A Algebra · graphs — 1MA1 revision notes

A14

Plot and interpret graphs (including reciprocal and exponential graphs) and graphs of non-standard functions in real contexts, to find approximate solutions e.g. simple kinematic problems

Notes

Evidence from your answers: none yet

Your confidence:

A self-report of how sure you feel. It does not measure mastery. Evidence from your answers reaches secure after the latest Tier 2/3 attempt is correct, with three correct distinct drills across at least two dates and two practice sources.

Explanation

A graph represents the ordered pairs satisfying a rule. Use intercepts, turning points and asymptotes to describe its behaviour.
Reciprocal graphs such as $y=k/x$ have two branches and approach the axes without meeting them.
Higher tier: exponential graphs change by a constant multiplier for equal changes in $x$ .
To solve equations graphically, plot both relations on the same axes and read every intersection accurately.
In a real context, the examiner expects the coordinate to be interpreted using the quantities and units on the axes, not reported as an unexplained pair of numbers.

For

y=k/x

with

k>0

, reciprocal branches lie in opposite quadrants and approach both axes.

Worked example

The journey time $t$ hours is modelled by $t=24/v$ , where $v$ is the average speed in km/h. Find and interpret the point when $v=16$ .

1.Substitute $v=16$ into the model: $t=24\div16$ .
2.Evaluate to obtain $t=1.5$ , so the coordinate is $(16,1.5)$ .
3.Interpret the axes: at $16$ km/h, the journey takes $1.5$ hours.

Answer: $(16,1.5)$ ; travelling at $16$ km/h gives a journey time of $1.5$ hours.

Common mistakes

Don't read the $y$ -coordinate when the question asks for the solution in $x$ .
Don't give an intersection such as $(16,1.5)$ without interpreting either coordinate or its unit.

Exam tip

For an 'estimate' question, show the plotted curves and read the intersection to a precision justified by the graph scale.

Tier 1 · Easy

ORIGINAL

1

The time

t

hours for a fixed journey is modelled by

t=24/v

, where

v

is the average speed in km/h. Work out the point on this graph when

v=16

and interpret it.

(2)

(Total for Question 1 is 2 marks)

Tier 2 · Standard

ORIGINAL

1

Higher only: Plot

y=2^x

and

y=9-2x

for

1\le x\le3

. Use the intersection to estimate the solution of

2^x=9-2x

to one decimal place.

(4)

(Total for Question 1 is 4 marks)

Tier 3 · Hard

ORIGINAL

1

For

0\le t\le5

, two moving objects have distances from a marker modelled by

d=3t^2+2

and

d=14t

, with

d

in metres and

t

in seconds. Draw both graphs and estimate the later time when the objects are equally far from the marker.

(5)

(Total for Question 1 is 5 marks)

A15

Calculate or estimate gradients of graphs and areas under graphs (incl. quadratic and other non-linear); interpret e.g. distance-time, velocity-time and financial graphs (not calculus) [Higher only]

Notes

Worked answers & exam appearances →

Evidence from your answers: none yet

Your confidence:

A self-report of how sure you feel. It does not measure mastery. Evidence from your answers reaches secure after the latest Tier 2/3 attempt is correct, with three correct distinct drills across at least two dates and two practice sources.

Explanation

Higher tier only. Gradient is change in the vertical coordinate divided by change in the horizontal coordinate, with units formed from the axis units.
For a curve, draw a tangent at the required point and use two well-separated points on that tangent, not two points on the curve.
Estimate an area under a non-linear graph by dividing it into strips and applying the trapezium rule.
On a distance-time graph gradient represents speed; on a velocity-time graph gradient represents acceleration and area represents displacement.
The examiner expects a numerical result, correct units and an interpretation of its sign where relevant.

Worked example

A distance-time graph is a straight line from $(12,150)$ to $(32,510)$ , with time in seconds and distance in metres. Calculate and interpret its gradient.

1.Use change in distance divided by change in time: $\dfrac{510-150}{32-12}$ .
2.Evaluate $\dfrac{360}{20}=18$ .
3.Attach the units metres per second and interpret the constant straight-line gradient.

Answer: $18$ m/s; the object travels at a constant speed of $18$ m/s.

Common mistakes

Don't use two points on the curve instead of two points on the tangent when estimating a gradient.
Don't add trapezium heights without multiplying by half the strip width.
Don't report a velocity-time area in m/s instead of metres.

Exam tip

For an estimate, leave the tangent or trapezia visible because the method marks depend on the construction.

Tier 1 · Easy

ORIGINAL

1

A distance-time graph is a straight line from

(12,150)

to

(32,510)

, where time is in seconds and distance is in metres. Calculate and interpret its gradient.

(3)

(Total for Question 1 is 3 marks)

Tier 2 · Standard

ORIGINAL

1

A velocity-time graph joins the points

(0,0)

,

(6,15)

,

(14,15)

and

(20,3)

with straight lines. Work out the distance travelled in the first

20

seconds.

(4)

(Total for Question 1 is 4 marks)

Tier 3 · Hard

ORIGINAL

1

A curved velocity-time graph passes through the values

v=18,25,29,24,10

m/s at

t=0,3,6,9,12

seconds. Use four trapezia to estimate the distance travelled. A tangent at

t=6

passes through

(3,38)

and

(11,14)

; estimate the acceleration then.

(6)

(Total for Question 1 is 6 marks)

A16

Recognise and use the equation of a circle with centre at the origin; find the equation of a tangent to a circle at a given point [Higher only]

Notes

Worked answers & exam appearances →

Evidence from your answers: none yet

Your confidence:

A self-report of how sure you feel. It does not measure mastery. Evidence from your answers reaches secure after the latest Tier 2/3 attempt is correct, with three correct distinct drills across at least two dates and two practice sources.

Explanation

Higher tier only. A circle centred at $(0,0)$ with radius $r$ has equation $x^2+y^2=r^2$ .
Check that a given point lies on the circle by substituting its coordinates. The radius from the origin to a point $(a,b)$ has gradient $b/a$ when $a\ne0$ .
The tangent at that point is perpendicular to the radius, so its gradient is the negative reciprocal, and a point-gradient equation gives the tangent.
The examiner expects an exact equation, normally simplified, and it should pass through the stated point.
Horizontal and vertical radii produce vertical and horizontal tangents respectively.

The tangent at a point on a circle is perpendicular to the radius.

Worked example

The point $P(5,12)$ lies on $x^2+y^2=169$ . Find the equation of the tangent at $P$ .

1.The radius $OP$ has gradient $12/5$ , so the tangent gradient is $-5/12$ .
2.Use point-gradient form: $y-12=-\dfrac{5}{12}(x-5)$ .
3.Multiply by $12$ and rearrange to obtain $5x+12y=169$ .

Answer: $5x+12y=169$ .

Common mistakes

Don't use the radius gradient $12/5$ as the tangent gradient.
Don't find the negative reciprocal correctly but write a line that does not pass through the given point.

Exam tip

Substitute the contact point into your final tangent equation for a quick accuracy check.

Tier 1 · Easy

ORIGINAL

1

Write down the equation of the circle with centre

(0,0)

and radius

7

.

(1)

(Total for Question 1 is 1 mark)

Tier 2 · Standard

ORIGINAL

1

The point

P(5,12)

lies on the circle

x^2+y^2=169

. Determine the tangent's equation there.

(4)

(Total for Question 1 is 4 marks)

Tier 3 · Hard

ORIGINAL

1

The tangent to

x^2+y^2=50

at

Q(1,7)

meets the positive coordinate axes. Find the exact area of the triangle enclosed by the tangent and the axes.

(5)

(Total for Question 1 is 5 marks)

A17

Solve linear equations in one unknown algebraically (including those with the unknown on both sides of the equation); find approximate solutions using a graph

Notes

Worked answers & exam appearances →

Evidence from your answers: none yet

Your confidence:

A self-report of how sure you feel. It does not measure mastery. Evidence from your answers reaches secure after the latest Tier 2/3 attempt is correct, with three correct distinct drills across at least two dates and two practice sources.

Definition

Specification A17: solve linear equations in one unknown algebraically, including those with the unknown on both sides of the equation; find approximate solutions using a graph.

Explanation

Do the same thing to both sides until the unknown stands alone. With a fraction, multiply every term on both sides by the denominator, writing the line $5-x=2(2x-7)$ before expanding; the method mark needs the whole of each side multiplied, not just one term.
Expand brackets, then collect the $x$ terms on one side and the numbers on the other in a single visible step such as $4x+x=14+5$ .
Divide last, and leave the answer as an exact fraction or decimal; $\dfrac{19}{5}$ and $3.8$ both score, but $\dfrac{-19}{-5}$ does not.
Check by substituting into the original equation.
A graphical solution is the $x$ -coordinate where the two graphs meet, read to the accuracy of the grid.

Another way to see this:StandardVisual

Worked example

1

Solve

\dfrac{3(x-2)}{4}-\dfrac{x+1}{3}=\dfrac{5}{6}

.

(3)

(Total for Question 1 is 3 marks)

Working

1.Multiply every term by $12$ , the lowest common multiple of $4$ , $3$ and $6$ : $9(x-2)-4(x+1)=10$ .
2.Expand the brackets: $9x-18-4x-4=10$ .
3.Collect the $x$ terms and the numbers: $5x=10+18+4=32$ .
4.Divide by $5$ : $x=\dfrac{32}{5}$ .

Answer: $x=\dfrac{32}{5}$

Mark scheme

Mark scheme for question 1
Question	Answer	Mark	Additional guidance
1	for a correct first step, eg $9(x-2)-4(x+1)=10$ or $9x-18-4x-4=10$	M1	Method must show every term multiplied by 12 (or a common multiple)
	(dep) for isolating terms in $x$ , eg $9x-4x=10+18+4$ or $5x=32$	M1
	for $\dfrac{32}{5}$	A1	oe; accept 6.4 or $6\dfrac{2}{5}$

Common mistakes

Don't multiply only the fractional terms, not every term, when clearing denominators.
Don't change the sign of a term when moving it without performing the same addition or subtraction on both sides.

Mark-scheme wording to use

“for a correct first step, eg 5 – x = 2(2x − 7) or 5 – x = 4x − 14 or 5/2 – x/2 = 2x – 7” — Pearson 1MA1, June 2018 Paper 3H, Q7
“(dep) for isolating terms in x … Accept 19/5 , 3 4/5 oe but not −19/−5 oe” — Pearson 1MA1, June 2018 Paper 3H, Q7
“for correct expansion of the bracket or dividing all terms by 3 as a first step … for isolating terms in x on one side of an equation eg 5x – 6 – 3x = −3” — Pearson 1MA1, November 2017 Paper 2H, Q1

What the examiner wants

Every term treated the same. The June 2018 Paper 3H scheme states that the method must show the left side and both terms on the right multiplied by $2$ ; the November 2024 Paper 3H report says many knew to multiply both sides by the denominator but too often did not carry it out correctly, and then lost the marks for isolating the terms.

Exam tip

For a multi-mark 'solve' question, keep each balancing step visible so an arithmetic slip does not lose the method marks.

Tier 1 · Easy

ORIGINAL

1

Solve

7x+5=33

.

(2)

(Total for Question 1 is 2 marks)

Tier 2 · Standard

ORIGINAL

1

Draw

y=4.5x-2

and

y=9.2-x

on the same axes. Use the intersection to solve

4.5x-2=9.2-x

, giving an estimate to one decimal place.

(4)

(Total for Question 1 is 4 marks)

Tier 3 · Hard

ORIGINAL

1

Solve

\frac{3x+1}{5}+\frac{13-2x}{7}=\frac{5x+9}{8}

.

(4)

(Total for Question 1 is 4 marks)

Plot and interpret graphs (including reciprocal and exponential graphs) and graphs of non-standard functions in real contexts, to find approximate solutions e.g. simple kinematic problems

Tier 1 · Easy

Tier 2 · Standard

Tier 3 · Hard

Calculate or estimate gradients of graphs and areas under graphs (incl. quadratic and other non-linear); interpret e.g. distance-time, velocity-time and financial graphs (not calculus) [Higher only]

Tier 1 · Easy

Tier 2 · Standard

Tier 3 · Hard

Recognise and use the equation of a circle with centre at the origin; find the equation of a tangent to a circle at a given point [Higher only]

Tier 1 · Easy

Tier 2 · Standard

Tier 3 · Hard

Solve linear equations in one unknown algebraically (including those with the unknown on both sides of the equation); find approximate solutions using a graph

Tier 1 · Easy

Tier 2 · Standard

Tier 3 · Hard

Want help turning these notes into marks?