Mathematics Advanced • Year 11 • Module 1 • Lesson 1
Functions & Relations
Build fluency in the function/relation distinction, the vertical line test, and evaluating $f(x)$ for simple inputs.
1. Quick recall
Answer in the space provided. 1 mark each
Q1.1 Complete the definition:
A function is a relation in which each input has ______________ output.
Q1.2 State the rule of the vertical line test.
A graph represents a function if and only if every vertical line crosses the graph __________________________.
Q1.3 The notation f(x) is read as ______________ of ______________. The letter f is the ______________ of the function and x is the ______________.
2. Worked example — testing a graph with the VLT
Follow each line. Every step has a reason on the right.
Problem. Does the equation x = y² define y as a function of x?
Step 1 — Solve for y to see how many outputs each x produces.
x = y² ⇒ y = ±√x
Reason: square-rooting both sides gives two possible y values whenever x > 0.
Step 2 — Pick a test value, e.g. x = 4.
x = 4 ⇒ y = +2 or y = −2
Reason: one input (x = 4) gives two outputs (2 and −2).
Step 3 — Apply the vertical line test graphically.
The vertical line x = 4 crosses the sideways parabola twice — at (4, 2) and (4, −2).
Conclusion. The graph fails the vertical line test, so x = y² is not a function of x.
3. Faded example — fill in the missing steps
Test whether the relation defined by the set of ordered pairs
S = { (0, 1), (1, 3), (2, 5), (1, −2) } is a function. Fill in each blank. 3 marks
Step 1 — List the inputs (x-values) that appear:
Inputs used: ____, ____, ____, ____.
Step 2 — Check whether any input appears more than once with different outputs.
The input x = ______ appears with outputs y = ______ and y = ______.
Step 3 — Apply the definition.
A function requires exactly one output per input. The repeated input has _________ different outputs, so the rule is violated.
Conclusion. S is __________________________ (function / not a function).
4. Graduated practice
For Foundation, classify each as function or not a function. For Standard, also show the algebra step. For Extension, justify in full.
Foundation — quick classification (4 questions)
| Q | Relation | Function or not? |
|---|---|---|
| 4.1 1 | y = 3x + 2 | |
| 4.2 1 | { (1, 4), (2, 5), (3, 6) } | |
| 4.3 1 | x² + y² = 9 (a circle) | |
| 4.4 1 | y = |x| |
Standard — evaluate / explain (6 questions)
Show at least one line of working for each.
4.5 If f(x) = x² − 3x + 5, find f(2). 2 marks
4.6 If f(x) = x² − 3x + 5, find f(−1). 2 marks
4.7 The graph of a sideways "V" opens to the right with vertex at the origin (i.e. the graph of x = |y|). Apply the vertical line test to decide whether it is a function. 2 marks
4.8 A taxi charge is modelled by C(d) = 5 + 2d, where C is the cost in dollars and d is the distance in km. Identify the independent variable, the dependent variable, and find C(8). 2 marks
4.9 Classify the set { (−2, 4), (−1, 1), (0, 0), (1, 1), (2, 4) } as a function or not, and explain in one line. Is the rule one-to-one? 2 marks
4.10 The temperature T (in °C) of a cup of coffee t minutes after pouring is given by T(t) = 90 e−0.05 t. Without a calculator, state what T(0) represents physically. 2 marks
Extension — combine concepts (2 questions)
4.11 A graph consists of the upper semicircle x² + y² = 25, y ≥ 0. Decide whether the graph defines y as a function of x. Justify using the vertical line test and state the natural domain of the resulting function. 3 marks
4.12 A student claims: "Every graph that is symmetric about the x-axis fails the vertical line test." Decide whether the claim is true or false. If true, explain why. If false, give a graph that is symmetric about the x-axis but still represents a function. 3 marks
5. Self-check the easy 3
Tick the first three once you've checked your method works.
How did this worksheet feel?
What I'll revisit before next class:
Q1.1 — Definition
A function is a relation in which each input has exactly one output.
Q1.2 — Vertical line test
Every vertical line crosses the graph at most once.
Q1.3 — Function notation
f(x) is read as "f of x". f is the name of the function and x is the input (independent variable).
Q3 — Faded example, S = {(0,1), (1,3), (2,5), (1,−2)}
Inputs used: 0, 1, 2, 1. The input x = 1 appears with outputs y = 3 and y = −2. That gives 2 different outputs for the same input. Not a function.
Q4.1 — y = 3x + 2
Function. The graph is a straight non-vertical line; every vertical line meets it exactly once.
Q4.2 — {(1, 4), (2, 5), (3, 6)}
Function. Each input (1, 2, 3) appears once, with a single output.
Q4.3 — x² + y² = 9
Not a function. Solving for y gives y = ±√(9 − x²), so each x in (−3, 3) produces two y-values. The vertical line x = 0 cuts the circle at y = 3 and y = −3.
Q4.4 — y = |x|
Function. The V-graph: each x has exactly one y because |x| is single-valued.
Q4.5 — f(2) for f(x) = x² − 3x + 5
f(2) = (2)² − 3(2) + 5 = 4 − 6 + 5 = 3.
Q4.6 — f(−1) for f(x) = x² − 3x + 5
f(−1) = (−1)² − 3(−1) + 5 = 1 + 3 + 5 = 9. (Brackets keep the minus sign safe inside the squared term.)
Q4.7 — x = |y|
Not a function. For any x > 0, y can be either +x or −x. A vertical line at x = 3 cuts the graph at (3, 3) and (3, −3).
Q4.8 — Taxi C(d) = 5 + 2d
Independent variable: d (distance in km). Dependent variable: C (cost in $). C(8) = 5 + 2(8) = $21.
Q4.9 — {(−2, 4), (−1, 1), (0, 0), (1, 1), (2, 4)}
Function — every input appears once. However it is not one-to-one: inputs −2 and 2 share output 4; inputs −1 and 1 share output 1. (This is the parabola y = x² sampled at integer points.)
Q4.10 — T(t) = 90 e−0.05 t
T(0) = 90 e⁰ = 90 × 1 = 90 °C. Physically this is the temperature of the coffee at the moment it is poured (t = 0).
Q4.11 — Upper semicircle x² + y² = 25, y ≥ 0
Solving for y under the constraint y ≥ 0 gives the single-valued y = +√(25 − x²). Every vertical line in the strip −5 ≤ x ≤ 5 meets the graph exactly once, so this is a function. Natural domain: [−5, 5].
Q4.12 — Symmetry about the x-axis claim
Almost true, with one exception. If a graph is symmetric about the x-axis and contains a point (a, b) with b ≠ 0, then it also contains (a, −b), so the vertical line x = a hits the graph twice — VLT fails. The only graph symmetric about the x-axis that does represent a function is one that lies entirely on the x-axis (e.g. y = 0), because then b = −b = 0. So the claim is false in general, but the counter-example is the trivial constant function y = 0.