Mathematics Extension 1: should I choose it?

What will I learn?

How will I be assessed?

Progression and companion subjects

No complete companion or exclusion review is attached to this profile. Check the official course-entry conditions, language eligibility and your school’s sequence before enrolling. Links between resource hubs show related learning; they do not establish a compulsory prerequisite.

Will I find it hard?

Difficulty depends on your experience, interests, teaching and assessment. These are our reflection prompts, not an official difficulty rating.

• How do you feel about explaining a method and working through unfamiliar multi-step problems?
• Try the sample without looking at a worked answer, then check the steps you need to practise.

Does it scale up or down?

Mathematics Extension 1, 2025. UAC Table A3 reports separate means for reported HSC marks and scaled marks per unit, with candidature for each course in 2025. These are cohort statistics. They do not supply individual raw exam marks or a conversion between the two columns. Historical cohort data cannot predict your result or a future scaling direction.

Scaling is not a measure of difficulty or a guaranteed bonus. Prioritise content you can engage with and required subjects for courses you are considering.

Try a little of the subject

Further work with functions

Extension 1 asks you to build new graphs out of a graph you already know. The first and most important transformation is the reciprocal : given y = f(x), sketch y = 1/f(x). Rather than plotting points, you read features straight off the original curve. Zeros become vertical asymptotes. Wherever f(x) = 0, the reciprocal is undefined and its magnitude grows without bound, so draw a dashed vertical line there. Large values become small values. If f(x) → ±∞, then 1/f(x) → 0. In particular, if f(x) grows without bound as x → ±∞, the reciprocal has the horizontal asymptote y = 0. If f(x) → k (k ≠ 0), then 1/f(x) → 1/k. Sign is preserved. 1/f(x) is positive exactly where f(x) is positive, so the reciprocal stays on the same side of the x-axis. The values 1 and −1 are fixed. Where f(x) = ±1, the two graphs meet. These are useful anchor points. Turning points stay at the same x-value. A local maximum of f with a non-zero value becomes a local minimum of 1/f, and vice versa, because a larger positive number has a smaller reciprocal. Increasing becomes decreasing on any interval where f does not change sign. Worked example 1. Sketch y = 1/(x² − 4). Start with f(x) = x² − 4, a concave-up parabola with zeros at x = ±2 and minimum value −4 at x = 0. Step 1: zeros x = ±2 become vertical asymptotes x = −2 and x = 2. Step 2: as x → ±∞, f(x) → ∞, so 1/f(x) → 0⁺; the horizontal asymptote is y = 0, approached from above. Step 3: the minimum (0, −4) of f becomes a local maximum (0, −1/4) of the reciprocal. Step 4: signs. For |x| > 2, f > 0, so the outer branches lie above the x-axis. For −2 < x < 2, f < 0, so the middle branch is below the axis, an upside-down ‘cap’ peaking at (0, −1/4) and plunging to −∞ at each asymptote. Domain: x ≠ ±2. Range: y ≤ −1/4 or y > 0. Reciprocal tr

Original practice question: Let f(x) = x and g(x) = (x − 3)². For which values of x is the product y = f(x)g(x) negative?

This is ATARMAxxing’s original sample learning material. It is not an official assessment task or a prediction of your school’s workload.

Explore the resources

Open the Mathematics Extension 1 subject hub and free preview (opens in a new tab)

• Mathematics Extension 1 — HSC Year 12 (opens in a new tab)

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