◆ Slope formula
m = (y₂ − y₁) / (x₂ − x₁)
Slope measures steepness: vertical change divided by horizontal change.
Example — how to apply it
- Points (2, 3) and (5, 9): m = (9 − 3) / (5 − 2) = 6 / 3.
Answer: m = 2
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💎 Formula Treasure Vault
Every essential formula for Algebra I through Calculus BC, each one a gem in the vault — with a worked example showing exactly how to use it. Pick your subject and start collecting.
◆ Algebra I · Ruby · 10 essential formulas
The foundation: lines, slopes, and quadratics. Master these and every later course gets easier.
◆ Slope-intercept form
y = mx + b
m is the slope, b is where the line crosses the y-axis.
Example — how to apply it
- For y = 3x − 2, the slope is 3 and the y-intercept is −2.
- At x = 4: y = 3(4) − 2 = 10.
Answer: y = 10 when x = 4
◆ Point-slope form
y − y₁ = m(x − x₁)
Build a line from its slope and one point on it.
Example — how to apply it
- Slope 2 through (1, 5): y − 5 = 2(x − 1).
- Simplify: y = 2x + 3.
Answer: y = 2x + 3
◆ Distance formula
d = √((x₂ − x₁)² + (y₂ − y₁)²)
The straight-line distance between two points — Pythagoras in disguise.
Example — how to apply it
- From (0, 0) to (3, 4): d = √(3² + 4²) = √25.
Answer: d = 5
◆ Midpoint formula
((x₁ + x₂)/2, (y₁ + y₂)/2)
Average the x-coordinates and the y-coordinates to find the middle point.
Example — how to apply it
- Midpoint of (2, 4) and (6, 10): ((2 + 6)/2, (4 + 10)/2).
Answer: (4, 7)
◆ Quadratic formula
x = (−b ± √(b² − 4ac)) / 2a
Solves any quadratic ax² + bx + c = 0, even when factoring fails.
Example — how to apply it
- For x² − 5x + 6 = 0: a = 1, b = −5, c = 6.
- x = (5 ± √(25 − 24)) / 2 = (5 ± 1) / 2.
Answer: x = 2 or x = 3
◆ Difference of squares
a² − b² = (a − b)(a + b)
A squared minus a squared always factors into a sum times a difference.
Example — how to apply it
- x² − 9 = x² − 3², so it factors at once.
Answer: (x − 3)(x + 3)
◆ Perfect square trinomial
(a + b)² = a² + 2ab + b²
Squaring a sum gives the two squares plus twice the product.
Example — how to apply it
- (x + 4)² = x² + 2(x)(4) + 4² = x² + 8x + 16.
Answer: x² + 8x + 16
◆ Exponent rules
aᵐ · aⁿ = aᵐ⁺ⁿ, (aᵐ)ⁿ = aᵐⁿ
Multiply powers by adding exponents; raise a power by multiplying exponents.
Example — how to apply it
- x³ · x⁴ = x³⁺⁴.
Answer: x⁷
◆ Parallel and perpendicular slopes
Parallel lines share m; perpendicular lines use −1/m
Same slope means never meeting; a negative reciprocal means meeting at a right angle.
Example — how to apply it
- A line perpendicular to y = 2x + 1 has slope −1/2.
Answer: m = −1/2
◆ Algebra II · Sapphire · 10 essential formulas
Quadratics go deeper: complex numbers, logarithms, exponentials, and sequences.
◆ Vertex form
y = a(x − h)² + k, vertex (h, k)
The vertex form shows the turning point of a parabola at a glance.
Example — how to apply it
- For y = 2(x − 3)² + 1, h = 3 and k = 1.
Answer: Vertex (3, 1)
◆ Discriminant
D = b² − 4ac
D > 0: two real roots. D = 0: one root. D < 0: two complex roots.
Example — how to apply it
- For x² + 2x + 5: D = 4 − 20 = −16 < 0.
Answer: Two complex roots
◆ Complex numbers
i² = −1
Complex numbers follow the same algebra rules, with i² replaced by −1.
Example — how to apply it
- (3 + 2i)(1 − i) = 3 − 3i + 2i − 2i² = 3 − i + 2.
Answer: 5 − i
◆ Logarithm product rule
logb(xy) = logbx + logby
A logarithm turns multiplication into addition.
Example — how to apply it
- log₂8 + log₂4 = log₂(8 · 4) = log₂32.
Answer: 5
◆ Exponential growth
y = a(1 + r)t
a is the starting amount, r the growth rate per period, t the number of periods.
Example — how to apply it
- $1000 at 5% annual interest for 3 years: 1000 · 1.05³ = 1157.625.
Answer: $1157.63
◆ Simplifying rational expressions
(x² − 9) / (x − 3) = x + 3, x ≠ 3
Factor first, then cancel — but keep the excluded value.
Example — how to apply it
- x² − 9 = (x − 3)(x + 3), so the (x − 3) factors cancel.
- The original is undefined at x = 3, so the simplified form keeps x ≠ 3.
Answer: x + 3, x ≠ 3
◆ Remainder and factor theorems
Remainder of P(x) ÷ (x − a) is P(a); if P(a) = 0, (x − a) is a factor
Evaluate instead of dividing: the value at a tells you the remainder.
Example — how to apply it
- P(x) = x³ − 2x + 1: P(1) = 1 − 2 + 1 = 0.
Answer: (x − 1) is a factor
◆ Sum and difference of cubes
a³ ± b³ = (a ± b)(a² ∓ ab + b²)
Cubes factor into a linear factor times a quadratic — note the sign flip.
Example — how to apply it
- x³ + 8 = (x + 2)(x² − 2x + 4).
Answer: (x + 2)(x² − 2x + 4)
◆ Arithmetic sequences
aₙ = a₁ + (n − 1)d
Each term adds the same difference d to the previous term.
Example — how to apply it
- For 3, 7, 11, …: d = 4, so a₁₀ = 3 + 9 · 4.
Answer: 39
◆ Geometric sequences
aₙ = a₁ · rn−1
Each term multiplies the previous term by the same ratio r.
Example — how to apply it
- For 2, 6, 18, …: r = 3, so a₅ = 2 · 3⁴.
Answer: 162
◆ Geometry · Emerald · 10 essential formulas
Shapes, measurement, and proof: the formulas that turn pictures into numbers.
◆ Pythagorean theorem
a² + b² = c²
In a right triangle, the squares on the legs add to the square on the hypotenuse.
Example — how to apply it
- Legs 3 and 4: c² = 9 + 16 = 25.
Answer: c = 5
◆ Area of a triangle
A = (1/2)bh
Half the base times the height — half of the enclosing rectangle.
Example — how to apply it
- Base 8, height 5: A = (1/2)(8)(5).
Answer: 20
◆ Circle: circumference and area
C = 2πr, A = πr²
Both circle formulas grow from the radius; area grows with the square.
Example — how to apply it
- Radius 3: A = π · 3².
Answer: 9π
◆ Volume of a cylinder
V = πr²h
Base area times height — a stack of circles.
Example — how to apply it
- Radius 2, height 5: V = π · 4 · 5.
Answer: 20π
◆ Triangle angle sum
Angles of a triangle add to 180°
Know two angles and the third is forced.
Example — how to apply it
- Angles 50° and 60° leave 180° − 110°.
Answer: 70°
◆ Similar triangles
Corresponding sides are proportional
Same shape means every side scales by the same factor.
Example — how to apply it
- If 6/9 = 8/x, then 6x = 72.
Answer: x = 12
◆ Special right triangles
30°-60°-90°: 1 : √3 : 2; 45°-45°-90°: 1 : 1 : √2
Memorize the side ratios and skip the calculator.
Example — how to apply it
- 30°-60°-90° with hypotenuse 10: short leg = 10/2.
Answer: 5
◆ Arc length
s = rθ (θ in radians)
An arc is a fraction of the circumference, measured by its angle.
Example — how to apply it
- Radius 6, angle π/3: s = 6 · π/3.
Answer: 2π
◆ Area of a trapezoid
A = (1/2)(b₁ + b₂)h
Average the two parallel bases, then multiply by the height.
Example — how to apply it
- Bases 6 and 10, height 4: A = (1/2)(16)(4).
Answer: 32
◆ Surface area of a rectangular prism
SA = 2(lw + lh + wh)
Add the three pairs of opposite faces.
Example — how to apply it
- Sides 3, 4, 5: SA = 2(12 + 15 + 20).
Answer: 94
◆ Trigonometry · Amethyst · 10 essential formulas
Triangles, circles, and waves: the ratios and identities that unlock angles.
◆ Right-triangle ratios
sin θ = opp/hyp, cos θ = adj/hyp, tan θ = opp/adj
SOHCAHTOA: each ratio pairs two sides of a right triangle.
Example — how to apply it
- Opposite 5, hypotenuse 13: sin θ = 5/13.
Answer: sin θ = 5/13
◆ Pythagorean identity
sin²θ + cos²θ = 1
Sine and cosine are the legs of a unit right triangle.
Example — how to apply it
- If sin θ = 3/5 and θ is acute: cos²θ = 1 − 9/25 = 16/25.
Answer: cos θ = 4/5
◆ Special angles
sin 30° = 1/2, cos 45° = √2/2, tan 60° = √3
The 30°-60°-90° and 45°-45°-90° triangles give these by heart.
Example — how to apply it
- sin 150° = sin(180° − 150°) = sin 30°.
Answer: 1/2
◆ Law of sines
a / sin A = b / sin B = c / sin C
Sides and sines of opposite angles share one constant ratio.
Example — how to apply it
- a = 6, A = 30°, B = 45°: 6 / (1/2) = b / (√2/2), so b = 6√2.
Answer: b = 6√2
◆ Law of cosines
c² = a² + b² − 2ab cos C
Pythagoras with a correction term for non-right triangles.
Example — how to apply it
- a = 5, b = 7, C = 60°: c² = 25 + 49 − 2 · 5 · 7 · (1/2) = 39.
Answer: c = √39
◆ Angle-sum identity
sin(A + B) = sin A cos B + cos A sin B
Split an awkward angle into two friendly ones.
Example — how to apply it
- sin 75° = sin(45° + 30°) = (√2/2)(√3/2) + (√2/2)(1/2).
Answer: (√6 + √2)/4
◆ Double-angle identity
sin 2θ = 2 sin θ cos θ
Doubling the angle doubles the product of sine and cosine.
Example — how to apply it
- sin θ = 3/5 (acute) gives cos θ = 4/5, so sin 2θ = 2 · (3/5)(4/5).
Answer: 24/25
◆ Reciprocal identities
csc θ = 1/sin θ, sec θ = 1/cos θ, cot θ = 1/tan θ
Each reciprocal ratio is one divided by its partner.
Example — how to apply it
- If cos θ = 1/2, then sec θ = 1 / (1/2).
Answer: 2
◆ Degrees to radians
180° = π radians
Multiply degrees by π/180 to switch to radian measure.
Example — how to apply it
- 120° · π/180 = 2π/3.
Answer: 2π/3
◆ Inverse trigonometric functions
sin θ = 1/2 (acute) → θ = 30°
Inverse trig recovers the angle from the ratio.
Example — how to apply it
- sin θ = 1/2 with θ acute: θ = arcsin(1/2).
Answer: 30°
◆ Precalculus · Topaz · 10 essential formulas
The bridge to calculus: functions, logarithms, vectors, and conics.
◆ Function composition
(f ∘ g)(x) = f(g(x))
Apply g first, then feed its output into f.
Example — how to apply it
- f(x) = x², g(x) = x + 1: (f ∘ g)(x) = (x + 1)².
Answer: (x + 1)²
◆ Inverse functions
Swap x and y, then solve for y
An inverse undoes the original function, step by step in reverse.
Example — how to apply it
- f(x) = 2x − 3: x = 2y − 3 gives y = (x + 3)/2.
- Check: f((x + 3)/2) = x.
Answer: f⁻¹(x) = (x + 3)/2
◆ Change of base
logba = ln a / ln b
Any logarithm can be computed with natural logs.
Example — how to apply it
- log₂10 = ln 10 / ln 2 ≈ 3.32.
Answer: ≈ 3.32
◆ Continuous growth
A = Pert
Interest or growth compounded continuously, not once per period.
Example — how to apply it
- $1000 at 4% continuously for 5 years: 1000 · e0.04 · 5 = 1000e0.2.
Answer: ≈ $1221.40
◆ Binomial theorem
(a + b)ⁿ = Σ C(n, k) an−kbk
Each term picks how many times b appears; the coefficient counts the choices.
Example — how to apply it
- x² term of (x + 2)⁵: C(5, 2) · x² · 2³ = 10 · 8.
Answer: 80x²
◆ Rational asymptotes
Vertical: denominator zeros; horizontal: compare degrees
Asymptotes describe where a graph blows up or levels off.
Example — how to apply it
- For (2x + 1)/(x − 3): vertical asymptote x = 3.
- Equal degrees, so horizontal asymptote y = 2/1 = 2.
Answer: x = 3, y = 2
◆ Vector magnitude
|a| = √(x² + y²)
A vector's length is the distance formula from the origin.
Example — how to apply it
- |(3, 4)| = √(9 + 16) = 5.
Answer: 5
◆ Parametric elimination
x = f(t), y = g(t) → eliminate t
Parametric curves trace a path; eliminating t reveals the plain equation.
Example — how to apply it
- x = 2t, y = t²: t = x/2, so y = (x/2)².
Answer: y = x²/4
◆ Polar coordinates
x = r cos θ, y = r sin θ, r² = x² + y²
Describe a point by distance and angle instead of x and y.
Example — how to apply it
- Point (3, 4): r = 5 and θ = arctan(4/3).
Answer: r = 5, θ = arctan(4/3)
◆ Circle in standard form
(x − h)² + (y − k)² = r²
Complete the square to read the center and radius off a circle equation.
Example — how to apply it
- x² + y² − 6x + 4y − 3 = 0 → (x − 3)² + (y + 2)² = 16.
Answer: Center (3, −2), radius 4
◆ Calculus AB · Teal · 10 essential formulas
Limits, derivatives, and integrals: the core ideas of first-year calculus.
◆ Definition of the derivative
f′(x) = limh→0 [f(x + h) − f(x)] / h
A derivative is the limit of average rates of change — the slope of the tangent line.
Example — how to apply it
- For f(x) = x²: [(x + h)² − x²] / h = 2x + h, and h → 0.
Answer: f′(x) = 2x
◆ Power rule
d/dx [xⁿ] = n xn−1
Bring the exponent down, then subtract one from it.
Example — how to apply it
- d/dx [x⁵] = 5x⁴.
Answer: 5x⁴
◆ Product rule
(fg)′ = f′g + fg′
Differentiate each factor in turn, keeping the other fixed.
Example — how to apply it
- (x² · sin x)′ = 2x · sin x + x² · cos x.
Answer: 2x sin x + x² cos x
◆ Quotient rule
(f/g)′ = (f′g − fg′) / g²
Low d-high minus high d-low, over low squared.
Example — how to apply it
- (x/(x + 1))′ = [(x + 1) − x] / (x + 1)².
Answer: 1/(x + 1)²
◆ Chain rule
d/dx [f(g(x))] = f′(g(x)) · g′(x)
Differentiate the outside, then multiply by the derivative of the inside.
Example — how to apply it
- d/dx [(3x + 1)⁴] = 4(3x + 1)³ · 3.
Answer: 12(3x + 1)³
◆ Trigonometric derivatives
(sin x)′ = cos x, (cos x)′ = −sin x, (tan x)′ = sec²x
Sine and cosine trade places, with a sign flip on cosine.
Example — how to apply it
- d/dx [sin 2x] = cos 2x · 2 (chain rule).
Answer: 2 cos 2x
◆ Exponential and logarithm derivatives
(eˣ)′ = eˣ, (ln x)′ = 1/x
eˣ is its own derivative; ln x differentiates to a reciprocal.
Example — how to apply it
- d/dx [ln(x²)] = (1/x²) · 2x (chain rule).
Answer: 2/x
◆ Fundamental theorem of calculus
∫ab f(x) dx = F(b) − F(a)
Integration undoes differentiation: evaluate the antiderivative at the ends.
Example — how to apply it
- ∫₀¹ 3x² dx = [x³]₀¹ = 1 − 0.
Answer: 1
◆ Mean value theorem
f′(c) = [f(b) − f(a)] / (b − a) for some c in (a, b)
Somewhere inside, the instantaneous rate equals the average rate.
Example — how to apply it
- f(x) = x² on [0, 2]: average rate = 4/2 = 2, so f′(c) = 2c = 2.
Answer: c = 1
◆ First and second derivative tests
f′ = 0 marks candidates; f″ > 0 → minimum, f″ < 0 → maximum
The first derivative finds flat points; the second tells their shape.
Example — how to apply it
- f(x) = x³ − 3x: f′ = 3x² − 3 = 0 at x = ±1.
- f″ = 6x: f″(−1) < 0 (maximum), f″(1) > 0 (minimum).
Answer: Local max at x = −1, local min at x = 1
◆ Calculus BC · Diamond · 10 essential formulas
Beyond AB: advanced integration, series, parametric and polar calculus.
◆ Integration by parts
∫ u dv = uv − ∫ v du
Trade a hard integral for an easier one by splitting the integrand.
Example — how to apply it
- ∫ x eˣ dx: let u = x, dv = eˣ dx. Then uv − ∫v du = x eˣ − ∫ eˣ dx.
- So the answer is x eˣ − eˣ + C.
Answer: (x − 1)eˣ + C
◆ Partial fractions
1/(x² − x) = 1/(x − 1) − 1/x
Split a hard rational function into simple fractions you can integrate.
Example — how to apply it
- ∫ dx/(x² − x) = ∫ [1/(x − 1) − 1/x] dx.
Answer: ln|x − 1| − ln|x| + C
◆ Improper integrals
∫1∞ f(x) dx = limb→∞ ∫1b f(x) dx
An infinite bound is handled as a limit of finite ones.
Example — how to apply it
- ∫₁∞ x−2 dx = lim [−1/x]₁b = 0 − (−1).
Answer: 1
◆ Euler's method
yn+1 = yn + h · f(xn, yn)
Follow the tangent line in small steps to approximate a solution curve.
Example — how to apply it
- y′ = y, y(0) = 1, h = 0.5: y(0.5) ≈ 1 + 0.5 · 1.
Answer: y(0.5) ≈ 1.5
◆ Logistic growth
dP/dt = kP(1 − P/M), P(t) = M / (1 + Ae−kt)
Growth that slows as it approaches the carrying capacity M.
Example — how to apply it
- A population with M = 1000 follows P = 1000 / (1 + Ae−kt).
- As t → ∞, P → 1000, the carrying capacity.
Answer: P → M as t → ∞
◆ Geometric series
Σ arⁿ = a / (1 − r) when |r| < 1
A geometric series converges exactly when its ratio is smaller than 1 in size.
Example — how to apply it
- Σ (1/2)ⁿ from n = 0: a = 1, r = 1/2, so the sum is 1 / (1 − 1/2).
Answer: 2
◆ Taylor series for eˣ
eˣ = Σ xⁿ / n! = 1 + x + x²/2! + x³/3! + …
A function rebuilt from its derivatives at a single point.
Example — how to apply it
- First three terms: eˣ ≈ 1 + x + x²/2.
Answer: 1 + x + x²/2 + …
◆ Ratio test
If lim |an+1/an| < 1, Σaₙ converges absolutely
Compare each term with the next: a shrinking ratio means convergence.
Example — how to apply it
- For Σ xⁿ/n!: the ratio is |x|/(n + 1) → 0 for every x.
- So the series converges for all x — infinite radius.
Answer: Radius of convergence ∞
◆ Polar area
A = (1/2) ∫ r² dθ
Sweep a radius around: each slice contributes a thin sector.
Example — how to apply it
- Circle r = 2: A = (1/2) ∫₀2π 4 dθ = (1/2) · 8π.
Answer: 4π
◆ p-series test
Σ 1/np converges if and only if p > 1
The exponent decides: just above 1 converges, at 1 or below diverges.
Example — how to apply it
- Σ 1/n² converges (p = 2), while the harmonic series Σ 1/n diverges.
Answer: Converges for p > 1
Every formula above is one students are expected to know cold — the example under each one shows exactly how it is used. Work the example yourself first, then open it to check.