Recognising and Sketching Linear and Quadratic Graphs Revision Notes | GCSE AQA Higher Maths

1. Why Graphs Matter: The Big Picture

Every equation tells a story, and a graph is how that story looks. When you can glance at $y = 3x - 2$ and immediately picture a rising straight line, or see $y = x^2 - 4$ and know it dips below the x-axis, you have a powerful tool for the whole of GCSE Maths — from solving equations graphically to interpreting real-world models.

examiners use the command word sketch deliberately. A sketch is not a precise plot — it is a diagram showing the correct shape with key features labelled. That distinction saves you time and earns you marks.

This lesson covers two graph families: linear (straight lines) and quadratic (parabolas). Master these and you have the foundation for every other graph type at GCSE.

2. Linear Graphs: Straight Lines and Their Equations

A linear function produces a straight-line graph. The equation is always written in the form $y = mx + c$, where two numbers control everything about the line: how steep it is and where it crosses the y-axis.

Think of gradient as the slope of a hill. A gradient of $2$ means for every $1$ unit you move right, you move $2$ units up. A gradient of $-3$ means for every $1$ unit right, you drop $3$ units down.

Once you know $m$ and $c$, you can sketch any straight line in seconds. Plot the y-intercept first, then use the gradient to find a second point, and draw the line through both.

Not every linear equation arrives in $y = mx + c$ form. You might see $2x + 3y = 6$ or $y - 4 = 2(x - 1)$. Rearranging to $y = mx + c$ is always your first move — it unlocks the gradient and intercept immediately.

Worked example: Sketching a Linear Graph from $y = mx + c$

Sketch the graph of $y = 2x - 3$, labelling the y-intercept and one other point. (Without a Calculator)

1. $m = 2$, $c = -3$ — Read directly from $y = mx + c$. The line crosses the y-axis at $(0, -3)$. [M1 for identifying both values]
2. Mark the point $(0, -3)$ on the y-axis. — This is your anchor point. Every sketch starts here.
3. From $(0, -3)$: move $1$ right, $2$ up $\Rightarrow$ $(1, -1)$ — Gradient $= 2 = \frac{2}{1}$, so rise $= 2$, run $= 1$. [M1 for correct use of gradient]
4. Set $y = 0$: $0 = 2x - 3 \Rightarrow 2x = 3 \Rightarrow x = 1.5$ So x-intercept is $(1.5, 0)$. — Labelling the x-intercept earns an extra mark on sketch questions.
5. Draw a straight line through $(0, -3)$ and $(1, -1)$, extending in both directions. Label both intercepts. — A sketch must show the correct direction (positive gradient = rising line). [A1 for correct sketch with labels]

Answer: A straight line with gradient $2$, crossing the y-axis at $(0, -3)$ and the x-axis at $(1.5, 0)$.

Special cases are worth knowing cold. A horizontal line like $y = 4$ has gradient $0$ — it runs perfectly flat. A vertical line like $x = 3$ is not a function and has an undefined gradient. These appear in GCSE Higher exam questions more often than students expect.

The graph above shows two linear functions. Notice how $y = 2x - 3$ (steeper, positive gradient) and $y = -x + 2$ (negative gradient, falling left to right) look completely different despite both being straight lines. The gradient is the single biggest visual clue.

3. Quadratic Graphs: Parabolas and Their Key Features

A quadratic function contains an $x^2$ term, and its graph is always a smooth, symmetrical curve called a parabola. The shape is either a U (when the $x^2$ coefficient is positive) or an n (when it is negative). Recognising this immediately from the equation is a core exam skill.

For a sketch, you need four key features: the shape (U or n), the y-intercept, the x-intercepts (roots, if they exist), and the turning point. You do not need to plot every point — the examiner wants to see these features correctly labelled.

The y-intercept is always the easiest to find: substitute $x = 0$ and you get $y = c$. The x-intercepts require solving $ax^2 + bx + c = 0$, usually by factorising at GCSE. The turning point sits exactly halfway between the two roots — that midpoint is the line of symmetry.

For $y = x^2 - 2x - 3$: the y-intercept is $(0, -3)$, the roots are at $x = -1$ and $x = 3$, and the line of symmetry is $x = 1$ (the midpoint of $-1$ and $3$). Substituting $x = 1$ gives $y = 1 - 2 - 3 = -4$, so the turning point is $(1, -4)$.

Worked example: Sketching a Quadratic Graph by Factorising

Sketch the graph of $y = x^2 - x - 6$, labelling all intercepts and the turning point. (Without a Calculator)

1. Coefficient of $x^2$ is $+1 > 0$, so the graph is U-shaped. — Always state the shape first — it earns a mark and guides your sketch. [M1]
2. Substitute $x = 0$: $y = 0 - 0 - 6 = -6$ Y-intercept: $(0, -6)$ — The constant term $c = -6$ gives the y-intercept directly. [A1]
3. $x^2 - x - 6 = 0$ $(x - 3)(x + 2) = 0$ $x = 3$ or $x = -2$ — Need two numbers that multiply to $-6$ and add to $-1$: those are $-3$ and $+2$. Roots are $(3, 0)$ and $(-2, 0)$. [M1 for factorisation, A1 for both roots]
4. $x = \frac{3 + (-2)}{2} = \frac{1}{2} = 0.5$ — The line of symmetry is the midpoint of the two roots. [M1]
5. Substitute $x = 0.5$: $y = (0.5)^2 - (0.5) - 6 = 0.25 - 0.5 - 6 = -6.25$ Turning point: $(0.5, -6.25)$ — Verify: $0.5 \times 0.5 = 0.25$. Then $0.25 - 0.5 - 6 = -6.25$. This is the minimum point. [A1]
6. Draw a U-shaped curve through $(-2, 0)$, $(0, -6)$, $(3, 0)$ with minimum at $(0.5, -6.25)$. — Label all four key points. The curve should be smooth and symmetrical about $x = 0.5$.

Answer: U-shaped parabola with roots at $(-2, 0)$ and $(3, 0)$, y-intercept $(0, -6)$, and minimum turning point at $(0.5, -6.25)$.

Notice the pattern: finding the turning point always follows the same sequence — roots first, midpoint second, substitute third. Practise this sequence until it is automatic.

4. Comparing Linear and Quadratic: Recognising Graph Types

GCSE Higher exam questions often show you a graph and ask you to identify the type, or give you an equation and ask you to match it to a sketch. The key is knowing the visual signature of each family at a glance.

The table above is your rapid-recognition toolkit. When you see a curved graph with one turning point and symmetry, it is quadratic. When you see a straight line, it is linear. These distinctions appear in matching questions, graph interpretation, and multi-part problems.

Interpreting a graph means reading meaning from its features. The gradient of a linear graph tells you the rate of change — in a distance-time graph, that is speed. The x-intercepts of a quadratic tell you when $y = 0$ — in a projectile problem, those are the moments the object is at ground level.

Worked example: Recognising Graph Type from an Equation

State the type of graph for each equation and give one key feature: (a) $y = 5 - 3x$ (b) $y = x^2 - 9$ (c) $y = -x^2 + 4x$

1. $y = 5 - 3x = -3x + 5$ Highest power of $x$ is $1$. — This is linear. Gradient $m = -3$ (falling line), y-intercept $(0, 5)$.
2. $y = x^2 - 9$ Highest power of $x$ is $2$, coefficient $a = +1 > 0$. — This is quadratic, U-shaped. Y-intercept $(0, -9)$. Roots: $x^2 = 9 \Rightarrow x = \pm 3$.
3. $y = -x^2 + 4x$ Highest power of $x$ is $2$, coefficient $a = -1 < 0$. — This is quadratic, n-shaped. Y-intercept $(0, 0)$ — the curve passes through the origin. Roots: $x(-x + 4) = 0 \Rightarrow x = 0$ or $x = 4$.

Answer: (a) Linear, falling straight line, y-intercept $(0, 5)$. (b) Quadratic U-shape, roots at $x = \pm 3$, y-intercept $(0, -9)$. (c) Quadratic n-shape, roots at $x = 0$ and $x = 4$, passes through origin.

5. Deliberate Practice: Try These Yourself

The best way to consolidate these methods is to work through short problems independently before checking. Attempt each one fully — including labelling all key features — before revealing the solution.

6. Exam Traps and How to Avoid Them

This is where marks are won and lost. Sketch questions are generous with method marks — but only if you show the right features. Knowing the traps in advance puts you ahead of most students in the room.

The most common error on linear graph questions is misreading a rearranged equation. If the equation is $y = 4 - 3x$, the gradient is $-3$, not $4$. Students who do not rewrite it as $y = -3x + 4$ often swap the two values and lose both marks.

On quadratic sketches, the turning point is the most frequently dropped mark. Many students find the roots correctly but then forget to substitute back to find the y-coordinate of the vertex. The examiner expects a labelled coordinate pair — not just the x-value.