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v0.1.1506
NotesMath AATopic 2.6
Unit 2 Β· Functions Β· Topic 2.6

IB Math AA β€” Quadratic functions

Topic 2.6 of IB Mathematics: Analysis and Approaches covers Quadratic functions, which is part of Unit 2: Functions. Students explore key concepts including Quadratic graphs, Vertex & axis of symmetry, Maximum, minimum & range. A strong understanding of quadratic functions is essential for IB Math AA exams and builds the foundation for connected topics across the syllabus.

Exam technique guidePractice questions

Key concepts in Quadratic functions

Key Idea: A quadratic graph is a parabola, and you can write it three ways β€” each hands you a different feature for free. It's mostly Paper 1 (by hand): sketch it, find the roots, or find the turning point.

πŸ“ The three forms β€” what each reveals

FormLooks likeRead off instantly
Standardy = axΒ² + bx + cDirection (sign of a) and the y-intercept c, i.e. (0, c).
Vertexy = a(x βˆ’ h)Β² + kThe turning point (h, k) β€” and k is the max/min value.
Factoredy = a(x βˆ’ p)(x βˆ’ q)The x-intercepts x = p and x = q (set each bracket to 0).
a > 0 β†’ opens up β†’ minimum. a < 0 β†’ opens down β†’ maximum. Watch the bracket signs: (x + 1) gives the root x = βˆ’1, and (x βˆ’ 3)Β² gives h = +3.

🎯 Vertex & axis of symmetry

x=βˆ’b2ax = -\frac{b}{2a}x=βˆ’2ab​
a,ba, ba,b
the coefficients in y = axΒ² + bx + c
x=βˆ’b2ax = -\frac{b}{2a}x=βˆ’2ab​
the axis of symmetry β€” also midway between the two x-intercepts; the vertex sits on it
Tip: To reach a(x βˆ’ h)Β² + k: halve the x-coefficient, square it for the bracket, then add/subtract to fix the constant. The vertex is (h, k), and k is the min (a > 0) or max (a < 0) value, since the squared part is never negative.

✏️ IB-style worked examples

IB-style question β€” direction & y-intercept from standard form

For y = βˆ’3xΒ² + 4x βˆ’ 7, state the direction it opens and its y-intercept.

Step by step:

  1. The sign of a sets the direction.

    a=βˆ’3<0β‡’opensΒ downa = -3 < 0 \Rightarrow \text{opens down}a=βˆ’3<0β‡’opensΒ down
  2. The constant c is the y-intercept.

    c=βˆ’7β‡’(0,β€‰βˆ’7)c = -7 \Rightarrow (0,\, -7)c=βˆ’7β‡’(0,βˆ’7)
Final answer:

Opens downward (so the vertex is a maximum); y-intercept (0, βˆ’7).

IB-style question β€” find the x-intercepts (factored form)

Find the x-intercepts of y = (x βˆ’ 5)(x + 2).

Step by step:

  1. Set each factor equal to zero.

    xβˆ’5=0β€…β€Šorβ€…β€Šx+2=0x - 5 = 0 \;\text{or}\; x + 2 = 0xβˆ’5=0orx+2=0
  2. Solve each β€” mind the sign on the second bracket.

    x=5β€…β€Šorβ€…β€Šx=βˆ’2x = 5 \;\text{or}\; x = -2x=5orx=βˆ’2
Final answer:

x-intercepts at (5, 0) and (βˆ’2, 0).

IB-style question β€” write in the form (x βˆ’ h)Β² + k

Write xΒ² βˆ’ 8x + 19 in the form (x βˆ’ h)Β² + k, and state the vertex.

Step by step:

  1. Halve the x-coefficient (βˆ’8 β†’ βˆ’4) and square it (16).

    (xβˆ’4)2=x2βˆ’8x+16(x - 4)^2 = x^2 - 8x + 16(xβˆ’4)2=x2βˆ’8x+16
  2. Fix the constant: 19 = 16 + 3.

    x2βˆ’8x+19=(xβˆ’4)2+3x^2 - 8x + 19 = (x - 4)^2 + 3x2βˆ’8x+19=(xβˆ’4)2+3
Final answer:

(x βˆ’ 4)Β² + 3, so the vertex is (4, 3).

IB-style question β€” build a quadratic from its vertex

A parabola has vertex (3, βˆ’4) and passes through (1, 4). Find a in y = a(x βˆ’ 3)Β² βˆ’ 4.

Step by step:

  1. Substitute the known point (1, 4).

    4=a(1βˆ’3)2βˆ’44 = a(1 - 3)^2 - 44=a(1βˆ’3)2βˆ’4
  2. Simplify and solve for a.

    4=4aβˆ’4β‡’a=24 = 4a - 4 \Rightarrow a = 24=4aβˆ’4β‡’a=2
Final answer:

a = 2, so y = 2(x βˆ’ 3)Β² βˆ’ 4.


Important: In a(x βˆ’ h)Β² + k, the vertex x-coordinate is +h: (x βˆ’ 3)Β² means h = 3, not βˆ’3. And re-read what's asked: the min/max value is k (a number); the min/max point is (h, k).

Tap each card to reveal the answer.

Which form gives the y-intercept at a glance? Standard form y = axΒ² + bx + c β€” the y-intercept is (0, c).

Roots of y = (x βˆ’ 6)(x + 4)? x = 6 and x = βˆ’4 β€” set each bracket to zero.

Axis of symmetry of y = 2xΒ² βˆ’ 12x + 1? x = 3 β€” use x = βˆ’b/(2a) = βˆ’(βˆ’12)/(2Β·2).

Vertex of y = (x + 5)Β² βˆ’ 2? (βˆ’5, βˆ’2) β€” (x + 5)Β² gives h = βˆ’5, and k = βˆ’2.

Does y = βˆ’xΒ² + 6x βˆ’ 1 have a max or a min? A maximum β€” a = βˆ’1 < 0, so the parabola opens down.

Exam Tips

  • Pick the form that answers the question: factored β†’ roots, vertex β†’ turning point, standard β†’ y-intercept & direction.
  • Sign of a: a > 0 opens up (minimum), a < 0 opens down (maximum).
  • Axis of symmetry x = βˆ’b/(2a) is also exactly midway between the two x-intercepts.
  • Complete the square: halve b, square it, then add/subtract to keep the constant correct.
  • (x βˆ’ h)Β² gives h = +h; the min/max value is k, the min/max point is (h, k).

What you'll learn in Topic 2.6

  • 2.6.1 Quadratic graphs
  • 2.6.2 Vertex & axis of symmetry
  • 2.6.3 Maximum, minimum & range
Suggested study order: Read the notes for each sub-topic below β†’ test yourself with flashcards β†’ attempt practice questions β†’ review exam technique.

Study resources β€” 2.6 Quadratic functions

2.6.1

Quadratic graphs

Notes
2.6.2

Vertex & axis of symmetry

Notes
2.6.3

Maximum, minimum & range

Notes

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Topic 2.6 Quadratic functions forms a core part of Unit 2: Functions in IB Math AA. Mastering these concepts will strengthen your understanding of connected topics across the syllabus and prepare you for exam questions that require analysis, evaluation, and real-world application.

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