Lesson 1 of 12
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1. Quadratic Functions P1
Quadratic Function: A function of the form f(x) = ax² + bx + c where a ≠ 0. The graph is a parabola — opening upward (a > 0) or downward (a < 0). The vertex (turning point) is found by completing the square.
Completing the Square
Completing the Square — Standard Form
ax² + bx + c = a(x + b/2a)² + (c − b²/4a)
Vertex: (−b/2a, c − b²/4a)
Axis of symmetry: x = −b/2a
📐 Worked Example 1 — Completing the Square
Express f(x) = 2x² − 8x + 5 in the form a(x + p)² + q. State the vertex and the range of f.
1
Factor out the coefficient of x²:
f(x) = 2(x² − 4x) + 5
f(x) = 2(x² − 4x) + 5
2
Complete the square inside the bracket — add and subtract (−4/2)² = 4:
= 2(x² − 4x + 4 − 4) + 5 = 2(x − 2)² − 8 + 5
f(x) = 2(x − 2)² − 3
= 2(x² − 4x + 4 − 4) + 5 = 2(x − 2)² − 8 + 5
f(x) = 2(x − 2)² − 3
3
Vertex: (2, −3) — minimum point since a = 2 > 0.
Range: f(x) ≥ −3, i.e. f ∈ [−3, ∞)
Range: f(x) ≥ −3, i.e. f ∈ [−3, ∞)
The Discriminant
Discriminant Δ = b² − 4ac
Δ > 0 : Two distinct real roots (graph crosses x-axis twice)
Δ = 0 : One repeated real root (graph touches x-axis)
Δ < 0 : No real roots (graph does not meet x-axis)
📐 Worked Example 2 — Discriminant and Conditions
Find the values of k for which x² + kx + (k+3) = 0 has two distinct real roots.
1
For two distinct real roots: Δ > 0
b² − 4ac > 0 → k² − 4(1)(k+3) > 0
b² − 4ac > 0 → k² − 4(1)(k+3) > 0
2
k² − 4k − 12 > 0 → (k − 6)(k + 2) > 0
Roots of (k−6)(k+2)=0 are k = 6 and k = −2.
Roots of (k−6)(k+2)=0 are k = 6 and k = −2.
3
Since a = 1 > 0 (upward parabola in k), the quadratic in k is positive outside the roots:
k < −2 or k > 6
k < −2 or k > 6
Quadratic Inequalities
Method for Quadratic Inequalities:
1. Rearrange so one side is 0. 2. Factorise to find roots p and q (p < q).
3. Sketch the parabola (know which way it opens).
4. For ax²+bx+c > 0 (a>0): solution is x < p or x > q (outside roots).
5. For ax²+bx+c < 0 (a>0): solution is p < x < q (between roots).
1. Rearrange so one side is 0. 2. Factorise to find roots p and q (p < q).
3. Sketch the parabola (know which way it opens).
4. For ax²+bx+c > 0 (a>0): solution is x < p or x > q (outside roots).
5. For ax²+bx+c < 0 (a>0): solution is p < x < q (between roots).
📐 Worked Example 3 — Quadratic Inequality
Solve 3x² − x − 10 ≤ 0.
1
Factorise: 3x² − x − 10 = (3x + 5)(x − 2)
Roots: x = −5/3 and x = 2.
Roots: x = −5/3 and x = 2.
2
Since a = 3 > 0, parabola opens upward. The expression ≤ 0 between the roots:
−5/3 ≤ x ≤ 2
−5/3 ≤ x ≤ 2
2. Functions P1
Function: A rule that maps each element of the domain to exactly one element of the range. Written f: x ↦ f(x) or f(x) = ... The domain is the set of valid input values; the range is the set of actual output values.
Domain and Range
| Function | Natural Domain | Range |
|---|---|---|
| f(x) = x² | x ∈ ℝ | f(x) ≥ 0 |
| f(x) = √x | x ≥ 0 | f(x) ≥ 0 |
| f(x) = 1/x | x ∈ ℝ, x ≠ 0 | f(x) ∈ ℝ, f(x) ≠ 0 |
| f(x) = ln x | x > 0 | f(x) ∈ ℝ |
| f(x) = eˣ | x ∈ ℝ | f(x) > 0 |
| f(x) = sin x | x ∈ ℝ | −1 ≤ f(x) ≤ 1 |
| f(x) = a(x+p)²+q, a>0 | x ∈ ℝ | f(x) ≥ q |
Composite Functions
Composite function fg(x) = f(g(x)): Apply g first, then f. The domain of fg is the set of x in the domain of g such that g(x) lies in the domain of f.
📐 Worked Example 4 — Composite Functions
f(x) = x² + 1 for x ∈ ℝ and g(x) = 2x − 3 for x ∈ ℝ. Find fg(x) and gf(x). Are they equal?
1
fg(x) = f(g(x)) = f(2x−3) = (2x−3)² + 1
= 4x²−12x+9+1 = 4x²−12x+10
= 4x²−12x+9+1 = 4x²−12x+10
2
gf(x) = g(f(x)) = g(x²+1) = 2(x²+1)−3 = 2x²−1
3
fg(x) ≠ gf(x) — composite functions are not generally commutative.
Inverse Functions
Finding the Inverse f⁻¹(x):
1. Write y = f(x). 2. Swap x and y. 3. Solve for y. 4. Write as f⁻¹(x).
Key properties: ff⁻¹(x) = x and f⁻¹f(x) = x.
Domain of f⁻¹ = Range of f. Range of f⁻¹ = Domain of f.
Graph of f⁻¹ is the reflection of f in the line y = x.
Condition: Inverse exists only if f is one-to-one (strictly monotonic) on its domain.
1. Write y = f(x). 2. Swap x and y. 3. Solve for y. 4. Write as f⁻¹(x).
Key properties: ff⁻¹(x) = x and f⁻¹f(x) = x.
Domain of f⁻¹ = Range of f. Range of f⁻¹ = Domain of f.
Graph of f⁻¹ is the reflection of f in the line y = x.
Condition: Inverse exists only if f is one-to-one (strictly monotonic) on its domain.
📐 Worked Example 5 — Inverse of a Restricted Quadratic
f(x) = (x−2)² + 3 for x ≥ 2. Find f⁻¹(x) and state its domain.
1
The domain x ≥ 2 restricts f to the right branch (one-to-one, increasing). Range of f: f ≥ 3.
2
y = (x−2)²+3 → swap: x = (y−2)²+3 → x−3 = (y−2)² → y−2 = ±√(x−3)
Since x ≥ 2 (domain of f) maps to y ≥ 2 (range of f⁻¹), take positive root:
f⁻¹(x) = √(x−3) + 2
Since x ≥ 2 (domain of f) maps to y ≥ 2 (range of f⁻¹), take positive root:
f⁻¹(x) = √(x−3) + 2
3
Domain of f⁻¹: x ≥ 3 (= Range of f)
Graph Transformations
| Transformation | Effect on y = f(x) | Example |
|---|---|---|
| y = f(x) + a | Translate a units upward | y=x²+3 shifts x² up 3 |
| y = f(x + a) | Translate a units to the left | y=(x+2)² shifts x² left 2 |
| y = af(x) | Vertical stretch by factor a | y=3x² stretches x² ×3 vertically |
| y = f(ax) | Horizontal stretch by factor 1/a | y=(2x)² squeezes x² horizontally ×½ |
| y = −f(x) | Reflection in x-axis | y=−x² reflects x² in x-axis |
| y = f(−x) | Reflection in y-axis | y=(−x)²=x² (symmetric) |
| y = |f(x)| | Reflect negative parts upward | y=|x²−4| — negative arch flipped up |
📐 Worked Example 6 — Chain of Transformations
Describe the transformations that map y = x² to y = 3(x−1)² + 2.
1
x² → (x−1)²: translate 1 unit to the right. Translation vector: (1, 0).
2
(x−1)² → 3(x−1)²: vertical stretch by factor 3 from the x-axis.
3
3(x−1)² → 3(x−1)²+2: translate 2 units upward. Translation vector: (0, 2).
3. Coordinate Geometry P1
Core Coordinate Geometry Formulae
Gradient: m = (y₂ − y₁)/(x₂ − x₁)
Distance: d = √[(x₂−x₁)² + (y₂−y₁)²]
Midpoint: M = ((x₁+x₂)/2, (y₁+y₂)/2)
Line equation (gradient-intercept): y = mx + c
Line equation (point-slope): y − y₁ = m(x − x₁)
Perpendicular gradient: m₂ = −1/m₁ (product = −1)
Circles
Circle Equations
Standard form: (x − a)² + (y − b)² = r²
Centre (a, b), radius r
General form: x² + y² + 2gx + 2fy + c = 0
Centre (−g, −f), radius = √(g² + f² − c)
Key Circle Properties:
• The angle in a semicircle is 90° — if AB is a diameter, then ∠APB = 90° for any point P on the circle.
• The tangent to a circle at point P is perpendicular to the radius at P.
• The perpendicular from the centre to a chord bisects the chord.
• To find the tangent at point P: find the radius OP gradient, then use perpendicular gradient.
• The angle in a semicircle is 90° — if AB is a diameter, then ∠APB = 90° for any point P on the circle.
• The tangent to a circle at point P is perpendicular to the radius at P.
• The perpendicular from the centre to a chord bisects the chord.
• To find the tangent at point P: find the radius OP gradient, then use perpendicular gradient.
📐 Worked Example 7 — Circle: Centre, Radius, and Tangent
Find the centre and radius of x² + y² − 6x + 4y − 12 = 0. Find the equation of the tangent at the point (7, 1).
1
Complete the square in x and y:
(x²−6x+9) + (y²+4y+4) = 12+9+4 = 25
(x−3)² + (y+2)² = 25
Centre = (3, −2), radius = 5.
(x²−6x+9) + (y²+4y+4) = 12+9+4 = 25
(x−3)² + (y+2)² = 25
Centre = (3, −2), radius = 5.
2
Verify (7,1) is on circle: (7−3)²+(1+2)²=16+9=25 ✓
3
Gradient of radius from centre (3,−2) to (7,1):
m_radius = (1−(−2))/(7−3) = 3/4
Gradient of tangent (perpendicular): m_tangent = −4/3
m_radius = (1−(−2))/(7−3) = 3/4
Gradient of tangent (perpendicular): m_tangent = −4/3
4
Tangent at (7,1): y−1 = −4/3(x−7)
3y−3 = −4x+28 →
4x + 3y = 31
3y−3 = −4x+28 →
4x + 3y = 31
📐 Worked Example 8 — Intersection of Line and Circle
Find the coordinates where y = 2x − 1 meets the circle (x−1)² + (y−3)² = 20.
1
Substitute y = 2x−1 into the circle equation:
(x−1)² + (2x−1−3)² = 20
(x−1)² + (2x−4)² = 20
(x−1)² + (2x−1−3)² = 20
(x−1)² + (2x−4)² = 20
2
x²−2x+1 + 4x²−16x+16 = 20
5x²−18x+17 = 20
5x²−18x−3 = 0
5x²−18x+17 = 20
5x²−18x−3 = 0
3
x = (18 ± √(324+60))/10 = (18 ± √384)/10 = (18 ± 8√6)/10 = (9 ± 4√6)/5
Corresponding y values: y = 2x−1 substituted for each x.
x = (9+4√6)/5 or x = (9−4√6)/5
Corresponding y values: y = 2x−1 substituted for each x.
x = (9+4√6)/5 or x = (9−4√6)/5
Coordinate Geometry — Harder Results
Perpendicular Bisector
The locus of points equidistant from A and B. Steps:
- Find midpoint M of AB.
- Find gradient of AB: m.
- Perpendicular gradient: −1/m.
- Line through M with gradient −1/m.
Angle Between Line and Circle
To determine if a line y=mx+c intersects, touches, or misses circle (x−a)²+(y−b)²=r²:
- Substitute line into circle → quadratic in x.
- Δ > 0: two intersections.
- Δ = 0: tangent (touches).
- Δ < 0: no intersection.
4. Polynomials and the Factor Theorem P1
Remainder and Factor Theorems
Remainder Theorem: When f(x) is divided by (x−a), the remainder = f(a)
Factor Theorem: (x−a) is a factor of f(x) ⟺ f(a) = 0
For (ax−b): substitute x = b/a
📐 Worked Example 9 — Factor Theorem
f(x) = 2x³ + x² − 5x + 2. Factorise completely.
1
Try f(1)=2+1−5+2=0 ✓ → (x−1) is a factor.
2
Divide: 2x³+x²−5x+2 ÷ (x−1) = 2x²+3x−2
(by inspection or long division)
(by inspection or long division)
3
Factorise 2x²+3x−2 = (2x−1)(x+2)
f(x) = (x−1)(2x−1)(x+2)
f(x) = (x−1)(2x−1)(x+2)
5. Simultaneous Equations — One Linear, One Quadratic P1
Method: Substitute the linear equation into the quadratic, form a single quadratic equation, solve using factorisation or the quadratic formula. The number of solutions depends on the discriminant of the resulting quadratic.
📐 Worked Example 10 — Simultaneous Equations
Solve: y = 3x − 2 and y = x² + x − 4.
1
Set equal: 3x−2 = x²+x−4
0 = x²−2x−2
0 = x²−2x−2
2
x = (2 ± √(4+8))/2 = (2 ± 2√3)/2 = 1 ± √3
3
y = 3x−2:
x = 1+√3: y = 3(1+√3)−2 = 1+3√3
x = 1−√3: y = 3(1−√3)−2 = 1−3√3
Solutions: (1+√3, 1+3√3) and (1−√3, 1−3√3)
x = 1+√3: y = 3(1+√3)−2 = 1+3√3
x = 1−√3: y = 3(1−√3)−2 = 1−3√3
Solutions: (1+√3, 1+3√3) and (1−√3, 1−3√3)
📐 Worked Example 11 — Tangency Condition
Find the values of c for which y = 2x + c is a tangent to y = x² − 3.
1
Substitute: 2x + c = x²−3 → x²−2x−(c+3) = 0
2
For tangency: Δ = 0
4 + 4(c+3) = 0 → 4 + 4c + 12 = 0 → 4c = −16
c = −4
4 + 4(c+3) = 0 → 4 + 4c + 12 = 0 → 4c = −16
c = −4
📝 Exam Practice Questions
Q1 [4 marks] — Express f(x) = 3x² + 12x − 1 in the form a(x+p)²+q. Hence state the minimum value of f and the value of x at which it occurs. State also the range of f.
f(x) = 3(x²+4x)−1 = 3(x²+4x+4−4)−1 = 3(x+2)²−12−1
f(x) = 3(x+2)²−13
Minimum value = −13 at x = −2.
Range: f(x) ≥ −13
f(x) = 3(x+2)²−13
Minimum value = −13 at x = −2.
Range: f(x) ≥ −13
Q2 [3 marks] — Find the set of values of k for which the equation 2x² − 3x + k = 0 has no real roots.
For no real roots: Δ < 0
b²−4ac < 0 → 9−8k < 0 → 8k > 9
k > 9/8
b²−4ac < 0 → 9−8k < 0 → 8k > 9
k > 9/8
Q3 [4 marks] — f(x) = 2x+1 for x ∈ ℝ and g(x) = x²−2 for x ∈ ℝ.
(a) Find gf(x). (b) Find f⁻¹(x). (c) Solve gf(x) = f⁻¹(x).
(a) gf(x) = g(2x+1) = (2x+1)²−2 = 4x²+4x−1
(b) y=2x+1 → x=(y−1)/2 → f⁻¹(x)=(x−1)/2
(c) 4x²+4x−1 = (x−1)/2
8x²+8x−2 = x−1 → 8x²+7x−1=0 → (8x−1)(x+1)=0
x = 1/8 or x = −1
(b) y=2x+1 → x=(y−1)/2 → f⁻¹(x)=(x−1)/2
(c) 4x²+4x−1 = (x−1)/2
8x²+8x−2 = x−1 → 8x²+7x−1=0 → (8x−1)(x+1)=0
x = 1/8 or x = −1
Q4 [5 marks] — A circle has equation x²+y²−4x+6y−3=0.
(a) Find the centre and radius. (b) Find the equation of the tangent at (5,−2). (c) Determine whether the point (−1, 1) lies inside, on, or outside the circle.
(a) Complete the square: (x−2)²+(y+3)²=3+4+9=16
Centre = (2, −3), radius = 4
(b) Gradient of radius to (5,−2): m=(−2+3)/(5−2)=1/3
Tangent gradient = −3
Equation: y+2=−3(x−5) → 3x+y=13
(c) Distance from (−1,1) to centre (2,−3):
d=√(9+16)=√25=5 > 4 = radius
Outside the circle
Centre = (2, −3), radius = 4
(b) Gradient of radius to (5,−2): m=(−2+3)/(5−2)=1/3
Tangent gradient = −3
Equation: y+2=−3(x−5) → 3x+y=13
(c) Distance from (−1,1) to centre (2,−3):
d=√(9+16)=√25=5 > 4 = radius
Outside the circle
Exam Tip: For part (c), compare the distance from the point to the centre with the radius. No need to find the full equation — just compute d² and compare with r². Using d² avoids the square root: d²=9+16=25 > 16=r², so outside.
Q5 [4 marks] — f(x) = x³ − 4x² + x + 6. Show that (x−2) is a factor, factorise completely, and solve f(x) = 0.
f(2) = 8−16+2+6 = 0 ✓ → (x−2) is a factor.
Dividing: x³−4x²+x+6 = (x−2)(x²−2x−3) = (x−2)(x−3)(x+1)
Roots: x = −1, x = 2, x = 3
Dividing: x³−4x²+x+6 = (x−2)(x²−2x−3) = (x−2)(x−3)(x+1)
Roots: x = −1, x = 2, x = 3
Q6 [4 marks] — The line y = mx + 4 is a tangent to the curve y = x² + 2x + 5. Find the two possible values of m and the corresponding points of tangency.
Substitute: mx+4 = x²+2x+5
x²+(2−m)x+1=0. For tangency Δ=0:
(2−m)²−4=0 → (2−m)²=4 → 2−m=±2
m=0 or m=4.
m=0: x²+2x+1=0 → (x+1)²=0 → x=−1, y=4. Point: (−1, 4)
m=4: x²−2x+1=0 → (x−1)²=0 → x=1, y=8. Point: (1, 8)
x²+(2−m)x+1=0. For tangency Δ=0:
(2−m)²−4=0 → (2−m)²=4 → 2−m=±2
m=0 or m=4.
m=0: x²+2x+1=0 → (x+1)²=0 → x=−1, y=4. Point: (−1, 4)
m=4: x²−2x+1=0 → (x−1)²=0 → x=1, y=8. Point: (1, 8)