Grade by Grade
A. Characteristics of Functions · Functions and Equivalent Expressions

Stretches, Compressions and Reflections: y = af(kx)

A1.8 A1.9About 70 minutes
Learning goals
  • I can describe the effect of a in y = af(x) as a vertical stretch or compression, with a reflection in the x-axis when a < 0.
  • I can describe the effect of k in y = f(kx) as a horizontal stretch or compression, with a reflection in the y-axis when k < 0.
  • I can use the mapping rules (x, y) → (x, ay) and (x, y) → (x/k, y) to sketch transformed parent functions.
  • I can state the domain and range after a stretch or reflection.

Words to know

vertical stretch / compression
y = af(x) with |a| > 1 (stretch) or 0 < |a| < 1 (compression); every y-value is multiplied by a.
horizontal stretch / compression
y = f(kx) with 0 < |k| < 1 (stretch) or |k| > 1 (compression), by a factor of 1/|k|; every x-value is divided by k.
reflection in the x-axis
y = −f(x); (x, y) → (x, −y).
reflection in the y-axis
y = f(−x); (x, y) → (−x, y).
invariant point
A point that does not move under a transformation, such as a point on the x-axis under a vertical stretch.

Stretching the stencil

Translations slide a graph. Now we change its shape: pull it taller or squash it flatter (vertical), pull it wider or squeeze it narrower (horizontal), or flip it over an axis. In Grade 10 you saw that y = ax² is narrower when a is large and opens down when a is negative. Here you will meet two different numbers, a outside the function and k inside it:

y = af(x) and y = f(kx)

Vertical stretches and compressions: y = af(x)

The number a multiplies every output: (x, y) → (x, ay).

  • |a| > 1: vertical stretch by a factor of |a| (taller).
  • 0 < |a| < 1: vertical compression by a factor of |a| (flatter).
  • a < 0: also a reflection in the x-axis (upside down).

Points on the x-axis (y = 0) do not move, because a × 0 = 0. That is why we say the stretch is "from the x-axis".

xy−3−2−1132468y = af(x)multiply every y by athick: y = 2x²thin: y = x²dashed: y = 0.5x²at x = 2: heights 8, 4, 2a = 2: vertical stretch,factor 2 (from the x-axis)a = 0.5: verticalcompression, factor 0.5
Every y-value of y = x² is doubled for y = 2x² and halved for y = 0.5x². At x = 2 the heights are 8, 4 and 2. The vertex (0, 0) does not move.

Horizontal stretches and compressions: y = f(kx)

The number k multiplies the input, and again the effect feels backwards: (x, y) → (x/k, y).

  • |k| > 1: horizontal compression by a factor of 1/|k| (narrower).
  • 0 < |k| < 1: horizontal stretch by a factor of 1/|k| (wider).
  • k < 0: also a reflection in the y-axis (left-right flip).

Why divide? For y = √(2x) to output 2, we need 2x = 4, so x = 2, not 4. Each output arrives at half the x-value. Points on the y-axis (x = 0) do not move.

xy246810−11234y = f(kx): divide x by kthick: y = √(2x)thin: y = √xdashed: y = √(0.5x)k = 2: compression, factor 1/2k = 0.5: stretch, factor 2height 2 at x = 2, 4, 8
To reach the height 2: √x needs x = 4, √(2x) needs only x = 2 (compressed), and √(0.5x) needs x = 8 (stretched).

Outside affects y; inside affects x

a is outside the function, so it acts on y the way you expect (multiply). k is inside with x, so it acts on x the opposite way (divide). The same was true for c and d in translations.

Worked example

Example 1: Vertical stretches of a parabola

Using the points (−2, 4), (−1, 1), (0, 0), (1, 1), (2, 4) of y = x², find points on (a) y = 3x² (b) y = (1/2)x² (c) y = −2x². Describe each transformation and give the range.

  1. (a) (x, y) → (x, 3y): (−2, 12), (−1, 3), (0, 0), (1, 3), (2, 12). Vertical stretch by a factor of 3. R: y ≥ 0.
  2. (b) (x, y) → (x, y/2): (−2, 2), (−1, 0.5), (0, 0), (1, 0.5), (2, 2). Vertical compression by a factor of 1/2. R: y ≥ 0.
  3. (c) (x, y) → (x, −2y): (−2, −8), (−1, −2), (0, 0), (1, −2), (2, −8). Vertical stretch by a factor of 2 and a reflection in the x-axis. R: y ≤ 0.

Only the reflection changed the range; the stretch and the compression kept it as y ≥ 0.

Worked example

Example 2: Horizontal changes to the square root function

Using the points (0, 0), (1, 1), (4, 2), (9, 3) of y = √x, find points on (a) y = √(3x) (b) y = √((1/2)x). Describe each.

  1. (a) k = 3: (x, y) → (x/3, y): (0, 0), (1/3, 1), (4/3, 2), (3, 3). Horizontal compression by a factor of 1/3.
  2. (b) k = 1/2: (x, y) → (x ÷ 1/2, y) = (2x, y): (0, 0), (2, 1), (8, 2), (18, 3). Horizontal stretch by a factor of 2.
  3. Check (b) with the equation: √((1/2)(8)) = √4 = 2 ✓.

√(3x) is narrower (compressed towards the y-axis); √(x/2) is wider.

Reflections

  • y = −f(x): reflection in the x-axis, (x, y) → (x, −y).
  • y = f(−x): reflection in the y-axis, (x, y) → (−x, y).

For y = √x, a reflection in the x-axis changes the range to y ≤ 0, while a reflection in the y-axis changes the domain to x ≤ 0. Some graphs look the same after a reflection: x² is symmetric in the y-axis, so (−x)² = x².

xy−9−669−3−113(4, 2)(4, −2)(−4, 2)y = √xy = −√xy = √(−x)reflect in x-axisreflect in y-axis
y = √x (solid), its reflection in the x-axis y = −√x (dashed), and its reflection in the y-axis y = √(−x) (faint).
Worked example

Example 3: Transforming the reciprocal function

Describe and sketch (a) y = 4/x (b) y = 1/(2x) (c) y = −1/x. Do the asymptotes move?

  1. (a) 4/x = 4(1/x): a = 4, vertical stretch by 4. (1, 1) → (1, 4); (2, 0.5) → (2, 2); (−1, −1) → (−1, −4).
  2. (b) 1/(2x) = f(2x): k = 2, horizontal compression by 1/2. (1, 1) → (0.5, 1); (2, 0.5) → (1, 0.5); (−1, −1) → (−0.5, −1).
  3. Notice 1/(2x) also equals (1/2)(1/x), so it can be described as a vertical compression by 1/2 as well. Both descriptions give the same graph.
  4. (c) a = −1: reflection in the x-axis. (1, 1) → (1, −1); (−1, −1) → (−1, 1). The branches are now in the second and fourth quadrants.
  5. Stretches, compressions and reflections in the axes do not move the asymptotes x = 0 and y = 0, because each of these transformations maps each axis onto itself.

Asymptotes stay at x = 0 and y = 0 in all three.

Worked example

Example 4: Mapping a single point

The point (6, −4) is on y = f(x). Find its image on (a) y = 3f(x) (b) y = f(2x) (c) y = −(1/2)f(x) (d) y = f(−(1/3)x).

  1. (a) (x, 3y): (6, −12).
  2. (b) (x/2, y): (3, −4).
  3. (c) (x, −y/2): (6, 2).
  4. (d) k = −1/3, so x ÷ (−1/3) = −3x: (−18, −4). (Horizontal stretch by 3 and a reflection in the y-axis.)

(6, −12); (3, −4); (6, 2); (−18, −4).

Just for fun

A parabola walks into a gym. The trainer says, "Let's work on your stretches." The parabola asks, "Vertical from the x-axis, or horizontal from the y-axis?" The trainer quietly updates her rates.

Hands-on activity

Stretch it with a rubber band

You need: grid paper, a pencil, a ruler and a wide rubber band (or a strip of elastic). Take care not to snap the rubber band towards anyone's face.

  1. Draw y = x² from x = −3 to 3 on grid paper, and the points (±1, 1), (±2, 4).
  2. On the rubber band, mark dots 1 cm apart. Lay it along the y-axis and stretch it to twice its length: each dot moves twice as far from the end at the x-axis. That is a vertical stretch by a factor of 2.
  3. Use the idea to plot y = 2x²: keep each x, double each y.
  4. Now plot y = (2x)² by halving each x of the original points. Compare it with your graph of y = 4x² (keep x, multiply y by 4).

What to look for: y = 2x² passes through (±1, 2) and (±2, 8). y = (2x)² passes through (±0.5, 1) and (±1, 4), exactly on y = 4x², because (2x)² = 4x². A horizontal compression of a parabola by 1/2 looks the same as a vertical stretch by 4.

Your turn

1

Describe the transformation: y = 5√x.

2

Describe the transformation: y = (1/4)x².

3

Describe the transformation: y = √(4x).

4

Describe the transformation: y = 1/(−x).

5

The point (8, 2) is on y = f(x). Find its image on y = f(4x).

6

The point (8, 2) is on y = f(x). Find its image on y = −3f(x).

7

State the domain and range of y = −√x.

More practice: warm-up

1

Which parameter, a or k, causes a reflection in the x-axis when negative?

2

Write the mapping rule for y = af(x).

3

Write the mapping rule for y = f(kx).

4

Is y = 0.2x² wider or narrower than y = x²?

5

Which points do not move under a vertical stretch?

More practice: on your own

1

Find four points on y = −2√x and state its domain and range.

2

Find four points on y = √(−2x) and state its domain.

3

Find the images of (1, 1), (2, 0.5) and (−1, −1) on y = 6/x.

4

Show that y = (3x)² is the same as a vertical stretch of y = x² by a factor of 9.

5

The graph of y = f(x) has domain {x ∈ ℝ | −4 ≤ x ≤ 6}. Find the domain of y = f(2x) and of y = f(−x).

Challenge

1

Show that √(4x) = 2√x for x ≥ 0, and explain what this means about two different transformations of y = √x.

2

For f(x) = 1/x, show that f(kx) = (1/k)f(x). What does this say about stretches of the reciprocal function?

3

A graph passes through (2, 5). After the transformation y = af(kx), the image of that point is (−1, 15). Find a and k.

Check your understanding

Quiz

1.

How does y = 4f(x) change the graph of y = f(x)?

2.

How does y = f(3x) change the graph of y = f(x)?

3.

Which equation is y = √x reflected in the y-axis?

4.

The point (−6, 2) is on y = f(x). Which point is on y = f((1/2)x)?

5.

What is the range of y = −3x²?

6.

Which transformation does NOT change the asymptotes of y = 1/x?

For parents and teachers

Big idea: Expectations A1.8 and A1.9 ask students to investigate the roles of a and k in y = af(kx) (stretches, compressions and reflections, "to and from the x- and y-axes") for the parent functions, and to sketch the results with domain and range.

How to help: Have your teen say the mapping rule out loud before they plot anything: "a multiplies y; k divides x." Then ask which points stay fixed (points on the x-axis for a vertical stretch; points on the y-axis for a horizontal one).

Common difficulties:

  • Multiplying x by k instead of dividing.
  • Calling y = f(2x) a "stretch by 2" (it is a compression by 1/2).
  • Forgetting that a negative a or k also reflects the graph.
  • Changing the domain of √x for a vertical stretch (only a horizontal reflection changes it).