Graphs & transformations — notes, question types & practice · Functions, Graphs & Logarithms

01

A graph is a picture of the rule

Every point on the curve $y = f(x)$ says: at input $x$ , the output is $f(x)$ .

For $f(x) = 2x + 1$ the points $(0, 1)$ , $(1, 3)$ , $(2, 5)$ line up into a straight line.

Graph questions are scoring. The paper shows a shifted parabola or a V shape and asks where it peaks, where it cuts an axis, or how many times it crosses a line.

02

Sliding up and down

Adding outside the bracket lifts or lowers the whole picture.

$y = f(x) + 3$ moves every point up by $3$ . $y = f(x) - 2$ drops it by $2$ .

The parabola $y = x^2$ has its lowest point at $(0, 0)$ . So $y = x^2 + 4$ has its lowest point at $(0, 4)$ .

Rule: Changes made outside the rule (added after $f$ ) move the picture up or down, exactly as written.

03

Sliding left and right

Changes inside the bracket work in reverse.

$y = f(x - 2)$ moves the picture $2$ units right. $y = f(x + 3)$ moves it $3$ units left.

Why backwards? At $x = 2$ , the rule $f(x-2)$ computes $f(0)$ — the old picture starts where the new zero sits.

The parabola $y = (x-2)^2$ has its vertex at $x = 2$ : the old vertex at $0$ has walked to $2$ .

Watch: Inside changes are opposite. $(x - a)$ shifts right, $(x + a)$ shifts left. Under pressure, this is the one students flip.

04

Flips and mirrors

$y = -f(x)$ : outputs multiplied by $-1$ . The picture flips over the $x$ -axis — a water image. Point $(3, -2)$ becomes $(3, 2)$ .
$y = f(-x)$ : inputs replaced by their negatives. The picture flips over the $y$ -axis — a mirror image. Point $(2, 5)$ becomes $(-2, 5)$ .

The point itself tells you the answer: flip the sign of the $y$ for the first, flip the sign of the $x$ for the second.

05

The modulus folds the graph up

$y = |f(x)|$ keeps everything on or above the axis. Outputs below zero are reflected up; the rest stays put.

$y = |x - 3|$ is a V with its corner at $(3, 0)$ and arms climbing at $45^\circ$ . It cuts the $y$ -axis where $|0 - 3| = 3$ , at $(0, 3)$ .

Only the below-axis part folds. The above-axis part never changes.

Tip: For $y = |x - a|$ , draw a V with corner at $(a, 0)$ . Read crossings straight off the picture.

06

The parabola and its vertex

For $f(x) = ax^2 + bx + c$ :

$a > 0$ : opens up, vertex is the minimum.
$a < 0$ : opens down, vertex is the maximum.
Vertex input: $x = -\dfrac{b}{2a}$ .

Worked: $f(x) = -2x^2 + 12x - 11$ . Here $a = -2 < 0$ , so a maximum exists. $x = -\dfrac{12}{-4} = 3$ . Then $f(3) = -18 + 36 - 11 = 7$ . Maximum value $7$ .

Where the graph meets the $x$ -axis, put $f(x) = 0$ . For $(x-1)^2 - 4 = 0$ : $(x-1)^2 = 4$ , so $x = 3$ or $x = -1$ .

Example: $f(x) = x^2 - 6x + 8$ has vertex at $x = 3$ and cuts the axis where $(x-2)(x-3) = 0$ , at $x = 2$ and $x = 3$ .

07

Counting solutions from a picture

"How many solutions?" means "how many times do the two pictures cross?"

$|2x - 5| = 3$ : the V with corner at $x = 2.5$ meets the flat line $y = 3$ twice. Answer: $2$ solutions.

A horizontal line above a V's corner cuts it twice; exactly at the corner, once; below the corner, never.

Rule: Rough-sketch the V or parabola with its corner and one point on each arm. Counting crossings beats algebra when only the number of solutions is asked.

08

Question types you will see

Each type: how to recognise it, the method step by step, and one question to try.

Type 1very common4 practice Q

Vertex: maximum or minimum value

How to spot it:

A quadratic is given and 'the maximum/minimum value of $f(x)$ ' or the input where it occurs is asked.

x = -\frac{b}{2a}

Method

Check the sign of $a$ for min or max.
Compute $x = -b/2a$ .
Substitute back for the value.
State both the input and the value if asked.

Why it works:

The parabola turns around at its vertex, so the vertex carries the peak or the valley.

Try this

Find the minimum value of $f(x) = x^2 - 8x + 17$ .

Show solution

$a = 1 > 0$ , so a minimum.
$x = 8/2 = 4$ .
$f(4) = 16 - 32 + 17 = 1$ .

Answer

1

Type 2common2 practice Q

Shift a graph

How to spot it:

' $y = f(x)$ is shifted...' or a new rule like $f(x - 3) + 1$ is given and the movement is asked.

y = f(x - a) + k

Method

Look for changes inside the bracket: they move opposite.
Look for changes outside: they move as written.
Combine both moves.
State right/left and up/down.

Why it works:

Inputs feed the rule from inside, so an inside change delays or advances the picture — the reverse of its sign.

Try this

How does $y = (x - 3)^2 + 1$ relate to $y = x^2$ ?

Show solution

Inside $-3$ : shift $3$ right.
Outside $+1$ : shift $1$ up.

Answer

3 right and 1 up

Type 3common2 practice Q

Graph of a modulus

How to spot it:

A rule with $|\;|$ , or 'the graph of $y = |x - a|$ ...' with an axis crossing asked.

|x - a| = \begin{cases} x - a, & x \ge a \\ a - x, & x < a \end{cases}

Method

Mark the corner at $(a, 0)$ .
Draw two rising arms at $45^\circ$ .
For $|f(x)|$ , fold the below-axis part up.
Read axis crossings from the sketch.

Why it works:

A modulus erases negative signs, so the picture is whatever remains above the axis.

Try this

Where does $y = |x - 3|$ cut the $y$ -axis?

Show solution

Put $x = 0$ .
$y = |0 - 3| = 3$ .

Answer

At (0, 3)

Type 4common2 practice Q

Count solutions from the picture

How to spot it:

'How many solutions does ... have?' — usually a modulus or parabola against a constant.

Method

Sketch the modulus V or the parabola roughly.
Draw the horizontal line on the other side.
Count the crossing points.
Answer with the count only.

Why it works:

Each crossing is one solution, so a rough sketch answers a counting question without solving.

Try this

How many solutions does $|2x - 5| = 3$ have?

Show solution

V with corner at $x = 2.5$ .
Line $y = 3$ is above the corner value $0$ .
Two crossings.

Answer

2

Type 5occasional2 practice Q

Reflect a graph or a point

How to spot it:

' $(a, b)$ lies on $y = f(x)$ . Which point lies on $y = -f(x)$ or $y = f(-x)$ ?'

y = -f(x): (a, b) \to (a, -b); \quad y = f(-x): (a, b) \to (-a, b)

Method

Identify which rule changed: outside minus flips outputs.
Inside minus flips inputs.
Change the sign of the matching coordinate.
Leave the other coordinate alone.

Why it works:

Multiplying outputs by $-1$ reflects over the $x$ -axis; negating inputs reflects over the $y$ -axis.

Try this

$(3, -2)$ lies on $y = f(x)$ . Which point lies on $y = -f(x)$ ?

Show solution

Outside minus: outputs flip.
$(3, -2) \to (3, 2)$ .

Answer

(3, 2)

Type 6common2 practice Q

Where a graph meets the axes

How to spot it:

'The graph of ... cuts the x-axis at' — find roots, or the y-axis crossing.

Method

For the $x$ -axis, put $y = 0$ and solve.
For the $y$ -axis, put $x = 0$ .
Factor the quadratic if needed.
List all crossing points.

Why it works:

Crossings are exactly the points where one coordinate is zero, so one substitution does it.

Try this

Where does $y = (x - 1)^2 - 4$ cut the $x$ -axis?

Show solution

$(x-1)^2 - 4 = 0$ .
$(x-1)^2 = 4$ , so $x - 1 = \pm 2$ .

Answer

At x = -1 and x = 3

09

Formula sheet

Vertical shift

y = f(x) + k

$k > 0$ lifts the picture, $k < 0$ lowers it.

Horizontal shift

y = f(x - a)

Moves right by $a$; $f(x+a)$ moves left.

Reflections

y = -f(x), \quad y = f(-x)

First flips over the $x$-axis, second over the $y$-axis.

Modulus graph

y = |f(x)|

The part below the axis is reflected up.

Vertex of a parabola

x = -\frac{b}{2a}, \quad \text{value} = f\left(-\frac{b}{2a}\right)

Minimum if $a > 0$, maximum if $a < 0$.

10

Shortcuts that save time

⚡ Vertex in one line

For $ax^2 + bx + c$ , the turning input is $-b/2a$ . Substitute it back for the peak or valley value. Check the sign of $a$ first.

Example

Find the maximum value of $f(x) = -2x^2 + 12x - 11$ .

Show solution

$a = -2 < 0$ , so a maximum exists.
$x = -12/(-4) = 3$ .
$f(3) = -18 + 36 - 11 = 7$ .

Answer

7

⚡ Inside moves opposite

For shifts, look at where the rule sits: inside the bracket means the move is opposite to the sign; outside means the move is as written.

Example

How does $y = (x - 3)^2$ relate to $y = x^2$ ?

Show solution

Inside change $x - 3$ .
Opposite direction: right.

Answer

Shifted 3 units right

⚡ Count crossings, not algebra

For 'how many solutions', sketch the V or parabola roughly and count crossings with the horizontal line. No solving needed.

Example

How many solutions does $|x + 2| = 5$ have?

Show solution

V with corner at $x = -2$ , opening up.
Line $y = 5$ sits above the corner.

Answer

2 solutions

11

Mistakes to avoid

Where most students lose marks on this subtopic.

Mistake 01

Reading $f(x + 3)$ as a shift right — inside changes work in reverse: $f(x+3)$ is $3$ left.

Mistake 02

Taking $y = f(x) + 2$ down — an outside $+k$ lifts the picture up.

Mistake 03

For $y = |f(x)|$ , mirroring the above-axis part too — only the part below the axis folds up.

Mistake 04

Calling the vertex a minimum for every quadratic — if $a < 0$ the parabola opens down and the vertex is a maximum.

Mistake 05

Reflecting the wrong coordinate: $y = -f(x)$ flips $y$ -values, $y = f(-x)$ flips $x$ -values.

Mistake 06

Answering '2 solutions' for $|x - a| = 0$ — at the corner the V meets the line once, not twice.

12

Quick revision

Read this the night before the exam.

$f(x) + k$ : up $k$ . $f(x - a)$ : right $a$ .
Inside changes work opposite to their sign.
$-f(x)$ : water image. $f(-x)$ : mirror image.
$|f(x)|$ : fold everything above the axis.
Vertex at $x = -b/2a$ ; $a > 0$ minimum, $a < 0$ maximum.
Axis crossings: put the other coordinate equal to $0$ .

13

Practice: 14 questions

Sets of 10, mixed across the question types above. Every answer has a step-by-step explanation.

Topic test · 10 questions

Suggested time 7 min · wrong answers go to your mistake notebook automatically.