Unit 2 · 11th Grade Mathematics
Quadratic
Functions.
Features, forms, complex numbers, and solving quadratic equations — all in one guide.
Topic A: Features of Quadratic Functions
Lessons 1–4 · Vertex, axis of symmetry, zeros, and forms of quadratics.
01 — Lesson 1
Key Features of Parabolas
A quadratic function creates a U-shaped parabola with distinct features.
- Vertex: the highest or lowest point — (h, k).
- Axis of symmetry: vertical line x = h through the vertex.
- Opens up (a > 0): minimum at vertex. Opens down (a < 0): maximum.
- Zeros/roots: x-intercepts where f(x) = 0.
- Y-intercept: f(0) = the constant term c in ax² + bx + c.
f(x) = −(x−3)² + 9 → vertex (3, 9) is a maximum; axis: x = 3; opens down.
Practice — Lesson 1
Q1 · Multiple Choice
For f(x) = 2(x−1)² − 4, what is the vertex?
Q2 · Multiple Choice
f(x) = −3x² + 6x − 1. Does this parabola have a maximum or minimum?
02–03 — Lessons 2–3
Forms of Quadratic Functions
Quadratics can be written in three equivalent forms, each revealing different features.
- Standard form: f(x) = ax² + bx + c → y-intercept = c.
- Vertex form: f(x) = a(x−h)² + k → vertex = (h, k).
- Factored/Intercept form: f(x) = a(x−p)(x−q) → zeros at x = p and x = q.
- Convert between forms by expanding or completing the square.
f(x) = (x−2)(x−6) → zeros: x = 2, x = 6; axis: x = 4; vertex: (4, −4)
Practice — Lessons 2–3
Q3 · Multiple Choice
Which form of a quadratic most easily reveals its x-intercepts?
Q4 · Short Answer
What is the y-intercept of f(x) = 3x² − 5x + 7?
04 — Lesson 4
Transformations of Quadratics
Transformations shift, stretch, or flip the parent function f(x) = x².
- f(x) = x² + k: shifts up k units (down if k < 0).
- f(x) = (x − h)²: shifts right h units (left if h < 0).
- f(x) = ax²: a > 1 narrows; 0 < a < 1 widens; a < 0 flips.
f(x) = −2(x+1)² + 5 → flipped (a=−2), shifted left 1, up 5. Vertex: (−1, 5).
Practice — Lesson 4
Q5 · Multiple Choice
Compared to f(x) = x², how is g(x) = (x−4)² + 3 transformed?
Q6 · Multiple Choice
Which quadratic is narrower than f(x) = x²?
Topic B: Complex Numbers
Lessons 5–7 · Imaginary numbers, complex number operations, and complex solutions.
05 — Lesson 5
Imaginary & Complex Numbers
When a quadratic has no real zeros, its solutions involve the imaginary unit i.
- i = √(−1), so i² = −1.
- Complex number: a + bi, where a is real and bi is imaginary.
- Pure imaginary: no real part, e.g. 3i.
- √(−9) = √9 · √(−1) = 3i
√(−25) = 5i · (2 + 3i) + (1 − i) = 3 + 2i
Practice — Lesson 5
Q7 · Short Answer
Simplify √(−16). Write in the form bi.
Q8 · Multiple Choice
What is i²?
06–07 — Lessons 6–7
Operating with Complex Numbers
Add, subtract, multiply, and divide complex numbers — treat i like a variable, but replace i² with −1.
- Addition/Subtraction: combine real parts and imaginary parts separately.
- Multiplication: FOIL, then replace i² = −1.
- Conjugate of a + bi: a − bi. Used for division.
- Division: multiply numerator and denominator by the conjugate.
(2 + 3i)(1 − 2i) = 2 − 4i + 3i − 6i² = 2 − i + 6 = 8 − i
Practice — Lessons 6–7
Q9 · Short Answer
Multiply: (3 + i)(3 − i). (Hint: it's a difference of squares.)
Q10 · Multiple Choice
What is the complex conjugate of 5 − 2i?
Topic C: Solving Quadratic Equations
Lessons 8–11 · Factoring, completing the square, and the quadratic formula.
08–09 — Lessons 8–9
Solving by Factoring
Set the quadratic equal to zero, factor it, then use the zero product property.
- Zero product property: if AB = 0, then A = 0 or B = 0.
- Factor: find two binomials whose product equals the quadratic.
- GCF first: always factor out the greatest common factor first.
- Works cleanly when solutions are rational numbers.
x² − 5x + 6 = 0 → (x−2)(x−3) = 0 → x = 2 or x = 3
Practice — Lessons 8–9
Q11 · Short Answer
Solve: x² + 7x + 12 = 0. Give the smaller solution.
Q12 · Multiple Choice
The zero product property says: if (x−5)(x+2) = 0, then...
10–11 — Lessons 10–11
Quadratic Formula & Discriminant
When factoring is difficult, use the quadratic formula to find exact solutions.
- Formula: x = [−b ± √(b²−4ac)] / 2a
- Discriminant (b²−4ac): determines the number of solutions.
- > 0: two real solutions. = 0: one real solution. < 0: two complex solutions.
x² − 4x + 1 = 0 → discriminant = 16−4 = 12 > 0 → two real solutions:
x = (4 ± √12)/2 = 2 ± √3
Practice — Lessons 10–11
Q13 · Multiple Choice
For 2x² + 3x + 5 = 0, the discriminant is 9 − 40 = −31. What does this tell us?
Q14 · Multiple Choice
Using the quadratic formula on x² − 6x + 9 = 0, what is x?