Grade 12 Functions Practice Questions for Paper 1 (2026)

Practise Grade 12 functions for NSC Paper 1 on 23 October 2026: parabolas, hyperbolas, exponentials, inverses and transformations, solved step by step.

· 7 min read · Snap&Learn

Functions and Graphs is worth ~35 marks in Paper 1, and almost every sub-question follows one of five patterns. Drill the patterns, not random examples.

How functions are examined in Paper 1 (23 October 2026)

NSC Mathematics Paper 1 is written on Friday 23 October 2026 at 09:00: 3 hours for 150 marks. Functions and graphs carry about 35 of those marks, more than any other Paper 1 topic except differential calculus, so this is where steady practice pays most.

  • Sketching parabolas, hyperbolas and exponential graphs with their intercepts, asymptotes and turning points.
  • Finding a function's equation from a given graph.
  • Inverses of linear, quadratic and exponential functions, including the logarithmic inverse.
  • Transformations, and reading the domain and range from a graph.

Practice 1 — Parabola turning point

Given f(x) = −2x² + 8x − 3. Find the turning point. Use x = −b/2a = 2, then f(2) = 5. Turning point: (2, 5).

Practice 2 — Hyperbola asymptotes

Given g(x) = 2/(x − 3) + 1. Vertical asymptote: x = 3. Horizontal asymptote: y = 1. Domain: x ≠ 3. Range: y ≠ 1.

Practice 3 — Exponential function

Given h(x) = 3·2^x − 6. Find the x-intercept: 3·2^x = 6 → 2^x = 2 → x = 1. Horizontal asymptote: y = −6.

Practice 4 — Inverse function

Find the inverse of f(x) = 2x + 3. Swap and solve: x = 2y + 3 → y = (x − 3)/2. So f⁻¹(x) = (x − 3)/2.

Practice 5 — Transformations

If f(x) = x², describe f(x + 2) − 3. The graph shifts 2 units left and 3 units down. New turning point: (−2, −3).

Practice 6 — Average gradient

For f(x) = x² − 4x − 5, find the average gradient between x = 1 and x = 4. f(1) = −8 and f(4) = −5, so the average gradient is (−5 − (−8)) / (4 − 1) = 1.

Practice 7 — A hyperbola's equation

A hyperbola has asymptotes x = 2 and y = −1 and passes through (3; 1). Then y = a/(x − 2) − 1, and substituting the point gives 1 = a − 1, so a = 2 and y = 2/(x − 2) − 1.

Practice 8 — A logarithmic inverse

For g(x) = (½)ˣ, swap x and y: x = (½)ʸ, so g⁻¹(x) = log_½ x, defined for x > 0. Both graphs are decreasing, and they are reflections of each other in the line y = x.

Get full worked solutions on any question

When a question stops you, snap it into Snap&Learn — you'll get a CAPS-aligned step-by-step explanation in seconds, and your first three AI solutions are free.

Frequently asked questions

Which functions are examined in CAPS Grade 12?

Parabolas, hyperbolas, exponentials, logarithmic functions (as inverses of exponentials), and their transformations.

Do I need to memorise transformation rules?

Yes — horizontal shifts, vertical shifts, reflections about the x- and y-axes, and stretches. They appear in nearly every Paper 1.

How do I find an inverse function?

Swap x and y in the equation, then solve for y. Check the domain restriction — many inverses (like √x) require x ≥ 0.

Snap&Learn gives South African Matric learners instant, CAPS-aligned, step-by-step Mathematics solutions. Snap or upload a question and the AI shows every method mark the way the NSC memo awards them. Your first three AI solutions are free.