Sketching Functions: Linear and Quadratic Revision Notes | GCSE AQA Higher Further Maths

1. Why Sketching Graphs Matters

Graphs are one of the most powerful tools in GCSE Maths — they turn equations into pictures you can reason about. When an exam question says sketch, it is not asking for a perfect plot on graph paper. It is asking you to show the shape and label the features that matter.

The two graph families you need to master at this level are linear (straight lines) and quadratic (parabolas). Once you know what to look for in an equation, sketching becomes a checklist — not a guessing game.

2. Sketching Linear Functions

A linear function has the form $y = mx + c$. The letter $m$ tells you the gradient — how steeply the line rises or falls — and $c$ tells you where the line crosses the $y$-axis, called the $y$-intercept.

To sketch a linear function, you need two things: where the line crosses the $y$-axis (set $x = 0$) and where it crosses the $x$-axis (set $y = 0$). Mark both intercepts, then draw a straight line through them.

Worked example: Sketching a Linear Function

Sketch the graph of $y = 2x - 6$, labelling both intercepts.

1. $m = 2$, $\quad c = -6$ — Comparing with $y = mx + c$: gradient is 2 (line rises), $y$-intercept is $-6$. [M1]
2. Set $x = 0$: $\quad y = 2(0) - 6 = -6$ $\quad \Rightarrow$ point $(0, -6)$ — This is where the line crosses the $y$-axis.
3. Set $y = 0$: $\quad 0 = 2x - 6$ $\quad \Rightarrow 2x = 6$ $\quad \Rightarrow x = 3$ $\quad \Rightarrow$ point $(3, 0)$ — This is where the line crosses the $x$-axis. [M1]
4. Plot $(0, -6)$ and $(3, 0)$. Draw a straight line through both points. Label each intercept. — A positive gradient means the line slopes upward from left to right. [A1]

Answer: Straight line through $(0, -6)$ and $(3, 0)$ with positive gradient.

Sketching a straight line is a two-intercept job. Now that you have the linear case, the next step is the curve that appears on almost every GCSE paper: the parabola.

3. Sketching Quadratic Functions

A quadratic function has the form $y = ax^2 + bx + c$. The key difference from a linear function is the $x^2$ term, which gives the graph its characteristic curved, symmetric shape — the parabola.

The sign of $a$ is the first thing to check. When $a > 0$, the parabola opens upwards (a U-shape). When $a < 0$, it opens downwards (an n-shape). This single observation earns you a mark in a sketch question.

• Shape: U-shaped if $a > 0$; n-shaped if $a < 0$
• $y$-intercept: set $x = 0$, giving $y = c$, so the point is $(0, c)$
• $x$-intercepts (roots): set $y = 0$ and solve $ax^2 + bx + c = 0$ by factorising or using the quadratic formula
• Turning point (vertex): lies on the axis of symmetry, which is at $x = -\dfrac{b}{2a}$
• Axis of symmetry: a vertical line halfway between the two roots

Worked example: Sketching a Quadratic Function

Sketch the graph of $y = x^2 - 4x + 3$, labelling the intercepts and turning point.

1. $a = 1 > 0$, so the parabola opens upwards (U-shape). — Always state the shape first — it earns a mark. [M1]
2. Set $x = 0$: $\quad y = (0)^2 - 4(0) + 3 = 3$ $\quad \Rightarrow$ point $(0, 3)$ — The constant term $c = 3$ gives the $y$-intercept directly.
3. Set $y = 0$: $\quad x^2 - 4x + 3 = 0$ $\quad (x-1)(x-3) = 0$ $\quad \Rightarrow x = 1$ or $x = 3$ — Two roots: $(1, 0)$ and $(3, 0)$. [M1 for factorisation, A1 for both roots]
4. $x = -\dfrac{b}{2a} = -\dfrac{-4}{2 \times 1} = \dfrac{4}{2} = 2$ — The axis of symmetry is $x = 2$, halfway between the roots $x=1$ and $x=3$. This is a useful cross-check.
5. Substitute $x = 2$: $\quad y = (2)^2 - 4(2) + 3 = 4 - 8 + 3 = -1$ $\quad \Rightarrow$ turning point $(2, -1)$ — This is the minimum point of the parabola. [A1]
6. Draw a U-shaped curve through $(1,0)$, $(3,0)$, $(0,3)$, with minimum at $(2,-1)$. Label all four points. — The curve must be smooth and symmetric about $x = 2$.

Answer: U-shaped parabola with roots at $(1, 0)$ and $(3, 0)$, $y$-intercept $(0, 3)$, and minimum turning point $(2, -1)$.

4. Interpreting Graphs

Sketching is only half the skill. Interpreting a graph means reading information from it — finding values, describing behaviour, and connecting the graph back to its equation or a real-world context.

For a linear graph, the gradient tells you the rate of change. If the graph shows distance against time, the gradient is speed. For a quadratic, the turning point tells you the maximum or minimum value of the quantity being modelled.

• Read a $y$-value: draw a vertical line from the $x$-value to the curve, then read across to the $y$-axis
• Read an $x$-value: draw a horizontal line from the $y$-value to the curve, then read down to the $x$-axis
• Find where two graphs intersect: the $x$-value at the intersection is the solution to the equation formed by setting the two expressions equal
• Describe the gradient of a straight line: state whether it is positive, negative, zero, or undefined, and give its numerical value if possible
• Identify roots from a graph: the $x$-intercepts of $y = f(x)$ are the solutions to $f(x) = 0$

Worked example: Interpreting a Quadratic Graph

The graph of $y = x^2 - 2x - 8$ is shown. Use the graph to find the roots and the minimum value of $y$.

1. The curve crosses the $x$-axis where $y = 0$. Read off: $x = -2$ and $x = 4$. — These are the solutions to $x^2 - 2x - 8 = 0$. You can verify: $(-2)^2 - 2(-2) - 8 = 4 + 4 - 8 = 0$ ✓ and $(4)^2 - 2(4) - 8 = 16 - 8 - 8 = 0$ ✓
2. Midpoint of roots: $x = \dfrac{-2 + 4}{2} = \dfrac{2}{2} = 1$ — The turning point lies on the axis of symmetry $x = 1$.
3. Substitute $x = 1$: $\quad y = (1)^2 - 2(1) - 8 = 1 - 2 - 8 = -9$ — The minimum value of $y$ is $-9$, occurring at the point $(1, -9)$. [A1]

Answer: Roots at $x = -2$ and $x = 4$; minimum value of $y$ is $-9$ at the point $(1, -9)$.

5. Functions Defined Over Restricted Domains

Sometimes a function is only defined for a specific range of $x$-values — this is called a restricted domain. The graph only exists within that interval, and you must show this clearly by stopping the curve or line at the boundary values.

At GCSE, you may encounter up to three separate domain restrictions on a single graph. Each piece is sketched independently using the same techniques you have already practised, then placed side by side on the same axes.

Worked example: Sketching a Function with a Restricted Domain

Sketch $y = 2x + 1$ for $-1 \leq x \leq 3$. Label the endpoints.

1. $y = 2(-1) + 1 = -2 + 1 = -1$ $\quad \Rightarrow$ point $(-1, -1)$ — This is the left boundary of the domain.
2. $y = 2(3) + 1 = 6 + 1 = 7$ $\quad \Rightarrow$ point $(3, 7)$ — This is the right boundary of the domain.
3. Draw a straight line segment from $(-1, -1)$ to $(3, 7)$. Mark both endpoints with filled circles (since both boundary values are included). — Filled circles show the endpoints are included ($\leq$). Open circles would show they are excluded ($<$). [A1]

Answer: Line segment from $(-1, -1)$ to $(3, 7)$, both endpoints included.

6. Practice Problems

The best way to consolidate these techniques is to apply them yourself. Work through each problem using the checklist from the previous sections before checking the answers.

7. Exam Technique and Common Mistakes

Sketch questions are mark-rich if you follow a method, and mark-poor if you rush. Examiners award marks for labelled intercepts, the correct shape, and the turning point — not for artistic accuracy. A rough curve with all features labelled beats a beautiful curve with nothing marked.

One of the most common errors is confusing the $y$-intercept with the $x$-intercept. The $y$-intercept is always found by setting $x = 0$; the $x$-intercept is found by setting $y = 0$. Keep these distinct in your working.