Transformations: translations

7 minAS Level1.2.5

AI-drafted, independently checked · not yet teacher-reviewed

By the end of this lesson

By the end you can sketch and describe the graphs of $y = f(x) + a$ and $y = f(x + a)$ as translations, find the new equation, and track where points move.

Slide, don't bend

A translation picks the whole graph up and slides it, without stretching, flipping or bending it. Every point moves the same distance in the same direction.

How a slide is written. Cambridge describes a translation with a column vector $\begin{pmatrix} h \\ k \end{pmatrix}$ : two numbers stacked in a bracket. The top number $h$ is how far every point moves right (a negative number means left). The bottom number $k$ is how far it moves up (negative means down). So $\begin{pmatrix} 4 \\ 0 \end{pmatrix}$ is "4 right, no vertical move", and $\begin{pmatrix} 0 \\ -2 \end{pmatrix}$ is "2 down".

Below, $f(x)$ is the original function, $a$ is a number, and $(p, q)$ is any point on the original graph $y = f(x)$ .

The one rule to tattoo on your brain:

A change outside the bracket acts on $y$ and does what it says.
A change inside the bracket acts on $x$ and does the opposite.

The two translations

Equation	Translation by	In words	Point $(p, q)$ goes to
$y = f(x) + a$	$\begin{pmatrix} 0 \\ a \end{pmatrix}$	up $a$ (down if $a < 0$ )	$(p,\ q + a)$
$y = f(x + a)$	$\begin{pmatrix} -a \\ 0 \end{pmatrix}$	left $a$ (right if $a < 0$ )	$(p - a,\ q)$

So $y = f(x - 4)$ is a translation by $\begin{pmatrix} 4 \\ 0 \end{pmatrix}$ : four to the right.

Why does + inside move it left?

Try a number. Suppose $f(3) = 10$ , so $(3, 10)$ is on $y = f(x)$ . Where does height 10 appear on $y = f(x + 2)$ ? We need the bracket to be 3, so $x + 2 = 3$ , which means $x = 1$ . The point has moved from $x = 3$ to $x = 1$ : 2 to the left.

Every point does the same, so the new graph shows everything $f$ did, 2 units earlier, and the whole picture appears 2 units further left.

Think of a song that starts 2 seconds early: every beat lands 2 seconds sooner on the timeline. The analogy breaks a bit for $f(x) + a$ , which isn't about timing at all: it simply adds $a$ to every height.

Quick check

Multiple choiceQuestion 1 · Recall

Answer the 2 questions above to continue.