Edexcel IGCSE Math A Higher 3.3 Graphs Questions

Use this graphs page to read coordinates, plot points, work with straight-line equations and interpret graph features in real data contexts.

Exam points

Plot and read coordinates across quadrants, including points from geometric diagrams.
Find midpoint, gradient or straight-line equations from coordinates or graph features.
Use plotted curves, tangents or intersections to estimate solutions and gradients.

Covered syllabus points

3.3.A Interpreting linear and non-linear graphs 3.3.B Cartesian coordinate conventions 3.3.C Points (x, y) in any of the four quadrants or locate 3.3.D The coordinates of points identified by geometrical 3.3.E Midpoint coordinates 3.3.F Straight-line conversion graphs 3.3.G Gradient of a straight line 3.3.H Straight-line graphs in the form y = mx + c 3.3.I Linear and quadratic function graphs 3.3.HA Polynomial, reciprocal and trigonometric graphs 3.3.HB Transformations of graphs 3.3.HC Analysing transformations of functions 3.3.HD The gradients of non-linear graphs By drawing a tangent 3.3.HE Intersections of linear and non-linear graphs 3.3.HF Gradient from two points 3.3.HG Parallel and perpendicular line equations

[Maximum number: 3]

Jalina left her home at 1000 to cycle to a park.
On her way to the park, she stopped at a friend's house and then continued her journey to the park.
Here is the distance-time graph for her journey to the park.

Figure for Question 1 — Edexcel IGCSE Math A Higher

(a)

On her journey to the park, did Jalina cycle at a faster speed before or after she stopped at her friend's house?
Give a reason for your answer.

Jalina stayed at the park for 45 minutes.
She then cycled, without stopping, at a constant speed of $16 \mathrm{~km} / \mathrm{h}$ from the park back to her home.

[ 1 ]

$15 \mathrm{~km} / \mathrm{h}$ or $\frac{25}{6} \mathrm{~m} / \mathrm{sec}$ or $0.25 \mathrm{~km} / \mathrm{min}$ or $\frac{15}{4}$
oe
$12 \mathrm{~km} / \mathrm{h}$ or $\frac{10}{3} \mathrm{~m} / \mathrm{sec}$ or $0.2 \mathrm{~km} / \mathrm{min}$ or $\frac{9}{3}$ oe

'before' with
reason

1
e.g. before as gradient is steeper
or before as speed before is 15
km / h speed after is $12 \mathrm{~km} / \mathrm{h}$ or
before as she goes over 11(allow
11-12 ) km in $\frac{3}{4}$ hour but only
goes 9 km in $\frac{3}{4}$ hour after oe
NB: any figures used for the
reason must be accurate if they
haven't used 'gradient is steeper'
oe

(b)

Show all this information on the distance-time graph.

[ 2 ]

Line from:
$(12{:}00,24)\to(12{:}45,24)\to(14{:}15,0)$

Marking guidance:
- B2 for the complete line
- If not B2, award B1 for either line segment, or for a time of 1.5 hours seen

[Maximum number: 1]

Use your answer to part (a) to write down the coordinates of A
(...

From:
$19-3(x-2)^2$

Maximum point:
(2,19)

[Maximum number: 6]

25 A B C D is a trapezium with AB parallel to DC
A is the point with coordinates (-4,6)
B is the point with coordinates (2,3)
D is the point with coordinates (-1,8)
The trapezium has one line of symmetry.
The line of symmetry intersects CD at the point E
Work out the coordinates of the point E

Midpoint of AB:
$\left(\frac{-4+2}{2},\frac{6+3}{2}\right)=(-1,4.5)$
Gradient of AB:
$\frac{6-3}{-4-2}=-\frac12$
Gradient of the line of symmetry:
2
Equation of the line of symmetry through (-1,4.5):
y=2x+6.5
Equation of CD through D(-1,8), parallel to AB:
y=-0.5x+7.5
Intersection:
2x+6.5=-0.5x+7.5 $x=0.4,\qquad y=7.3$

Final answer:
(0.4,7.3)

M marks for midpoint, gradient of AB, perpendicular gradient, equations, and solving the intersection. A1 oe.

[Maximum number: 2]

The point A has coordinates (5,-4)
The point B has coordinates (13,1)

Work out the coordinates of the midpoint of AB.
(, )

Line L has equation y=2-3 x

$\left(\frac{5+13}{2},\frac{-4+1}{2}\right) = (9,-1.5)$

Final answer:
(9,-1.5)

[Maximum number: 2]

The diagram shows the graph of y=f(x) for $-4 \leqslant x \leqslant 12$

Figure for Question 5 — Edexcel IGCSE Math A Higher

The point P on the curve has x coordinate 2

Hence find an equation of the tangent to the curve at P.

Give your answer in the form y=m x+c

The equation f(x)=k has exactly two different solutions for $-4 \leqslant x \leqslant 12$

[Maximum number: 3]

Figure for Question 6 — Edexcel IGCSE Math A Higher

On the grid, draw and label with its equation the straight line with equation

\begin{aligned}(i)\;& y=1 \\ (ii)\;& x=2 \\ (iii)\;& x+y=7\end{aligned}

[Maximum number: 1]

Here are six graphs.
A

Figure for Question 7 — Edexcel IGCSE Math A Higher

Write down the letter of the graph that could have the equation

$y=-\frac{1}{x}$

[Maximum number: 2]

The graph of $y=\sin x^{\circ}$ for $0 \leqslant x \leqslant 360$ is drawn on the grid.

Figure for Question 8 — Edexcel IGCSE Math A Higher

On the grid, draw the graph of $y=2 \sin (x+30)^{\circ}$ for $0 \leqslant x \leqslant 360$

Correct graph: must go through
$(60,2),\quad (150,0),\quad (240,-2),\quad (330,0)$
and not through (0,0).

Marking guidance:
- B2 for the correct graph
- If not B2, award B1 for the correct shape through 2 or 3 given points, a clear stretch of scale factor 2, or a clear translation of $\binom{-30}{0}$

[Maximum number: 2]

The graph of $y=\sin x^{\circ}$ for $0 \leqslant x \leqslant 360$ is drawn on the grid.

Figure for Question 9 — Edexcel IGCSE Math A Higher

Hence, describe fully the single transformation that maps the curve with equation $y=x^{2}$ onto the curve with equation $y=x^{2}-6 x+10$

Translation by:
$\begin{pmatrix} 3\\ 1 \end{pmatrix}$

Marking guidance:
- B1 for translation
- B1 for $\begin{pmatrix}3\\1\end{pmatrix}$ , ft from part (b)(i);
must be a column vector

[Maximum number: 3]

The diagram shows the graph of y=f(x) for $-4 \leqslant x \leqslant 12$

Figure for Question 10 — Edexcel IGCSE Math A Higher

The point P on the curve has x coordinate 2

Use the graph to find an estimate for the gradient of the curve at P.

[Maximum number: 2]

The diagram shows the graph of y=f(x) for $-4 \leqslant x \leqslant 12$

Figure for Question 11 — Edexcel IGCSE Math A Higher

The point P on the curve has x coordinate 2

Use the graph to find the two possible values of k.

Final answers:
$k=3,\qquad k=-1$

[Maximum number: 2]

Point A has coordinates (5,8)
Point B has coordinates (9,-4)

Work out the gradient of AB.

The straight line L has equation y=-4 x+5

$\frac{8-(-4)}{5-9}=\frac{12}{-4}=-3$

Final answer:
-3

[Maximum number: 1]

Point A has coordinates (5,8)
Point B has coordinates (9,-4)

Write down the gradient of a straight line that is perpendicular to L.

A line perpendicular to y=-4x+5 has gradient:
$\frac14$

Equivalent answers include 0.25 or -1/-4.

All question bank results loaded

Covered syllabus points

Question 1

Question (a)

Question (b)

Question 2

Question 3

Question 4

Question 5

Question 6

Question 7

Question 8

Question 9

Question 10

Question 11

Question 12

Question 13