Loav · VCE Mathematical Methods · Units 3 & 4
F02 · 2026 curriculum · Original teaching guide
1. Read inside and outside changes differently
A transformation changes a graph in a controlled way. Translations shift it, reflections reverse a coordinate’s sign, and dilations change distances from an axis. Rather than memorising a separate slogan for every formula, track an arbitrary point. If (u, v) lies on y = f(x), then v = f(u). Ask which new input makes the argument of f equal to u, and then calculate the new output from v.
For non-zero a and b, write a transformed rule in the form g(x) = a f(b(x − h)) + k. To recover the old input u, solve b(x − h) = u, obtaining x = h + u/b. Its new output is av + k. This gives the point mapping:
(u, v) → (h + u/b, av + k).
The inverse transformation takes a point (X, Y) on the transformed graph back to the original graph. Solve the coordinate equations X = h + u/b and Y = av + k for the old coordinates: u = b(X − h) and v = (Y − k)/a. Thus the inverse map is (X, Y) → (b(X − h), (Y − k)/a). Subtract each translation before undoing its dilation or reflection. The conditions a ≠ 0 and b ≠ 0 ensure these steps can be reversed.
The factor outside f acts directly on outputs: vertical scale factor |a|, with reflection in the x-axis when a is negative. The factor inside acts reciprocally on input coordinates: horizontal scale factor 1/|b|, with reflection in the y-axis when b is negative, before the horizontal translation. The values h and k then shift the transformed graph horizontally and vertically.
2. Factor the argument before naming a translation
In f(2x − 6), the argument is 2(x − 3), so the horizontal translation is three units right after a horizontal compression by factor 1/2. It is not six units right. Similarly, f(−x + 4) = f(−(x − 4)) can be described as a reflection in the y-axis followed by a translation four units right. The point map x = 4 − u confirms the description without relying on a visual guess.
Transformation order matters when operations affect the same coordinate. Translating a graph up three and then reflecting it in the x-axis produces y = −f(x) − 3. Reflecting first and then translating up three produces y = −f(x) + 3. A vertical operation and a horizontal operation can often be interchanged, but there is no universal licence to reorder every instruction. Preserve the stated sequence or calculate its coordinate map.
Be precise about dilation language. A dilation by factor 3 from the x-axis multiplies y-coordinates by 3. A dilation by factor 3 from the y-axis multiplies x-coordinates by 3, giving y = f(x/3). Naming the reference axis removes the ambiguity in phrases such as “stretch by three”. A negative coefficient combines a positive scale factor with a reflection; a dilation factor itself is usually described as positive.
Worked example 1: transform a square-root graph
Sketch g(x) = −3√(2(x − 1)) + 6. Use the parent f(x) = √x, with domain [0, ∞) and range [0, ∞). Here a = −3, b = 2, h = 1 and k = 6, giving (u, v) → (1 + u/2, 6 − 3v).
The parent points (0, 0), (1, 1) and (4, 2) become (1, 6), (3/2, 3) and (3, 0). The endpoint moves to (1, 6), and the curve decreases to the right. The radicand condition 2(x − 1) ≥ 0 gives domain [1, ∞). Since √(2(x − 1)) ≥ 0, the range is (−∞, 6]. There is no y-intercept because zero is outside the domain. Setting the output to zero confirms the x-intercept (3, 0). As an inverse-transformation check, the image point (3, 0) returns to u = 2(3 − 1) = 4 and v = (0 − 6)/(−3) = 2, recovering the parent point (4, 2).
3. Transform sets as well as points
A transformed graph inherits its domain through the inside expression. If the parent domain is D, require b(x − h) ∈ D and solve that condition. This is safer than applying a remembered shift to endpoint numbers because a negative b reverses their order. Endpoint inclusion is preserved under a non-zero linear coordinate change, but which endpoint is left or right can swap.
The transformed range is obtained by applying y → ay + k to every original output. If a is negative, the original largest output becomes the new smallest output. Again, include an endpoint exactly when the original corresponding output was attained. For example, if the original range is [−1, 5], multiplying by −2 and adding 1 gives [−9, 3], not [3, −9]. Intervals are always written from smaller to larger.
Restrictions travel with the graph. Transforming a full parabola and then displaying only half of it is a different task from transforming an already restricted quadratic. Map all included and excluded endpoints, and retain any gaps in the original domain. A smooth-looking formula cannot restore inputs that the function’s definition excluded.
4. Asymptotes and intercepts need different treatment
Asymptotes transform as lines. A vertical asymptote x = c maps to x = h + c/b, and a horizontal asymptote y = d maps to y = ad + k. These lines are not ordinarily graph points, so do not mark them as intercepts. For a transformed reciprocal curve, they identify the centre and the branches’ limiting behaviour. Draw them dashed and label their equations.
Intercepts, by contrast, depend on the new coordinate axes. An old x-intercept can move away from the x-axis after a vertical translation. To find new x-intercepts, solve g(x) = 0 subject to the domain. To find a new y-intercept, evaluate g(0) only if zero belongs to the domain. Mapping useful old points and calculating the new intercepts are complementary steps.
Worked example 2: a shifted reciprocal
Consider r(x) = 2/(x + 3) − 1. Starting from y = 1/x, apply a vertical dilation by factor 2, translate three units left, then one unit down. The asymptotes become x = −3 and y = −1. The domain is ℝ \ {−3}, and the range is ℝ \ {−1}.
For the x-intercept, solve 2/(x + 3) = 1. This gives x = −1, which is allowed. For the y-intercept, r(0) = 2/3 − 1 = −1/3. The parent points (1, 1) and (−1, −1) map to (−2, 1) and (−4, −3). These points locate the branches relative to the two asymptotes. The graph approaches y = −1 for large positive or negative x but never attains that value.
5. Recover a rule from graph features
Choose a form that displays the supplied features. A quadratic vertex suggests a(x − h)² + k. A translated cubic’s centre of rotational symmetry suggests a(x − h)³ + k. A square-root endpoint suggests a√(b(x − h)) + k, together with its allowed direction. Reciprocal asymptotes suggest a/(x − h) + k. Substitute an additional point to determine any remaining scale factor.
Not every possible description is unique. For example, 4x² can result from a vertical dilation of y = x² by factor 4 or a horizontal compression by factor 1/2. Both yield the same graph because (2x)² = 4x². If a question asks for one sequence, provide a valid sequence with a clear order. If it gives a particular transformation family, use that family rather than assuming its parameters are interchangeable.
Worked example 3: determine a transformed cubic
A graph is known to have form y = a(x − 2)³ + 1 and passes through (4, 17). Substitute that point: 17 = a(4 − 2)³ + 1 = 8a + 1, so a = 2. The rule is y = 2(x − 2)³ + 1.
One valid construction from y = x³ is a vertical dilation by factor 2, followed by translations two units right and one unit up. The central point is (2, 1); the y-intercept is (0, −15). To find the x-intercept, solve (x − 2)³ = −1/2, so x = 2 − ∛(1/2). Both domain and range are ℝ. The single supplied point determines a only because the other parameters were already specified.
Practise, then check
- A point (6, −1) lies on y = f(x). Where does it move on y = −2f(3(x − 4)) + 5?
- Suppose f has domain [−3, 6] and range [−2, 4). Find the domain and range of g(x) = −2f(3(x − 4)) + 5.
- Describe y = f(x/4). Where does the original point (2, 7) move?
- Starting from f(x) = x² − 1, compare “translate up 3, then reflect in the x-axis” with the reversed order. Give both resulting rules.
- Find the endpoint, domain, range and x-intercept of y = 4 − √(x + 5).
- A reciprocal graph has vertical asymptote x = 2 and horizontal asymptote y = 3, and passes through (4, 8). Find its rule in the form a/(x − h) + k.
- Let f have domain [−2, 4) and range [−1, 5]. Find the domain and range of g(x) = −2f(3 − x) + 1. Explain the endpoint reversal.
Full reasoned answers
- Use x = 4 + u/3 and y = −2v + 5. With u = 6 and v = −1, the image is (6, 7). Substituting x = 6 into the new argument gives 6, which checks the horizontal coordinate.
- Require −3 ≤ 3(x − 4) ≤ 6, giving 3 ≤ x ≤ 6. For outputs, −2 ≤ v < 4 becomes −8 < −2v ≤ 4, then −3 < −2v + 5 ≤ 9. Thus the domain is [3, 6] and range is (−3, 9].
- This is a horizontal dilation by factor 4 from the y-axis. Solve x/4 = u to obtain x = 4u, leaving the output unchanged. Thus (2, 7) moves to (8, 7).
- Up then reflect gives −(f(x) + 3) = −x² − 2. Reflect then up gives −f(x) + 3 = 4 − x². The outputs differ by 6, demonstrating that these operations do not commute.
- The radicand requires x ≥ −5, and the endpoint is (−5, 4). The range is (−∞, 4]. Setting √(x + 5) = 4 gives x = 11, so the x-intercept is (11, 0). Squaring is safe because both sides are non-negative.
- The asymptotes give h = 2 and k = 3. Substituting (4, 8) yields 8 = a/2 + 3, so a = 10. Therefore y = 10/(x − 2) + 3, with x ≠ 2. The extra point sets the branch scale and sign.
- Require −2 ≤ 3 − x < 4. Solving gives −1 < x ≤ 5, so domain (−1, 5]. The range transforms from [−1, 5] to [−9, 3]. The negative horizontal coefficient sends the original excluded right endpoint to the excluded left endpoint; both original range endpoints are included.
Exam habits and technology checks
Before sketching, list the parent graph, point mapping, domain, range and any asymptotes. Map two or three distinctive points, then calculate intercepts separately. Label open endpoints clearly and draw the correct branch direction. A sketch communicates structure; many unlabelled plotted points do not replace those features.
Technology can check a proposed rule against a graph, but an automatic window may make a compression look like a dilation or conceal a translated asymptote. Use equal or clearly labelled axis scales when comparing shapes. If a calculator simplifies two expressions to the same formula, explain why the stated transformation sequence still gives that expression.
Curriculum and sources
Scope: VCE Mathematical Methods Units 3 and 4, current 2023–2027 study design, Functions, relations and graphs and Algebra, number and structure. See the official VCAA Mathematical Methods curriculum and VCAA Mathematical Methods FAQs, updated February 2026. This guide covers translations, dilations, reflections and transformed graph features.
Further reading: Michael Evans, Kay Lipson and David Greenwood, Cambridge Senior Mathematics VCE Mathematical Methods Units 3 & 4, second edition (2023), ISBN 9781009110495. Chapter 3, pp. 90–136. Explanations, diagrams, worked examples and practice questions in this guide are original. This is an independent Loav learning resource, not a VCAA publication.
