Three forms, one parabola.

General form ax² + bx + c makes the coefficients clear. Factorised form shows real zeros, while vertex form a(x − h)² + k exposes symmetry and the turning point. Completing the square converts between these forms.

For a ≠ 0, the discriminant $b^{2}$ − 4ac counts real roots: positive gives two distinct roots, zero gives a repeated root and negative gives none. The quadratic formula works when simple factorisation is unavailable. Interpret any roots within the domain of the model.

INTERACTIVE MODEL

Try it. Watch it change.

y = 1(x − 0)² + 0

Vertical scale / gradient a1 Horizontal shift h0 Vertical shift k0Input x1

f(1) = 1 · Vertex (0, 0)

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

Explore: Move a positive-opening parabola above, onto and below the x-axis. Count its real roots.

Quadratic formula

x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}

WORKED EXAMPLE

Solve $2x^{2}$ − 4x − 1 = 0 exactly.

Discriminant = (−4)² − 4(2)(−1) = 24.
x = (4 ± $\sqrt{24}$ )/4 = 1 ± $\sqrt{6}$ /2.

Assumed knowledge

Year 10 factorisation and index notation.

Learning checkpoints & source

YOUR LEARNING CHECKPOINT

Solve quadratics and use the discriminant.
Choose a form that reveals roots or a turning point.

QCAA Mathematical Methods 2025 v1.3 · p. 16 ↗

READY TO TRY IT?

Three forms, one parabola.

Solve 2x22x^{2}2x2 − 4x − 1 = 0 exactly.

Put the idea to work.

Solve $2x^{2}$ − 4x − 1 = 0 exactly.