Transformations with Asymptotes

How horizontal and vertical asymptotes move under each transformation

In the exam for Edexcel International AS Maths · 7 real past-paper questions below

Asymptotes move too

You can already move a point with $f (x) + a$ , $f (x + a)$ , $a f (x)$ , $f (a x)$ and $- f (x)$ , $f (- x)$ . AS questions add asymptotes, full labelled sketches, and where two curves meet.

An asymptote is a line the curve gets closer and closer to but never reaches. Move it with the same rules as a point.

You see	Asymptote $y = c$	Asymptote $x = d$
$f (x) + a$	$y = c + a$	stays $x = d$
$a f (x)$	$y = a c$	stays $x = d$
$- f (x)$	$y = - c$	stays $x = d$
$f (x + a)$	stays $y = c$	$x = d - a$
$f (a x)$	stays $y = c$	$x = \frac{d}{a}$
$f (- x)$	stays $y = c$	$x = - d$

Memory trick

A line $y = c$ is a y value, so it only hears changes outside the bracket. A line $x = d$ is an x value, so it only hears changes inside.

Worked example

1
A curve $y = f (x)$ has a minimum at $(- 2, 0)$ , a maximum at $(3, 4)$ and one asymptote, $y = 2$ . Sketch $y = f (x) - 5$ and $y = f (- x)$ .
2
$f (x) - 5$ — The −5 is outside, so every y value drops 5. x values stay.
3
$(- 2, 0) \to (- 2, - 5), (3, 4) \to (3, - 1)$ — Move both turning points down 5.
4
$y = 2 \to y = - 3$ — The asymptote is a y value, so it drops 5 as well.
5
$f (- x)$ — The minus is inside, so every x value changes sign. y values stay.
6
$(- 2, 0) \to (2, 0), (3, 4) \to (- 3, 4)$ — Swap the sign of each x. The minimum now sits on the positive x-axis.
7
$y = 2 stays y = 2$ — A horizontal line looks the same in a mirror on the y-axis.

Rule

Sketch marks usually go one each to: the right shape in the right place, the asymptote written as an equation ( $y = - 3$ ), and the turning points or axis crossings labelled with coordinates.

Example

Reciprocal curves too. $y = \frac{2}{x} - 4$ is $y = \frac{2}{x}$ moved down 4. Asymptotes: $x = 0$ and $y = - 4$ . It crosses the x-axis where $\frac{2}{x} = 4$ , so at $(\frac{1}{2}, 0)$ .

Common mistake

Writing just −3 next to a dashed line. The question asks for the equation, so write $y = - 3$ .

Key idea

The word asymptote comes from Greek for “not falling together”: the curve and the line never meet.

Send a point to the origin

Some questions turn it round: they say where the new curve goes, and ask for the number in the transformation.

Take $f (x) = (x + 2) (x - 1) (x - 5)$ . It crosses the x-axis at $- 2$ , $1$ and $5$ , and the y-axis at $f (0) = 10$ .

Worked example

1
The curve $y = f (x) - p$ passes through $(0, 4)$ . Find $p$ .
2
$f (0) - p = 4$ — Put $x = 0$ into the new curve. Its y value must be 4.
3
$10 - p = 4$ — Use $f (0) = 10$ : $2 \times (- 1) \times (- 5) = 10$ .
4
$p = 6$ — Solve. The curve drops 6, so $(0, 10)$ lands on $(0, 4)$ .

Worked example

1
The curve $y = f (x + q)$ passes through the origin. Find the possible values of $q$ .
2
$f (0 + q) = 0$ — At the origin $x = 0$ and $y = 0$ .
3
$f (q) = 0$ — So $q$ must be a value that makes $f$ zero: a root.
4
$q = - 2, 1 or 5$ — Check one: $q = 5$ gives $f (x + 5)$ , which moves the root at 5 left by 5, onto the origin.

Common mistake

Writing $q = 2, - 1, - 5$ . Inside works the opposite way, so test it: $f (0 + q)$ must be $f$ of a root.

Key idea

To pass through a point, put its x into the new equation and make it equal its y. Then solve for the unknown.

f(−ax): two moves at once

In $f (- 2 x)$ the bracket holds a stretch and a reflection. Treat it as one rule: the new x value is the old one divided by $- 2$ .

Formula

f (- a x) : (x, y) \to (\frac{x}{- a}, y)

Worked example

1
$y = f (x)$ crosses the x-axis at $(- 4, 0)$ , touches it at $(2, 0)$ and crosses the y-axis at $(0, 3)$ . Sketch $y = f (- 2 x)$ .
2
$- 2 x = - 4 \Rightarrow x = 2$ — The bracket must equal the old x, −4. Divide by −2.
3
$(- 4, 0) \to (2, 0)$ — This root jumps to the other side and comes closer to O.
4
$- 2 x = 2 \Rightarrow x = - 1$ — Same for the point where the curve touches.
5
$(2, 0) \to (- 1, 0)$ — It still touches the axis here; only its place changes.
6
$(0, 3) \to (0, 3)$ — 0 ÷ (−2) = 0, so the y-axis crossing does not move.

Rule

“Describe fully”: a stretch parallel to the x-axis, scale factor $\frac{1}{2}$ , and a reflection in the y-axis. Either order gives the same curve.

Example

Solving with the bracket. If $f (x) = 5$ at $x = - 6$ and $x = 4$ , then $f (2 x) = 5$ when $2 x = - 6$ or $2 x = 4$ , so at $x = - 3$ or $x = 2$ .

Common mistake

Moving the points but forgetting the reflection: $(- 4, 0)$ going to $(- 2, 0)$ . The minus sign sends every x to the other side.

Key idea

Points on the y-axis never move in a stretch or reflection inside the bracket. Use them to check your sketch.

Where two curves meet

The last part often asks where two transformed curves cross. Set their equations equal and solve with algebra. These parts usually say “you must show all stages of your working”: a calculator answer scores nothing.

Take $f (x) = x$ . Then $y = f (3 x)$ is $y = 3 x$ , and $y = f (x) + 2$ is $y = x + 2$ .

Worked example

1
Find where $y = 3 x$ meets $y = x + 2$ . Give exact answers.
2
$3 x = x + 2$ — Where they meet, the y values are equal.
3
$3 x = x + 2$ — Split the root: $3 x = 3 x$ .
4
$(3 - 1) x = 2$ — Collect the $x$ terms and take out the factor.
5
$x = \frac{2}{3 - 1} \times \frac{3 + 1}{3 + 1}$ — Rationalise: multiply top and bottom by $3 + 1$ .
6
$x = \frac{2 ( 3 + 1 )}{3 - 1} = 3 + 1$ — The bottom is $3 - 1 = 2$ , which cancels the 2 on top.
7
$x = (3 + 1)^{2} = 4 + 23$ — Square: $3 + 23 + 1$ .
8
$y = x + 2 = 3 + 3$ — Use $x = 3 + 1$ from before. No need to square root again.

Rule

Marks for this type: B1 for setting the two equations equal, M1 for collecting the $x$ terms, A1 for the rationalised answer. A “show that” must reach the given form line by line.

Common mistake

Squaring each term: $(x + 2)^{2}$ is not $x + 4$ . Collect the roots first and you never need to square a sum.

Key idea

$3 x = 3 x$ , so $y = f (3 x)$ is also a stretch parallel to the y-axis, scale factor $3$ . One curve, two correct descriptions.

Which method? Read the question

The wording tells you what to do.

You see	You do
“State the equation of the asymptote”	Move it like a point: $y = c$ hears outside, $x = d$ hears inside
“Sketch … show clearly”	Shape, asymptotes as equations, every key point as coordinates
New curve passes through a point	Put that x into the new equation, set it equal to that y
$f (x + q)$ through the origin	$q$ is a root of $f$
$f (- a x)$	Divide each x by $- a$ ; y-axis points stay
Two transformed curves meet	Set equal, collect, rationalise, exact answer

Key idea

Asymptotes, roots and crossings all move by the same rules as points. Label every one of them.

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Real past-paper questions on transformations with asymptotes

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