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AP Precalculus Unit 4 Visual ReviewA topic-by-topic visual walkthrough of Unit 4: Functions Involving Parameters, Vectors, and Matrices — parametric functions, implicit functions, conics, vectors, and matrices.
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TOPIC 4.1
Parametric Functions
x = f(t), y = g(t) — both coordinates depend on a parameter t
A parameter drives both x and y
As t runs over an interval, the point
(x, y) traces a curve in the plane.
t often represents TIME, so the curve
has a DIRECTION (orientation).
Can trace curves that fail the vertical line test.
Worked example
x = t, y = t²
t=0 → (0,0), t=1 → (1,1),
t=2 → (2,4)
Eliminate t: y = x² (a parabola),
traced left → right as t increases.
x=f(t), y=g(t) traces a directed curve; eliminate t to get a relation in x, y.The Review Hub · AP Precalculus Unit 4
TOPIC 4.2
Parametric Functions: Planar Motion
Position over time
Model an object moving in a plane with
(x(t), y(t)) = position at time t
x(t) handles horizontal motion;
y(t) handles vertical motion — separately.
The orientation shows the path's direction.
Example: projectile
x(t) = v₀cos(θ)·t
y(t) = v₀sin(θ)·t − ½gt²
Horizontal is linear; vertical is quadratic
→ together they give a parabolic path.
Find when it lands by solving y(t) = 0.
Read position, and where the object is at a given time
Plug in t to get the location; the curve's shape shows the trajectory, and the arrows show direction of travel.
Model motion with (x(t), y(t)) — horizontal and vertical components separately.
The Review Hub · AP Precalculus Unit 4
TOPIC 4.3
Parametric Functions & Rates of Change
Separate rates for x and y
The rate of change of x and of y with
respect to t are found independently.
Δx/Δt and Δy/Δt
Positive Δx/Δt → moving right;
positive Δy/Δt → moving up.
Direction of travel
The two rates together give the direction
the point is moving.
slope of path = (Δy/Δt)/(Δx/Δt)
Where Δx/Δt = 0 the motion is vertical;
where Δy/Δt = 0 it's horizontal.
Estimate rates from a table of (t, x, y)
Compute average rates over intervals of t to describe how fast and in which direction the object is moving.
Analyze Δx/Δt and Δy/Δt separately — together they give speed and direction.
The Review Hub · AP Precalculus Unit 4
TOPIC 4.4
Parametric Circles & Lines
Circle of radius r, center (h, k)
x = h + r cos t
y = k + r sin t
t from 0 to 2π traces one full circle,
counterclockwise.
Swap sin/cos or negate to change direction.
Line through (x₀, y₀)
x = x₀ + at
y = y₀ + bt
Direction ⟨a, b⟩; slope = b/a.
At t = 0 you're at the start point;
t moves you along the line steadily.
Ellipse: different radii
x = h + a cos t, y = k + b sin t traces an ELLIPSE with horizontal radius a and vertical radius b.
Circle: x=h+r cos t, y=k+r sin t ; line: x=x₀+at, y=y₀+bt.
The Review Hub · AP Precalculus Unit 4
TOPIC 4.5
Implicitly Defined Functions
x² + y² = 25 — a relation, not solved for y
Implicit vs. explicit
EXPLICIT: y is written alone, y = f(x).
IMPLICIT: x and y appear in an equation
that isn't solved for either variable.
The graph can fail the vertical line test
(a circle has two y's per x).
Solve for y in pieces
x² + y² = 25 splits into two functions:
y = √(25 − x²) (top half)
y = −√(25 − x²) (bottom half)
Each branch is a function on its own.
Domain here: −5 ≤ x ≤ 5.
An implicit relation (like x²+y²=25) isn't solved for y — split into function branches.
The Review Hub · AP Precalculus Unit 4
TOPIC 4.6
Conic Sections
Ellipse
x²/a² + y²/b² = 1
circle if a = b
two foci; sum of distances constant
Parabola
y = a(x − h)² + k
one focus & a directrix
vertex at (h, k)
Hyperbola
x²/a² − y²/b² = 1
two branches & asymptotes
difference of distances constant
Conics: ellipse (+), parabola (one squared), hyperbola (−) in x², y².
The Review Hub · AP Precalculus Unit 4
TOPIC 4.7
Parametrizing Implicit Functions
Turn a relation into (x(t), y(t))
Rewrite an implicit curve as parametric
equations, adding a direction & speed.
Circle x²+y²=r²:
x = r cos t, y = r sin t
Verify: (r cos t)² + (r sin t)² = r².
Many parametrizations exist
The same curve can be parametrized in
different ways (different speed/direction).
A simple trick: let x = t, then solve the
relation for y in terms of t.
Ellipse: x = a cos t, y = b sin t.
Check by eliminating the parameter
Substitute your x(t), y(t) back into the original equation — it should simplify to a true statement.
Parametrize a relation as (x(t), y(t)) — e.g. a circle as (r cos t, r sin t).The Review Hub · AP Precalculus Unit 4
TOPIC 4.8
Vectors
v = ⟨4, 3⟩
4 3
|v| = √(a² + b²) = magnitude
⟨4,3⟩: |v| = √(16+9) = 5
direction θ = arctan(b/a)
Vector operations
add: ⟨a,b⟩ + ⟨c,d⟩ = ⟨a+c, b+d⟩
scale: k⟨a,b⟩ = ⟨ka, kb⟩
A vector has both magnitude AND direction.
A vector ⟨a,b⟩ has magnitude √(a²+b²) and direction; add componentwise.
The Review Hub · AP Precalculus Unit 4
TOPIC 4.9
Vector-Valued Functions
p(t) = ⟨ x(t), y(t) ⟩ — position as a vector of time
Output is a vector
A vector-valued function maps a number t
to a VECTOR (a position in the plane).
Equivalent to parametric equations —
it packages x(t) and y(t) as one object.
Traces the same directed path as t varies.
Displacement between times
The change in position from t₁ to t₂ is a
vector (displacement):
p(t₂) − p(t₁)
Its direction is the direction of travel;
its length is the straight-line distance.
p(t)=⟨x(t), y(t)⟩ outputs a position vector; displacement = p(t₂) − p(t₁).The Review Hub · AP Precalculus Unit 4
TOPIC 4.10
Matrices
A rectangular array of numbers
An m × n matrix has m rows, n columns.
A = [ a b ]
[ c d ] (2×2)
Add/subtract element-by-element (same size);
scalar multiply scales every entry.
Matrix multiplication
Multiply ROWS of A by COLUMNS of B
(dot product of each pair).
Columns of A must match rows of B.
Order matters: AB ≠ BA in general.
Identity I: AI = IA = A.
Matrices act on vectors
A 2×2 matrix times a 2×1 vector gives a new vector — the basis for transformations (Topic 4.12).
A matrix is a number array; multiply row × column , and AB ≠ BA.
The Review Hub · AP Precalculus Unit 4
TOPIC 4.11
Inverse & Determinant of a Matrix
A = [a b; c d] → det(A) = ad − bc
The 2×2 inverse
A⁻¹ = 1/det(A) · [ d −b ]
[ −c a ]
Swap a & d, negate b & c, divide by det.
A·A⁻¹ = I (identity)
Use A⁻¹ to solve AX = B → X = A⁻¹B.
What the determinant tells you
det(A) = the AREA scaling factor of the
transformation A applies.
det(A) = 0 → NO inverse exists
(the matrix is "singular").
A negative det flips orientation.
det = ad − bc ; if det = 0 there's no inverse. A⁻¹ scales the adjugate by 1/det.The Review Hub · AP Precalculus Unit 4
TOPIC 4.12
Linear Transformations & Matrices
A · v transforms the vector v (rotate, scale, reflect, shear)
A matrix moves points
Multiplying a vector by a matrix maps it
to a new point — a linear transformation.
The columns of A are where the basis
⟨1,0⟩ and ⟨0,1⟩ land.
Lines stay lines; the origin stays fixed.
Common transformations
rotate 90°: [0 −1; 1 0]
scale by k: [k 0; 0 k]
reflect over x: [1 0; 0 −1]
Compose transforms by MULTIPLYING
their matrices (apply right-to-left).
A·v is a linear transformation — its columns show where the basis vectors land.The Review Hub · AP Precalculus Unit 4
TOPIC 4.13
Matrices as Functions
Input vector → output vector
Think of A as a FUNCTION: it takes an
input vector and returns an output vector.
T(v) = A·v
The inverse matrix A⁻¹ UNDOES the
transformation, like an inverse function.
Composition = multiplication
Doing transformation B then A is the
single matrix A·B.
A(B v) = (AB) v
Just like composing functions (f∘g),
applied from the inside out.
Invertible ⟺ det ≠ 0
The matrix-function has an inverse exactly when det(A) ≠ 0 — the same one-to-one condition as any function.
A matrix is a function T(v)=A·v ; A⁻¹ undoes it, and composition is A·B.
The Review Hub · AP Precalculus Unit 4
TOPIC 4.14
Matrices Modeling Contexts
State transitions
A TRANSITION matrix models how a system
moves between states each step.
next state = A · current state
Apply A repeatedly (Aⁿ) to project
several steps into the future.
Example
Populations, market share, or ecosystems
that shift between categories over time.
Entries are proportions moving from one
state to another.
A long-run steady state may be reached.
Solve systems with the inverse
A system of linear equations AX = B is solved as X = A⁻¹B when A is invertible (det ≠ 0).
Transition matrices model state changes (next = A·current); solve AX=B with X=A⁻¹B.The Review Hub · AP Precalculus Unit 4
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How to use the visual review
Spend 30 seconds per slide before clicking next. Look at the diagram, then ask yourself: "Could I sketch this from memory and explain what it represents?"
Use the fullscreen button () on desktop for the best experience. Use arrow keys to navigate. Tap "Show all slides" to jump around.
Unit 4 is not tested on the AP Exam , but these visuals are a strong bridge to calculus and linear algebra.