Graphs of Functions and Sketching Curves

Syllabus coverage

Construct tables of values, and draw, recognise and interpret graphs of functions of the form $a x + b$ , $\pm x^{2} + a x + b$ and $\frac{a}{x}$ where $x$ is not zero, with $a$ and $b$ integers
Solve associated equations graphically, including finding and interpreting roots by graphical methods, for example the intersection of a line and a curve
Recognise, sketch and interpret graphs of linear and quadratic functions, where knowledge of symmetry and roots is required and turning points are not
Extended only: draw, recognise and interpret graphs of the form $a x^{n}$ , including sums of no more than three of these, and of the form $a b^{x} + c$
Extended only: draw and interpret graphs representing exponential growth and decay problems
Extended only: recognise, sketch and interpret cubic, reciprocal and exponential graphs, where knowledge of turning points, roots, symmetry, and vertical and horizontal asymptotes is required

The graph of y = x^2 - 2x - 3. The roots are where the curve crosses the x-axis, the y-intercept is the value at x = 0, and the turning point lies on the line of symmetry halfway between the roots.

From an equation to a table to a graph

To draw a graph from an equation, build a table of values first. Choose a set of $x$ -values that covers the range the question asks about, work out each matching $y$ -value, then plot the pairs and join them.

Substitute carefully, especially with negative numbers. For $y = x^{2} - 3 x + 1$ at $x = - 2$ the first term is $(- 2)^{2} = 4$ , not $- 4$ , because squaring a negative number gives a positive one. Most of the marks lost in this topic come from a sign error in the table rather than from the plotting.

Five or more points are enough to show the shape of a curve, and two are enough for a straight line. Join the points of a curve with a smooth curve drawn in one movement: a ruler between plotted points turns a parabola into a chain of straight lines and loses the shape mark.

Take $x$ -values across the range the question asks for
Use brackets when you substitute a negative number
Plot with small crosses so the points stay visible under the curve
Join a curve smoothly and a straight line with a ruler

$x$	-2	-1	0	1	2
$y = x^{2} - 2 x - 3$	5	0	-3	-4	-3

Show the table even when the question only asks for the graph. A correct table earns the method marks if the curve goes wrong, and it makes a plotting error obvious before you draw anything.

The straight line: $y = m x + c$

Every equation that can be written as $y = m x + c$ is a straight line. The number $m$ is the gradient, how many units the line rises for each 1 across, and $c$ is the $y$ -intercept, the value of $y$ when $x = 0$ . To sketch one, mark the intercept on the vertical axis and step across 1 and up $m$ to find a second point.

An equation such as $2 x + 3 y = 12$ is still a straight line. Rearrange it into the form $y = m x + c$ before reading anything off it: subtracting $2 x$ from both sides and dividing by 3 gives $y = - \frac{2}{3} x + 4$ , so the line falls from left to right with gradient $- \frac{2}{3}$ and crosses the $y$ -axis at 4.

The sign of the gradient tells you the direction: a positive gradient rises from left to right, a negative gradient falls, and a gradient of zero gives a horizontal line. Two lines with the same gradient are parallel.

$m > 0$ : the line rises from left to right
$m < 0$ : the line falls from left to right
$m = 0$ : a horizontal line
Rearrange $a x + b y = c$ into $y = m x + c$ before reading off $m$ and $c$

The quadratic: $y = a x^{2} + b x + c$

A quadratic graph is a parabola, a symmetric U-shaped curve. If $a$ is positive the arms open upwards and the turning point is a minimum; if $a$ is negative the arms open downwards and the turning point is a maximum. The value of $a$ controls how narrow the curve is, and $c$ is where it crosses the $y$ -axis.

The roots are the values of $x$ where the curve crosses the $x$ -axis, and they are the solutions of $a x^{2} + b x + c = 0$ . A quadratic has two roots, one repeated root when the curve just touches the axis, or none at all when it stays clear of it.

The curve is symmetric about the vertical line through its turning point, so that line sits exactly halfway between the two roots. For $y = x^{2} - 2 x - 3$ the roots are $- 1$ and $3$ , so the line of symmetry is $x = \frac{- 1 + 3}{2} = 1$ , the minimum is at $(1, - 4)$ , and the curve crosses the $y$ -axis at $(0, - 3)$ .

$a > 0$ : a minimum, with the arms opening upwards
$a < 0$ : a maximum, with the arms opening downwards
Roots: where the curve crosses the $x$ -axis
Line of symmetry: the vertical line through the turning point, halfway between the roots

Core candidates are asked about the roots and the symmetry; turning points are Extended content. Either way, the turning point is the easiest feature to find once you have the roots: average them.

The graph of y = x^2 - 2x - 3. The roots are where the curve cuts the x-axis, the y-intercept is the value at x = 0, and the turning point sits on the line of symmetry halfway between the roots.

Cubic, reciprocal and exponential shapes

The cubic $y = a x^{3} + b$ is an S-shaped curve. When $a$ is positive it runs from bottom-left to top-right and passes through $(0, b)$ ; when $a$ is negative it runs from top-left to bottom-right. A cubic of the form $y = a x^{3} + b x^{2} + c x$ has the same overall shape but may turn twice on the way.

The reciprocal graph $y = \frac{a}{x} + b$ has two separate branches and never touches either axis. The vertical axis is a vertical asymptote: as $x$ gets close to zero the curve shoots off to a very large value. As $x$ becomes large the curve flattens towards the horizontal line $y = b$ . For $a > 0$ one branch lies in the region where $x$ and $y$ are both large and positive, and the other where both are negative.

The exponential graph $y = a b^{x} + c$ rises or falls faster and faster. If $b > 1$ the curve grows steeply; if $0 < b < 1$ it decays towards the horizontal asymptote $y = c$ . It crosses the $y$ -axis at $(0, a + c)$ , because $b^{0} = 1$ , and it has no vertical asymptote.

Function	Shape	Features to label on a sketch
$y = m x + c$	a straight line	the gradient and the $y$ -intercept
$y = a x^{2} + b x + c$	a parabola	the roots, the $y$ -intercept, the turning point and the line of symmetry
$y = a x^{3} + b$	an S-shaped curve	where it crosses the axes, and which way up it is; no asymptote
$y = \frac{a}{x} + b$	two separate branches	the asymptotes $x = 0$ and $y = b$
$y = a b^{x} + c$	a growth or decay curve	the asymptote $y = c$ and the intercept $(0, a + c)$

Draw an asymptote as a dashed line and label it with its equation before you draw the curve. An unlabelled asymptote scores nothing, and a curve that touches its own asymptote is wrong however good the shape looks.

Reading roots, values and intersections from a graph

Every equation with $y$ on one side can be solved from its graph. To find a value of $y$ , go up from the $x$ -value to the curve and across to the vertical axis. To find an $x$ from a $y$ , go the other way. Read to the accuracy the grid allows and give the answer to the decimal place the question asks for.

To solve $f (x) = 0$ , read where the curve crosses the $x$ -axis: those $x$ -values are the roots. To solve $f (x) = k$ , draw the horizontal line $y = k$ and read the $x$ -coordinates of the points where it meets the curve.

To solve $f (x) = g (x)$ , draw both graphs on the same axes and read the $x$ -coordinates of the intersections. The example the syllabus gives is finding the intersection of a line and a curve, which is the same instruction: draw the line, mark where it crosses the curve, and read off the $x$ -values.

Roots: where the curve crosses the $x$ -axis
The $y$ -intercept: the value of $y$ at $x = 0$
$f (x) = k$ : draw the line $y = k$ and read the crossings
$f (x) = g (x)$ : read the $x$ -coordinates of the intersections

The answer to a graphical solution is always the

x

-coordinate of the crossing, not the

y

-coordinate. If the question asks for the value of

x

where the line meets the curve, the

y

-value of the crossing is not part of the answer.

Sketching a curve

A sketch is not a plot. It shows the shape of the curve and its key features in roughly the right positions, without graph paper and without plotting a table of points. The marks are for the shape and for the labelled features: the examiner is testing what you know about the curve, not how neatly you can draw.

Label every point the question mentions, and always show the $y$ -intercept, the roots, and the turning point or asymptote if the curve has one. A cubic sketch should show which way up it is and where it crosses the axes; a reciprocal sketch should show both branches and both asymptotes.

Before you draw, decide three things: which way up the curve is, from the sign of the highest power; where it crosses the $y$ -axis, by putting $x = 0$ ; and whether it has an asymptote, by looking for a zero denominator or a value the curve settles towards. Those three facts are enough to fix a full-mark sketch.

Quadratic: a U shape opening up if $a > 0$ , down if $a < 0$
Cubic: bottom-left to top-right if $a > 0$
Reciprocal: two branches with the asymptotes labelled
Exponential: passes through $(0, a + c)$ with asymptote $y = c$

On a sketch, mark the coordinates of the features rather than the values of the axes.

(0, - 3)

and

x = - 1

tell the examiner what you know; a scale marked 0 to 10 tells them nothing.

The constants move and stretch the curve

The Extended syllabus gives these shapes in equivalent forms, and the constants tell you where the curve has moved to. $y = x^{2} + b$ is the basic parabola moved up by $b$ , and $y = (x - 3)^{2}$ is the basic parabola moved 3 to the right, so its turning point is at $(3, 0)$ . The shape is unchanged; only the position is.

For the reciprocal, $y = \frac{a}{x + b}$ is the curve $y = \frac{a}{x}$ moved $b$ to the left, so its vertical asymptote is the line $x = - b$ . The quick way to find a vertical asymptote is always the same: ask which value of $x$ makes the denominator zero.

For the exponential, $y = a b^{x} + c$ is the curve $y = a b^{x}$ moved up by $c$ , so its horizontal asymptote is $y = c$ and it still crosses the $y$ -axis at $(0, a + c)$ . A vertical asymptote comes from a denominator that can be zero; a horizontal asymptote comes from the value the curve settles towards as $x$ becomes very large.

$+ c$ at the end: the whole curve moves up by $c$
$(x - b)$ inside a bracket: the curve moves $b$ to the right
Denominator zero: a vertical asymptote
The value the curve approaches for large $x$ : a horizontal asymptote

Every one of these curves has the same shape as the simple version, so you never have to plot points to sketch one. Work out the intercept, the asymptotes and which way up it is, then draw the shape you already know.

Key terms you must be able to define

Function: A rule that gives exactly one output for each input. The graph of a function shows every input paired with its output.

Linear function: A function whose graph is a straight line, written in the form $y = m x + c$ .

Gradient: The steepness of a line: the change in $y$ divided by the change in $x$ .

$y$ -intercept: The value of $y$ where a graph crosses the vertical axis, found by putting $x = 0$ .

Quadratic function: A function in which the highest power of $x$ is $x^{2}$ . Its graph is a parabola.

Parabola: The symmetric U-shaped curve of a quadratic function.

Root: A value of $x$ for which $y = 0$ : where the graph crosses the $x$ -axis.

Turning point: The point where a curve changes direction, either a minimum or a maximum.

Minimum: The lowest point of a curve, where it stops falling and starts rising.

Maximum: The highest point of a curve, where it stops rising and starts falling.

Line of symmetry: The vertical line through the turning point of a parabola. The curve is a mirror image either side of it.

Cubic function: A function in which the highest power of $x$ is $x^{3}$ . Its graph is an S-shaped curve.

Reciprocal function: A function of the form $y = \frac{a}{x} + b$ , whose graph has two separate branches that never touch the axes.

Asymptote: A line that a curve approaches but never reaches. A vertical asymptote comes from a zero denominator; a horizontal one is the value the curve settles towards.

Exponential function: A function of the form $y = a b^{x} + c$ , where the variable is the power. It grows when $b > 1$ and decays when $0 < b < 1$ .

Table of values: A table of $x$ -values with the matching $y$ -values, used to plot the graph of a function.

Sketch: A drawing that shows the shape of a graph and its key features without plotting points to scale.

In context

Cambridge sets 0580 in real situations - a recipe, a bank statement, a map, a garden, a graph of a car's journey. These show how the mathematics appears in a question, then you write your own working for the numbers you are given.

A table of values

For $y = x^{2} - 3 x + 1$ : at $x = - 1$ , $y = 1 + 3 + 1 = 5$ ; at $x = 0$ , $y = 1$ ; at $x = 1$ , $y = - 1$ ; at $x = 2$ , $y = - 1$ ; at $x = 3$ , $y = 1$ . The values 5, 1, -1, -1, 1 repeat either side of the middle, which shows the line of symmetry is at $x = 1.5$ .

Reading the roots

The graph of $y = x^{2} - 2 x - 3$ crosses the $x$ -axis at $x = - 1$ and $x = 3$ , so those are the roots and the solutions of $x^{2} - 2 x - 3 = 0$ . Factorising checks it: $(x + 1) (x - 3) = x^{2} - 2 x - 3$ , and each bracket is zero at one of the two roots.

Solving graphically

To solve $x^{2} - 2 x - 3 = 2$ from the graph, draw the line $y = 2$ and read where it cuts the curve, at about $- 1.45$ and $3.45$ . Rearranging to $x^{2} - 2 x - 5 = 0$ and using the quadratic formula gives $x = 1 \pm \sqrt{6}$ , which is $- 1.45$ and $3.45$ to 2 decimal places.

Reciprocal asymptotes

For $y = \frac{3}{x} + 1$ : when $x = 3$ , $y = 2$ ; when $x = 0.5$ , $y = 7$ ; when $x = 30$ , $y = 1.1$ . As $x$ grows the curve flattens towards the horizontal asymptote $y = 1$ , and the vertical asymptote is the $y$ -axis, $x = 0$ , because the denominator is zero there.

Exponential growth

For $y = 2 \times 3^{x} + 1$ : at $x = 0$ , $y = 3$ ; at $x = 1$ , $y = 7$ ; at $x = 2$ , $y = 19$ . Each step multiplies the growing part by 3, so the values rise faster and faster, and for large negative $x$ the curve flattens towards the horizontal asymptote $y = 1$ .

Flashcards

Click a card to flip it. Mark honestly — the tracker uses this.

Q1What shape is the graph of a quadratic function?

A parabola: a symmetric U-shaped curve, opening upwards when the coefficient of

x^{2}

is positive and downwards when it is negative.

Q2How do you find the

y

-intercept of a graph?

Put

x = 0

and work out

y

. For

y = x^{2} - 2 x - 3

that gives

(0, - 3)

Q3What is a root of

y = f (x)

A value of

x

where

y = 0

: a point where the graph crosses the

x

-axis.

Q4How do you find the line of symmetry of a quadratic from its roots?

It is halfway between the roots:

x = \frac{r_{1} + r_{2}}{2}

. For roots

- 1

and

3

the line is

x = 1

Q5What is the turning point of

y = x^{2} - 2 x - 3

(1, - 4)

, a minimum. The line of symmetry

x = 1

gives the

x

-coordinate, and substituting gives the

y

-coordinate.

Q6What shape is the graph of

y = a x^{3} + b

An S-shaped curve: bottom-left to top-right when

a > 0

, top-left to bottom-right when

a < 0

Q7How many branches does the graph of

y = \frac{a}{x}

have?

Two, one each side of the vertical asymptote

x = 0

. The curve never touches either axis.

Q8What are the asymptotes of

y = \frac{a}{x} + b

x = 0

and

y = b

. The curve approaches both lines but never meets them.

Q9Where does the graph of

y = a b^{x} + c

cross the

y

-axis?

(0, a + c)

, because

b^{0} = 1

and

a \times 1 + c = a + c

Q10What is the horizontal asymptote of

y = a b^{x} + c

y = c

: the value the curve settles towards as

x

becomes very large, or very negative for a growth curve.

Q11How do you solve

f (x) = k

from a graph?

Draw the line

y = k

and read the

x

-coordinates of the points where it crosses the curve.

Q12How do you solve

f (x) = g (x)

from a graph?

Read the

x

-coordinates of the points where the two graphs intersect. The

y

-coordinate is not the answer.

0 of 12 marked as known

Exam-style questions

7 questions, 15 marks. Write your answer, then reveal the mark scheme and tick each point you actually made. Answering honestly here is what makes the rest of the site useful.

Write downAO12 marks

The graph of

y = x^{2} - 2 x - 3

is shown. Write down the two values of

x

for which

x^{2} - 2 x - 3 = 0

Your answer

Give every part of the answer, separated by commas or the word 'or'.

Mark scheme — tick each point you actually made

B1 $x = - 1$ not checked yet
B1 $x = 3$ not checked yet

Examiner tip: These are the values where the curve crosses the

x

-axis. Check the signs against the factors

(x + 1) (x - 3)

, which give

- 1

and

3

Write downAO11 mark

The graph of

y = x^{2} - 2 x - 3

is shown. Write down the value of

y

at the point where the graph crosses the

y

-axis.

Your answer

Write a number, a fraction such as 3/4, or an expression such as 3x+2.

Mark scheme — tick each point you actually made

B1 $y = - 3$ when $x = 0$ not checked yet

Examiner tip: Put

x = 0

into the equation:

y = 0 - 0 - 3

. The point is

(0, - 3)

, and the value is the constant term of the quadratic.

Work outAO13 marks

A table of values for

y = x^{2} - 3 x + 1

is being constructed. Work out the value of

y

when

x = 0

, the value when

x = 2

and the value when

x = 4

Your answer

Write a number, a fraction such as 3/4, or an expression such as 3x+2.

Mark scheme — tick each point you actually made

M1substitutes correctly, for example $2^{2} - 3 \times 2 + 1$
A1two of the three values correct
A1 $y = 1$ , $y = - 1$ and $y = 5$ not checked yet

Examiner tip: Substitute one value at a time and keep the signs. At

x = 4

the value is

16 - 12 + 1 = 5

, and at

x = 0

the

x^{2}

and

x

terms vanish and leave 1.

DetermineAO22 marks

By drawing the line

y = 2

on the graph of

y = x^{2} - 2 x - 3

, determine the larger value of

x

for which

x^{2} - 2 x - 3 = 2

. Give your answer correct to 1 decimal place.

Your answer

Write a number, a fraction such as 3/4, or an expression such as 3x+2.

Mark scheme — tick each point you actually made

M1draws the line $y = 2$ and reads where it crosses the curve
A1 $x = 3.4$ not checked yet

Examiner tip: The two crossings are at about

- 1.4

and

3.4

. Checking with algebra:

x^{2} - 2 x - 3 = 2

rearranges to

x^{2} - 2 x - 5 = 0

, so

x = 1 \pm \sqrt{6}

, and the larger value is the one the question wants.

Work outAO23 marks

The graph of

y = \frac{12}{x} + 2

is drawn. Work out the value of

y

when

x = 3

, and write down the equation of the horizontal asymptote.

Your answer

Write a number, a fraction such as 3/4, or an expression such as 3x+2.

Mark scheme — tick each point you actually made

M1substitutes $x = 3$ : $\frac{12}{3} + 2$
A1 $y = 6$
B1the horizontal asymptote is $y = 2$ not checked yet

Examiner tip: The horizontal asymptote is the number added on the end, because as

x

becomes large

\frac{12}{x}

shrinks towards zero and

y

settles at 2. Answer with an equation,

y = 2

, not just the number 2.

SketchAO12 marks

Sketch the graph of

y = \frac{3}{x} + 1

for

x > 0

, showing the horizontal asymptote and labelling it with its equation.

Mark scheme — tick each point you actually made

B1a decreasing curve in the region where $x > 0$ and $y > 1$ , getting close to the asymptote without touching it
B1the horizontal asymptote drawn as a dashed line and labelled $y = 1$

Examiner tip: Two marks: one for the shape and one for the labelled asymptote. Start the curve high up on the left of the first quadrant, flatten it towards

y = 1

, and never let it touch the line.

ExplainAO22 marks

Explain why the graph of

y = 2 \times 3^{x} + 1

never crosses the line

y = 1

Mark scheme — tick each point you actually made

B1 $3^{x}$ is positive for every value of $x$ , so $2 \times 3^{x}$ is positive and $y$ is always greater than 1
B1as $x$ becomes large and negative, $3^{x}$ gets closer and closer to zero, so $y$ gets closer and closer to 1 without ever reaching it

Examiner tip: Two ideas, two marks: the value is always above the line, and it approaches the line without reaching it. The line

y = 1

is the horizontal asymptote of the curve.

Exam technique for this topic

Show the table of values even when the question does not ask for it. It is where the method marks sit, and it turns a plotting mistake into something you can see before you draw the curve.
On a sketch, label the coordinates of every feature you show: intercepts as $(0, c)$ , roots as $x = \dots$ , the turning point, and each asymptote with its equation. An unlabelled curve is a shape rather than an answer.
Check the sign of the coefficient of the highest power before you draw. $y = - x^{2} + 4$ is a downward parabola with a maximum, and $y = - x^{3}$ runs from top-left to bottom-right.
For a quadratic, find the line of symmetry as the mean of the two roots, $x = \frac{r_{1} + r_{2}}{2}$ . It gives the $x$ -coordinate of the turning point in one line of working.
When a question asks you to solve an equation using a graph, rearrange it so that the curve you have been given is on one side and a simple line is on the other. Solving $x^{2} - 2 x - 3 = 2$ means drawing the line $y = 2$ , and solving $x^{2} - 2 x - 3 = x$ means drawing $y = x$ .

Why students lose marks here

A student wroteThe graph of

y = x^{2} - 2 x - 3

has roots at

x = - 3

and

x = 1

Why it loses marksThe factors are

(x + 1) (x - 3)

, and a root is the value that makes a bracket zero. Reading the numbers straight out of the brackets without changing the sign gives the wrong pair.

The answer that scoresFactorise to

(x + 1) (x - 3) = 0

, so the roots are

- 1

and

3

, and the curve crosses the

x

-axis at those two values.

A student wroteThe turning point of

y = x^{2} - 2 x - 3

is at

x = 3

, because that is the largest root.

Why it loses marksThe turning point lies on the line of symmetry, which is halfway between the roots:

x = \frac{- 1 + 3}{2} = 1

. The value

x = 3

is where the curve crosses the

x

-axis, not where it turns.

The answer that scoresThe line of symmetry is

x = 1

, so the turning point is

(1, - 4)

, the minimum value of

x^{2} - 2 x - 3

A student wroteThe graph of

y = \frac{3}{x}

crosses the

x

-axis when

x

is very large.

Why it loses marksIt never crosses. As

x

becomes large the curve gets closer and closer to the axis without reaching it, which is what an asymptote means. The same is true of

y = \frac{a}{x} + b

and the line

y = b

The answer that scoresThe axes are asymptotes of

y = \frac{3}{x}

: the curve approaches them but never touches them, so this graph has no roots.

A student wroteA sketch has to be accurate, so I will draw a full table of values and plot it on graph paper.

Why it loses marksA sketch shows the shape and the labelled features, not a scaled plot. Time spent plotting to scale is time not spent labelling the intercepts, the turning point and the asymptotes, and those labels are where the marks are.

The answer that scoresDraw the shape freehand in roughly the right position, label the intercepts, roots, turning point and asymptotes, and save accurate plotting for questions that ask for a graph from a table of values.

Graphs of Functions and Sketching Curves

Syllabus coverage

From an equation to a table to a graph

The straight line: y=mx+c

The quadratic: y=ax2+bx+c

Cubic, reciprocal and exponential shapes

Reading roots, values and intersections from a graph

Sketching a curve

The constants move and stretch the curve

Key terms you must be able to define

In context

Flashcards

Exam-style questions

Mark scheme — tick each point you actually made

Mark scheme — tick each point you actually made

Mark scheme — tick each point you actually made

Mark scheme — tick each point you actually made

Mark scheme — tick each point you actually made

Mark scheme — tick each point you actually made

Mark scheme — tick each point you actually made

Exam technique for this topic

Why students lose marks here

The straight line: $y = m x + c$

The quadratic: $y = a x^{2} + b x + c$