Mathematics 9709 · Pure Mathematics 1

Transformations of functions

You can move, flip and stretch $y=\mathrm{f}(x)$ without plotting any points again.
One simple rule: a change outside the bracket does what it says. A change inside does the opposite.

Outside the bracket: vertical, as written. Inside the bracket: horizontal, the opposite way (so $\mathrm{f}(x+3)$ moves left).

The four transformations

For each one, first ask: is it outside or inside the bracket?

$\mathrm{f}(x)+a$ : translate up by $a$ , vector $\binom{0}{a}$ . Outside, as written.
$\mathrm{f}(x-a)$ : translate right by $a$ , vector $\binom{a}{0}$ (the textbook's primary form). The sign flips, so $\mathrm{f}(x+a)$ goes left, vector $\binom{-a}{0}$ . Inside, the opposite way.
$a\,\mathrm{f}(x)$ : stretch factor $a$ parallel to the $y$ -axis (vertical). Outside, as written.
$\mathrm{f}(ax)$ : stretch factor $\tfrac{1}{a}$ parallel to the $x$ -axis (horizontal). Inside, the reciprocal.

Two reflections are just stretches with a minus sign. $-\mathrm{f}(x)$ flips in the $x$ -axis. $\mathrm{f}(-x)$ flips in the $y$ -axis.
Same rule: the minus outside changes the output, the minus inside changes the input.

Drag it: watch a, b and c reshape the curve

Drag the three sliders in $y=a\,\mathrm{f}(x-b)+c$ (here $\mathrm{f}$ is a parabola) and watch the gold curve change; the vertex moves straight to $(b,c)$ .

a (stretch)b (left/right)c (up/down)

y = (x + 0)2 + 0·the parent curve, untouched

Notice: the b inside the bracket moves the curve the opposite way you might expect, while a and c outside act exactly as written.

Drag a (vertical stretch), b (left and right, the inside one) and c (up and down): the curve changes as you drag, and the vertex moves to (b, c).

Notice this: move $b$ to $+2$ and the curve moves right, not left.
The bracket holds $x-b$ . So a positive $b$ moves the graph the opposite way to the sign. This is the inside-the-bracket rule, and now you can see it.

Before and after: one worked transformation

Here is the starting curve $y=x^2$ and the curve it becomes after a move right, a vertical stretch and a move up, where the vertex moves from $(0,0)$ to $(2,6)$ .

y = x² (vertex at the origin) and its image 2(x − 2)² + 6 (vertex at (2, 6)): right 2, stretch ×2, up 6.

Why inside the bracket runs backwards

The inside rule often seems backwards — here is the short reason.

Why does f(x + 3) move LEFT, not right?

$y=\mathrm{f}(x+3)$ reaches a height when the bracket equals the value that gave that height in $\mathrm{f}$ . To make $x+3$ equal that value, $x$ must be 3 smaller. So the new graph reaches the same height 3 steps sooner. That is a move left. Same idea for $\mathrm{f}(2x)$ : the bracket reaches each value when $x$ is half as big, so the graph is squashed left to right by factor $\tfrac12$ .

Describing a transformation in the exam

A common question gives you a quadratic and asks how $y=x^2$ maps onto it.
What to do: complete the square first (see Quadratics §1.2). Then read the bracket as horizontal and everything outside as vertical.

Worked example

Real exam: describe how $y=x^2$ maps onto $\mathrm{g}(x)=2x^2-8x+14$ (9709/12 March 2022 Q5b)

Complete the square so the bracket is clean: $2x^2-8x+14=2\big[(x-2)^2+3\big]=2(x-2)^2+6$ .
The $(x-2)$ is inside, so it is horizontal: translate right by 2, vector $\tbinom{2}{0}$ .
The $2\times$ is outside: stretch factor 2 parallel to the $y$ -axis.
The $+6$ is outside: translate up by 6, vector $\tbinom{0}{6}$ .

Answer

\text{right }2,\ \ \text{stretch }{\times}2\ (\parallel y),\ \ \text{up }6

(9709/12 March 2022 Q5(b)) The mark scheme grades the order here, because the stretch and the up 6 are both vertical. See the Deeper below.

When the order matters (and when it does not)

One horizontal + one vertical: order does not matter. A sideways move and an up-or-down move do not affect each other. So you may write them in either order and both score. (9709/12 March 2022 Q5)

Two moves in the same direction: order does matter. Above, the stretch and the up 6 are both outside the bracket. “Stretch then up 6” gives $2(x-2)^2+6$ . But “up 6 then stretch” gives $2(x-2)^2+12$ , a different curve. So for two moves in the same direction, you cannot use any order. Use the order that gives the right answer.

When a single transformation is really two

The hardest version writes a combined transformation as one expression and asks you to split it into two single moves, so read inside and outside separately.

Worked example

Real exam: describe fully the two transformations in $y=\mathrm{f}(2x)-3$ (9709/12 Oct/Nov 2021 Q2a)

The $2x$ is inside the bracket, so it is horizontal and you take the reciprocal factor: a stretch parallel to the $x$ -axis, factor $\tfrac12$ .
The $-3$ is outside, so it is vertical and as written: a translation $\tbinom{0}{-3}$ (down 3).

Answer

\text{stretch }{\times}\tfrac12\ (\parallel x),\ \ \text{then }\tbinom{0}{-3}

(9709/12 Oct/Nov 2021 Q2(a)) Here the two moves point in different directions, so the mark scheme accepts either order.

Worked example

Same paper, the follow-up: $P(5,6)$ lies on $y=\mathrm{f}(2x)-3$ , find the matching point on $y=\mathrm{f}(x)$ (Q2b)

The new curve is made from $\mathrm{f}$ by changing $x$ to $2x$ and taking away 3, so to get back to $\mathrm{f}$ , undo each step on the coordinates of $P$ .

At $P$ , $6=\mathrm{f}(2\cdot 5)-3$ , so $\mathrm{f}(10)=6+3=9$ (undo the down 3 by adding it back).
The input that $\mathrm{f}$ actually receives is $2x=2\cdot 5=10$ , so the original point is $(10,9)$ .

Answer

(10,\ 9)

(9709/12 Oct/Nov 2021 Q2(b)) A point on the new curve matches the original only after you undo both moves: the horizontal squash and the vertical shift.

Where the marks go

Common mistake

\mathrm{f}(x+3)

moves the graph left 3, not right. Inside the bracket is always the opposite direction. This sign mistake is the most common in the topic. (9709/12 Oct/Nov 2021 Q2)

Common mistake

“Describe fully” wants the type (translation, stretch, reflection), the direction or axis (say “parallel to the

y

-axis” for a stretch), and the factor or vector. Order is only graded when the two moves go the same way. A horizontal and a vertical move can be written in either order and still score. (9709/12 March 2022 Q5)

Now you try

The graph of $y=\mathrm{f}(x)$ is stretched by factor 3 parallel to the $y$ -axis and then translated by $\binom{0}{-2}$ . Write down the equation of the resulting graph in terms of $\mathrm{f}$ . (textbook 2.5-style)

Your turn— tap to reveal the worked answer (textbook 2.5-style)

Both moves are outside the bracket (vertical). A stretch of factor 3 multiplies the output: $3\mathrm{f}(x)$ . The move down 2 then takes away 2:

Answer

y=3\mathrm{f}(x)-2

Order matters here because both are vertical. Stretching then moving down 2 gives $3\mathrm{f}(x)-2$ , the answer asked for.

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Transformations of functions

You can move, flip and stretch $y=\mathrm{f}(x)$ without plotting any points again.
One simple rule: a change outside the bracket does what it says. A change inside does the opposite.

Outside the bracket: vertical, as written. Inside the bracket: horizontal, the opposite way (so $\mathrm{f}(x+3)$ moves left).

The four transformations

For each one, first ask: is it outside or inside the bracket?

$\mathrm{f}(x)+a$ : translate up by $a$ , vector $\binom{0}{a}$ . Outside, as written.
$\mathrm{f}(x-a)$ : translate right by $a$ , vector $\binom{a}{0}$ (the textbook's primary form). The sign flips, so $\mathrm{f}(x+a)$ goes left, vector $\binom{-a}{0}$ . Inside, the opposite way.
$a\,\mathrm{f}(x)$ : stretch factor $a$ parallel to the $y$ -axis (vertical). Outside, as written.
$\mathrm{f}(ax)$ : stretch factor $\tfrac{1}{a}$ parallel to the $x$ -axis (horizontal). Inside, the reciprocal.

Two reflections are just stretches with a minus sign. $-\mathrm{f}(x)$ flips in the $x$ -axis. $\mathrm{f}(-x)$ flips in the $y$ -axis.
Same rule: the minus outside changes the output, the minus inside changes the input.

Drag it: watch a, b and c reshape the curve

Drag the three sliders in $y=a\,\mathrm{f}(x-b)+c$ (here $\mathrm{f}$ is a parabola) and watch the gold curve change; the vertex moves straight to $(b,c)$ .

a (stretch)b (left/right)c (up/down)

y = (x + 0)2 + 0·the parent curve, untouched

Notice: the b inside the bracket moves the curve the opposite way you might expect, while a and c outside act exactly as written.

Drag a (vertical stretch), b (left and right, the inside one) and c (up and down): the curve changes as you drag, and the vertex moves to (b, c).

Notice this: move $b$ to $+2$ and the curve moves right, not left.
The bracket holds $x-b$ . So a positive $b$ moves the graph the opposite way to the sign. This is the inside-the-bracket rule, and now you can see it.

Before and after: one worked transformation

Here is the starting curve $y=x^2$ and the curve it becomes after a move right, a vertical stretch and a move up, where the vertex moves from $(0,0)$ to $(2,6)$ .

y = x² (vertex at the origin) and its image 2(x − 2)² + 6 (vertex at (2, 6)): right 2, stretch ×2, up 6.

Why inside the bracket runs backwards

The inside rule often seems backwards — here is the short reason.

Why does f(x + 3) move LEFT, not right?

Describing a transformation in the exam

A common question gives you a quadratic and asks how $y=x^2$ maps onto it.
What to do: complete the square first (see Quadratics §1.2). Then read the bracket as horizontal and everything outside as vertical.

Worked example

Real exam: describe how $y=x^2$ maps onto $\mathrm{g}(x)=2x^2-8x+14$ (9709/12 March 2022 Q5b)

Complete the square so the bracket is clean: $2x^2-8x+14=2\big[(x-2)^2+3\big]=2(x-2)^2+6$ .
The $(x-2)$ is inside, so it is horizontal: translate right by 2, vector $\tbinom{2}{0}$ .
The $2\times$ is outside: stretch factor 2 parallel to the $y$ -axis.
The $+6$ is outside: translate up by 6, vector $\tbinom{0}{6}$ .

Answer

\text{right }2,\ \ \text{stretch }{\times}2\ (\parallel y),\ \ \text{up }6

(9709/12 March 2022 Q5(b)) The mark scheme grades the order here, because the stretch and the up 6 are both vertical. See the Deeper below.

When the order matters (and when it does not)

When a single transformation is really two

The hardest version writes a combined transformation as one expression and asks you to split it into two single moves, so read inside and outside separately.

Worked example

Real exam: describe fully the two transformations in $y=\mathrm{f}(2x)-3$ (9709/12 Oct/Nov 2021 Q2a)

The $2x$ is inside the bracket, so it is horizontal and you take the reciprocal factor: a stretch parallel to the $x$ -axis, factor $\tfrac12$ .
The $-3$ is outside, so it is vertical and as written: a translation $\tbinom{0}{-3}$ (down 3).

Answer

\text{stretch }{\times}\tfrac12\ (\parallel x),\ \ \text{then }\tbinom{0}{-3}

(9709/12 Oct/Nov 2021 Q2(a)) Here the two moves point in different directions, so the mark scheme accepts either order.

Worked example

Same paper, the follow-up: $P(5,6)$ lies on $y=\mathrm{f}(2x)-3$ , find the matching point on $y=\mathrm{f}(x)$ (Q2b)

The new curve is made from $\mathrm{f}$ by changing $x$ to $2x$ and taking away 3, so to get back to $\mathrm{f}$ , undo each step on the coordinates of $P$ .

At $P$ , $6=\mathrm{f}(2\cdot 5)-3$ , so $\mathrm{f}(10)=6+3=9$ (undo the down 3 by adding it back).
The input that $\mathrm{f}$ actually receives is $2x=2\cdot 5=10$ , so the original point is $(10,9)$ .

Answer

(10,\ 9)

(9709/12 Oct/Nov 2021 Q2(b)) A point on the new curve matches the original only after you undo both moves: the horizontal squash and the vertical shift.

Where the marks go

Common mistake

\mathrm{f}(x+3)

moves the graph left 3, not right. Inside the bracket is always the opposite direction. This sign mistake is the most common in the topic. (9709/12 Oct/Nov 2021 Q2)

Common mistake

“Describe fully” wants the type (translation, stretch, reflection), the direction or axis (say “parallel to the

y

Now you try

Your turn— tap to reveal the worked answer (textbook 2.5-style)

Both moves are outside the bracket (vertical). A stretch of factor 3 multiplies the output: $3\mathrm{f}(x)$ . The move down 2 then takes away 2:

Answer

y=3\mathrm{f}(x)-2

Order matters here because both are vertical. Stretching then moving down 2 gives $3\mathrm{f}(x)-2$ , the answer asked for.