1The idea in one paragraph

Every point on the graph of y = f(x) is a pair (x, f(x)). Change the equation and every one of those points moves in the same way, so the whole graph moves as one piece. There are only three kinds of movement at this level: a translation (a slide), a reflection (a flip) and a stretch (a pull away from an axis). Where you make the change decides which way the graph moves. A change outside the function, to f(x) itself, acts on the y-coordinates, so the graph moves vertically and it does exactly what the algebra says. A change inside the bracket, to x before f sees it, acts on the x-coordinates, so the graph moves horizontally and it does the opposite of what the algebra seems to say. Learn that one sentence and the rest of this subtopic is detail.

2The graph we will move

To see what each change does, take one function and keep it fixed through the page:

f(x) = ½(x − 1)²(x + 2)

It has four points worth watching: A(−2, 0), B(−1, 2), which is its local maximum, C(0, 1), the y-intercept, and D(1, 0), where it touches the x-axis. It is lopsided on purpose, so the two reflections look different. In every figure the original graph is teal and the transformed graph is amber.

3Translations

Vertical translation: y = f(x) + b. The new function takes each output of f and adds b. The x-coordinates are untouched; every y-coordinate goes up by b. So the graph moves b units up (down if b is negative). Figure 1(a) shows y = f(x) + 2: B(−1, 2) moves to (−1, 4), and every other point moves up 2 with it.

Horizontal translation: y = f(x − a). This is the one that feels backwards. The graph moves a units to the right, even though the bracket shows a minus. Here is why. The new function g(x) = f(x − 3) gives, at x = 3, the value f(0). So whatever f did at x = 0, g does at x = 3: three units later. Every feature arrives 3 units further right. Figure 1(b) shows y = f(x − 3): C(0, 1) moves to (3, 1). And y = f(x + 3) is f(x − (−3)), a translation 3 units to the left.

Both are translations by a vector. The IB writes the vector as a column, the horizontal move on top and the vertical move below. In these notes it is written on one line, so "translation by the vector (3, 0)" means 3 right and 0 up. The two combine without any fuss: y = f(x − 3) + 2 is a translation by the vector (3, 2), and every point (x, y) goes to (x + 3, y + 2).

4Reflections

y = −f(x): reflection in the x-axis. Every output changes sign, so every point (x, y) goes to (x, −y). Points above the x-axis go below it and the other way round. Points on the x-axis stay where they are: in Figure 2(a), A(−2, 0) and D(1, 0) do not move, and the maximum B(−1, 2) becomes a minimum at (−1, −2).

y = f(−x): reflection in the y-axis. Now the input changes sign. The new function at x = 2 gives f(−2), so what f did at −2, the new graph does at +2. Every point (x, y) goes to (−x, y). Points on the y-axis stay put: C(0, 1) is fixed in Figure 2(b), while B(−1, 2) moves to (1, 2).

Fixed points check a sketch: a reflection leaves every point on its mirror line where it was.

5Stretches

y = p·f(x): vertical stretch with scale factor p. Every output is multiplied by p, so every point (x, y) goes to (x, py). Each point moves away from the x-axis to p times its distance. With p = 2 in Figure 3(a), B(−1, 2) goes to (−1, 4) and C(0, 1) goes to (0, 2). The points on the x-axis, A and D, have distance zero, and twice zero is zero, so they stay put. If 0 < p < 1 the "stretch" squashes the graph towards the x-axis, and it is still called a stretch, with a scale factor less than 1.

y = f(qx): horizontal stretch with scale factor 1/q. Every point (x, y) goes to (x/q, y). With q = 2, the new function reaches at x = 0.5 the value f had at x = 1: it gets there twice as fast, so the graph is squeezed towards the y-axis to half its width. The scale factor is ½, not 2. In Figure 3(b), B(−1, 2) moves to (−0.5, 2) and D(1, 0) moves to (0.5, 0), while C(0, 1) on the y-axis stays where it is. By the same reasoning y = f(x/3), which is f(⅓x), is a horizontal stretch with scale factor 3: three times as wide.

Every stretch has an invariant line, the axis it stretches away from. Say which in your description: "a vertical stretch, scale factor 3" is enough, because vertical stretches always pull away from the x-axis, and "a stretch parallel to the y-axis with scale factor 3" means the same thing and is also accepted.

6The whole set on one table

Every standard level transformation, what it does to a point, and P(4, −3) taken through each one. Memorise this table.

Figure 4 is the reason behind the whole table. Read y = p·f(x − a) + b as a machine. The input x is changed first, before f sees it: that is the inside, and it moves the graph sideways, undoing what it appears to do. Then f acts. Then the output is multiplied and added to: that is the outside, and it moves the graph up and down, doing exactly what it says.

This is composite functions (2.5) in disguise. y = f(x) + b is g ∘ f with g(x) = x + b: the extra function acts after f, on the output. y = f(x − a) is f ∘ h with h(x) = x − a: the extra function acts before f, on the input. Which side of f the new function sits on decides which coordinate it changes.

7Using a graph you know: y = x² to y = 3x² + 2

The guide's own example is to sketch y = 3x² + 2 from y = x². Read the machine: x is squared, then multiplied by 3, then 2 is added. So the transformations are, in that order:

1. a vertical stretch with scale factor 3, giving y = 3x²;
2. a translation by the vector (0, 2), giving y = 3x² + 2.

Figure 5 draws both steps and follows the point (1, 1): the stretch takes it to (1, 3), the translation to (1, 5). The vertex (0, 0) is on the x-axis, so the stretch leaves it alone, and the translation lifts it to (0, 2). The sketch needs the vertex (0, 2), the axis of symmetry x = 0, and a parabola narrower than x², since every height has been tripled.

The vertex form of a quadratic from 2.6 is exactly this idea. y = 2(x − 3)² − 5 is y = x² translated 3 right, stretched vertically by factor 2, then translated 5 down. The vertex (0, 0) goes to (3, 0), stays at (3, 0) under the stretch, and ends at (3, −5), which is why the vertex of a(x − h)² + k is (h, k). Every parabola is one parabola, transformed.

8Two or more transformations, and when the order matters

A question may give you a sequence of transformations and ask for the new equation, or give you the equation and ask for the sequence. Both need one rule.

Vertical changes happen in the order they are applied to f(x). Horizontal changes happen to x. A horizontal change and a vertical change never interfere with each other.

Two vertical changes: the order matters. Start from y = f(x) and take two changes: a vertical stretch by 2 and a translation up 1.

• Stretch first, then translate: f(x) becomes 2f(x), then 2f(x) + 1.
• Translate first, then stretch: f(x) becomes f(x) + 1, then 2(f(x) + 1) = 2f(x) + 2.

The second is a different graph, because the stretch has stretched the translation as well. Figure 6 shows both, following B(−1, 2): it ends at (−1, 5) one way and (−1, 6) the other.

The same goes for a reflection and a translation: reflect-then-add-3 turns 2ˣ into −2ˣ + 3, while add-3-then-reflect gives −(2ˣ + 3) = −2ˣ − 3.

A stretch and a reflection in the same direction can be done in either order: −2f(x) is both 2(−f(x)) and −(2f(x)). A horizontal and a vertical change can also be done in either order, because one works on x and the other on y. In y = 3f(x − 2) + 1 the translation of 2 to the right can come first, last or in the middle; only the stretch by 3 must come before the +1.

Worked example 1 · from words to an equation (Paper 1). Let f(x) = x² − 4x. The graph of f is translated by the vector (−1, 3) and then stretched vertically with scale factor 2 to give the graph of g. Find g(x).

Had the stretch come first, the answer would be 2f(x + 1) + 3 = 2x² − 4x − 3.

Worked example 2 · from an equation to words. Describe a sequence of transformations that maps y = f(x) onto y = 5 − 2f(x).

Rewrite it in the machine order: y = −2f(x) + 5. The outputs of f are multiplied by −2, then 5 is added. So:

1. a vertical stretch with scale factor 2,
2. a reflection in the x-axis (these first two in either order),
3. a translation by the vector (0, 5).

The point (1, 4) on f goes to (1, 8), then (1, −8), then (1, −3). Check with the equation: 5 − 2 × 4 = −3.

Worked example 3 · one change inside, one outside. The point (4, −3) lies on y = f(x). Find its image on y = 2f(x + 1) − 3.

The inside +1 moves every point 1 to the left, so x = 4 becomes 3. The outside doubles and then subtracts 3, so y = −3 becomes 2(−3) − 3 = −9. The image is (3, −9).

9Asymptotes, intercepts and turning points move too

An asymptote is a line the graph approaches, and the transformation moves it exactly as it moves the graph. So when you sketch, transform the asymptote first and the curve second.

Take y = 2ˣ, with horizontal asymptote y = 0 and y-intercept (0, 1). Reflect it in the x-axis to get y = −2ˣ: the asymptote y = 0 lies on the mirror line, so it does not move, and (0, 1) becomes (0, −1). Now translate by the vector (1, 3) to get y = −2ˣ⁻¹ + 3. Figure 7 shows both stages.

• The horizontal asymptote moves up 3, to y = 3. A horizontal line is not affected by the move to the right.
• The y-intercept does not come from moving (0, −1), because (0, −1) moves to (1, 2), which is not on the y-axis. Always find a new intercept from the new equation: at x = 0, y = −2⁻¹ + 3 = 2.5.
• The x-intercept: −2ˣ⁻¹ + 3 = 0 gives 2ˣ⁻¹ = 3, so x − 1 = log₂ 3 and x = 1 + log₂ 3 = log₂ 6 ≈ 2.58.

That second bullet is the trap. Transformations carry points to points, but an intercept is defined by an axis, and the axes do not move. Turning points, on the other hand, are carried faithfully: a local maximum of f at (−1, 2) becomes a local maximum of f(x − 3) + 2 at (2, 4), and a local minimum of −f(x), because a reflection in the x-axis turns maxima into minima.

The same thinking handles 2.8: y = 6/(x + 2) + 1 is y = 1/x stretched vertically by 6 and translated by (−2, 1), so its asymptotes are x = −2 and y = 1, the axes moved.

10With technology

On Paper 2, graph f and the transformed function on the same axes and check where the key points went. Better still, use the GDC once to find a feature of f, such as a turning point, and then write down the matching feature of the new function by transforming it. That second step is what a "hence" question rewards, and it is Q5 below.