the right-hand limit: lim(x→a⁻)f(x)=lim(x→a⁺)f(x).
- Indeterminate form 0/0
- A signal to factor and simplify before evaluating a limit — it does NOT mean the limit fails to exist.
- Average rate of change
- [f(b)−f(a)]/(b−a) — the slope of the secant line through (a,f(a)) and (b,f(b)).
- Difference quotient
- [f(a+h)−f(a)]/h — the average rate of change over an interval of width h starting at a.
- Derivative f'(a)
- lim(h→0) [f(a+h)−f(a)]/h — the instantaneous rate of change; the slope of the tangent line at x=a.
- Secant line vs tangent line
- A secant line crosses the curve at two points (average rate); a tangent line touches at one point (instantaneous rate).
- Continuity and direct substitution
- If f is continuous at x=a, then lim(x→a) f(x)=f(a), so you can find the limit by direct substitution.
- Jump discontinuity
- Occurs when the left-hand and right-hand limits at a point exist but are NOT equal — the overall limit does not exist there.