3 Sequences, functions and graphs

Exam analysis

Most tested topics

3.1 Sequences

A sequence is an ordered list. A term-to-term rule tells how to get from one term to the next; a position-to-term rule gives any term directly from its position $n$ .

Rule type	Example	How to generate
term-to-term	start at 3, then multiply by 2	$3,6,12,24,\ldots$
position-to-term	$u_n=n^2$	$1,4,9,16,\ldots$
named pattern	odd numbers	$1,3,5,7,\ldots$
named pattern	powers of 2	$1,2,4,8,\ldots$

Identify the starting position, apply the stated rule exactly, and label the terms. For a position rule, substitute $n=1,2,3,\ldots$ rather than repeatedly changing the previous value.

Check that every generated term obeys the same definition. A list may look familiar but still follow a different rule after its displayed terms.

A finite list alone does not determine one unique sequence. Use the stated rule or context; do not invent a pattern only because it fits the first few terms.

To continue a sequence, compare consecutive terms and find an operation that is applied consistently. State both the operation and any starting information needed.

Consecutive comparison	Likely rule
constant difference	add or subtract that value
constant ratio	multiply or divide by that value
alternating changes	repeat the change cycle
changing first differences	inspect second differences or another structure

For $5,9,13,17,\ldots$ , each difference is $4$ , so the next terms are $21,25$ and the term-to-term rule is ‘add 4’.

For $1,2,4,8,\ldots$ , each term is twice the previous term, so the next terms are $16,32$ and the rule is ‘multiply by 2’.

Do not describe $1,2,4,8$ as ‘add 1, add 2, add 4’ if the intended invariant is multiply by 2. Prefer the simplest consistent rule supported by the task.

An arithmetic sequence has a constant difference $d$ . Its nth term is linear: $u_n=dn+c$ , where $c$ is chosen so that the formula reproduces the first term.

Step	Action
difference	calculate $d$ from consecutive terms
first draft	write $dn$
adjustment	compare $d$ with the first term to find $c=a-d$
verify	substitute $n=1,2,3$
use	substitute a position, or solve the formula for $n$

For $7,11,15,19,\ldots$ , $d=4$ . The sequence $4n$ begins $4,8,12,16$ , so add 3: $u_n=4n+3$ .

To test whether 83 is a term, solve $4n+3=83$ , giving $n=20$ . Because $20$ is a positive integer, 83 is a term.

The common difference is the coefficient of $n$ , not usually the complete nth term. Always include the adjustment and verify at least the first term.

Every arithmetic term satisfies $u_n=a+(n-1)d$ , where $a$ is the first term and $d$ is the common difference. Two independent term facts can determine both unknowns.

Given fact	Equation
second term is 7	$a+d=7$
fifth term is 19	$a+4d=19$
terms $u_p$ and $u_q$ known	subtract to get $(q-p)d=u_q-u_p$

If $u_2=7$ and $u_5=19$ , subtract $a+d=7$ from $a+4d=19$ : $3d=12$ , so $d=4$ and then $a=3$ .

Rebuild the stated terms: $u_2=3+4=7$ and $u_5=3+16=19$ . This catches an off-by-one error in the multiplier of $d$ .

The nth term contains $(n-1)d$ , not $nd$ , because the first term is reached before any common-difference steps have been taken.

The formula $u_n=a+(n-1)d$ gives any term of an arithmetic sequence directly from its first term $a$ , common difference $d$ , and position $n$ .

Goal	Set up
find a term	substitute its position for $n$
find a position	set $a+(n-1)d$ equal to the given value
compare terms	use $u_q-u_p=(q-p)d$
write linear form	expand to $u_n=dn+(a-d)$

For $a=4$ and $d=3$ , $u_{50}=4+49(3)=151$ . The multiplier is 49 because there are 49 steps from the first term to the 50th.

If $4+(n-1)3=100$ , then $3n+1=100$ and $n=33$ , so 100 is the 33rd term. A non-positive or non-integer result would mean it is not a term.

This formula applies to arithmetic sequences only. A constant ratio or changing difference requires a different model.

An arithmetic series is the sum of the terms of an arithmetic sequence. Pairing first and last terms gives $S_n=\frac n2(a+l)=\frac n2[2a+(n-1)d]$ .

Information available	Efficient form
first term $a$ and last term $l$	$S_n=\frac n2(a+l)$
first term $a$ and difference $d$	$S_n=\frac n2[2a+(n-1)d]$
a range of terms	subtract cumulative sums or re-index carefully

For $4+7+10+13+\cdots$ with 50 terms, $a=4$ , $d=3$ , so $S_{50}=\frac{50}{2}[2(4)+49(3)]=25(155)=3875$ .

The last term is $l=a+(n-1)d$ . Writing the series forwards and backwards makes each pair total $a+l$ ; there are $n$ such pairs across two copies of the series.

Do not substitute the last term as $n$ . First determine how many terms are being summed, especially when the sum starts after the first term.

3.3 Graphs

A graph tells a story about how one quantity changes with another. Read the axis labels, units and scale before interpreting its shape.

Feature	Distance–time meaning	Speed–time meaning
horizontal segment	stopped	constant speed
steeper segment	faster travel	faster acceleration or deceleration
rising segment	moving away	speed increasing
falling segment	returning	speed decreasing

To answer a question, locate the given value on one axis, move to the graph, then read the corresponding value from the other axis. For an interval, compare its two endpoints.

A negative gradient on a distance-from-home graph means returning towards home; distance travelled itself has not become negative.

Rectangular Cartesian coordinates use two perpendicular number lines: the horizontal $x$ -axis and vertical $y$ -axis. Their intersection is the origin $(0,0)$ .

Convention	Meaning
$(x,y)$	horizontal coordinate first, vertical coordinate second
positive $x$	right of the origin
negative $x$	left of the origin
positive $y$	above the origin
negative $y$	below the origin

Check the scale on each axis separately: one square need not represent one unit, and the two axes may use different scales.

$(3,-2)$ and $(-2,3)$ are different points. Never reverse the coordinate order to match the direction you move.

To plot $(x,y)$ , start at the origin, move horizontally to $x$ , then vertically to $y$ . To read a point, project it to the $x$ -axis first and the $y$ -axis second.

Quadrant	Sign of $x$	Sign of $y$
I	+	+
II	−	+
III	−	−
IV	+	−

A point on the $x$ -axis has $y=0$ ; a point on the $y$ -axis has $x=0$ . Such points are not in any quadrant.

After plotting, read the point back from the axes. This catches swapped coordinates and incorrect signs.

Geometrical facts can fix a point's horizontal coordinate, vertical coordinate or both. Translate each fact into a coordinate constraint before calculating.

Geometrical fact	Coordinate consequence
vertical alignment	same $x$ -coordinate
horizontal alignment	same $y$ -coordinate
reflection in $x$ -axis	$(x,y)\to(x,-y)$
reflection in $y$ -axis	$(x,y)\to(-x,y)$
translation by $\binom{a}{b}$	$(x,y)\to(x+a,y+b)$

Mark known coordinates, use shape properties such as equal sides, parallel lines or perpendicular diagonals, and solve only for coordinates not already fixed.

A diagram may not be drawn to scale. Coordinates must follow stated properties and axis values, not visual appearance.

For endpoints $A(x_1,y_1)$ and $B(x_2,y_2)$ , the midpoint is $M=((x_1+x_2)/2,(y_1+y_2)/2)$ .

For $A(-5,4)$ and $B(3,-2)$ , $M=((-5+3)/2,(4-2)/2)=(-1,1)$ .

The midpoint lies halfway in both directions, so average the two $x$ -coordinates and independently average the two $y$ -coordinates.

Do not average an $x$ -coordinate with a $y$ -coordinate, and keep negative values inside brackets when adding.

A straight-line conversion graph represents a constant rate between two quantities. If zero converts to zero, the line passes through the origin.

Task	Action
draw	choose accurate conversion pairs, plot them, join with a straight line
convert across	start at the known axis, move to the line, then to the other axis
check	confirm direction, units and sensible magnitude
extend	use the same constant gradient only while the rate remains fixed

If 10 dollars converts to 60 krone, the rate is 6 krone per dollar. Thus 40 dollars corresponds to 240 krone.

A conversion graph can be read in either direction, but the numerical rate must be inverted when the direction is reversed.

For a straight line, $\text{gradient}=\Delta y/\Delta x=\text{rise}/\text{run}$ . Use two points on the line and measure changes in the same direction.

Line direction left to right	Gradient
rises	positive
falls	negative
horizontal	zero
vertical	undefined

Choose two clear, widely separated points on the line. Form a right-angled gradient triangle and divide the vertical change by the horizontal change.

Gradient is not $y/x$ unless one chosen point is the origin. In general it is a change divided by a change.

In $y=mx+c$ , $m$ is the gradient and $c$ is the $y$ -coordinate where the line crosses the $y$ -axis, so the intercept is $(0,c)$ .

Task	Method
read equation	identify coefficient $m$ and constant $c$
write from graph	find gradient, then read the $y$ -intercept
write from gradient and intercept	substitute directly into $y=mx+c$
rearrange	make $y$ the subject before reading $m,c$

For $2y=-7x+10$ , divide every term by 2: $y=-3.5x+5$ . The gradient is $-3.5$ and the intercept is $(0,5)$ .

The $x$ -intercept is not $c$ . Set $y=0$ and solve separately if an $x$ -intercept is required.

Choose suitable $x$ -values, calculate each $y$ -value accurately, plot the coordinate pairs, then join them with the correct shape.

Function	Expected graph
$y=mx+c$	straight line
$x=k$	vertical line
$y=c$	horizontal line
$y=ax^2+bx+c$	smooth parabola
$ax+by=c$	rearrange to straight-line form when useful

For a quadratic, use enough points around the turning point and look for symmetry. For a linear graph, two correct points determine the line, but a third point checks arithmetic.

Join quadratic points with a smooth curve, not straight line segments. Do not force a straight line through points from a non-linear rule.

A graph family is recognised from its intercepts, symmetry, turning points, end behaviour, asymptotes and periodicity—not from one isolated point.

Family	Signature features
quadratic	one turning point; parabolic shape
cubic	opposite end directions; up to two turning points
reciprocal $k/x$	two branches; axes are asymptotes
sine/cosine	smooth periodic waves
tangent	repeating branches with vertical asymptotes

For polynomial graphs, calculate a table including intercept regions and turning behaviour. For trigonometric graphs in degrees, mark key angles and repeat using the correct period.

A reciprocal or tangent graph must not be drawn through a vertical asymptote. Separate branches never join across an undefined input.

A transformation changes known points of $y=f(x)$ without rebuilding the whole value table. Changes outside $f$ act vertically; changes inside $f$ act horizontally.

New graph	Point mapping from $(x,y)$	Effect
$y=f(x)+a$	$(x,y)\to(x,y+a)$	up by $a$
$y=af(x)$	$(x,y)\to(x,ay)$	vertical scale factor $a$
$y=f(x+a)$	$(x,y)\to(x-a,y)$	left by $a$
$y=f(ax)$	$(x,y)\to(x/a,y)$	horizontal scale factor $1/a$

Transform several defining points—intercepts, vertices and turning points—then preserve the original curve's connections and shape.

Inside changes act in the opposite horizontal direction: $f(x+3)$ moves the graph 3 units left, not right.

Compare a distinctive point on the original and transformed graphs. Decide whether $x$ -coordinates or $y$ -coordinates changed, then test the corresponding transformation rule on another point.

Observation	Likely algebra
every $y$ increases by $a$	$f(x)+a$
every $x$ decreases by $a$	$f(x+a)$
every $y$ is multiplied by $a$	$af(x)$
every $x$ is divided by $a$	$f(ax)$

Use invariant features: vertical translations keep $x$ -coordinates of turning points; horizontal transformations keep their $y$ -coordinates.

Describe exactly one coherent transformation unless the graph genuinely shows a combination. A visual guess must be verified with coordinates.

A curve has a changing gradient. Its gradient at one point is estimated by the gradient of the tangent touching the curve there.

Step	Action
1	draw a tangent that matches the curve's local direction
2	choose two well-separated points on the tangent
3	calculate $\Delta y/\Delta x$
4	include a sign and sensible precision

Use points on the tangent, not necessarily points on the curve. A large gradient triangle reduces the effect of reading error.

A chord through two curve points gives an average gradient over an interval; it is not automatically the gradient at the named point.

At an intersection, two graphs have the same $x$ and $y$ . Therefore the intersection $x$ -coordinates solve $y_2-y_1=0$ .

Task	Action
prepare	write each side as a graph $y=y_1$ and $y=y_2$
draw	plot both on the same axes
solve	read every intersection's $x$ -coordinate
report	give estimates to precision supported by the scale

To solve $x^2-5x=x-7$ , plot $y=x^2-5x$ and $y=x-7$ , or rearrange consistently to use an already drawn curve and a suitable line.

Do not report the intersection's $y$ -coordinate when the equation asks for $x$ . Check for all intersections in the shown domain.

For distinct points $(x_1,y_1)$ and $(x_2,y_2)$ , $m=(y_2-y_1)/(x_2-x_1)$ .

Between $(9,-4)$ and $(5,8)$ , $m=(8-(-4))/(5-9)=12/(-4)=-3$ .

Subtract coordinates in the same order on top and bottom. Reversing both orders gives the same gradient.

If $x_2=x_1$ , the denominator is zero and the vertical line has undefined gradient—not gradient zero.

Parallel lines have equal gradients. For two non-vertical perpendicular lines, their gradients satisfy $m_1m_2=-1$ .

Step	Action
1	rearrange the given line to identify its gradient
2	keep that gradient for parallel, or use the negative reciprocal for perpendicular
3	substitute the given point into $y=mx+c$ to find $c$
4	verify both the point and gradient relationship

A line perpendicular to $y=-4x+5$ has gradient $1/4$ . Through $(3,7)$ it satisfies $7=(1/4)(3)+c$ , so $c=25/4$ .

Changing only the sign is not enough for a perpendicular gradient. Take the reciprocal and change the sign.

3 Sequences, functions and graphs

Exam analysis

Most tested topics

3.1 Sequences

Generate sequence terms from two kinds of rule

Continue an integer sequence and state its rule

Find and use the nth term of an arithmetic sequence

Recover the first term and common difference

Use the arithmetic nth-term formula

Sum the first n terms of an arithmetic series

3.3 Graphs

Interpret journeys and other real-world graphs

Use Cartesian coordinates consistently

Plot and read points in all four quadrants

Deduce coordinates from geometrical information

Find the midpoint of a line segment

Draw and use straight-line conversion graphs

Understand gradient as a rate of change

Use the straight-line form y = mx + c

Generate and plot linear and quadratic graphs

Recognise polynomial, reciprocal and trigonometric graphs

Transform graphs from y = f(x)

Infer the algebra behind a graph transformation

Estimate a non-linear gradient using a tangent

Solve equations using graph intersections

Calculate gradient from two coordinates

Find equations of parallel and perpendicular lines