Mathematics · AQA 8300 §A9

Graphs

Spec: AQA 8300 §A9 ⭐⭐⭐⭐ ⌛ 55 mins AQA · Edexcel · OCR Grade 9

Find gradients and equations of straight lines using $y = mx + c$ and point-gradient form
Identify parallel lines (equal gradients) and perpendicular lines ($m_1 m_2 = -1$)
Sketch quadratic, cubic and reciprocal curves with key features labelled
Describe and apply all four graph transformations in function notation
Interpret and calculate from distance-time and velocity-time graphs

🔑 Core Concepts

1. Gradient and the Equation $y = mx + c$

Every straight line can be described completely by two values: its gradient $m$ (how steep it is) and its y-intercept $c$ (where it crosses the y-axis). The standard form $y = mx + c$ encodes both values directly.

📖

DEFINITION — Gradient

The gradient of a straight line is the change in $y$ per unit increase in $x$. Formally: $$m = \frac{\text{rise}}{\text{run}} = \frac{\Delta y}{\Delta x}$$ A positive gradient ($m > 0$) slopes upward left-to-right; negative ($m < 0$) slopes downward; zero gradient ($m = 0$) is horizontal.

Gradient Between Two Points

$$m = \frac{y_2 - y_1}{x_2 - x_1}$$

$m$ = gradient of the line $(x_1,\, y_1)$ = first point $(x_2,\, y_2)$ = second point

Standard Form of a Straight Line

$$y = mx + c$$

$m$ = gradient (coefficient of $x$) $c$ = y-intercept (value of $y$ when $x=0$)

🎯

EXAM TIP — Reading Off $m$ and $c$

Always rearrange to $y = mx + c$ before reading gradient and y-intercept. For example, $3y - 6x = 9$ divides by $3$ to give $y = 2x + 3$, so $m = 2$ and $c = 3$. A common trap is reading from an unrearranged equation.

✗

COMMON MISTAKE — Sign of the y-intercept

In $y = 5x - 7$, students sometimes write $c = 7$. The y-intercept is $-7$ (negative). The sign is part of the value and must be included.

2. Finding the Equation of a Line

When you know a gradient and one point on the line, or any two points, you can determine the equation uniquely. The point-gradient form is the most efficient method.

Point-Gradient Form

$$y - y_1 = m(x - x_1)$$

$(x_1,\, y_1)$ = any known point on the line $m$ = gradient Rearrange to $y = mx + c$ as required

Find gradient $m = \dfrac{y_2-y_1}{x_2-x_1}$

→

Substitute $m$ and one point $(x_1, y_1)$

→

Write $y - y_1 = m(x - x_1)$

→

Expand and rearrange to required form

🎯

EXAM TIP — Always Verify

After finding the equation, substitute the other point to verify. If the point satisfies the equation, your answer is correct. This check takes 10 seconds and guarantees the mark.

3. Parallel Lines

📖

DEFINITION — Parallel Lines

Two distinct lines are parallel if and only if they have equal gradients. They never intersect and remain a constant distance apart. If $m_1 = m_2$ but the lines are different (i.e. $c_1 \neq c_2$), they are parallel.

To find the equation of a line through $(3, 7)$ parallel to $y = 4x - 1$: the gradient must also be $4$. Use $y - 7 = 4(x - 3)$, giving $y = 4x - 5$.

✗

COMMON MISTAKE — Same Line vs Parallel Line

Two lines with the same gradient and the same y-intercept are identical, not parallel. Parallel lines must have the same $m$ but different $c$.

4. Perpendicular Lines

📖

DEFINITION — Perpendicular Lines

Two lines are perpendicular if they intersect at exactly $90°$. Their gradients satisfy: $$m_1 \times m_2 = -1$$ Equivalently, $m_2 = -\dfrac{1}{m_1}$ (the negative reciprocal of $m_1$).

Perpendicular Gradient Rule

$$m_2 = -\frac{1}{m_1}$$

$m_1$ = gradient of original line $m_2$ = gradient of perpendicular line Check: $m_1 \times m_2 = -1$ always

Original gradient $m_1$	Perpendicular gradient $m_2 = -1/m_1$
$2$	$-\frac{1}{2}$
$-3$	$\frac{1}{3}$
$\frac{2}{3}$	$-\frac{3}{2}$
$-\frac{1}{4}$	$4$

🧠

MEMORY TRICK — Flip and Negate

To find a perpendicular gradient: (1) Write the gradient as a fraction, (2) Flip it (reciprocal), (3) Change the sign. Example: $m = \frac{2}{3}$ → flip to $\frac{3}{2}$ → negate to $-\frac{3}{2}$.

✗

COMMON MISTAKE — Only Negating

Given $m = 4$, many students write the perpendicular gradient as $-4$. This only negates — it does not reciprocate. The correct answer is $-\frac{1}{4}$. Check: $4 \times (-\frac{1}{4}) = -1$ ✓

5. Quadratic Graphs

A quadratic function has the form $y = ax^2 + bx + c$ where $a \neq 0$. Its graph is a parabola with a single turning point (vertex). The sign of $a$ determines the orientation.

📖

KEY FEATURES OF A QUADRATIC

Roots: values of $x$ where $y = 0$. Found by factorising, completing the square, or the quadratic formula. A quadratic has 0, 1, or 2 real roots.
y-intercept: the constant $c$ (substitute $x = 0$).
Turning point (vertex): minimum if $a > 0$; maximum if $a < 0$.
Line of symmetry: a vertical line $x = -\frac{b}{2a}$ passing through the vertex.
Shape: U-shape (smile) if $a > 0$; inverted U (frown) if $a < 0$.

Line of Symmetry and Vertex

$$x_{\text{vertex}} = -\frac{b}{2a}$$

For $y = ax^2 + bx + c$ Substitute $x_{\text{vertex}}$ back to find the $y$-coordinate If factorised as $(x-p)(x-q)$: $x_{\text{vertex}} = \dfrac{p+q}{2}$

🎯

EXAM TIP — Use Factorised Form for Vertex

For $y = (x-1)(x-5)$, the roots are $x = 1$ and $x = 5$ by inspection. The line of symmetry is $x = \frac{1+5}{2} = 3$ (average of roots), which is much faster than $-b/2a$. The vertex is at $x = 3$; substitute to find $y = (3-1)(3-5) = -4$. Turning point: $(3, -4)$.

6. Cubic Graphs

A cubic function has the form $y = ax^3 + \ldots$ with a non-zero $x^3$ term. Cubics have a characteristic shape that depends on the sign of the leading coefficient and the nature of the roots.

📖

KEY FEATURES OF A CUBIC

Roots: up to 3 real x-intercepts. A cubic always has at least one real root (it crosses the x-axis at least once).
End behaviour: if $a > 0$, the graph goes from bottom-left to top-right; if $a < 0$, top-left to bottom-right.
Turning points: 0 or 2 local turning points. $y = x^3$ has no turning points; $y = x(x-2)(x+3)$ has two.
y-intercept: the constant term $d$ (set $x = 0$).

🎯

EXAM TIP — Sketching a Factorised Cubic

For $y = x(x-2)(x+3)$: roots at $x = 0,\; 2,\; -3$; y-intercept at $0$; positive leading coefficient so starts bottom-left. Plot the three roots, mark the y-intercept, and draw a smooth S-curve through them.

✗

COMMON MISTAKE — Assuming All Cubics Have Two Turning Points

$y = x^3$ has no turning points — it passes through the origin with a point of inflection (it flattens momentarily). Only draw two turning points when the cubic has three distinct real roots or a specific structure that creates them.

7. Reciprocal Graphs

A reciprocal function has the form $y = \dfrac{k}{x}$ (equivalently $y = kx^{-1}$). It produces a hyperbola — two separate branches, one in each of two opposite quadrants.

📖

KEY FEATURES OF $y = k/x$

Asymptotes: $x = 0$ (the y-axis) and $y = 0$ (the x-axis). The graph never touches either axis.
If $k > 0$: branches in quadrants 1 (top-right) and 3 (bottom-left).
If $k < 0$: branches in quadrants 2 (top-left) and 4 (bottom-right).
No roots, no y-intercept, no turning points.
As $x \to 0^+$, $y \to +\infty$ (for $k>0$); as $x \to \infty$, $y \to 0^+$.

✗

COMMON MISTAKE — Touching the Axes

The branches of $y = k/x$ approach but never reach the coordinate axes. Drawing a branch that touches or crosses $x = 0$ or $y = 0$ is wrong and costs marks. Plot guide points such as $(0.5, 2k)$ and $(2, k/2)$ to ensure your curve stays away from the axes.

8. Graph Transformations

Graph transformations describe how the graph of $y = f(x)$ is repositioned or rescaled to produce a new graph. There are four standard transformations, each with a precise mathematical description.

Why the horizontal cases seem counter-intuitive: When you replace $x$ with $x + a$, you are asking "for which input does the transformed function give the same output as $f(x)$ did at $x$?" The answer is $x - a$, so the graph shifts by $-a$ (left if $a > 0$). The change happens inside the function, so it works horizontally and in the opposite direction.

Translation Upward: $y = f(x) + a$

$$y = f(x) + a$$

Translation: UP by $a$ units (DOWN if $a < 0$) Every point $(x, y) \mapsto (x,\; y + a)$ Change is outside $f$ → vertical, in the direction of $a$

Translation Sideways: $y = f(x + a)$

$$y = f(x + a)$$

Translation: LEFT by $a$ units (RIGHT if $a < 0$) Every point $(x, y) \mapsto (x - a,\; y)$ Change is inside $f$ → horizontal, opposite to sign of $a$

Vertical Stretch: $y = af(x)$

$$y = af(x)$$

Stretch by scale factor $a$ parallel to the y-axis Every point $(x, y) \mapsto (x,\; ay)$ $a = -1$: reflection in the x-axis

Horizontal Stretch: $y = f(ax)$

$$y = f(ax)$$

Stretch by scale factor $\dfrac{1}{a}$ parallel to the x-axis Every point $(x, y) \mapsto \!\left(\dfrac{x}{a},\; y\right)$ $a = -1$: reflection in the y-axis

⚠

IMPORTANT — Order of Multiple Transformations

When applying multiple transformations such as $y = 2f(3x - 6) + 1$, work from inside out:

Factorise the argument: $y = 2f(3(x-2)) + 1$
Apply horizontal stretch by $\frac{1}{3}$ (from $f(3\cdot)$)
Apply horizontal translation RIGHT 2 (from $f(\cdot - 2)$) — note: right because $-2$
Apply vertical stretch ×2
Apply vertical translation UP 1

The horizontal stretch is applied before the horizontal translation.

🧠

MEMORY TRICK — Inside vs Outside

INSIDE the function = HORIZONTAL = OPPOSITE direction
OUTSIDE the function = VERTICAL = EXPECTED direction
Outside: $+a$ goes up, $\times a$ stretches by $a$ upward.
Inside: $+a$ goes left, $\times a$ stretches by $\frac{1}{a}$ inward.

9. Distance-Time Graphs

📖

DEFINITION — Distance-Time Graph

A distance-time graph plots distance from a fixed reference point (usually the starting point) against time elapsed. The gradient at any point equals the speed at that moment. $$\text{Speed} = \frac{\text{change in distance}}{\text{change in time}} = \text{gradient of d-t graph}$$

Feature of D-T Graph	Physical Meaning
Steep positive gradient	Moving quickly away from starting point
Shallow positive gradient	Moving slowly away from starting point
Horizontal line (zero gradient)	Stationary — not moving
Negative gradient	Moving back towards the starting point
Curved line	Changing speed (acceleration or deceleration)

🎯

EXAM TIP — Units

If distance is in km and time in hours, speed is in km/h. If distance is in m and time in seconds, speed is in m/s. Always state units in your final answer.

10. Velocity-Time Graphs

📖

DEFINITION — Velocity-Time Graph

A velocity-time graph plots velocity against time. Two key relationships: $$\text{Acceleration} = \frac{\Delta v}{\Delta t} = \text{gradient of v-t graph}$$ $$\text{Distance travelled} = \text{area under v-t graph}$$

Feature of V-T Graph	Physical Meaning
Positive gradient	Accelerating (speeding up in positive direction)
Negative gradient	Decelerating (slowing down)
Zero gradient (horizontal)	Constant velocity (uniform motion)
Area above x-axis	Distance in positive direction
Area below x-axis	Distance in negative direction (returning)
Curved line	Non-uniform acceleration

Velocity-Time Key Relationships

$$a = \frac{v - u}{t} = \text{gradient} \qquad d = \text{area under graph}$$

$a$ = acceleration (m/s²); $v$ = final velocity; $u$ = initial velocity Area of triangle: $\frac{1}{2} \times \text{base} \times \text{height}$ Area of trapezium: $\frac{1}{2}(a+b) \times h$

✗

COMMON MISTAKE — Gradient vs Area Confusion

On a velocity-time graph, the gradient gives acceleration — NOT distance. The area gives distance — NOT acceleration. Confusing these two gives completely wrong answers. Write a mental note: "v-t: grad = acc, area = dist."

🎯

GRADE 9 TIP — Gradient of a Curved v-t Graph

For a curved velocity-time graph, instantaneous acceleration at a point is found by drawing a tangent line at that point and calculating its gradient. Use a large triangle (long base) to minimise reading errors. State that your answer is an estimate, since the tangent is drawn by eye.

⚠

IMPORTANT — Area Under a Curved v-t Graph

When the v-t graph is curved, you cannot find the exact distance with simple shapes. At GCSE, you can: (1) count squares under the curve, or (2) use the trapezium rule by dividing the region into strips. Both give estimates — acknowledge this in your answer.

🗺 Visual Notes

Graphs

Straight Lines

$y = mx + c$: gradient & y-intercept
Gradient: $m = (y_2-y_1)/(x_2-x_1)$
Parallel: $m_1 = m_2$, different $c$
Perpendicular: $m_1 m_2 = -1$

Curved Graphs

Quadratic: parabola, 0/1/2 roots
Cubic: S-shape, up to 3 roots
Reciprocal: $y = k/x$, hyperbola
Key: roots, intercepts, turning points

Transformations

$f(x)+a$: UP $a$ (vertical)
$f(x+a)$: LEFT $a$ (horizontal, opposite)
$af(x)$: vertical stretch ×$a$
$f(ax)$: horizontal stretch ÷$a$

Real-World Graphs

Distance-time: gradient = speed
Velocity-time: gradient = acceleration
Area under v-t graph = distance
Horizontal line = constant value

Grade 9 Skills

Tangent to curve → instantaneous rate
Multiple transformations in order
Connect equation features to graph
Estimate area under curved graph

Comparison: Types of Graph

Graph Type	General Form	Shape	Max Roots	Asymptotes?	Turning Points?
Linear	$y = mx + c$	Straight line	1	No	No
Quadratic	$y = ax^2 + bx + c$	Parabola (U or ∩)	2	No	1 (min or max)
Cubic	$y = ax^3 + \ldots$	S-shape or single curve	3	No	0 or 2
Reciprocal	$y = k/x$	Hyperbola (2 branches)	0	Yes ($x=0$, $y=0$)	No

Transformation Quick-Reference Table

Notation	Type	Direction	Effect on point $(x, y)$	Inside/Outside?
$f(x) + a$	Translation	UP by $a$	$(x,\; y+a)$	Outside
$f(x + a)$	Translation	LEFT by $a$	$(x-a,\; y)$	Inside
$af(x)$	Stretch	Vertical ×$a$	$(x,\; ay)$	Outside
$f(ax)$	Stretch	Horizontal ÷$a$	$(x/a,\; y)$	Inside
$-f(x)$	Reflection	In x-axis	$(x,\; -y)$	Outside
$f(-x)$	Reflection	In y-axis	$(-x,\; y)$	Inside

Decision Tree: How to Identify a Transformation

Look at the change from $f(x)$

→

Is change INSIDE $f(\ldots)$?

→

Horizontal effect
(OPPOSITE direction to number)

→

Is change OUTSIDE $f(\ldots)$?

→

Vertical effect
(SAME direction as number)

Comparison: Distance-Time vs Velocity-Time Graphs

Property	Distance-Time Graph	Velocity-Time Graph
y-axis quantity	Distance (m or km)	Velocity (m/s or km/h)
Gradient represents	Speed (m/s)	Acceleration (m/s²)
Area under graph	Not typically calculated	Distance travelled
Horizontal line	Stationary (not moving)	Constant velocity
Steep upward line	Fast motion away	Large acceleration
Downward slope	Returning to start	Decelerating
Curve	Changing speed	Changing acceleration

✏ Worked Examples

Grade 4–5 · Straight Lines

Find the equation of the line passing through the points $(1,\; 4)$ and $(3,\; 10)$. Give your answer in the form $y = mx + c$.

Calculate the gradient

Label the points: $(x_1, y_1) = (1, 4)$ and $(x_2, y_2) = (3, 10)$. $$m = \frac{y_2 - y_1}{x_2 - x_1} = \frac{10 - 4}{3 - 1} = \frac{6}{2} = 3$$

Use the point-gradient formula

Using $m = 3$ and point $(1, 4)$: $$y - 4 = 3(x - 1)$$ $$y - 4 = 3x - 3$$

Rearrange and verify

$$y = 3x - 3 + 4 = 3x + 1$$ Verify with $(3, 10)$: $3(3) + 1 = 10$ ✓

The equation of the line is $\boldsymbol{y = 3x + 1}$.

Grade 6–7 · Quadratic Graph

Sketch the graph of $y = x^2 - 4x - 5$. Label the roots, y-intercept, and turning point clearly.

Find the y-intercept

Set $x = 0$: $$y = (0)^2 - 4(0) - 5 = -5$$ y-intercept: $(0,\; -5)$

Find the roots by factorising

Set $y = 0$: $$x^2 - 4x - 5 = 0$$ Find two numbers that multiply to $-5$ and add to $-4$: these are $-5$ and $+1$. $$(x - 5)(x + 1) = 0 \implies x = 5 \text{ or } x = -1$$

Find the turning point

Line of symmetry = average of roots: $$x = \frac{-1 + 5}{2} = 2$$ Substitute $x = 2$: $$y = (2)^2 - 4(2) - 5 = 4 - 8 - 5 = -9$$ Minimum turning point: $(2,\; -9)$

Sketch the graph

Since $a = 1 > 0$, draw a U-shaped parabola through: root $(-1,\; 0)$, y-intercept $(0,\; -5)$, minimum $(2,\; -9)$, root $(5,\; 0)$. The curve is symmetric about $x = 2$.

Roots at $x = -1$ and $x = 5$; y-intercept at $(0, -5)$; minimum turning point at $(2, -9)$; line of symmetry $x = 2$.

Grade 9 · Combined Graph Transformations

The graph of $y = f(x)$ has a minimum point at $(4,\; -3)$ and passes through $(0,\; 5)$.
(a) Write down the coordinates of the minimum point on $y = f(2x) - 1$.
(b) Write down the coordinates of the image of $(0,\; 5)$ on $y = 3f(x + 2)$.
(c) Describe fully the two transformations that map $y = f(x)$ onto $y = 3f(x + 2)$.

Part (a): Analyse $y = f(2x) - 1$

Identify the two transformations:

$f(2x)$: horizontal stretch by scale factor $\frac{1}{2}$ (x-coordinates halved)
$f(\cdot) - 1$: vertical translation down 1 (y-coordinates reduced by 1)

Apply to $(4, -3)$:
After $f(2x)$: $(4 \div 2,\; -3) = (2,\; -3)$
After $-1$: $(2,\; -3 - 1) = (2,\; -4)$

Part (b): Analyse $y = 3f(x + 2)$

Identify the two transformations:

$f(x + 2)$: horizontal translation LEFT 2 (x-coordinates decrease by 2)
$3f(\cdot)$: vertical stretch by scale factor 3 (y-coordinates multiplied by 3)

Apply to $(0, 5)$:
After $f(x+2)$: $(0 - 2,\; 5) = (-2,\; 5)$
After $3f(\cdot)$: $(-2,\; 5 \times 3) = (-2,\; 15)$

Part (c): Describe both transformations fully

To earn full marks, state: type + direction + amount for each transformation.

A translation by the vector $\begin{pmatrix}-2\\0\end{pmatrix}$ (2 units to the left, horizontal)
A vertical stretch by scale factor 3 parallel to the y-axis (centred on the x-axis)

Note: both transformations must be stated; stating only one earns partial marks.

(a) Minimum at $\boldsymbol{(2,\; -4)}$.
(b) Image of $(0, 5)$ is $\boldsymbol{(-2,\; 15)}$.
(c) Translation by $\begin{pmatrix}-2\\0\end{pmatrix}$, then vertical stretch by scale factor 3 about the x-axis.

❓ Exam Questions

Q11 mark

The line $L$ has equation $y = 5x - 3$. Write down the gradient of a line perpendicular to $L$.

Answer: $-\dfrac{1}{5}$
Mark scheme: B1 for $-\frac{1}{5}$. Accept $-0.2$. Do not accept $\frac{1}{5}$ (must be negative reciprocal, not just reciprocal).

Q23 marks

Find the equation of the line perpendicular to $y = 2x + 1$ that passes through the point $(4,\; 3)$. Give your answer in the form $y = mx + c$.

Step 1: Gradient of $y = 2x + 1$ is $m_1 = 2$.
Perpendicular gradient: $m_2 = -\dfrac{1}{2}$ [M1]
Step 2: Equation through $(4, 3)$ with $m = -\frac{1}{2}$:
$y - 3 = -\dfrac{1}{2}(x - 4)$ [M1]
$y - 3 = -\dfrac{x}{2} + 2$
$y = -\dfrac{1}{2}x + 5$ [A1]
Answer: $\boldsymbol{y = -\frac{1}{2}x + 5}$
Verify: at $(4,3)$: $-\frac{1}{2}(4)+5 = -2+5 = 3$ ✓

Q33 marks

Describe fully the single transformation that maps the graph of $y = x^2$ onto the graph of $y = (x + 3)^2 - 4$.

Answer: A translation by the vector $\begin{pmatrix}-3\\-4\end{pmatrix}$.
Mark scheme:
B1: Identifies the transformation as a translation (not stretch, not reflection).
B1: Correct horizontal component: $-3$ (left 3). The $+3$ inside the function shifts left.
B1: Correct vertical component: $-4$ (down 4). The $-4$ outside the function shifts down.
Note: "Translation left 3 and down 4" is acceptable. Vector notation is preferred but not required.

Q44 marks

A car's velocity increases uniformly from $0\ \text{m/s}$ to $20\ \text{m/s}$ over the first 5 seconds, then remains constant for 7 seconds, then decreases uniformly to $0\ \text{m/s}$ over the next 4 seconds.
(a) Find the acceleration during the first 5 seconds. [1]
(b) Find the total distance travelled. [3]

(a) Acceleration $=$ gradient $= \dfrac{20 - 0}{5 - 0} = 4\ \text{m/s}^2$ [B1]
(b) Split the v-t graph into three regions:
Triangle (0 to 5s): $\frac{1}{2} \times 5 \times 20 = 50\ \text{m}$ [M1]
Rectangle (5 to 12s): $7 \times 20 = 140\ \text{m}$ [M1]
Triangle (12 to 16s): $\frac{1}{2} \times 4 \times 20 = 40\ \text{m}$
Total: $50 + 140 + 40 = \boldsymbol{230\ \text{m}}$ [A1]

Q54 marks

The graph of $y = f(x)$ has a maximum point at $(2,\; 5)$ and crosses the x-axis at $(-1,\; 0)$ and $(5,\; 0)$. Write down the coordinates of the corresponding point or intercept on each transformed graph:
(a) $y = f(x) - 3$: maximum point (b) $y = f(x - 4)$: maximum point
(c) $y = 2f(x)$: maximum point (d) $y = f(3x)$: both x-intercepts

(a) $f(x) - 3$: translate down 3 → max at $(2,\; 5-3) = \boldsymbol{(2,\; 2)}$ [B1]
(b) $f(x-4)$: translate right 4 → max at $(2+4,\; 5) = \boldsymbol{(6,\; 5)}$ [B1]
(c) $2f(x)$: vertical stretch ×2 → max at $(2,\; 2 \times 5) = \boldsymbol{(2,\; 10)}$ [B1]
(d) $f(3x)$: horizontal stretch ×$\frac{1}{3}$ → x-intercepts at $(-1 \div 3,\; 0)$ and $(5 \div 3,\; 0)$:
i.e. $\boldsymbol{\left(-\tfrac{1}{3},\; 0\right)}$ and $\boldsymbol{\left(\tfrac{5}{3},\; 0\right)}$ [B1 for both]

Q66 marks

Consider $y = x^3 - 3x^2 - 4x$.
(a) Show that $x = 4$ is a root of $x^3 - 3x^2 - 4x = 0$. [1]
(b) Hence fully factorise $x^3 - 3x^2 - 4x$. [3]
(c) State the coordinates of all three x-intercepts and the y-intercept of the graph. [2]

(a) Substitute $x = 4$:
$4^3 - 3(4)^2 - 4(4) = 64 - 48 - 16 = 0$ ✓ [B1]
(b) Factor $x$ from every term: $x(x^2 - 3x - 4)$ [M1]
Factorise the quadratic (need two numbers: product $-4$, sum $-3$, i.e. $-4$ and $+1$):
$x(x-4)(x+1)$ [M1 for method, A1 for correct answer]
(c) Set $y = 0$: $x = 0,\; x = 4,\; x = -1$
x-intercepts: $\boldsymbol{(0,0)},\; \boldsymbol{(4,0)},\; \boldsymbol{(-1,0)}$
y-intercept: set $x = 0 \Rightarrow y = 0$, so $\boldsymbol{(0, 0)}$ [B2: 1 mark for any two, 2 marks for all four]

⭐ Grade 9 Model Answers

Full annotated solution to Q6 (the hardest question above), demonstrating the techniques and communication that earn every mark at Grade 9.

Grade 9 · Cubic Factorisation · 6 marks

Show that $x = 4$ is a root of $x^3 - 3x^2 - 4x = 0$, hence fully factorise the expression, and state all intercepts.

Verify the root by substitution [B1]

Substitute $x = 4$ directly into the expression: $$4^3 - 3(4)^2 - 4(4) = 64 - 3(16) - 16 = 64 - 48 - 16 = 0 \checkmark$$ Grade 9 communication: Do not just state "it is a root" — you must show the arithmetic and conclude with $= 0$.

Extract the common factor of $x$ [M1]

Before using the known root $(x - 4)$, notice every term has a factor of $x$: $$x^3 - 3x^2 - 4x = x(x^2 - 3x - 4)$$ Grade 9 insight: Spotting this first simplifies the problem. Many students miss this and try to divide by $(x-4)$ immediately, which is harder.

Factorise the quadratic $x^2 - 3x - 4$ [M1 + A1]

Find two integers that multiply to $-4$ and add to $-3$: these are $-4$ and $+1$. $$x^2 - 3x - 4 = (x - 4)(x + 1)$$ Therefore the complete factorisation is: $$x^3 - 3x^2 - 4x = x(x-4)(x+1)$$ Grade 9 habit: Expand to verify: $x(x-4)(x+1) = x(x^2 - 3x - 4) = x^3 - 3x^2 - 4x$ ✓

State all intercepts [B2]

Setting $y = 0$: each factor gives a root: $x = 0$, $x = 4$, $x = -1$.
x-intercepts: $(0,\; 0)$, $(4,\; 0)$, $(-1,\; 0)$.
y-intercept: $x = 0 \Rightarrow y = 0$, so the origin $(0, 0)$ is both the y-intercept and an x-intercept.
Grade 9 communication: Write coordinates, not just $x$-values.

Full factorisation: $\boldsymbol{x(x-4)(x+1)}$
x-intercepts: $(0,\; 0),\; (4,\; 0),\; (-1,\; 0)$. y-intercept: $(0,\; 0)$.

🎯

Why Each Step Earns Marks

B1 (verification): Substitution shown with full arithmetic and "= 0 ✓" conclusion. Just stating "a root" without working scores 0.
M1 (common factor): Method mark for $x(\ldots)$ extraction — even if the quadratic inside is wrong, this mark can be earned.
M1 (quadratic method): Attempting to factorise $x^2 - 3x - 4$ into two brackets earns the method mark.
A1 (correct factorisation): The final complete expression $x(x-4)(x+1)$ earns this accuracy mark.
B2 (intercepts): One mark for any two correct intercepts as coordinates; both marks for all three/four.

📋 Revision Sheet

Key Definitions

Gradient	Rise over run; rate of change of $y$ with $x$
y-intercept	Point where graph crosses y-axis ($x = 0$)
Root	Value of $x$ where $y = 0$ (x-intercept)
Turning point	Local maximum or minimum of a curve
Asymptote	Line a graph approaches but never reaches
Parabola	U-shaped curve of a quadratic function
Perpendicular	Lines meeting at 90°; $m_1 m_2 = -1$
Acceleration	Rate of change of velocity (gradient on v-t)

Essential Formulae

$m = \dfrac{y_2-y_1}{x_2-x_1}$ (gradient)
$y - y_1 = m(x - x_1)$ (point-gradient)
$m_\perp = -\dfrac{1}{m}$ (perpendicular)
$x_{\text{vertex}} = -\dfrac{b}{2a}$ (line of symmetry)
$f(x)+a$: up $a$; $\;\;f(x+a)$: left $a$
$af(x)$: vert ×$a$; $\;\;f(ax)$: horiz ÷$a$
Acceleration $=$ gradient (v-t graph)
Distance $=$ area under v-t graph

Memory Hooks

"Flip and negate" — perpendicular gradient
"Inside = opposite" — horizontal transformations go in the opposite direction
"Outside = as stated" — vertical changes match the sign
Smile ($a>0$), frown ($a<0$) — quadratic orientation
S-shape — cubic with positive leading coefficient
Two branches, no touching — reciprocal graph rule
v-t: grad = acc, area = dist — velocity-time graph

Exam Tips

Rearrange to $y=mx+c$ before reading $m$ and $c$
Check perpendicular: $m_1 \times m_2$ must equal $-1$
Always label roots, y-intercept, turning point on sketches
Describe transformations with type + direction + size
Split v-t area into triangles and rectangles
Include correct units in motion graph answers
Verify line equations by substituting the second point
State "estimate" when gradient/area from a curved graph

🔄 Flashcards

Click each card to flip and reveal the answer. Test yourself before checking!

✗ Common Mistakes

✗

MISTAKE 1 — Perpendicular Gradient: Only Negating

What students do: Given gradient $m = 4$, write the perpendicular gradient as $-4$. They negate but forget to reciprocate.

Why marks are lost: The correct answer is $-\frac{1}{4}$. Writing $-4$ scores 0 on a "find perpendicular gradient" question.

How to avoid: Always do BOTH steps — flip the fraction AND change the sign. Verify: $4 \times (-\frac{1}{4}) = -1$ ✓. If the product is not $-1$, your gradient is wrong.

✗

MISTAKE 2 — Horizontal Translation Direction

What students do: Describe $y = f(x + 3)$ as "a translation 3 units to the right."

Why marks are lost: $f(x + 3)$ shifts the graph 3 units to the LEFT. The description "right" is the wrong direction and loses the mark for direction.

How to avoid: Memorise "inside = opposite." Test it: $y = (x+1)^2$ has its vertex where $x + 1 = 0$, i.e. $x = -1$. The vertex moved LEFT to $x = -1$, confirming the shift is left.

✗

MISTAKE 3 — Gradient vs Area on Velocity-Time Graphs

What students do: Calculate the gradient when asked for distance, or calculate the area when asked for acceleration.

Why marks are lost: On a v-t graph, gradient = acceleration and area = distance. Swapping these gives the completely wrong physical quantity and costs all the marks for that part.

How to avoid: Write a quick mental checklist every time you see a v-t graph: "Distance? → Area. Acceleration? → Gradient." Also check units: acceleration is m/s², distance is m.

✗

MISTAKE 4 — Turning Point Sign Error from Completed Square Form

What students do: For $y = (x - 3)^2 + 2$, write the turning point as $(-3,\; 2)$.

Why marks are lost: The turning point is $(3,\; 2)$. The vertex occurs where $x - 3 = 0$, so $x = +3$. Students apply the wrong sign to the $x$-coordinate.

How to avoid: In completed square form $y = (x - p)^2 + q$, solve $x - p = 0$ to get $x = p$ (NOT $-p$). The x-coordinate of the vertex is the value that makes the bracket zero.

✗

MISTAKE 5 — Reciprocal Graph Touching the Axes

What students do: Draw the branches of $y = 1/x$ so they touch or cross the x-axis or y-axis at the edges of the graph.

Why marks are lost: The axes are asymptotes: the graph gets closer and closer but never touches. Allowing the curve to touch an axis is incorrect and loses the accuracy mark for the sketch.

How to avoid: Plot explicit points: at $x = 0.1$, $y = 10$; at $x = 10$, $y = 0.1$. This shows the curve curving sharply towards (but away from) each axis. The curve should always be "pulling away" from both axes.

✗

MISTAKE 6 — Incomplete Transformation Descriptions

What students do: Write "a translation" or "a stretch" without specifying the direction, magnitude, or axis.

Why marks are lost: "Translation" alone scores 1 of 3 marks. A full description of a translation requires: (1) the word "translation," (2) the direction, and (3) the distance (or the vector). Missing any component loses marks.

How to avoid: Use a three-part checklist: (1) What type of transformation? (2) Which direction / which axis? (3) By how much / scale factor how large? Answer all three before moving on.

✅ Final Checklist

Click each item when you are confident with it. Your progress is saved in the browser.

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I can calculate the gradient between two points using $m = (y_2-y_1)/(x_2-x_1)$
I can identify gradient $m$ and y-intercept $c$ by rearranging to $y = mx + c$
I can write the equation of a line given a gradient and a point, using $y - y_1 = m(x-x_1)$
I can write the equation of a line through two given points
I can identify parallel lines (equal gradients) and write parallel line equations
I can find a perpendicular gradient by "flipping and negating," and verify it multiplies to $-1$
I can sketch a quadratic, labelling roots (by factorising), y-intercept, and turning point
I can sketch a cubic graph, identifying end behaviour and x-intercepts
I can sketch $y = k/x$, showing two branches that never touch the axes
I can apply and describe $f(x)+a$ (vertical shift) and $f(x+a)$ (horizontal, opposite direction)
I can apply and describe $af(x)$ (vertical stretch) and $f(ax)$ (horizontal stretch)
I can interpret a distance-time graph: gradient = speed, horizontal = stationary
I can calculate acceleration as the gradient of a velocity-time graph
I can calculate distance as the area under a velocity-time graph, splitting into triangles and rectangles