Higher Maths formula and reference guide

f(a) means substitute x = a

f(g(x)) means apply g first, then f.

For an inverse, swap x and y, rearrange, then check the domain and range.

y = f(x) + a moves the graph up a

y = f(x − a) moves the graph right a

y = −f(x) reflects in the x-axis; y = f(-x) reflects in the y-axis

x = -b ±√(b² − 4ac)2a

Discriminant Δ = b² − 4ac

Δ > 0: two real roots; Δ = 0: one repeated root; Δ < 0: no real roots

ax² + bx + c = a(x + b2a)² + c − b²/(4a)

Use the roots and the graph sign to solve a quadratic inequality.

Strict inequalities exclude boundary roots.

If f(a) = 0, then (x − a) is a factor.

The remainder when f(x) is divided by (x − a) is f(a).

After finding a factor, divide and solve every remaining factor.

a^x = b is equivalent to logₐ(b) = x.

log(ab) = log a + log b; log(ab) = log a − log b

log(aⁿ) = n log a; every logarithm argument must be positive

y = ab^x gives log y = log a + x log b

y = ax^b gives log y = log a + b log x

Compare transformed variables with Y = mX + c

m = y₂ − y₁x₂ − x₁

y − b = m(x − a)

Parallel gradients are equal; non-vertical perpendicular gradients multiply to −1.

m = tan θ

Distance = √((x₂ − x₁)² + (y₂ − y₁)²)

Midpoint = (x₁ + x₂2, y₁ + y₂2)

Median: vertex to opposite midpoint.

Altitude: through a vertex, perpendicular to the opposite side.

Perpendicular bisector: through the midpoint, perpendicular to the side.

(x − a)² + (y − b)² = r² has centre (a, b) and radius r

At a tangent point, the radius is perpendicular to the tangent.

Substitute a line into a circle; the discriminant counts intersections.

sin² x + cos² x = 1

tan x = sin xcos x

sin 30° = cos 60° = 12; sin 45° = cos 45° =√22

sin(A ± B) = sin A cos B ± cos A sin B

cos(A ± B) = cos A cos B −/+ sin A sin B

sin 2x = 2 sin x cos x

cos 2x = cos² x − sin² x = 2cos² x − 1 = 1 − 2sin² x

a cos x + b sin x = k cos(x − α).

k cos α = a; k sin α = b

k = √(a² + b²), k > 0

For a sin(bx + c) + d: amplitude = |a|; period = 2π|b| radians or 360°/|b|.

For a tan(bx + c) + d: period = π|b| radians or 180°/|b|.

Solve only in the stated degree or radian interval.

d/dx(xⁿ) = nx^{n − 1}

Constants differentiate to 0.

Rewrite roots and reciprocals as powers before differentiating where useful.

d/dx[sin(kx)] = k cos(kx)

d/dx[cos(kx)] = −k sin(kx)

Trigonometric calculus uses radians.

If y = f(g(x)), then dy/dx = f'(g(x))g'(x).

d/dx[(px + q)ⁿ] = np(px + q)^{n − 1}

Tangent gradient = dy/dx at the point

Normal gradient = −1m when tangent gradient m is non-zero

Use y − b = m(x − a) after finding the point and gradient.

Stationary point: dy/dx = 0

d²y/dx² > 0 gives a local minimum; d²y/dx² < 0 gives a local maximum

On a closed interval, compare every stationary value with both endpoint values.

f'(x) > 0: f is strictly increasing

f'(x) < 0: f is strictly decreasing

A sign change + to − gives a local maximum; − to + gives a local minimum.

∫xⁿ dx = x^{n + 1}n + 1 + C, n ≠ −1

Integrate each term and include + C for an indefinite integral.

∫(px + q)ⁿ dx = (px + q)^{n + 1}p(n + 1) + C, n ≠ −1

Differentiate the result to check the inner factor cancels.

Integral of p cos(qx + r) dx = (pq) sin(qx + r) + C

Integral of p sin(qx + r) dx = −(pq) cos(qx + r) + C

Integral from a to b of f(x) dx = F(b) − F(a)

An integral is signed area; geometrical area must be non-negative.

Area = integral from a to b of [top(x) − bottom(x)] dx

Find limits from intersections and split where the upper curve changes.

If dy/dx = f(x), then y = integral f(x) dx + C.

Use the initial condition to determine C and interpret the result in context.

For a = (a₁, a₂, a₃), |a| = √(a₁² + a₂² + a₃²).

The unit vector in the direction of non-zero a is a|a|.

ai + bj + ck corresponds to (a, b, c).

Add, subtract and scale matching components.

Collinearity: show two direction vectors sharing a point are scalar multiples.

If AP:PB = m:n, then p = n a + m bm + n.

a · b = a₁b₁ + a₂b₂ + a₃b₃.

a · b = |a||b|cos θ.

Non-zero vectors are perpendicular when a · b = 0

A linear model often has uₙ₊₁ = auₙ + b.

A candidate limit satisfies L = aL + b.

For this linear model, |a| < 1 gives convergence to L = b1 − a; interpret L in the model context.