Functions & Graphs

Functions and graphs covers linear, quadratic, exponential, logarithmic and hyperbolic functions, their inverses, transformations, intersections and key fe

Functions and graphs covers linear, quadratic, exponential, logarithmic and hyperbolic functions, their inverses, transformations, intersections and key features (asymptotes, intercepts, turning points).

Key concepts

  • f(x) = ax² + bx + c — turning point at x = −b/(2a)
  • Inverse: swap x and y, solve for y
  • Exponential f(x) = a·b^x and its inverse log_b
  • Hyperbola y = a/(x − p) + q has asymptotes x = p, y = q
  • Transformations: shifts, reflections, stretches

How it works

Identify the function type from its formula or shape. Find intercepts (set y=0 and x=0), asymptotes (denominator = 0 for hyperbolas), and turning points. For inverses, restrict the domain when needed so the inverse is also a function. Every graph question breaks into the same pieces: the equation, the intercepts, the turning point or asymptotes, and the domain and range. Find any unknowns in the equation first using the given points, because every later part of the question depends on them.

Step by step

1. Identify the family

Linear, quadratic, exponential, log, hyperbola.

2. Find intercepts

Set x=0 for y-intercept; set y=0 for x-intercepts.

3. Locate special features

Asymptotes, turning points, axes of symmetry.

4. Sketch with labels

Always label intercepts, asymptotes and key points.

Worked examples

Find the inverse of f(x) = 2x + 3.

  1. Let y = 2x + 3
  2. Swap: x = 2y + 3
  3. Solve: y = (x − 3)/2

Answer: f⁻¹(x) = (x − 3)/2

Find the equation of the hyperbola with asymptotes x = 2 and y = 1 that passes through (3; 4)

  1. Start with y = a/(x − 2) + 1
  2. Substitute (3; 4): 4 = a/1 + 1, so a = 3

Answer: y = 3/(x − 2) + 1

Find the turning point of y = 2x² − 8x + 3 by completing the square

  1. y = 2(x² − 4x) + 3
  2. y = 2(x − 2)² − 8 + 3

Answer: y = 2(x − 2)² − 5, so the turning point is (2; −5)

Find the x-intercepts of f(x) = −x² + 2x + 8

  1. −x² + 2x + 8 = 0, so x² − 2x − 8 = 0
  2. (x − 4)(x + 2) = 0

Answer: (4; 0) and (−2; 0)

For which values of x is f(x) = −x² + 2x + 8 ≥ 0?

  1. The parabola opens downwards, with roots −2 and 4
  2. It lies on or above the x-axis between the roots

Answer: −2 ≤ x ≤ 4

Common mistakes

  • Forgetting the domain restriction on inverses
  • Mislabelling asymptotes on hyperbolas
  • Sign errors when completing the square

Frequently asked questions

Do I need to memorise transformations?

Yes — shifts, reflections, stretches and compressions are tested in every paper.

How do I find the range of a parabola?

Use the y-value of the turning point, q. If a > 0 the range is y ≥ q; if a < 0 it is y ≤ q.

What are the axes of symmetry of a hyperbola?

The hyperbola y = a/(x − p) + q has two axes of symmetry through (p; q): y = x − p + q and y = −x + p + q.

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