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.