# Quadratics and Parabolas

> Solving quadratics three ways, sketching parabolas, the vertex, and the discriminant.

From HSC Maths Advanced on WhizzNotes, free study notes for NSW students: https://whizznotes.com/mathematics-advanced/quadratics/. Updated 3 November 2026.
Year 11 and Year 12 boxes go deeper than the Year 10 text around them.

### Story time: The basketball shot
Zara shoots a basketball from the free-throw line. The ball rises, slows down, reaches a highest point, then falls into the hoop. If you film it and trace the path, you get a smooth, symmetrical curve.

That curve is a **parabola**. It comes from a **quadratic**, an equation with an $x^2$ in it. Coaches, engineers who design bridges and anyone who launches anything use quadratics to find the highest point and where it lands.

**In short:** A quadratic is an expression like $ax^2 + bx + c$, and its graph is a parabola that opens up if $a > 0$ and down if $a < 0$. Solve a quadratic equation by factorising, completing the square or the formula $x = \dfrac{-b \pm \sqrt{b^2 - 4ac}}{2a}$.

## Starting point: Years 9 and 10

A **quadratic** has the form $y = ax^2 + bx + c$ with $a \neq 0$. Its graph is a parabola.

- If $a > 0$, the parabola opens **upward** (a smile) and has a minimum.
- If $a < 0$, it opens **downward** (a frown) and has a maximum.

### Solving quadratic equations

To solve $ax^2 + bx + c = 0$, get zero on one side first. Then use one of three methods.

**1. Factorise.** If two things multiply to zero, one of them must be zero.

$$x^2 - 2x - 3 = 0 \;\Rightarrow\; (x - 3)(x + 1) = 0 \;\Rightarrow\; x = 3 \text{ or } x = -1$$

**2. Quadratic formula.** It always works.

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

### Worked example: Using the formula
Solve $2x^2 + 3x - 4 = 0$. Here $a = 2$, $b = 3$, $c = -4$.

$$x = \frac{-3 \pm \sqrt{9 + 32}}{4} = \frac{-3 \pm \sqrt{41}}{4}$$

So $x \approx 0.85$ or $x \approx -2.35$.

**3. Completing the square.** Halve the middle number, square it, add and subtract it.

$$x^2 + 6x + 2 = (x + 3)^2 - 9 + 2 = (x + 3)^2 - 7$$

### Calculator tip: The quadratic formula without slips
Work out the discriminant $b^2 - 4ac$ first and write it down. If it is negative there are no real solutions, so stop there.

Then type the whole top and the whole bottom in brackets: `( -b + √(Δ) ) ÷ ( 2 × a )`, using your numbers, and repeat with minus. A common slip is `-b + √(Δ) ÷ 2a`, which divides only the root. Press `S⇔D` to see exact answers such as $\frac{3 + \sqrt{5}}{2}$.

### Sketching a parabola

Find these four things:

1. **Shape:** up or down from the sign of $a$.
2. **y-intercept:** put $x = 0$, so it is $c$.
3. **x-intercepts:** solve $y = 0$.
4. **Vertex:** the turning point. Its x-value is $x = -\dfrac{b}{2a}$, which is also the **axis of symmetry**.

*Diagram: Parabola y equals x squared minus 2x minus 3, crossing the x-axis at minus 1 and 3, with vertex at 1, minus 4*
*y = x² − 2x − 3: roots at −1 and 3, vertex halfway between them.*

### Key idea: The vertex sits halfway between the roots
The parabola is symmetrical. The roots $-1$ and $3$ average to $1$, which is the x-value of the vertex. Then $y = 1 - 2 - 3 = -4$.

### Watch out: Divide everything by 2a
In the formula, the whole top is divided by $2a$, not just the square root.

### Exam tip: Which method?
Try to factorise first because it is fastest. If the numbers do not work out in 30 seconds, use the formula.

**Mind map**

```text
Quadratics
  Form
    y = ax² + bx + c
    a > 0 smile, a < 0 frown
  Solve
    Factorise
    Formula
    Complete the square
  Sketch
    y-intercept is c
    Roots from y = 0
    Vertex at x = -b/2a
  Discriminant
    Δ = b² - 4ac
    Number of roots
```

### Year 11: The discriminant
The part under the root, $\Delta = b^2 - 4ac$, is the **discriminant**. It tells you how many real roots there are without solving.

- $\Delta > 0$: two different real roots (the parabola cuts the x-axis twice).
- $\Delta = 0$: one repeated root (it just touches the axis).
- $\Delta < 0$: no real roots (it misses the axis).

**Example.** For what values of $k$ does $x^2 + 4x + k = 0$ have two real roots? Need $16 - 4k > 0$, so $k < 4$.

A quadratic that is always positive (**positive definite**) has $a > 0$ and $\Delta < 0$.

If $a$ contains $k$ (as in $kx^2 - 4x + 1 = 0$), remember $k \neq 0$, or the equation is not a quadratic.

To show an equation always has two real roots, show $\Delta > 0$ for every $k$. For $x^2 + kx - 9 = 0$: $\Delta = k^2 + 36$, which is always positive because $k^2 \geq 0$.

**Sum and product of the roots.** If $\alpha$ and $\beta$ are the roots of $ax^2 + bx + c = 0$, then
$$\alpha + \beta = -\frac{b}{a} \qquad \alpha\beta = \frac{c}{a}$$
They answer questions without solving. Roots of opposite signs need $\alpha\beta < 0$, so $\frac{c}{a} < 0$. If the roots of $x^2 + kx - 4 = 0$ differ by 5, use $(\alpha - \beta)^2 = (\alpha + \beta)^2 - 4\alpha\beta$: $25 = k^2 + 16$, so $k = \pm 3$.

### Year 11: Turning point form and quadratic inequalities
Completing the square gives **turning point form** $y = a(x - h)^2 + k$ with vertex $(h, k)$.

To solve $x^2 - 2x - 3 < 0$, sketch the parabola. It is below the axis between its roots, so $-1 < x < 3$. For $> 0$, the answer is outside the roots: $x < -1$ or $x > 3$.

When $a \neq 1$, take $a$ out of the $x$ terms first, then complete the square inside the bracket:
$$3x^2 - 12x + 5 = 3(x^2 - 4x) + 5 = 3\big[(x - 2)^2 - 4\big] + 5 = 3(x - 2)^2 - 7$$
So the vertex is $(2, -7)$.

### Year 12: Maximising with a quadratic
A farmer has 40 m of fencing for a rectangle against a wall, so only three sides need fencing. With width $x$, the length is $40 - 2x$ and the area is

$$A = x(40 - 2x) = -2x^2 + 40x$$

This is a frown parabola, so the vertex is a maximum at $x = -\dfrac{40}{2(-2)} = 10$. The best area is $A = 10 \times 20 = 200 \text{ m}^2$. In Year 12 the same answer comes from setting the derivative to zero.

## Quick summary

- A quadratic $y = ax^2 + bx + c$ graphs as a parabola. The sign of $a$ decides up or down.
- Solve by factorising, the formula $x = \dfrac{-b \pm \sqrt{b^2 - 4ac}}{2a}$, or completing the square.
- The vertex is at $x = -\dfrac{b}{2a}$, halfway between the roots.
- The discriminant $\Delta = b^2 - 4ac$ counts the real roots.

## Quiz

Try these questions from the chapter quiz.

1. Solve $x^2 - 5x + 6 = 0$.
   - A. $x = -2$ or $x = -3$
   - B. $x = 2$ or $x = 3$
   - C. $x = 1$ or $x = 6$
   - D. $x = -1$ or $x = 6$
   - Answer: B. $x = 2$ or $x = 3$. $(x - 2)(x - 3) = 0$.

2. Solve $x^2 = 9x$.
   - A. $x = 9$ only
   - B. $x = 3$ or $x = -3$
   - C. $x = 0$ or $x = 9$
   - D. $x = 0$ only
   - Answer: C. $x = 0$ or $x = 9$. $x^2 - 9x = 0$, so $x(x - 9) = 0$. Never divide by $x$, or you lose $x = 0$.

3. The parabola $y = -2x^2 + 5$ opens:
   - A. upward
   - B. downward
   - C. to the right
   - D. it is a straight line
   - Answer: B. downward. $a = -2$ is negative, so it is a frown.

The whole chapter quiz is on the page, for signed-in students: https://whizznotes.com/mathematics-advanced/quadratics/#quiz