What is the importance of graphing functions in Cambridge IGCSE Mathematics?

Graphing functions is crucial in Cambridge IGCSE Mathematics as it helps students visually understand the behavior and properties of different functions. By plotting graphs, students can easily identify key features such as intercepts, slopes, and asymptotes, which are essential for solving equations and understanding real-world applications.

How do you determine the domain and range from a graph of a function?

To determine the domain from a graph, look at the set of all possible x-values that the function can take, which is typically represented by the horizontal extent of the graph. The range is found by identifying the set of all possible y-values, shown by the vertical extent of the graph. Understanding these concepts is fundamental in Cambridge IGCSE Mathematics for analyzing functions.

What are the common types of graphs of functions studied in the Cambridge IGCSE curriculum?

In the Cambridge IGCSE curriculum, students commonly study linear, quadratic, cubic, and reciprocal function graphs. Each type has distinct characteristics: linear graphs are straight lines, quadratic graphs form parabolas, cubic graphs have an S-shape, and reciprocal graphs have hyperbolic shapes. Recognizing these types helps in solving various mathematical problems.

How can you identify asymptotes in the graph of a function?

Asymptotes are lines that a graph approaches but never touches. In the graph of a function, vertical asymptotes occur where the function is undefined, often due to division by zero. Horizontal asymptotes are determined by the behavior of the function as x approaches infinity. Recognizing asymptotes is a key skill in Cambridge IGCSE Mathematics for understanding the limits of functions.

What role do intercepts play in the graph of a function?

Intercepts are points where the graph of a function crosses the axes. The x-intercept is where the graph crosses the x-axis, indicating the value of x when y is zero. The y-intercept is where the graph crosses the y-axis, showing the value of y when x is zero. Identifying intercepts is essential in Cambridge IGCSE Mathematics for solving equations and analyzing function behavior.

Why is "Reading $f(x - 3)$ as a left-shift" a common mistake in Graphs of Functions?

Why it happens: The minus sign is counter-intuitive. How to avoid it: f(x - 3) shifts RIGHT by 3 (because input x now has to be 3 bigger to give the same output). Source: 0580 examiner reports — recurring

Why is "Drawing the curve crossing or touching the asymptote" a common mistake in Graphs of Functions?

Why it happens: Sketch is rushed. How to avoid it: An asymptote is a line the curve APPROACHES but never reaches. Leave a clear gap.

Why is "Drawing $y = \frac{k}{x}$ branches in the wrong quadrants" a common mistake in Graphs of Functions?

Why it happens: Sign of k determines which quadrant pair the branches occupy. How to avoid it: k > 0 → 1st and 3rd quadrants. k < 0 → 2nd and 4th.

Why is "Forgetting that $a^{0} = 1$" a common mistake in Graphs of Functions?

Why it happens: Students compute a^0 as 0. How to avoid it: Any non-zero base raised to the 0 power is 1. So y = a^x always passes through (0, 1).

Cambridge IGCSE Mathematics (0580)

Graphs of Functions

Q: Why is "Forgetting that $a^{0} = 1$" a common mistake in Graphs of Functions?

Why it happens: Students compute a^0 as 0. How to avoid it: Any non-zero base raised to the 0 power is 1. So y = a^x always passes through (0, 1).

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Short Notes - Graphs of Functions

Detailed Study Notes

Detailed notes on Functions for Cambridge IGCSE Mathematics, covering key concepts, explanations, examples, and exam-focused revision points.

Graphs of Functions — Cambridge IGCSE 0580 Maths Extended (2026)

Recognise the shape of common functions on sight: linear, quadratic, cubic, reciprocal, exponential, trig. Sketch them, identify intercepts, asymptotes and turning points — Paper 4 grading depends on getting all anchor features right.

At a glance

Linear $y = mx + c$ : straight line.
Quadratic $y = ax^{2} + bx + c$ : parabola.
Cubic $y = ax^{3}$ : S-shape through origin.
Reciprocal $y = \dfrac{a}{x}$ : two-branch hyperbola, asymptotes at $x = 0$ and $y = 0$ .
Exponential $y = a^{x}$ : passes through $(0, 1)$ , asymptote $y = 0$ .
$y = \sin x$ , $y = \cos x$ : periodic, range $[-1, 1]$ .
$y = \tan x$ : periodic, asymptotes at $90\degree$ , $270\degree$ , etc.
Use intercepts and asymptotes as anchor points.

What you’ll learn

Mapped to the Cambridge IGCSE 0580 syllabus (2025-2027).

E2.10 — Construct tables of values and draw graphs for functions of the form $y = ax^{n}$ and exponentials $y = ab^{x}$ and combinations.
Sketch and recognise shapes of standard functions.

Standard shapes — recognise on sight

Memorise these eight families. Cambridge expects instant recognition.

Linear ( $y = mx + c$ ).

Straight line. Gradient $m$ , $y$ -intercept $c$ .
Domain and range: all reals.

Quadratic ( $y = ax^{2} + bx + c$ ).

Parabola. Opens up if $a > 0$ , down if $a < 0$ .
Vertex on axis of symmetry $x = -\dfrac{b}{2a}$ .

Cubic ( $y = ax^{3}$ basic form).

S-shape through the origin. Sweeps from $-\infty$ to $+\infty$ in the $y$ -direction.
$y = ax^{3} + bx^{2} + cx + d$ is the general cubic — has one or two turning points.

Reciprocal ( $y = \dfrac{a}{x}$ ).

Two-branch hyperbola.
Asymptotes: $x = 0$ (vertical) and $y = 0$ (horizontal).
$a > 0$ : branches in 1st and 3rd quadrants. $a < 0$ : 2nd and 4th.

Exponential ( $y = a^{x}$ with $a > 0, a \ne 1$ ).

Passes through $(0, 1)$ .
$a > 1$ : rises rapidly to the right, decays to $0$ on the left.
$0 < a < 1$ : mirror-image (decays to right, rises to left).
Horizontal asymptote: $y = 0$ .

Sine and cosine ( $y = \sin x$ , $y = \cos x$ ).

Period $360\degree$ . Range $-1 \le y \le 1$ .
$\sin$ passes through origin. $\cos$ starts at $(0, 1)$ .

Tangent ( $y = \tan x$ ).

Period $180\degree$ . Range: all reals.
Vertical asymptotes at $x = 90\degree, 270\degree, \ldots$ .

Inverse function ( $y = f^{-1}(x)$ ).

Reflection of $f$ in the line $y = x$ .

The reciprocal curve never crosses the axes — they are its asymptotes.

Exponential passes through (0, 1); sine waves between −1 and 1; tangent repeats with vertical asymptotes.

Linear → straight line.
Quadratic → parabola.
Cubic → S-shape.
Reciprocal → two-branch hyperbola, asymptotes at axes.
Exponential → through $(0, 1)$ , asymptote $y = 0$ .
Sin/cos → wave, range $[-1, 1]$ .
Tan → asymptotes every $180\degree$ .

Sketching: find anchor points first

$y$ -intercept, $x$ -intercepts, asymptotes, turning points. Connect with a smooth curve.

The five anchors.

$y$ -intercept (substitute $x = 0$ ).
$x$ -intercepts / roots (solve $y = 0$ ).
Turning points (set $\dfrac{dy}{dx} = 0$ if you've covered differentiation; otherwise use formula or symmetry).
Asymptotes (vertical and horizontal).
Behaviour as $x \to \pm \infty$ .

Plot the anchor points, mark the asymptotes (dashed), then draw a smooth curve through everything.

Worked. Sketch $y = \dfrac{2}{x - 3}$ .

$y$ -intercept: $y = \dfrac{2}{-3} = -\dfrac{2}{3}$ .
$x$ -intercept: $\dfrac{2}{x - 3} = 0$ has no solution (numerator never zero) → no $x$ -intercept.
Vertical asymptote: $x = 3$ (denominator zero there).
Horizontal asymptote: $y \to 0$ as $x \to \pm\infty$ .
Two branches: one in upper-right (when $x > 3$ , $y > 0$ ), one in lower-left.

Tip. Always test ONE point on each branch to confirm direction. Plug in $x = 4$ : $y = 2$ . Plug in $x = 2$ : $y = -2$ . Confirms branches.

The curve hugs the asymptotes x = 3 and y = 0 without ever touching them.

$y$ -intercept first.
Then $x$ -intercepts.
Then asymptotes (vertical from undefined points; horizontal from end behaviour).
Then turning points (from differentiation or symmetry).
Smooth curve through anchors, respecting asymptotes.

Transformations of graphs

Translations, stretches, reflections — recognise the effect of each modification to the formula.

Vertical translation. $y = f(x) + a$ shifts UP by $a$ (DOWN if $a < 0$ ).

Horizontal translation. $y = f(x - a)$ shifts RIGHT by $a$ (LEFT if $a < 0$ ). Counter-intuitive: the sign FLIPS.

Vertical stretch. $y = af(x)$ stretches by factor $a$ in the $y$ -direction. $a < 0$ : also reflects in the $x$ -axis.

Horizontal stretch. $y = f(ax)$ stretches by factor $\dfrac{1}{a}$ in the $x$ -direction (compresses if $a > 1$ ).

Reflection.

$y = -f(x)$ : reflect in the $x$ -axis.
$y = f(-x)$ : reflect in the $y$ -axis.

Worked. Sketch $y = (x - 2)^{2} + 3$ given the basic $y = x^{2}$ .

$-2$ inside: shift RIGHT by 2.
$+3$ outside: shift UP by 3.
New vertex: $(2, 3)$ .

Inside-bracket changes move the curve along x; outside-bracket changes move it along y.

Combined transformations. Apply in order. Inside-the-bracket changes affect $x$ , outside-the-bracket changes affect $y$ .

Inside-bracket changes move x (opposite sign); outside changes move or stretch y.

Inside the function ( $x \to x + c$ ): horizontal shift, opposite direction.
Outside the function ( $f(x) + c$ ): vertical shift, same direction.
$af(x)$ : vertical stretch.
$f(ax)$ : horizontal stretch (compress for $a > 1$ ).
Negatives: reflection in respective axis.

Quick recap

Memorise the eight standard shapes.
Anchor points: $y$ -intercept, $x$ -intercepts, asymptotes, turning points.
Reciprocal $y = a/x$ : two branches with asymptotes at the axes.
Exponential $y = a^{x}$ passes through $(0, 1)$ .
Translations: $f(x + h)$ shifts opposite direction; $f(x) + k$ shifts same direction.
Reflections: $-f(x)$ in $x$ -axis; $f(-x)$ in $y$ -axis.

Memorise this

Verbatim phrases and definitions Cambridge mark schemes credit.

Linear function — $y = mx + c$ , straight line.
Quadratic function — $y = ax^{2} + bx + c$ , parabola.
Reciprocal function — $y = a/x$ , two-branch hyperbola with asymptotes at $x = 0$ and $y = 0$ .
Exponential function — $y = a^{x}$ with $a > 0, a \ne 1$ .
Asymptote — a line the graph approaches but never touches.

How it’s examined

Graph sketching appears every Paper 4 as a 4-6 mark item — usually a quadratic, cubic or reciprocal. Cambridge wants ALL anchor points labelled. Paper 2 has simpler 2-3 mark identify-the-shape questions. Examiner reports flag missing asymptotes on reciprocal sketches and unlabelled axis-intercepts.

Sources: Cambridge IGCSE Mathematics 0580 syllabus 2025-2027 (E2.10); 0580/42 Oct/Nov 2024 — Q11 (sketch reciprocal); 0580 Examiner Reports 2022-2024. Last reviewed 2026-05-05.

Take this whole topic with you

Step-by-step worked examples — Graphs of Functions

Step-by-step solutions to past-paper-style questions on graphs of functions, written exactly the way a tutor would explain them at the board.

Question type:

Question patterns to master — Graphs of Functions

Almost every graphs of functions exam question is one of these shapes. Learn to spot each one and you will always know how to start.

Graph or diagram

Recognise it by

Instructions to sketch, complete the table of values, use the graph to solve / estimate, draw a tangent, or describe the transformation of a curve.

How to approach it

For sketches, mark intercepts and asymptotes and use the function family's known shape. To solve graphically, draw the relevant horizontal line and read off intersection $x$ -values. For transformations, match to the standard form.

Common trap

Drawing a reciprocal curve that touches its asymptote, or shifting a translation the wrong way. Examiner reports flag $f(x-3)$ being read as a left-shift — it moves right.

Direct calculation

Recognise it by

A single algebraic instruction — find the $y$ -intercept, find the equation of the tangent — answered without reading a diagram.

How to approach it

Substitute $x=0$ for a $y$ -intercept; for a tangent, find the point, compute the gradient (differentiate and evaluate), then use $y-y_1=m(x-x_1)$ .

Common trap

Computing $a^0$ as $0$ instead of $1$ for exponential intercepts, or panicking at a zero gradient — a tangent at a turning point is simply horizontal.

Multi-step problem

Recognise it by

Find the coordinates where two curves intersect — solved algebraically by combining two equations.

How to approach it

Set the two expressions equal, clear any fractions, rearrange to a single quadratic, solve it, then substitute each $x$ back to find its paired $y$ .

Common trap

Stopping after finding the $x$ -values and not pairing them with their $y$ -values, or losing a solution by dividing through by $x$ .

1Sketch a cubic with given roots
ExtendedGraph or diagram• cubic
Question
Sketch $y = (x + 2)(x - 1)(x - 3)$ , showing the intercepts.
Step-by-step solution
1. Step 1
  x-intercepts: $x = -2, 1, 3$ .
2. Step 2
  y-intercept: $y = (2)(-1)(-3) = 6$ at $(0, 6)$ .
3. Step 3
  Coefficient of $x^{3}$ is positive, so the curve goes from bottom-left to top-right (rises through three roots).
Answer
Cubic crossing x-axis at $(-2, 0), (1, 0), (3, 0)$ and y-axis at $(0, 6)$ , going from bottom-left to top-right.
2Sketch a reciprocal graph
ExtendedGraph or diagram• reciprocal, asymptote
Question
Sketch $y = \dfrac{6}{x}$ and state any asymptotes.
Step-by-step solution
1. Step 1
  Two branches in the first and third quadrants (because $6 > 0$ ).
2. Step 2
  As $x \to 0$ , $y \to \pm \infty$ — vertical asymptote at $x = 0$ .
3. Step 3
  As $x \to \pm \infty$ , $y \to 0$ — horizontal asymptote at $y = 0$ .
Answer
Hyperbola; asymptotes at $x = 0$ and $y = 0$ .
3Sketch an exponential
ExtendedGraph or diagram• Adapted from 0580/42 Oct/Nov 2023 Q11• exponential
Question
Sketch $y = 2^{x}$ and state the y-intercept and asymptote.
Step-by-step solution
1. Step 1
  y-intercept: $y = 2^{0} = 1$ at $(0, 1)$ .
2. Step 2
  As $x \to -\infty$ , $y \to 0^{+}$ — horizontal asymptote at $y = 0$ .
3. Step 3
  As $x \to +\infty$ , $y \to \infty$ — curve rises rapidly.
Answer
Exponential growth curve through $(0, 1)$ , asymptote $y = 0$ as $x \to -\infty$ .
4Apply a vertical translation
ExtendedGraph or diagram• transformations
Question
Describe the transformation from $y = x^{2}$ to $y = x^{2} - 4$ .
Step-by-step solution
1. Step 1
  Subtracting $4$ from $y$ moves the graph DOWN by $4$ units.
Answer
Translation by vector $\begin{pmatrix} 0 \\ -4 \end{pmatrix}$ — i.e. $4$ units downward.
5Plot a cubic from a table of values
CoreGraph or diagram• table of values, cubic
Question
Complete the table of values for $y = x^{3} - 6x$ for $x = -3, -2, -1, 0, 1, 2, 3$ .
Step-by-step solution
1. Step 1
  Substitute each $x$ into $y = x^{3} - 6x$ .
  $x = -3: y = -27 + 18 = -9$
2. Step 2
  Continue.
  $x = -2: y = -8 + 12 = 4;\ x = -1: y = -1 + 6 = 5$
3. Step 3
  Centre and positive side.
  $x = 0: y = 0;\ x = 1: y = -5;\ x = 2: y = -4;\ x = 3: y = 9$
Answer
$y$ -values: $-9,\ 4,\ 5,\ 0,\ -5,\ -4,\ 9$
Examiner tip
The examiner report flags candidates often slip on signs of $x^{3}$ when $x$ is negative. Use brackets: $(-2)^{3} = -8$ , not $8$ .
6Recognise the y-intercept of a curve
CoreDirect calculation• y-intercept
Question
Find the $y$ -intercept of $y = 3^{x} - 4$ and state the equation of its horizontal asymptote.
Step-by-step solution
1. Step 1
  $y$ -intercept: substitute $x = 0$ .
  $y = 3^{0} - 4 = 1 - 4 = -3$
2. Step 2
  As $x \to -\infty$ , $3^{x} \to 0$ , so $y \to -4$ .
Answer
$y$ -intercept at $(0, -3)$ ; horizontal asymptote $y = -4$ .
7Sketch $y = -8/x$
ExtendedGraph or diagram• reciprocal, negative coefficient
Question
Sketch $y = -\dfrac{8}{x}$ , stating the quadrants the branches occupy and the asymptotes.
Step-by-step solution
1. Step 1
  $k = -8 < 0$ , so branches are in the 2nd and 4th quadrants.
2. Step 2
  Sample point: $x = 2 \Rightarrow y = -4$ , confirming the lower-right branch.
3. Step 3
  Asymptotes are the axes (the curve never touches $x = 0$ or $y = 0$ ).
Answer
Hyperbola in 2nd and 4th quadrants; asymptotes $x = 0$ and $y = 0$ .
8Solve $f(x) = g(x)$ graphically
ExtendedGraph or diagram• Adapted from 0580/42 May/Jun 2023 Q5• intersection, simultaneous
Question
Use the graph of $y = x^{2} - 2$ and $y = x$ to estimate the solutions of $x^{2} - 2 = x$ . Then check algebraically.
Step-by-step solution
1. Step 1
  Read the intersection $x$ -coordinates from the graph: approximately $x \approx -1$ and $x \approx 2$ .
2. Step 2
  Rearrange to confirm.
  $x^{2} - 2 = x \implies x^{2} - x - 2 = 0$
3. Step 3
  Factor.
  $(x - 2)(x + 1) = 0 \implies x = 2 \text{ or } x = -1$
Answer
$x = -1$ and $x = 2$
Examiner tip
The mark scheme awards method marks for showing the rearrangement to $f(x) - g(x) = 0$ explicitly; pure graph-reading without algebra often loses the accuracy mark.
9Use a graph to solve a cubic
ExtendedGraph or diagram• Adapted from 0580/42 Oct/Nov 2022 Q10• cubic, graphical solution
Question
The graph of $y = x^{3} - 3x$ is drawn for $-2.5 \leq x \leq 2.5$ . Use it to solve $x^{3} - 3x = 1$ .
Step-by-step solution
1. Step 1
  Draw the horizontal line $y = 1$ on the same axes.
2. Step 2
  The solutions of $x^{3} - 3x = 1$ are the $x$ -coordinates of the points where the cubic meets the line $y = 1$ .
3. Step 3
  Reading the graph carefully: $x \approx -1.5$ , $x \approx -0.3$ and $x \approx 1.9$ (accept $\pm 0.1$ ).
Answer
$x \approx -1.5,\ -0.3,\ 1.9$
Examiner tip
The examiner report flags candidates who draw the wrong horizontal line, e.g. $y = -1$ instead of $y = 1$ . Always label the constant clearly on the diagram.
10Estimate the gradient at a point on a curve
ExtendedGraph or diagram• gradient, tangent
Question
On the graph of $y = x^{2} + 1$ , draw a tangent at $(2, 5)$ and use it to estimate the gradient at that point.
Step-by-step solution
1. Step 1
  Draw a straight line just touching the curve at $(2, 5)$ — the tangent.
2. Step 2
  Pick two clear points on the tangent (e.g. $(0, -3)$ and $(2, 5)$ ) and use rise / run.
  $\text{gradient} \approx \dfrac{5 - (-3)}{2 - 0} = 4$
Answer
Gradient $\approx 4$
Examiner tip
The mark scheme awards method marks when the tangent is clearly drawn and the two coordinates used are visibly on the tangent — not on the curve.
11Intersection of two non-linear graphs
ChallengeMulti-step problem• intersection, non-linear
Question
Find the coordinates of the points where $y = \dfrac{6}{x}$ and $y = x + 1$ intersect.
Step-by-step solution
1. Step 1
  Set the right-hand sides equal.
  $\dfrac{6}{x} = x + 1$
2. Step 2
  Multiply both sides by $x$ (note $x \neq 0$ ).
  $6 = x^{2} + x \implies x^{2} + x - 6 = 0$
3. Step 3
  Factor.
  $(x + 3)(x - 2) = 0 \implies x = -3 \text{ or } x = 2$
4. Step 4
  Find $y$ for each.
  $x = -3: y = -2;\ x = 2: y = 3$
Answer
$(-3, -2)$ and $(2, 3)$
12Reflections: $y = -f(x)$ versus $y = f(-x)$
ChallengeGraph or diagram• Adapted from 0580/42 May/Jun 2024 Q12• transformations, reflection
Question
The graph of $y = f(x)$ has a maximum at $(3, 5)$ and passes through $(0, 2)$ . State the new maximum and $y$ -intercept of (a) $y = -f(x)$ , (b) $y = f(-x)$ .
Step-by-step solution
1. Step 1
  (a) $y = -f(x)$ reflects in the $x$ -axis: each $y$ -value is negated. The maximum becomes a minimum.
  $\text{Min at } (3, -5);\ \text{passes through } (0, -2)$
2. Step 2
  (b) $y = f(-x)$ reflects in the $y$ -axis: each $x$ -value is negated. Maxima remain maxima.
  $\text{Max at } (-3, 5);\ \text{passes through } (0, 2)$
Answer
(a) Minimum $(3, -5)$ , $y$ -intercept $(0, -2)$ (b) Maximum $(-3, 5)$ , $y$ -intercept $(0, 2)$
Examiner tip
The examiner report flags candidates often swap the two rules. Memorise: minus OUTSIDE flips top/bottom; minus INSIDE flips left/right.
13Find the equation of a tangent
ChallengeDirect calculation• tangent, differentiation
Question
Find the equation of the tangent to $y = x^{2} - 2x + 3$ at the point $x = 1$ .
Step-by-step solution
1. Step 1
  Find $y$ at $x = 1$ .
  $y = 1 - 2 + 3 = 2.\ \text{Point: } (1, 2)$
2. Step 2
  Differentiate and evaluate the gradient at $x = 1$ .
  $\dfrac{dy}{dx} = 2x - 2 \implies m = 0$
3. Step 3
  Gradient $0$ means the tangent is horizontal through $(1, 2)$ .
Answer
$y = 2$
Examiner tip
Tangents to a turning point are horizontal. The examiner report flags candidates who panic at gradient $0$ and try to write something like $x = 1$ instead.

Key Formulae — Graphs of Functions

The formulae you need to memorise for graphs of functions on the Cambridge IGCSE 0580 paper, with every variable defined in plain English and a note on when to use it.

Function translations
$y = f(x) + a \to \text{up by } a;\ \ y = f(x - a) \to \text{right by } a$
When to use
Recognising shifts of standard graphs.
Function stretches
$y = kf(x) \to \text{vertical stretch by } k;\ \ y = f(kx) \to \text{horizontal stretch by } \tfrac{1}{k}$
When to use
Recognising how a coefficient outside or inside the function changes the graph.
Function reflections
$y = -f(x) \to \text{reflect in x-axis};\ \ y = f(-x) \to \text{reflect in y-axis}$
When to use
When the question describes a reflection.

Key Definitions and Keywords — Graphs of Functions

Definitions to memorise and the exact keywords mark schemes credit for graphs of functions answers — sharpened from recent examiner reports for the 2026 0580 sitting.

Asymptote
Examiner keyword
A line that the curve approaches but never touches. Reciprocal graphs have horizontal and vertical asymptotes.
Intercept
Examiner keyword
The point where a graph crosses an axis. y-intercept: $x = 0$ . x-intercept: $y = 0$ .
Translation
Examiner keyword
A rigid shift of the graph. Described as a column vector $\begin{pmatrix} a \\ b \end{pmatrix}$ .
Stretch
A scaling that pulls or compresses the graph along an axis. Described by a scale factor.

Common Mistakes and Misconceptions — Graphs of Functions

The traps other students keep falling into on graphs of functions questions — taken from recent Cambridge IGCSE 0580 examiner reports and mark schemes — and how to avoid them.

✕Reading $f(x - 3)$ as a left-shift
0580 examiner reports — recurring
Why it happens
The minus sign is counter-intuitive.
How to avoid it
$f(x - 3)$ shifts RIGHT by $3$ (because input $x$ now has to be $3$ bigger to give the same output).
✕Drawing the curve crossing or touching the asymptote
Why it happens
Sketch is rushed.
How to avoid it
An asymptote is a line the curve APPROACHES but never reaches. Leave a clear gap.
✕Drawing $y = \frac{k}{x}$ branches in the wrong quadrants
Why it happens
Sign of $k$ determines which quadrant pair the branches occupy.
How to avoid it
$k > 0$ → 1st and 3rd quadrants. $k < 0$ → 2nd and 4th.
✕Forgetting that $a^{0} = 1$
Why it happens
Students compute $a^{0}$ as $0$ .
How to avoid it
Any non-zero base raised to the $0$ power is $1$ . So $y = a^{x}$ always passes through $(0, 1)$ .

Graphs of Functions — frequently asked questions

The things students keep getting wrong in this sub-topic, answered.

Cambridge IGCSE Mathematics (0580)

Graphs of Functions

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Short Notes - Graphs of Functions

Detailed Study Notes

Detailed notes on Functions for Cambridge IGCSE Mathematics, covering key concepts, explanations, examples, and exam-focused revision points.

Graphs of Functions — Cambridge IGCSE 0580 Maths Extended (2026)

At a glance

Linear $y = mx + c$ : straight line.
Quadratic $y = ax^{2} + bx + c$ : parabola.
Cubic $y = ax^{3}$ : S-shape through origin.
Reciprocal $y = \dfrac{a}{x}$ : two-branch hyperbola, asymptotes at $x = 0$ and $y = 0$ .
Exponential $y = a^{x}$ : passes through $(0, 1)$ , asymptote $y = 0$ .
$y = \sin x$ , $y = \cos x$ : periodic, range $[-1, 1]$ .
$y = \tan x$ : periodic, asymptotes at $90\degree$ , $270\degree$ , etc.
Use intercepts and asymptotes as anchor points.

What you’ll learn

Mapped to the Cambridge IGCSE 0580 syllabus (2025-2027).

E2.10 — Construct tables of values and draw graphs for functions of the form $y = ax^{n}$ and exponentials $y = ab^{x}$ and combinations.
Sketch and recognise shapes of standard functions.

Standard shapes — recognise on sight

Memorise these eight families. Cambridge expects instant recognition.

Linear ( $y = mx + c$ ).

Straight line. Gradient $m$ , $y$ -intercept $c$ .
Domain and range: all reals.

Quadratic ( $y = ax^{2} + bx + c$ ).

Parabola. Opens up if $a > 0$ , down if $a < 0$ .
Vertex on axis of symmetry $x = -\dfrac{b}{2a}$ .

Cubic ( $y = ax^{3}$ basic form).

S-shape through the origin. Sweeps from $-\infty$ to $+\infty$ in the $y$ -direction.
$y = ax^{3} + bx^{2} + cx + d$ is the general cubic — has one or two turning points.

Reciprocal ( $y = \dfrac{a}{x}$ ).

Two-branch hyperbola.
Asymptotes: $x = 0$ (vertical) and $y = 0$ (horizontal).
$a > 0$ : branches in 1st and 3rd quadrants. $a < 0$ : 2nd and 4th.

Exponential ( $y = a^{x}$ with $a > 0, a \ne 1$ ).

Passes through $(0, 1)$ .
$a > 1$ : rises rapidly to the right, decays to $0$ on the left.
$0 < a < 1$ : mirror-image (decays to right, rises to left).
Horizontal asymptote: $y = 0$ .

Sine and cosine ( $y = \sin x$ , $y = \cos x$ ).

Period $360\degree$ . Range $-1 \le y \le 1$ .
$\sin$ passes through origin. $\cos$ starts at $(0, 1)$ .

Tangent ( $y = \tan x$ ).

Period $180\degree$ . Range: all reals.
Vertical asymptotes at $x = 90\degree, 270\degree, \ldots$ .

Inverse function ( $y = f^{-1}(x)$ ).

Reflection of $f$ in the line $y = x$ .

The reciprocal curve never crosses the axes — they are its asymptotes.

Exponential passes through (0, 1); sine waves between −1 and 1; tangent repeats with vertical asymptotes.

Linear → straight line.
Quadratic → parabola.
Cubic → S-shape.
Reciprocal → two-branch hyperbola, asymptotes at axes.
Exponential → through $(0, 1)$ , asymptote $y = 0$ .
Sin/cos → wave, range $[-1, 1]$ .
Tan → asymptotes every $180\degree$ .

Sketching: find anchor points first

$y$ -intercept, $x$ -intercepts, asymptotes, turning points. Connect with a smooth curve.

The five anchors.

$y$ -intercept (substitute $x = 0$ ).
$x$ -intercepts / roots (solve $y = 0$ ).
Turning points (set $\dfrac{dy}{dx} = 0$ if you've covered differentiation; otherwise use formula or symmetry).
Asymptotes (vertical and horizontal).
Behaviour as $x \to \pm \infty$ .

Plot the anchor points, mark the asymptotes (dashed), then draw a smooth curve through everything.

Worked. Sketch $y = \dfrac{2}{x - 3}$ .

$y$ -intercept: $y = \dfrac{2}{-3} = -\dfrac{2}{3}$ .
$x$ -intercept: $\dfrac{2}{x - 3} = 0$ has no solution (numerator never zero) → no $x$ -intercept.
Vertical asymptote: $x = 3$ (denominator zero there).
Horizontal asymptote: $y \to 0$ as $x \to \pm\infty$ .
Two branches: one in upper-right (when $x > 3$ , $y > 0$ ), one in lower-left.

Tip. Always test ONE point on each branch to confirm direction. Plug in $x = 4$ : $y = 2$ . Plug in $x = 2$ : $y = -2$ . Confirms branches.

The curve hugs the asymptotes x = 3 and y = 0 without ever touching them.

$y$ -intercept first.
Then $x$ -intercepts.
Then asymptotes (vertical from undefined points; horizontal from end behaviour).
Then turning points (from differentiation or symmetry).
Smooth curve through anchors, respecting asymptotes.

Transformations of graphs

Translations, stretches, reflections — recognise the effect of each modification to the formula.

Vertical translation. $y = f(x) + a$ shifts UP by $a$ (DOWN if $a < 0$ ).

Horizontal translation. $y = f(x - a)$ shifts RIGHT by $a$ (LEFT if $a < 0$ ). Counter-intuitive: the sign FLIPS.

Vertical stretch. $y = af(x)$ stretches by factor $a$ in the $y$ -direction. $a < 0$ : also reflects in the $x$ -axis.

Horizontal stretch. $y = f(ax)$ stretches by factor $\dfrac{1}{a}$ in the $x$ -direction (compresses if $a > 1$ ).

Reflection.

$y = -f(x)$ : reflect in the $x$ -axis.
$y = f(-x)$ : reflect in the $y$ -axis.

Worked. Sketch $y = (x - 2)^{2} + 3$ given the basic $y = x^{2}$ .

$-2$ inside: shift RIGHT by 2.
$+3$ outside: shift UP by 3.
New vertex: $(2, 3)$ .

Inside-bracket changes move the curve along x; outside-bracket changes move it along y.

Combined transformations. Apply in order. Inside-the-bracket changes affect $x$ , outside-the-bracket changes affect $y$ .

Inside-bracket changes move x (opposite sign); outside changes move or stretch y.

Inside the function ( $x \to x + c$ ): horizontal shift, opposite direction.
Outside the function ( $f(x) + c$ ): vertical shift, same direction.
$af(x)$ : vertical stretch.
$f(ax)$ : horizontal stretch (compress for $a > 1$ ).
Negatives: reflection in respective axis.

Quick recap

Memorise the eight standard shapes.
Anchor points: $y$ -intercept, $x$ -intercepts, asymptotes, turning points.
Reciprocal $y = a/x$ : two branches with asymptotes at the axes.
Exponential $y = a^{x}$ passes through $(0, 1)$ .
Translations: $f(x + h)$ shifts opposite direction; $f(x) + k$ shifts same direction.
Reflections: $-f(x)$ in $x$ -axis; $f(-x)$ in $y$ -axis.

Memorise this

Verbatim phrases and definitions Cambridge mark schemes credit.

Linear function — $y = mx + c$ , straight line.
Quadratic function — $y = ax^{2} + bx + c$ , parabola.
Reciprocal function — $y = a/x$ , two-branch hyperbola with asymptotes at $x = 0$ and $y = 0$ .
Exponential function — $y = a^{x}$ with $a > 0, a \ne 1$ .
Asymptote — a line the graph approaches but never touches.

How it’s examined

Sources: Cambridge IGCSE Mathematics 0580 syllabus 2025-2027 (E2.10); 0580/42 Oct/Nov 2024 — Q11 (sketch reciprocal); 0580 Examiner Reports 2022-2024. Last reviewed 2026-05-05.

Take this whole topic with you

Step-by-step worked examples — Graphs of Functions

Step-by-step solutions to past-paper-style questions on graphs of functions, written exactly the way a tutor would explain them at the board.

Question type:

Question patterns to master — Graphs of Functions

Almost every graphs of functions exam question is one of these shapes. Learn to spot each one and you will always know how to start.

Graph or diagram

Recognise it by

Instructions to sketch, complete the table of values, use the graph to solve / estimate, draw a tangent, or describe the transformation of a curve.

How to approach it

Common trap

Drawing a reciprocal curve that touches its asymptote, or shifting a translation the wrong way. Examiner reports flag $f(x-3)$ being read as a left-shift — it moves right.

Direct calculation

Recognise it by

A single algebraic instruction — find the $y$ -intercept, find the equation of the tangent — answered without reading a diagram.

How to approach it

Substitute $x=0$ for a $y$ -intercept; for a tangent, find the point, compute the gradient (differentiate and evaluate), then use $y-y_1=m(x-x_1)$ .

Common trap

Computing $a^0$ as $0$ instead of $1$ for exponential intercepts, or panicking at a zero gradient — a tangent at a turning point is simply horizontal.

Multi-step problem

Recognise it by

Find the coordinates where two curves intersect — solved algebraically by combining two equations.

How to approach it

Set the two expressions equal, clear any fractions, rearrange to a single quadratic, solve it, then substitute each $x$ back to find its paired $y$ .

Common trap

Stopping after finding the $x$ -values and not pairing them with their $y$ -values, or losing a solution by dividing through by $x$ .

1Sketch a cubic with given roots
ExtendedGraph or diagram• cubic
Question
Sketch $y = (x + 2)(x - 1)(x - 3)$ , showing the intercepts.
Step-by-step solution
1. Step 1
  x-intercepts: $x = -2, 1, 3$ .
2. Step 2
  y-intercept: $y = (2)(-1)(-3) = 6$ at $(0, 6)$ .
3. Step 3
  Coefficient of $x^{3}$ is positive, so the curve goes from bottom-left to top-right (rises through three roots).
Answer
Cubic crossing x-axis at $(-2, 0), (1, 0), (3, 0)$ and y-axis at $(0, 6)$ , going from bottom-left to top-right.
2Sketch a reciprocal graph
ExtendedGraph or diagram• reciprocal, asymptote
Question
Sketch $y = \dfrac{6}{x}$ and state any asymptotes.
Step-by-step solution
1. Step 1
  Two branches in the first and third quadrants (because $6 > 0$ ).
2. Step 2
  As $x \to 0$ , $y \to \pm \infty$ — vertical asymptote at $x = 0$ .
3. Step 3
  As $x \to \pm \infty$ , $y \to 0$ — horizontal asymptote at $y = 0$ .
Answer
Hyperbola; asymptotes at $x = 0$ and $y = 0$ .
3Sketch an exponential
ExtendedGraph or diagram• Adapted from 0580/42 Oct/Nov 2023 Q11• exponential
Question
Sketch $y = 2^{x}$ and state the y-intercept and asymptote.
Step-by-step solution
1. Step 1
  y-intercept: $y = 2^{0} = 1$ at $(0, 1)$ .
2. Step 2
  As $x \to -\infty$ , $y \to 0^{+}$ — horizontal asymptote at $y = 0$ .
3. Step 3
  As $x \to +\infty$ , $y \to \infty$ — curve rises rapidly.
Answer
Exponential growth curve through $(0, 1)$ , asymptote $y = 0$ as $x \to -\infty$ .
4Apply a vertical translation
ExtendedGraph or diagram• transformations
Question
Describe the transformation from $y = x^{2}$ to $y = x^{2} - 4$ .
Step-by-step solution
1. Step 1
  Subtracting $4$ from $y$ moves the graph DOWN by $4$ units.
Answer
Translation by vector $\begin{pmatrix} 0 \\ -4 \end{pmatrix}$ — i.e. $4$ units downward.
5Plot a cubic from a table of values
CoreGraph or diagram• table of values, cubic
Question
Complete the table of values for $y = x^{3} - 6x$ for $x = -3, -2, -1, 0, 1, 2, 3$ .
Step-by-step solution
1. Step 1
  Substitute each $x$ into $y = x^{3} - 6x$ .
  $x = -3: y = -27 + 18 = -9$
2. Step 2
  Continue.
  $x = -2: y = -8 + 12 = 4;\ x = -1: y = -1 + 6 = 5$
3. Step 3
  Centre and positive side.
  $x = 0: y = 0;\ x = 1: y = -5;\ x = 2: y = -4;\ x = 3: y = 9$
Answer
$y$ -values: $-9,\ 4,\ 5,\ 0,\ -5,\ -4,\ 9$
Examiner tip
The examiner report flags candidates often slip on signs of $x^{3}$ when $x$ is negative. Use brackets: $(-2)^{3} = -8$ , not $8$ .
6Recognise the y-intercept of a curve
CoreDirect calculation• y-intercept
Question
Find the $y$ -intercept of $y = 3^{x} - 4$ and state the equation of its horizontal asymptote.
Step-by-step solution
1. Step 1
  $y$ -intercept: substitute $x = 0$ .
  $y = 3^{0} - 4 = 1 - 4 = -3$
2. Step 2
  As $x \to -\infty$ , $3^{x} \to 0$ , so $y \to -4$ .
Answer
$y$ -intercept at $(0, -3)$ ; horizontal asymptote $y = -4$ .
7Sketch $y = -8/x$
ExtendedGraph or diagram• reciprocal, negative coefficient
Question
Sketch $y = -\dfrac{8}{x}$ , stating the quadrants the branches occupy and the asymptotes.
Step-by-step solution
1. Step 1
  $k = -8 < 0$ , so branches are in the 2nd and 4th quadrants.
2. Step 2
  Sample point: $x = 2 \Rightarrow y = -4$ , confirming the lower-right branch.
3. Step 3
  Asymptotes are the axes (the curve never touches $x = 0$ or $y = 0$ ).
Answer
Hyperbola in 2nd and 4th quadrants; asymptotes $x = 0$ and $y = 0$ .
8Solve $f(x) = g(x)$ graphically
ExtendedGraph or diagram• Adapted from 0580/42 May/Jun 2023 Q5• intersection, simultaneous
Question
Use the graph of $y = x^{2} - 2$ and $y = x$ to estimate the solutions of $x^{2} - 2 = x$ . Then check algebraically.
Step-by-step solution
1. Step 1
  Read the intersection $x$ -coordinates from the graph: approximately $x \approx -1$ and $x \approx 2$ .
2. Step 2
  Rearrange to confirm.
  $x^{2} - 2 = x \implies x^{2} - x - 2 = 0$
3. Step 3
  Factor.
  $(x - 2)(x + 1) = 0 \implies x = 2 \text{ or } x = -1$
Answer
$x = -1$ and $x = 2$
Examiner tip
The mark scheme awards method marks for showing the rearrangement to $f(x) - g(x) = 0$ explicitly; pure graph-reading without algebra often loses the accuracy mark.
9Use a graph to solve a cubic
ExtendedGraph or diagram• Adapted from 0580/42 Oct/Nov 2022 Q10• cubic, graphical solution
Question
The graph of $y = x^{3} - 3x$ is drawn for $-2.5 \leq x \leq 2.5$ . Use it to solve $x^{3} - 3x = 1$ .
Step-by-step solution
1. Step 1
  Draw the horizontal line $y = 1$ on the same axes.
2. Step 2
  The solutions of $x^{3} - 3x = 1$ are the $x$ -coordinates of the points where the cubic meets the line $y = 1$ .
3. Step 3
  Reading the graph carefully: $x \approx -1.5$ , $x \approx -0.3$ and $x \approx 1.9$ (accept $\pm 0.1$ ).
Answer
$x \approx -1.5,\ -0.3,\ 1.9$
Examiner tip
The examiner report flags candidates who draw the wrong horizontal line, e.g. $y = -1$ instead of $y = 1$ . Always label the constant clearly on the diagram.
10Estimate the gradient at a point on a curve
ExtendedGraph or diagram• gradient, tangent
Question
On the graph of $y = x^{2} + 1$ , draw a tangent at $(2, 5)$ and use it to estimate the gradient at that point.
Step-by-step solution
1. Step 1
  Draw a straight line just touching the curve at $(2, 5)$ — the tangent.
2. Step 2
  Pick two clear points on the tangent (e.g. $(0, -3)$ and $(2, 5)$ ) and use rise / run.
  $\text{gradient} \approx \dfrac{5 - (-3)}{2 - 0} = 4$
Answer
Gradient $\approx 4$
Examiner tip
The mark scheme awards method marks when the tangent is clearly drawn and the two coordinates used are visibly on the tangent — not on the curve.
11Intersection of two non-linear graphs
ChallengeMulti-step problem• intersection, non-linear
Question
Find the coordinates of the points where $y = \dfrac{6}{x}$ and $y = x + 1$ intersect.
Step-by-step solution
1. Step 1
  Set the right-hand sides equal.
  $\dfrac{6}{x} = x + 1$
2. Step 2
  Multiply both sides by $x$ (note $x \neq 0$ ).
  $6 = x^{2} + x \implies x^{2} + x - 6 = 0$
3. Step 3
  Factor.
  $(x + 3)(x - 2) = 0 \implies x = -3 \text{ or } x = 2$
4. Step 4
  Find $y$ for each.
  $x = -3: y = -2;\ x = 2: y = 3$
Answer
$(-3, -2)$ and $(2, 3)$
12Reflections: $y = -f(x)$ versus $y = f(-x)$
ChallengeGraph or diagram• Adapted from 0580/42 May/Jun 2024 Q12• transformations, reflection
Question
The graph of $y = f(x)$ has a maximum at $(3, 5)$ and passes through $(0, 2)$ . State the new maximum and $y$ -intercept of (a) $y = -f(x)$ , (b) $y = f(-x)$ .
Step-by-step solution
1. Step 1
  (a) $y = -f(x)$ reflects in the $x$ -axis: each $y$ -value is negated. The maximum becomes a minimum.
  $\text{Min at } (3, -5);\ \text{passes through } (0, -2)$
2. Step 2
  (b) $y = f(-x)$ reflects in the $y$ -axis: each $x$ -value is negated. Maxima remain maxima.
  $\text{Max at } (-3, 5);\ \text{passes through } (0, 2)$
Answer
(a) Minimum $(3, -5)$ , $y$ -intercept $(0, -2)$ (b) Maximum $(-3, 5)$ , $y$ -intercept $(0, 2)$
Examiner tip
The examiner report flags candidates often swap the two rules. Memorise: minus OUTSIDE flips top/bottom; minus INSIDE flips left/right.
13Find the equation of a tangent
ChallengeDirect calculation• tangent, differentiation
Question
Find the equation of the tangent to $y = x^{2} - 2x + 3$ at the point $x = 1$ .
Step-by-step solution
1. Step 1
  Find $y$ at $x = 1$ .
  $y = 1 - 2 + 3 = 2.\ \text{Point: } (1, 2)$
2. Step 2
  Differentiate and evaluate the gradient at $x = 1$ .
  $\dfrac{dy}{dx} = 2x - 2 \implies m = 0$
3. Step 3
  Gradient $0$ means the tangent is horizontal through $(1, 2)$ .
Answer
$y = 2$
Examiner tip
Tangents to a turning point are horizontal. The examiner report flags candidates who panic at gradient $0$ and try to write something like $x = 1$ instead.

Key Formulae — Graphs of Functions

The formulae you need to memorise for graphs of functions on the Cambridge IGCSE 0580 paper, with every variable defined in plain English and a note on when to use it.

Function translations
$y = f(x) + a \to \text{up by } a;\ \ y = f(x - a) \to \text{right by } a$
When to use
Recognising shifts of standard graphs.
Function stretches
$y = kf(x) \to \text{vertical stretch by } k;\ \ y = f(kx) \to \text{horizontal stretch by } \tfrac{1}{k}$
When to use
Recognising how a coefficient outside or inside the function changes the graph.
Function reflections
$y = -f(x) \to \text{reflect in x-axis};\ \ y = f(-x) \to \text{reflect in y-axis}$
When to use
When the question describes a reflection.

Key Definitions and Keywords — Graphs of Functions

Definitions to memorise and the exact keywords mark schemes credit for graphs of functions answers — sharpened from recent examiner reports for the 2026 0580 sitting.

Asymptote
Examiner keyword
A line that the curve approaches but never touches. Reciprocal graphs have horizontal and vertical asymptotes.
Intercept
Examiner keyword
The point where a graph crosses an axis. y-intercept: $x = 0$ . x-intercept: $y = 0$ .
Translation
Examiner keyword
A rigid shift of the graph. Described as a column vector $\begin{pmatrix} a \\ b \end{pmatrix}$ .
Stretch
A scaling that pulls or compresses the graph along an axis. Described by a scale factor.

Common Mistakes and Misconceptions — Graphs of Functions

The traps other students keep falling into on graphs of functions questions — taken from recent Cambridge IGCSE 0580 examiner reports and mark schemes — and how to avoid them.

✕Reading $f(x - 3)$ as a left-shift
0580 examiner reports — recurring
Why it happens
The minus sign is counter-intuitive.
How to avoid it
$f(x - 3)$ shifts RIGHT by $3$ (because input $x$ now has to be $3$ bigger to give the same output).
✕Drawing the curve crossing or touching the asymptote
Why it happens
Sketch is rushed.
How to avoid it
An asymptote is a line the curve APPROACHES but never reaches. Leave a clear gap.
✕Drawing $y = \frac{k}{x}$ branches in the wrong quadrants
Why it happens
Sign of $k$ determines which quadrant pair the branches occupy.
How to avoid it
$k > 0$ → 1st and 3rd quadrants. $k < 0$ → 2nd and 4th.
✕Forgetting that $a^{0} = 1$
Why it happens
Students compute $a^{0}$ as $0$ .
How to avoid it
Any non-zero base raised to the $0$ power is $1$ . So $y = a^{x}$ always passes through $(0, 1)$ .

Graphs of Functions — frequently asked questions

The things students keep getting wrong in this sub-topic, answered.

Graphs of Functions

Short Notes - Graphs of Functions

Detailed Study Notes

At a glance

What you’ll learn

Standard shapes — recognise on sight

Sketching: find anchor points first

Transformations of graphs

Quick recap

Memorise this

How it’s examined

Take this whole topic with you

Question patterns to master — Graphs of Functions

1Sketch a cubic with given roots

2Sketch a reciprocal graph

3Sketch an exponential

4Apply a vertical translation

5Plot a cubic from a table of values

6Recognise the y-intercept of a curve

7Sketch y=−8/xy = -8/xy=−8/x

8Solve f(x)=g(x)f(x) = g(x)f(x)=g(x) graphically

9Use a graph to solve a cubic

10Estimate the gradient at a point on a curve

11Intersection of two non-linear graphs

12Reflections: y=−f(x)y = -f(x)y=−f(x) versus y=f(−x)y = f(-x)y=f(−x)

13Find the equation of a tangent

Function translations

Function stretches

Function reflections

Asymptote

Intercept

Translation

Stretch

✕Reading f(x−3)f(x - 3)f(x−3) as a left-shift

✕Drawing the curve crossing or touching the asymptote

✕Drawing y=kxy = \frac{k}{x}y=xk​ branches in the wrong quadrants

✕Forgetting that a0=1a^{0} = 1a0=1

Graphs of Functions — frequently asked questions

Graphs of Functions

Short Notes - Graphs of Functions

Detailed Study Notes

At a glance

What you’ll learn

Standard shapes — recognise on sight

Sketching: find anchor points first

Transformations of graphs

Quick recap

Memorise this

How it’s examined

Take this whole topic with you

Question patterns to master — Graphs of Functions

1Sketch a cubic with given roots

2Sketch a reciprocal graph

3Sketch an exponential

4Apply a vertical translation

5Plot a cubic from a table of values

6Recognise the y-intercept of a curve

7Sketch y=−8/xy = -8/xy=−8/x

8Solve f(x)=g(x)f(x) = g(x)f(x)=g(x) graphically

9Use a graph to solve a cubic

10Estimate the gradient at a point on a curve

11Intersection of two non-linear graphs

12Reflections: y=−f(x)y = -f(x)y=−f(x) versus y=f(−x)y = f(-x)y=f(−x)

13Find the equation of a tangent

Function translations

Function stretches

Function reflections

Asymptote

Intercept

Translation

Stretch

✕Reading f(x−3)f(x - 3)f(x−3) as a left-shift

✕Drawing the curve crossing or touching the asymptote

✕Drawing y=kxy = \frac{k}{x}y=xk​ branches in the wrong quadrants

✕Forgetting that a0=1a^{0} = 1a0=1

Graphs of Functions — frequently asked questions

7Sketch $y = -8/x$

8Solve $f(x) = g(x)$ graphically

12Reflections: $y = -f(x)$ versus $y = f(-x)$

✕Reading $f(x - 3)$ as a left-shift

✕Drawing $y = \frac{k}{x}$ branches in the wrong quadrants

✕Forgetting that $a^{0} = 1$

7Sketch $y = -8/x$

8Solve $f(x) = g(x)$ graphically

12Reflections: $y = -f(x)$ versus $y = f(-x)$

✕Reading $f(x - 3)$ as a left-shift

✕Drawing $y = \frac{k}{x}$ branches in the wrong quadrants

✕Forgetting that $a^{0} = 1$