Year 10 Extended · Unit 3 Functions, quadratics and transformations · Lesson 7 of 7

Does the order matter?

A graph can be moved, flipped and stretched in one go. This lesson asks when the order of two steps changes the result, and when two different-looking transformations are really one.

FunctionsNo calculator

Warm-up

Three quick ones from earlier work. Try each in your head or on paper before you open the answer.

  1. From Year 9: expand 2(x + 3).
    Answer2x + 6. The 2 multiplies both terms.
  2. From the last lesson: the point (4, 1) is on y = f(x). Where is its image on y = f(x − 3) + 2?
    Answer(7, 3). Right 3 and up 2.
  3. From the last lesson: the point (6, 5) is on y = f(x). Where is its image on y = f(3x)?
    Answer(2, 5). The x-coordinate is divided by 3.
Words used in this lesson
combination
two or more transformations applied one after another. For example: stretch, then lift.
order of transformations
which transformation is done first. For example: f(2x − 6) versus f(2(x − 6)).
equivalent transformations
different transformations giving the same graph. For example: squash by 1 ÷ 3 and stretch by 81 for x⁴.
counterexample
one case showing a statement is not always true. For example: x² + 5 for f(bx) = bⁿf(x).
power function
a function f(x) = xⁿ with n a positive whole number. For example: x⁴.
commute
give the same result in either order. For example: a lift and a shift sideways.

Two ways to build the same ramp

A designer in Bogotá is drawing a skate ramp from a reference curve y = f(x). Her plan: make it twice as tall, and lift the whole ramp 4 units off the ground. Her colleague does the same two things in the other order. Each is sure her ramp will match her colleague's.

Predict first: will the two orders give the same ramp when the stretch is 2 and the lift is 4? When would they always agree? Then use the figure: change the order and compare the ghost.

1
0
y = f(x)rule in this order
(2, 2) → (2, 2)corner at x = 2
yesdo the two orders agree?

Stretch first, then translate: y = f(x). The corner (2, 2) goes to (2, 2). Both orders agree here.

Show me the steps
  1. Set a = 2 and k = −3 with Stretch first. The rule is y = 2f(x) − 3 and the corner (2, 2) goes to (2, 1): double the height to 4, then take 3 away.
  2. Press Translate first, then stretch. The rule is y = 2(f(x) − 3) and the corner goes to (2, −2): take 3 away to get −1, then double. The ghost shows the other graph: they are different.
  3. Set k = 0. Both orders give y = 2f(x), and the ghost lies exactly on the graph: with nothing to translate by, the order cannot matter.
  4. Set a = 1 with k = −3. Both orders give y = f(x) − 3: with no stretch, the order cannot matter either.

Order can matter only when two operations act on the same coordinate. Stretch by a then lift by k gives af(x) + k. Lift by k then stretch gives a(f(x) + k) = af(x) + ak. They agree only when a = 1 or k = 0. Two lifts, or two stretches, still agree in either order. A horizontal move and a vertical move never interfere: they act on different coordinates.

Writing the equation. Apply the operations to the rule in the order described. Each operation replaces x (or y) in the rule you have so far.

f(bx − h) is a shift of h ÷ b. f(bx − h) = f(b(x − h ÷ b)), so it is a squash by 1 ÷ b and then a shift of h ÷ b.

Two operations, one graph. For f(x) = xⁿ, f(bx) = (bx)ⁿ = bⁿxⁿ = bⁿf(x), so a horizontal squash by 1 ÷ b gives the same graph as a vertical stretch by bⁿ.

1
0
y = f(x)rule
noneshift
(2, 2) → (2, 2)corner at x = 2

y = f(x): no shift. The corner (2, 2) goes to (2, 2).

Show me the steps
  1. Set b = 2 and h = 0. The rule is y = f(2x), a squash towards the y-axis, and the corner (2, 2) goes to (1, 2). There is no shift.
  2. Now set h = 6. The rule y = f(2x − 6) is f(2(x − 3)): the graph is squashed and moved 3 to the right, not 6. The corner goes to (4, 2).
  3. Set b = 3 and keep h = 6. The rule y = f(3x − 6) is f(3(x − 2)), a shift of 2 to the right, and the corner goes to about (2.67, 2).
  4. Set b = 1 with h = 6. The rule is y = f(x − 6), a shift of 6: when b is 1, the number taken off is the whole shift.

f(4x − 12) shifts 3, not 12. The 12 is shared with the 4: f(4x − 12) = f(4(x − 3)). And “translate then squash” and “squash then translate” use different amounts: be sure which one the sentence says before writing a rule.

Worked examples

1. Describe y = −2(x + 1)² + 3 as transformations of y = x².

  1. (x + 1)²: translate 1 to the left.
  2. 2(x + 1)²: stretch vertically by factor 2. Then −2(x + 1)²: reflect in the x-axis.
  3. +3: translate 3 up, last. The vertex is (−1, 3) and it opens downward.

2. Write the rule for y = f(x) translated 3 to the right and then squashed horizontally by factor 1 ÷ 2.

  1. Translating 3 right: y = f(x − 3).
  2. Squashing by 1 ÷ 2 replaces x by 2x in the rule so far: y = f(2x − 3).
  3. The rule is y = f(2x − 3). (Squashing first and then translating 3 would give f(2(x − 3)) = f(2x − 6).)

3. Show that for f(x) = x⁴, a squash by 1 ÷ 3 is a stretch by 81, but for f(x) = x² + 5 it is not.

  1. f(3x) = (3x)⁴ = 81x⁴ = 81f(x). The same graph: squash by 1 ÷ 3 equals stretch by 81.
  2. For f(x) = x² + 5: f(3x) = 9x² + 5, but 9f(x) = 9x² + 45. These differ by 40.
  3. So the equivalence needs a pure power. It fails for x² + 5.

Your go. The first step is given. The graph of y = f(x) is squashed horizontally by factor 1 ÷ 3 and then translated 2 to the left. Write the rule.

  1. Squashing by 1 ÷ 3 replaces x by 3x: y = f(3x).
  2. Now translate 2 to the left in the rule you have so far, replacing x by (x + 2).
Check the answer

Replacing x by (x + 2) in f(3x) gives y = f(3(x + 2)) = f(3x + 6). The 2 is inside the 3, so the number added is 6.

Check before you go on

1. Which describes y = f(2x − 10)?

Your turn

Start at Level 1 and go as far as you can. Type negative numbers with a minus sign, for example -7. A wrong answer gets a reason, not just a cross: read it, then try again.

Level 1 · Getting started

2. A point on y = f(x) is (8, 6). On y = f(2x) it becomes (x, 6). Find x.

3. The point (2, 5) is on y = f(x). On y = 3f(x) + 2 it becomes (2, q). Find q.

Level 2 · Building

4. The graph of y = f(x) has a vertex at (1, 4). Find the y-coordinate of the image of the vertex under y = −2f(x − 3) + 1.

5. The graph of y = x² is translated 2 to the right and 5 up, and then stretched vertically by factor 3 from the x-axis. Its equation is y = 3(x − 2)² + c. Find c.

Level 3 · Stretch

6. Stretching a graph vertically by factor 2 and then translating it 3 up gives the same graph as translating it up by t first and then stretching by factor 2. Find t.

7. For f(x) = x³, the graph of y = f(2x) is the same as the graph of y = kf(x), a vertical stretch by factor k. Find k.

Something to think about

For f(x) = x² we have f(−x) = f(x), and for f(x) = x³ we have f(−x) = −f(x). Which powers xⁿ behave like the first and which like the second? What do the graphs of each kind look like, and what does f(−x) = −f(x) say about a point (p, q) on the graph?

One way to think about it

For even n, (−x)ⁿ = xⁿ, so f(−x) = f(x): the graph is symmetric in the y-axis, like a parabola. For odd n, (−x)ⁿ = −xⁿ, so f(−x) = −f(x): if (p, q) is on the graph then so is (−p, −q), so the graph turns into itself under a half turn about the origin, like y = x³.

Exit ticket

8. Which two steps change the graph of y = f(x) into the graph of y = f(2x) + 3?