Sketching and interpreting quadratic functions and their graphs, and solving quadratic equations by factorising, by the quadratic formula and graphically.

What this dot point is asking

The SQA wants you to recognise and sketch a quadratic function (a parabola), identify its key features, and solve a quadratic equation by factorising, by the quadratic formula when it does not factorise, and by reading roots from a graph.

The shape of a quadratic

The graph of $y = ax^2 + bx + c$ is a parabola. The sign of $a$ decides which way it opens: a positive $a$ gives a U-shape with a minimum turning point, and a negative $a$ gives an upside-down U with a maximum. The curve is symmetrical about a vertical line through its turning point, and it crosses the y-axis at $(0, c)$.

The roots are the x-values where $y = 0$, that is, where the curve meets the x-axis. A parabola can cross the x-axis twice, touch it once, or miss it entirely, which links directly to the discriminant.

Solving by factorising

If a quadratic factorises, solving is quick: factorise, then use the fact that a product equals zero only when one of its factors is zero.

Solving with the quadratic formula

When a quadratic does not factorise, the formula always gives the roots (if they exist). Identify $a$, $b$ and $c$ carefully, including signs.

Solving graphically

The roots of $ax^2 + bx + c = 0$ are the x-coordinates where the graph of $y = ax^2 + bx + c$ crosses the x-axis. Reading these from an accurate sketch gives approximate solutions, useful as a check on algebraic work.

Sketching a parabola

A good sketch needs three features: the y-intercept (at $x = 0$, the value $c$), the roots (where it crosses the x-axis, found by solving), and the turning point (often from completing the square, or by symmetry halfway between the roots). The line of symmetry passes vertically through the turning point.

Examples in context

Quadratics model anything that rises and falls, such as the height of a thrown ball. If a ball's height in metres after $t$ seconds is $h = 20t - 5t^2$, it hits the ground when $h = 0$: $20t - 5t^2 = 0$, so $5t(4 - t) = 0$, giving $t = 0$ (launch) and $t = 4$ seconds (landing). The turning point of the parabola gives the greatest height. Factorising and the formula turn such models into precise answers.

Try this

Q1. Solve $x^2 - 9 = 0$. [2 marks]

• Cue. $(x + 3)(x - 3) = 0$, so $x = \pm 3$.

Q2. Solve $x^2 + 5x + 6 = 0$. [2 marks]

• Cue. $(x + 2)(x + 3) = 0$, so $x = -2$ or $x = -3$.

Q3. Solve $x^2 + 6x + 2 = 0$ to 1 decimal place. [3 marks]

• Cue. $x = \dfrac{-6 \pm \sqrt{28}}{2} = -0.4$ or $-5.6$.