Translations — shifting graphs up down left right — Mathematics

Before this

Domain and range
Completing the square
Translation of a shape · coming soon

This leads to

A minus sign that moves a graph to the right

You know the graph of $y = x^{2}$ . It is a U-shaped curve, and its lowest point, the vertex, is at the origin.

Now change one thing. Replace $x$ with $x - 3$ , to get $y = (x - 3)^{2}$ . The new graph has the same shape, but it is somewhere else. The question is where.

A minus sign usually means "go back" or "go left". So most people expect the graph to move 3 to the left. Before you read on, make your own prediction.

Predict, then reveal — which way does the graph move?

Compared with the graph of y = x², where is the graph of y = (x − 3)²? Choose one answer.

The graph of $y = (x - 3)^{2}$ is the graph of $y = x^{2}$ moved 3 units to the right. Its vertex is at $(3, 0)$ , because $(3 - 3)^{2} = 0$ .

The graph of

y = x^{2}

(ink) and the graph of

y = (x - 3)^{2}

(teal). Every point has moved 3 units to the right:

(- 2, 4)

(1, 4)

(1, 1)

(4, 1)

(2, 4)

(5, 4)

, and the vertex from

(0, 0)

(3, 0)

If you chose "left", you are in good company. The instinct is a good one, used in the wrong place. This lecture shows exactly where it goes wrong. By the end, "right" will feel like the only thing that could happen.

Two places to change a function

A function is a machine. An input goes in, the function works on it, and an output comes out. A graph shows every pair: the input $x$ along the bottom, and the output $y$ up the side.

So there are only two places where you can make a change. You can change the output, after the function has worked. Or you can change the input, before the function works. These two changes move the graph in different ways. That difference is the whole of this lecture.

Changing the output: up and down

Start with the output. Compare $y = x^{2}$ with $y = x^{2} + 3$ .

$x$	$- 2$	$- 1$	$0$	$1$	$2$
$x^{2}$	4	1	0	1	4
$x^{2} + 3$	7	4	3	4	7

The inputs are the same. Each output is 3 bigger. So each point $(x, y)$ moves to $(x, y + 3)$ . It goes straight up by 3.

The graph of

y = x^{2}

(ink) and the graph of

y = x^{2} + 3

(teal). The arrows show three points moving straight up by 3. The shape does not change.

Every point moves the same distance in the same direction, so the shape does not change at all. A move like this is called a translation. It is the same idea as a translation in geometry, where a whole shape slides without turning or changing size. Here the shape is a graph.

In the same way, $y = x^{2} - 5$ is the graph of $y = x^{2}$ moved down by 5. Adding to every height moves the graph up. Taking away from every height moves it down.

For any function $f$ , the graph of $y = f (x) + 3$ is the graph of $y = f (x)$ moved up by 3.

Changing the input: left and right

Now the change inside the bracket. Compare $y = x^{2}$ with $y = (x - 3)^{2}$ .

$x$	$- 1$	$0$	$1$	$2$	$3$	$4$	$5$	$6$
$x^{2}$	1	0	1	4	9	16	25	36
$(x - 3)^{2}$	16	9	4	1	0	1	4	9

Look for outputs you recognise. The old function gives 0 at $x = 0$ . The new one gives 0 at $x = 3$ . The old function gives 4 at $x = 2$ . The new one gives 4 at $x = 5$ . Every output of the old function appears again in the new row, 3 places further right.

Now ask why. Look at what the new function does with an input. It takes $x$ , subtracts 3, and only then squares. So at $x = 5$ , it squares 2. That is exactly what the old function did at $x = 2$ .

The new function always works on a number 3 smaller than the input you give it. So to get the same output as before, you must give it an input 3 bigger. Bigger inputs are further to the right. This means that the whole graph moves right.

This is the hard part of the lecture. Spend time here. The same argument works for any function $f$ , so let us write it once in general.

Take any point $(p, q)$ on the graph of $y = f (x)$ . This means that $q = f (p)$ . Where does $y = f (x - 3)$ give the same output $q$ ? It needs $f (x - 3) = f (p)$ . That happens when $x - 3 = p$ , so when $x = p + 3$ .

If (p, q) is on y = f (x), then (p + 3, q) is on y = f (x - 3) .

Every point has moved 3 to the right, and nothing else has changed.

That is it. That is the whole idea. The "− 3" is not an instruction to the graph. It is something the function does to the input before it works. To cancel it, the input must be 3 bigger. That is why the graph moves right.

A picture to hold in your mind

Think of a song that starts at time $t = 0$ . Let $f (t)$ be the sound at time $t$ seconds. Now start the same song 3 seconds late. At time $t$ , you hear what the song would normally play at time $t - 3$ . So the late song is $f (t - 3)$ . The sound you used to hear at time 0 now arrives at time 3. On a graph against time, later means further to the right. Subtracting 3 from the clock makes everything happen 3 seconds later.

Now you can see why the two kinds of change feel so different. A change to the output acts on the height you see, so it moves the graph the way it looks. A change to the input acts before the function works. The graph must move the opposite way to undo it. You will meet the same "opposite" effect again for stretches, later in this topic.

Translations of graphs

$y = f (x) + b$ is the graph of $y = f (x)$ moved $b$ units up. If $b$ is negative, it moves down.

$y = f (x - a)$ is the graph of $y = f (x)$ moved $a$ units to the right. If $a$ is negative, it moves left. For example, $f (x + 2) = f (x - (- 2))$ , so it moves 2 to the left.

$y = f (x - a) + b$ is a translation by the vector $(a b)$ . Every point $(p, q)$ moves to $(p + a, q + b)$ .

In a column vector, the top number is the move to the right and the bottom number is the move up. Negative numbers mean left and down.

See it on four different graphs below. Each graph has one marked point. Watch where it goes.

Graph builder — translate a graph by a vector

Choose a graph. Move a (right or left) and b (up or down). Watch the marked point, and read the new equation. For 1/x, watch the two dashed lines as well.

Starting graph:

a, the move to the right: 0

b, the move up: 0

The graph of $y = \frac{1}{x}$ translated 3 units right and 2 units up becomes $y = \frac{1}{x - 3} + 2$ . The point $(1, 1)$ moves to $(4, 3)$ . The two lines that the graph approaches, the axes $x = 0$ and $y = 0$ , move to $x = 3$ and $y = 2$ .

The graph of

y = \frac{1}{x}

(ink) and its translation by

(32)

(teal). The point

(1, 1)

moves to

(4, 3)

. The dashed lines

x = 3

and

y = 2

are where the axes, which

y = \frac{1}{x}

approaches, have moved to.

Starting graph	Move	New equation	Marked point
y = x²	2 right, 3 up	y = (x − 2)² + 3	(0, 0) to (2, 3)
y = x³	1 left, 2 down	y = (x + 1)³ − 2	(1, 1) to (0, −1)
y = √x	2 left, 1 up	y = √(x + 2) + 1	(0, 0) to (−2, 1)
y = 1/x	3 right, 2 up	y = 1/(x − 3) + 2	(1, 1) to (4, 3)

What should you have seen? Whatever the shape, every point moves by the same vector. The marked point always goes from $(p, q)$ to $(p + a, q + b)$ . The special features of each graph move with it: the vertex of $x^{2}$ , the starting point of $x$ , and the two dashed lines of $\frac{1}{x}$ .

A quick check that comes from the reason

Here is a fast check for the left-right move. Ask: which input makes the inside of the bracket zero?

For $f (x - 4)$ , the bracket is zero when $x = 4$ . At that input, the new function works on 0, exactly as the old function did at $x = 0$ . So whatever the old graph did at $x = 0$ , the new graph does at $x = 4$ . The graph has moved 4 to the right. For $f (x + 4)$ , the bracket is zero at $x = - 4$ , so the graph has moved 4 to the left.

Use this as a check, not as a replacement for the reason. When a bracket is less simple, it is the reason that will help you.

Stop and think

Move the line $y = 2 x$ 3 units to the right. Then, separately, move the same line 6 units down. Do you get two different lines?

Show the answer

No. The same line. Moving right by 3 gives $y = 2 (x - 3)$ , which is $y = 2 x - 6$ . Moving down by 6 also gives $y = 2 x - 6$ . The reason is the gradient. On this line, every 3 steps right go with 6 steps up. So the move "3 right" and the move "6 down" differ by "3 right and 6 up", which is a move along the line. Sliding a line along itself changes nothing. A curved graph like $y = x^{2}$ has no such direction, so for a curve, moving right and moving down always give different graphs.

Worked examples

Worked Example 1 — Following the vertex

Describe how to get the graph of $y = (x - 4)^{2} + 3$ from the graph of $y = x^{2}$ . Find its vertex and where it crosses the $y$ -axis.

There are two changes. Inside the bracket, $x - 4$ changes the input, so the graph moves 4 to the right. Outside, $+ 3$ changes the output, so it moves 3 up. Together, this is a translation by $(43)$ .

The vertex $(0, 0)$ moves to $(0 + 4, 0 + 3) = (4, 3)$ .

The graph crosses the $y$ -axis where $x = 0$ : $y = (0 - 4)^{2} + 3 = 16 + 3 = 19$ . So it crosses at $(0, 19)$ .

Check with another point. $(1, 1)$ is on $y = x^{2}$ , so $(5, 4)$ should be on the new graph. Put $x = 5$ : $(5 - 4)^{2} + 3 = 1 + 3 = 4$ . ✓

Worked Example 2 — Tracking one point

The graph of $y = f (x)$ passes through $(2, 5)$ . Find the matching point on the graph of $y = f (x + 1) - 2$ .

You do not need to know what $f$ is. Follow the point.

Inside the bracket, $x + 1 = x - (- 1)$ , so the graph moves 1 to the left. The $x$ -coordinate becomes $2 - 1 = 1$ .

Outside, $- 2$ moves the graph 2 down. The $y$ -coordinate becomes $5 - 2 = 3$ .

The matching point is $(1, 3)$ .

Check by substituting. At $x = 1$ , the new function gives $f (1 + 1) - 2 = f (2) - 2 = 5 - 2 = 3$ . ✓ Notice how the check works: the input 1 is changed back to 2 inside the bracket. That is the "opposite" effect again.

Worked Example 3 — A translation hidden in a completed square

Show that the graph of $y = x^{2} - 6 x + 11$ is a translation of $y = x^{2}$ . Find its lowest point and its range.

Complete the square. Half of $- 6$ is $- 3$ , and $(x - 3)^{2} = x^{2} - 6 x + 9$ . So:

x^{2} - 6 x + 11 = (x - 3)^{2} - 9 + 11, so x^{2} - 6 x + 11 = (x - 3)^{2} + 2 .

So the graph is $y = x^{2}$ translated by $(32)$ . The lowest point moves from $(0, 0)$ to $(3, 2)$ .

The range of $y = x^{2}$ is $y \geq 0$ . Every height has gone up by 2, so the range is now $y \geq 2$ .

Check at $x = 0$ : the original form gives $11$ , and the completed form gives $(0 - 3)^{2} + 2 = 11$ . ✓

This is what completing the square really does. It shows that every graph $y = x^{2} + p x + q$ is the same parabola as $y = x^{2}$ , moved to a new place.

Worked Example 4 — Why g never gives the output 2

Translate the graph of $y = \frac{7}{x}$ by $(32)$ . Find the new equation, its domain and its range.

Replace $x$ with $x - 3$ , then add 2:

y = \frac{7}{x - 3} + 2

The graph of $y = \frac{7}{x}$ approaches the two axes, $x = 0$ and $y = 0$ , but never meets them. After the translation, these lines are $x = 3$ and $y = 2$ . So the domain is $x \in R, x \neq = 3$ , and the range is $y \neq = 2$ .

In the domain and range lecture, we wrote $g (x) = \frac{2 x + 1}{x - 3}$ as $2 + \frac{7}{x - 3}$ . So the graph of $g$ is just the graph of $\frac{7}{x}$ , moved 3 right and 2 up. The function $\frac{7}{x}$ can never give 0. Move every output up by 2, and $g$ can never give 2. The strange missing output is the $x$ -axis, moved up.

Check: $g (4) = \frac{9}{1} = 9$ , and $\frac{7}{4 - 3} + 2 = 9$ . ✓

Worked Example 5 — The roots move too

$f (x) = x^{3} - 4 x$ . Find the roots of $f (x) = 0$ , then the roots of $f (x - 1) = 0$ .

Factorise: $x^{3} - 4 x = x (x^{2} - 4) = x (x - 2) (x + 2)$ . So the roots are $- 2$ , $0$ and $2$ . These are where the graph of $y = f (x)$ crosses the $x$ -axis.

The graph of $y = f (x - 1)$ is the same graph moved 1 to the right. A move to the right does not change any heights, so the crossing points simply move right by 1. The new roots are $- 1$ , $1$ and $3$ .

Check by substituting $x = 3$ : $f (3 - 1) = f (2) = 8 - 8 = 0$ . ✓ In factor form, $f (x - 1) = (x - 1) (x - 3) (x + 1)$ : each factor has had its $x$ replaced by $x - 1$ .

Three classic mistakes

Trap 1 — Moving f(x − 3) to the left

The minus sign inside the bracket does not mean "left". It is done to the input, so the graph moves the opposite way: $f (x - 3)$ moves 3 to the right, and $f (x + 3)$ moves 3 to the left. If you are unsure, find the input that makes the bracket zero, and check one point.

Trap 2 — Confusing f(x) − 3 with f(x − 3)

$x^{2} - 3$ is the graph of $x^{2}$ moved 3 down. $(x - 3)^{2}$ is the graph of $x^{2}$ moved 3 right. The bracket decides which: outside the function changes the output, inside changes the input. Always write the bracket. Without it, $x - 3^{2}$ means $x - 9$ , which is a straight line.

Trap 3 — Forgetting that the domain moves

The graph of $y = x$ starts at $x = 0$ . The graph of $y = x - 2$ starts at $x = 2$ , not at 0. Every feature of the graph moves with it: starting points, vertices, roots and the lines it approaches. So when you translate a graph, translate its domain and range as well.

Try it yourself

Work each one on paper first. Open a hint only when you are stuck.

Describe the translation that takes $y = x^{2}$ to $y = (x + 5)^{2} - 1$ , as a column vector. Give the vertex of the new graph and the point where it crosses the $y$ -axis.

Hint 1
Inside the bracket changes $x$ ; outside changes $y$ . Write $x + 5$ as $x - (- 5)$ .

Hint 2
The move is 5 left and 1 down. For the $y$ -axis, put $x = 0$ .

Full answer
$x + 5 = x - (- 5)$ , so the graph moves 5 to the left. The $- 1$ outside moves it 1 down. The vector is $(- 5 - 1)$ . The vertex moves from $(0, 0)$ to $(- 5, - 1)$ . At $x = 0$ : $y = 5^{2} - 1 = 24$ , so the graph crosses the $y$ -axis at $(0, 24)$ . Check: at $x = - 5$ , $y = 0^{2} - 1 = - 1$ , the vertex. ✓
The graph of $y = f (x)$ has a highest point at $(- 2, 7)$ . Find the highest point of (a) $y = f (x - 3) + 1$ and (b) $y = f (x + 4)$ .

Hint 1
A translation moves every point, including the highest point, by the same vector.

Hint 2
(a) is 3 right and 1 up. (b) is 4 left, with no vertical move.

Full answer
(a) The vector is $(31)$ , so $(- 2, 7)$ moves to $(1, 8)$ . (b) The vector is $(- 4 0)$ , so $(- 2, 7)$ moves to $(- 6, 7)$ . Check (a): at $x = 1$ , the new function gives $f (1 - 3) + 1 = f (- 2) + 1 = 7 + 1 = 8$ . ✓ A translation does not change the shape, so the highest point is still the highest point.
Translate the graph of $y = x$ by $(- 2 1)$ . Write the new equation, its domain and its range. Which point does $(4, 2)$ move to?

Hint 1
Replace $x$ with $x - (- 2)$ , then add 1.

Hint 2
The starting point $(0, 0)$ moves to $(- 2, 1)$ . The domain and range start there.

Full answer
The new equation is $y = x + 2 + 1$ . The number under the root must be 0 or more, so $x + 2 \geq 0$ , and the domain is $x \geq - 2$ . The smallest output is $0 + 1 = 1$ , so the range is $y \geq 1$ . The point $(4, 2)$ moves to $(4 - 2, 2 + 1) = (2, 3)$ . Check: $2 + 2 + 1 = 2 + 1 = 3$ . ✓
A ball is thrown upwards. Its height in metres after $t$ seconds is $h (t) = 20 t - 5 t^{2}$ . A second ball is thrown in exactly the same way, but 1.5 seconds later and from a platform 2 m higher. (a) Write the second ball's height $H (t)$ using $h$ . (b) Find the greatest height of each ball, and when it happens. (c) At $t = 3$ , which ball is higher? (d) When does the second ball come back down to the height it was thrown from?

Hint 1
"1.5 seconds later" is a move to the right on a time graph. "2 m higher" is a move up.

Hint 2
$H (t) = h (t - 1.5) + 2$ . Complete the square on $h (t)$ to find its highest point, then translate it.

Full answer
(a) Later in time means a move right on the time axis, and higher means a move up. So $H (t) = h (t - 1.5) + 2$ . (b) $h (t) = - 5 (t^{2} - 4 t) = - 5 ((t - 2)^{2} - 4) = 20 - 5 (t - 2)^{2}$ . So the first ball is highest at $t = 2$ , at 20 m. The second graph is translated by $(1.5 2)$ , so the second ball is highest at $t = 3.5$ , at 22 m. (c) $h (3) = 60 - 45 = 15$ m. $H (3) = h (1.5) + 2 = 30 - 11.25 + 2 = 20.75$ m. The second ball is higher. (d) The first ball is back at its starting height (0 m) when $20 t - 5 t^{2} = 0$ , at $t = 0$ and $t = 4$ . Move these 1.5 to the right: the second ball is back at 2 m at $t = 5.5$ . Check: $H (5.5) = h (4) + 2 = 0 + 2 = 2$ . ✓

The bigger idea

A translation is the simplest way to change a graph, but the idea behind it reaches far. A change to the input moves a graph the opposite way, because the function must undo it. You will use that sentence again for reflections, for stretches, and for every combined transformation in this topic.

It also explains how waves move. In physics, a wave of fixed shape $f$ that travels to the right at speed $v$ is written $y = f (x - v t)$ . After $t$ seconds the whole shape has moved $v t$ to the right, exactly as $f (x - 3)$ moves 3 to the right. The minus sign that looked backwards is what makes a wave go forwards.

Key concepts in this lesson

A translation moves every point of a graph the same distance in the same direction, so the shape does not change.
The graph of $y = f (x) + b$ is the graph of $y = f (x)$ moved $b$ units up, because every output increases by $b$ .
The graph of $y = f (x - a)$ is the graph of $y = f (x)$ moved $a$ units right, because the input must be $a$ bigger to give the same output.
The graph of $y = f (x - a) + b$ is a translation by the vector $(a b)$ , which moves each point $(p, q)$ to $(p + a, q + b)$ .
Vertices, roots, starting points and the lines a graph approaches all move with the graph, so the domain and range move too.