A-Level Maths (Edexcel 9MA0) · Pure Year 1 · Chapter 4: Graphs and transformations · Lesson 3

Transforming graphs

70 marks

Hints online, answers free with an account:
mathswariz.uk/worksheets/a-level-transforming-graphs/

Date

Do Now

Quick questions to start.

A1
Needed today
Factorise fully $x^{3} - 7 x^{2} + 12 x$ .
[2 marks]
Answer
A2
Last lesson
Find the coordinates of the points where $y = \frac{3}{x}$ meets $y = x + 2$ .
[2 marks]
Answer
A3
From chapter 2
Write $x^{2} + 14 x + 43$ in the form $(x + p)^{2} + q$ .
[2 marks]
Answer
A4
Needed today
$f (x) = 7 x - x^{2}$ . Write $f (x + 1)$ in the form $a x^{2} + b x + c$ .
[2 marks]
Answer

Example, then your turn

A-Level

Your teacher works through each example with you. Then try the one beside it.

Example 1

$f (x) = (x + 1) (x - 3)^{2}$ . On the same axes sketch $y = f (x)$ and $y = f (x + 2)$ , showing where each curve meets the axes.

Hint

f (x + 2)

moves the curve

2

to the left. Write it as

(x + 3) (x - 1)^{2}

to find the new

y

-intercept.

B1
$g (x) = (2 - x) (x + 3)^{2}$ . Sketch $y = g (x - 1)$ , showing where the curve meets the axes.
[4 marks]

Example 2

On separate axes sketch $y = \frac{4}{x} + 2$ and $y = \frac{4}{x + 2}$ . Give the equations of the asymptotes and the coordinates of any point where each curve meets an axis.

Hint

+ 2

outside moves

y = \frac{4}{x}

2

;

x + 2

inside moves it

2

left. Each asymptote moves with the curve.

B2
On separate axes sketch $y = \frac{9}{x ^{2}} - 1$ and $y = \frac{9}{( x - 1 ) ^{2}}$ . Give the equations of the asymptotes and the coordinates of any point where each curve meets an axis.
[5 marks]

Example 3

$f (x) = (x + 4) (x - 2)$ , whose minimum point is $(- 1, - 9)$ . On separate axes sketch $y = f (2 x)$ and $y = - f (x)$ , showing where each meets the axes and its turning point.

Hint

f (2 x)

halves every

x

-coordinate and keeps every

y

- f (x)

keeps every

x

and changes the sign of every

y

B3
$g (x) = (x - 1) (x + 7)$ , whose minimum point is $(- 3, - 16)$ . On separate axes sketch $y = 3 g (x)$ and $y = g (- x)$ , showing where each meets the axes and its turning point.
[4 marks]

Example 4

The diagram shows $y = f (x)$ , which has a maximum at $A (- 1, 7)$ and a minimum at $B (3, - 5)$ . Find where $A$ and $B$ go on (a) $y = f (\frac{1}{2} x)$ (b) $y = 3 - f (x)$ , saying which is now the maximum.

Hint Write

3 - f (x)

- f (x) + 3

: reflect in the

x

-axis first, then move up

3

B4
$P (- 3, 4)$ and $Q (1, - 2)$ lie on $y = f (x)$ . Find where $P$ and $Q$ go on (a) $y = f (3 x)$ (b) $2 y = f (x) - 6$ .
[4 marks]
Answer

Practice

A-Level

C1
The point $(6, - 12)$ on $y = f (x)$ goes to $(2, - 4)$ on $y = p f (q x)$ and to $(9, - 7)$ on $y = f (x - r) + s$ . Find $p$ , $q$ , $r$ and $s$ .
[4 marks]
Answer

C2
The curve $y = \frac{a}{x + b} + c$ has asymptotes $x = - 3$ and $y = 2$ , and passes through the point $(- 1, 6)$ . Find the values of $a$ , $b$ and $c$ .
[3 marks]
Answer
C3
The diagram shows the curve $y = f (x)$ . It has asymptotes $x = - 1$ and $y = 2$ , and it passes through $(1, 0)$ and $(0, - 2)$ . On separate axes sketch $y = f (2 x)$ and $y = - f (x)$ . Give the equations of the asymptotes and the coordinates of the points where each curve meets the axes.
[5 marks]

Exam-style questions

Edexcel exam style

D1
$f (x) = x^{2} - 16 x + 48$ .
(a)
Write $f (x)$ in the form $(x - p)^{2} - q$ .
[2 marks]
Answer
(b)
Hence describe fully the single transformation that maps the curve $y = x^{2}$ onto the curve $y = f (x)$ .
[2 marks]
Answer
(c)
Hence sketch $y = f (2 x)$ , showing the coordinates of its minimum point and of the points where it meets the axes.
[3 marks]
(d)
The curve $y = f (x) + k$ meets the $x$ -axis at exactly one point. Find the value of $k$ .
[2 marks]
Answer
Total for D1: 9 marks

D2
The curve $y = f (x)$ has one turning point, a minimum at $(4, - 3)$ . Josh says: " $y = 2 f (x) + 5$ is $y = f (x)$ moved up $5$ and then stretched by scale factor $2$ parallel to the $y$ -axis, so its minimum point is $(4, 4)$ ."
Josh is wrong.
(a)
Write down the equation of the curve that Josh's description gives, and find its minimum point.
[2 marks]
Answer
(b)
Hence explain Josh's mistake, and find the minimum point of $y = 2 f (x) + 5$ .
[3 marks]
Answer
Total for D2: 5 marks
D3
$f (x) = (x - 4) (x + 2)^{2}$ , and $y = f (x)$ has a minimum point at $(2, - 32)$ . The curve $y = f (x + p) + q$ touches the $x$ -axis at the origin. Find the two possible pairs of values of $p$ and $q$ .
[5 marks]
Answer
D4
The diagram shows the curve $y = f (x)$ . It has exactly two turning points: a maximum at $B (- 2, 4)$ and a minimum at $C (3, - 6)$ . As $x \to \infty$ , $y \to \infty$ , and as $x \to - \infty$ , $y \to - \infty$ .
(a)
Find where $B$ and $C$ go on $y = - 2 f (x)$ , saying which is now the maximum.
[2 marks]
Answer
(b)
Find where $B$ and $C$ go on $y = f (3 x) + 2$ .
[2 marks]
Answer
(c)
Find the set of values of $k$ for which $f (x) = k$ has exactly three real solutions.
[2 marks]
Answer
(d)
Hence find the set of values of $c$ for which $2 f (x) + c = 0$ has exactly three real solutions.
[2 marks]
Answer
Total for D4: 8 marks

Extension

Stretch

No route is given. The hints are at the end of the sheet. Use one at a time.

E1
Describe a sequence of transformations that maps the curve $y = \frac{1}{x}$ onto the curve $y = \frac{3 x - 1}{x + 2}$ . Hence sketch $y = \frac{3 x - 1}{x + 2}$ , giving its asymptotes and the points where it meets the axes.
[6 marks]
Hint 1 · What to try
Write the top as $3 (x + 2) - 7$ .
Hint 2 · The first line
So $y = 3 - \frac{7}{x + 2}$ . Build it from $\frac{1}{x}$ one change at a time: $\frac{1}{x + 2}$ , then $\frac{7}{x + 2}$ , then $- \frac{7}{x + 2}$ , then add $3$ .
Hint 3 · The full method
$3 x - 1 = 3 (x + 2) - 7$ , so $y = 3 - \frac{7}{x + 2}$ . A translation by $(- 2 0)$ gives $\frac{1}{x + 2}$ ; a stretch parallel to the $y$ -axis, scale factor $7$ , gives $\frac{7}{x + 2}$ ; a reflection in the $x$ -axis gives $- \frac{7}{x + 2}$ ; a translation by $(03)$ gives the curve. The asymptotes move with it, to $x = - 2$ and $y = 3$ . Put $y = 0$ : $x = \frac{1}{3}$ . Put $x = 0$ : $y = - \frac{1}{2}$ .

How did it go?

Colour one circle on each line.

	Not yet	Nearly	Yes
I can sketch y = f(x + a), y = f(x) + a, y = af(x) and y = f(ax).
I can find where a point goes when a graph is transformed.
I can find the new asymptotes of a transformed graph.
I can find the constants of a transformation.

Answers and mark scheme

The worked answers and the full mark scheme are free with a Mathswariz account. Teachers and students can both sign up.

Get the answers