Relations & Mappings
Q1. What is the DOMAIN of the relation {(2, 5), (4, 7), (6, 9)}?
Domain = {2, 4, 6}
Q1. Given f(x) = 4x − 1, find f(3).
f(3) = 4(3) − 1 = 11
Q1. State the gradient and y-intercept of y = 5x − 3.
Gradient = 5, y-intercept = −3
Q1. Write the equation of the axis of symmetry for y = x² − 6x + 2.
x = −b/2a = 6/2 = 3 → Axis of symmetry: x = 3
Q1. Describe the transformation that maps y = f(x) onto y = f(x) − 4.
Translation 4 units DOWNWARD (vertical shift down by 4)
Q2. State the RANGE of the relation {(1, 3), (2, 3), (5, 8)}.
Range = {3, 8}
$200
Easy-Medium
Q2. Given g(x) = x² − 3, evaluate g(−2).
g(−2) = (−2)² − 3 = 4 − 3 = 1
Q2. Find the x-intercept and y-intercept of the line 2x + 3y = 12.
x-intercept: set y=0 → x = 6 → (6, 0)
y-intercept: set x=0 → y = 4 → (0, 4)
Find the roots of y = x² − x − 6 by factoring.
(x − 3)(x + 2) = 0 → x = 3 or x = −2
Q2. The point (5, 3) lies on y = f(x). State the image of this point under the transformation y = f(x + 2).
Horizontal shift LEFT by 2 → image point: (3, 3)
Q3. Given the domain {−2, −1, 0, 1, 2} and the rule y = x + 3, list ALL ordered pairs.
{(−2, 1), (−1, 2), (0, 3), (1, 4), (2, 5)}
Q3. If h(x) = 2x + 5, find the inverse function h⁻¹(x).
Let y = 2x + 5 → x = (y − 5)/2 ∴ h⁻¹(x) = (x − 5)/2
Q3. Find the equation of the line passing through (3, 7) and (7, 15).
m = (15−7)/(7−3) = 8/4 = 2
y − 7 = 2(x − 3) → y = 2x + 1
Q3. Express y = x² − 8x + 10 in the form (x − h)² + k and state the minimum value.
y = (x − 4)² − 6
Minimum value = −6 at x = 4
Q3. Sketch the graphs of y = x² and y = (x − 2)² + 1 on the same axes. Describe the transformation.
y = (x−2)²+1 is y = x² translated 2 units RIGHT and 1 unit UP.
Vertex moves from (0,0) to (2, 1).
Q4. An arrow diagram shows: 1→1, 2→4, 3→9, 4→16. Write the algebraic rule and identify the type of mapping (one-to-one, many-to-one, one-to-many).
Rule: y = x² Type: One-to-one mapping (each input maps to a unique output)
Q4. Given f(x) = 3x − 1 and g(x) = x + 2, find fg(x) and gf(x). Are they equal?
fg(x) = f(x + 2) = 3(x + 2) − 1 = 3x + 5
gf(x) = g(3x − 1) = (3x − 1) + 2 = 3x + 1
They are NOT equal.
Q4. Line L₁ passes through (0, 4) and (6, 0). Find the equation of L₂, which is perpendicular to L₁ and passes through (3, 1).
Gradient of L₁ = (0−4)/(6−0) = −2/3
Gradient of L₂ = 3/2 (negative reciprocal)
y − 1 = (3/2)(x − 3) → y = (3/2)x − 7/2
Q4. Sketch the graph of f(x) = −x² + 2x + 3, clearly showing the vertex, axis of symmetry, x-intercepts and y-intercept.
Roots: −(x²−2x−3)=0 → (x−3)(x+1)=0 → x=3, x=−1
y-intercept: (0, 3)
Axis of symmetry: x = 1
Vertex: f(1) = −1+2+3 = 4 → V(1, 4) [maximum]
Q4. The graph of y = f(x) passes through (−1, 4), (0, 2), and (2, −3). State the coordinates of the corresponding points on y = −f(x) + 1.
Apply y → −y + 1 to each point:
(−1, 4) → (−1, −3)
(0, 2) → (0, −1)
(2, −3) → (2, 4)
Q5. The relation R = {(x, y) : y = |x| − 1, x ∈ {−3, −1, 0, 1, 3}}. List all ordered pairs and state the range.
Pairs: {(−3, 2), (−1, 0), (0, −1), (1, 0), (3, 2)} Range = {−1, 0, 2}
Q5. f(x) = (2x + 1)/(x − 3), x ≠ 3. Find f⁻¹(x) and state the value of x for which f⁻¹(x) is undefined.
Let y = (2x+1)/(x−3) → y(x−3) = 2x+1 → xy − 3y = 2x + 1 → x(y−2) = 3y+1 → f⁻¹(x) = (3x+1)/(x−2)
Undefined when x = 2
Q5. Lines L₁: 3x − 2y = 8 and L₂: 6x + ky = 15 are perpendicular. Find k and the coordinates of their point of intersection.
Gradient of L₁ = 3/2 → Gradient of L₂ must = −2/3
From L₂: gradient = −6/k = −2/3 → k = 9
Solve 3x−2y=8 and 6x+9y=15 simultaneously:
From L₁: y = (3x−8)/2; substitute: 6x + 9(3x−8)/2 = 15 → 12x + 27x − 72 = 30 → x = 102/39 = 34/13
y = (3(34/13)−8)/2 = −1/13
Point of intersection: (34/13, −1/13)
Q5. The graph of f(x) = ax² + bx + c has a maximum point at (2, 9) and passes through (0, 1). Find a, b and c. Hence state the range of f(x).
Vertex (2, 9): f(x) = a(x−2)²+9
f(0)=1: a(0−2)²+9=1 → 4a=−8 → a=−2
Expand: f(x)=−2(x−2)²+9=−2x²+8x+1
a=−2, b=8, c=1
Since a<0 (maximum), Range: f(x) ≤ 9
Q5. Describe fully ALL transformations that map y = x² onto y = 3(x + 1)² − 2. In what order must the transformations be applied? State the new vertex.
Starting from y = x²:
1. Translate 1 unit LEFT: y = (x+1)²
2. Vertical stretch by factor 3: y = 3(x+1)²
3. Translate 2 units DOWN: y = 3(x+1)²−2
Order: horizontal shift FIRST, then stretch, then vertical shift.
New vertex: (−1, −2)