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NotesMath AA HLTopic 2.11Stretches & reflections
Back to Math AA HL Topics
2.11.22 min read

Stretches & reflections

IB Mathematics: Analysis and Approaches • Unit 2

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Contents

  • Vertical stretch: a·f(x)
  • Horizontal stretch: f(bx)
  • Reflections
  • Effect on points
  • Reflecting to find a range — the exam question
Multiply outside → stretch the y's: y = a·f(x) stretches the graph vertically by factor a — every y-coordinate is multiplied by a.

(x-intercepts stay put, since 0 × a = 0.)

IB-style question — vertical stretch

y = f(x) passes through (2, 3).

Where does y = 4f(x) pass through (at x = 2)?

Step by step

  1. Multiply the y-coordinate by 4.

Final answer

(2, 12).

x-intercepts don't move: A vertical stretch leaves the x-intercepts fixed (their y is 0) but moves every other point away from the x-axis.
Multiply inside → stretch the x's by 1/b: y = f(bx) stretches the graph horizontally by factor 1/b — the reciprocal.

So f(2x) squashes by ½ (factor 1/2), and f(x/2) stretches by 2.

Inside changes are counterintuitive again.

IB-style question — horizontal stretch

Describe the transformation y = f(x) → y = f(2x).

Step by step

  1. Inside factor 2 → horizontal stretch by 1/2.

Final answer

Horizontal stretch with scale factor ½ (the graph is squashed toward the y-axis).

Reciprocal factor: f(bx) stretches by 1/b, not b. f(3x) → factor 1/3 (squash); f(x/3) → factor 3 (stretch).

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Minus outside / inside flips an axis: y = −f(x) reflects the graph in the x-axis (flip the y-coordinates).

y = f(−x) reflects it in the y-axis (flip the x-coordinates).

y = −f(x)

  • Reflect in the x-axis
  • (x, y) → (x, −y)
  • minus is OUTSIDE

y = f(−x)

  • Reflect in the y-axis
  • (x, y) → (−x, y)
  • minus is INSIDE
Outside flips y, inside flips x: Same theme: outside the function changes y; inside changes x.

[Diagram: math-transformation] - Available in full study mode

Track one point through the change: Apply the rule to a point's coordinates: a·f(x) multiplies y by a; f(bx) divides x by b; −f(x) negates y; f(−x) negates x.

IB-style question — image under a reflection

The point (2, 5) lies on y = f(x).

Find its image on y = −f(x).

Step by step

  1. −f(x) negates the y-coordinate.

Final answer

(2, −5).

Combine carefully: For 2f(x) the point (2, 5) → (2, 10); for f(−x) it → (−2, 5).

Apply each rule to the correct coordinate.

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Reflect in the x-axis → the range flips: g(x) = −f(x) reflects f in the x-axis: every output y becomes −y.

So take the range of f and negate both ends — which swaps ≤ and ≥.

An open bound (<) stays open and a closed bound (≤) stays closed; the symbol just moves to the other end of the range.

[Diagram: math-transformation] - Available in full study mode

IB-style question — reflect, then find the range

The graph of y = f(x), for x ≥ 0, has range −2 < f(x) ≤ 4.

The function g is defined by g(x) = −f(x) for x ≥ 0.

Find the range of g.

Step by step

  1. g = −f reflects f in the x-axis, so every output changes sign.
  2. The closed end: f's maximum 4 becomes g's minimum −4 (still reached).
  3. The open end: f's lower bound −2 becomes g's upper bound 2 (still not reached).
  4. Combine, smaller bound first.

Final answer

Range of g is −4 ≤ g(x) < 2.

Don't lose the open/closed end: The easy mark to drop is the inequality type.

Reflecting keeps each bound open or closed, but the direction flips (≤ ↔ ≥) because the values change sign: f's closed top (≤ 4) must become g's closed bottom (≥ −4).

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fully the transformation that maps y = f(x) onto y = 5f(x). [2 marks]

Related Math AA HL Topics

Continue learning with these related topics from the same unit:

2.1.1Equations of lines
2.1.2Parallel lines
2.1.3Perpendicular lines
2.1.4Perpendicular bisector
View all Math AA HL topics

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