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VCE Mathematical Methods (Year 11) · Unit 1 — Functions & Graphs · 12m read

Quadratic Functions and Graphs

Ever wondered why a basketball's path or a water fountain's arc looks the same shape? That's a parabola, and it comes from quadratic functions — equations with an x² term. You'll learn to sketch these curves and find their key features (like the vertex and roots), switch between different forms of the equation, and solve real problems from physics to design. It's essential for Methods because parabolas model everything from projectile motion to finding maximum profit, making this one of your most practical tools.

Learning objective: Graph and solve quadratic functions in different forms.

By the end of this lesson you can:

  • I convert between general, turning-point and factorised forms.
  • I find roots using the quadratic formula.
  • I sketch parabolas showing key features.

A quadratic is a fountain arc. Water rises, peaks at the turning point and falls symmetrically. The axis of symmetry runs through the peak.

Three forms, one parabola

THE GENERAL FORM:
• y = ax² + bx + c.
• a > 0 → opens up; a < 0 → opens down.
• y-intercept = c.

THE TURNING-POINT (VERTEX) FORM:
• y = a(x − h)² + k → turning point at (h, k).
• h shifts horizontally, k shifts vertically.
• Complete the square to convert from general.

THE FACTORISED FORM:
• y = a(x − p)(x − q) → x-intercepts at x = p, x = q.
• Axis of symmetry: x = (p + q)/2.

ROOTS (x-INTERCEPTS):
• Solve ax² + bx + c = 0.
• Factorise if possible.
• QUADRATIC FORMULA: x = (−b ± √Δ) / 2a where Δ = b² − 4ac.

DISCRIMINANT (Δ = b² − 4ac):
• Δ > 0 → 2 distinct real roots.
• Δ = 0 → 1 repeated root (graph touches x-axis).
• Δ < 0 → no real roots (graph doesn't cross x-axis).

AXIS OF SYMMETRY:
• x = −b / 2a (from general form).
• Turning point = (−b/2a, f(−b/2a)).

COMPLETING THE SQUARE:
• ax² + bx + c = a[(x + b/2a)² − b²/4a² + c/a].
• Used to find vertex, solve equations, integrate later.

GRAPHING CHECKLIST:
• Orientation (a > 0 or < 0).
• y-intercept (x = 0).
• x-intercepts (factorise / formula).
• Turning point (vertex form / −b/2a).
• Axis of symmetry.
• Domain = ℝ; range depends on vertex & orientation.

Quadratic shortcuts

  • Discriminant Δ: b² − 4ac predicts root count.
  • Vertex x: x = −b/2a always.
  • Sum & product: Sum of roots = −b/a, product = c/a.

Quadratic functions

  • Quadratic formula: x = \dfrac{-b \pm \sqrt{b^2 - 4ac}}{2a}
  • Discriminant: \Delta = b^2 - 4ac
  • Vertex form: y = a(x - h)^2 + k

Quick reference

| Form | Shows |
|---|---|
| y = ax² + bx + c | y-intercept c |
| y = a(x − h)² + k | Turning point (h, k) |
| y = a(x − p)(x − q) | x-intercepts p, q |
| Quadratic formula | x = (−b ± √Δ)/2a |
| Discriminant Δ = b² − 4ac | >0 two, =0 one, <0 none |

Worked example (Medium)

Sketch y = x² − 4x − 5 showing intercepts and turning point.

  1. Factorise. x² − 4x − 5 = (x − 5)(x + 1). Roots: x = 5, x = −1.
  2. y-intercept. Sub x = 0 → y = −5.
  3. Vertex. x = −(−4)/2 = 2. y = 4 − 8 − 5 = −9. Vertex (2, −9).

Answer: Intercepts (−1, 0), (5, 0), (0, −5). Turning point (2, −9), axis x = 2.

Worked example (Hard)

Find the values of k such that $2x^2 + kx + 8 = 0$ has exactly one real solution.

  1. Condition for one root. Discriminant $\Delta = 0$.
  2. Compute Δ. $\Delta = k^2 - 4(2)(8) = k^2 - 64$.
  3. Solve Δ = 0. $k^2 = 64 \Rightarrow k = \pm 8$.
  4. Verify. For $k = 8$: $2x^2 + 8x + 8 = 0 \Rightarrow x = -2$ (repeated). For $k = -8$: $2x^2 - 8x + 8 = 0 \Rightarrow x = 2$ (repeated). Both valid.

Answer: k = 8 or k = −8.

Common mistakes

  • Confusing the sign of h in turning-point form $y = a(x - h)^2 + k$ Students write $y = (x + 3)^2$ and claim the vertex is at $x = 3$ instead of $x = -3$ Fix: Remember: whatever makes the bracket zero is h. $(x + 3) = 0 \Rightarrow x = -3$.
  • Dropping the negative when using the quadratic formula Writing $x = b \pm \sqrt{\Delta}$ instead of $x = -b \pm \sqrt{\Delta}$ over $2a$ Fix: Memorise the formula as a chant: "negative b, plus or minus, root of b-squared minus 4ac, all over 2a".
  • Forgetting to check if the question asks for y-values or coordinates Students solve for x-intercepts but don't substitute back to verify or find other requested values Fix: After finding x, always re-read the question: does it want just x, or the full point (x, y)?

The discriminant as a detective tool

The discriminant $\Delta = b^2 - 4ac$ is far more than a formula to memorise — it tells you the shape story of a parabola before you sketch it.

When $\Delta > 0$, the parabola crosses the x-axis twice, creating two distinct roots. The larger $\Delta$ is, the further apart those roots lie. When $\Delta = 0$, the parabola just touches the axis at its vertex — a single repeated root. When $\Delta < 0$, the parabola never reaches the axis and sits entirely above (if $a > 0$) or below (if $a < 0$) it.

VCAA loves "find the values of $k$" problems where a parameter makes $\Delta = 0$, $\Delta > 0$, or $\Delta < 0$. These test your ability to set up an inequality involving the discriminant and solve it — combining algebra with quadratic theory.

A more advanced use appears in optimisation: the minimum or maximum of $f(x) = ax^2 + bx + c$ occurs at $x = -b/(2a)$, and its value is $c - b^2/(4a) = c - \Delta/(4a)$. So the discriminant directly relates to the extreme value.

In Year 12 Methods, the discriminant reappears in curve-fitting, intersection problems, and proof questions. Investing time now in understanding why $\Delta$ works — by completing the square inside the quadratic formula derivation — pays dividends across the entire course.

Practice questions

  1. The discriminant of y = 2x² + 3x + 5 is:

    1. −31
    2. 31
    3. −1
    4. 1

    (1 marks)

  2. A ball thrown from a 1 m cliff follows h(t) = −5t² + 20t + 1 metres. Find the maximum height and when the ball hits the ground.

    (4 marks)

Summary

  • Convert between general, turning-point and factorised forms.
  • Find roots using the quadratic formula.
  • Sketch parabolas showing key features.

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