weirdfacts
Year 11 / Applications of differential calculus

Use the slope to read the curve.

At x = a, the tangent has gradient f(a)f'(a) and passes through (a, f(a)f(a)). If that gradient is non-zero, the normal gradient is −1/f′(a). A horizontal tangent has a vertical normal.

Stationary points have f(x)f'(x) = 0. A change from positive to negative derivative gives a local maximum; negative to positive gives a local minimum. No sign change can give a stationary point of inflection. A global extremum on a closed interval also requires checking endpoints.

INTERACTIVE MODEL

Try it. Watch it change.

Function x³ − 3x, tangent and derivative-2.5-1.88-1.25-0.6300.631.251.882.5-5-1.751.54.758xy
f(x)f(x) = x3x^{3} − 3xtangentf(x)f'(x)f(x)f''(x)
f′(1) = 0 · f″(1) = 6

The graph is calculated from the displayed rule. Drag the highlighted point or use the sliders.

Explore: Use the cubic model and locate where the tangent becomes horizontal. Compare the gradient on each side.

Tangent at x = a
yf(a)=f(a)(xa)y-f(a)=f'(a)(x-a)
WORKED EXAMPLE

Find the tangent to y = x2x^{2} at x = 3.

  1. f(3)f(3) = 9 and f(3)f'(3) = 6.
  2. y − 9 = 6(x − 3), so y = 6x − 9.
Assumed knowledge

Differentiation of polynomials and line equations.

Learning checkpoints & sourceYOUR LEARNING CHECKPOINT
  • Find tangent and normal equations.
  • Classify stationary points using gradient signs.
QCAA Mathematical Methods 2025 v1.3 · p. 22
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