Reflections — flipping graphs across axes — Mathematics

Before this

Translations of graphs
Inverse functions
Reflection of a shape · coming soon

This leads to

Stretches of graphs
Combined transformations
Graphs of sin, cos and tan · coming soon

The square root of a negative number

You know that $- 4$ has no real value. No real number multiplied by itself gives a negative answer.

So what does the graph of $y = - x$ look like? Many students say it has no graph at all, because the square root of a negative is not allowed.

Try one input before you decide. Put $x = - 4$ . Then $- x = 4$ , and $4 = 2$ . So the point $(- 4, 2)$ is on the graph. The input was negative, but the number under the root was positive.

In fact, every negative input works, and every positive input fails. The graph of $y = - x$ is the graph of $y = x$ flipped to the other side of the $y$ -axis, like a reflection in a mirror. A single minus sign has flipped a whole graph. This lecture shows exactly why, and what happens when the minus sign goes in a different place.

Two places for a minus sign

The last lecture showed that there are only two places to change a function. You can change the output, after the function has worked. Or you can change the input, before the function works.

A minus sign can go in either place. In $y = - f (x)$ , the function works first, and then its output is made negative. In $y = f (- x)$ , the input is made negative first, and then the function works. These two give two different reflections.

Changing the sign of the output

Take $f (x) = x^{2} + 2 x - 3$ . Compare $y = f (x)$ with $y = - f (x)$ .

$x$	$- 4$	$- 3$	$- 2$	$- 1$	$0$	$1$	$2$
$f (x)$	5	0	−3	−4	−3	0	5
$- f (x)$	−5	0	3	4	3	0	−5

The inputs are the same. Each output has changed sign. A height of 5 above the $x$ -axis has become a depth of 5 below it. A depth of 4 below has become a height of 4 above. So each point $(p, q)$ moves to $(p, - q)$ .

Every point stays at the same distance from the $x$ -axis, on the other side of it, exactly as in a mirror. The graph of $y = - f (x)$ is the reflection of the graph of $y = f (x)$ in the $x$ -axis. The $x$ -axis is the mirror line.

The graph of

y = f (x)

, where

f (x) = x^{2} + 2 x - 3

(ink), and its reflection

y = - f (x)

(teal). Each arrow joins a point to its image:

(- 1, - 4)

(- 1, 4)

(2, 5)

(2, - 5)

and

(- 4, 5)

(- 4, - 5)

. The roots

- 3

and

1

do not move, because they lie on the mirror line.

The two roots did not move, because the negative of 0 is 0. And the lowest point of $f$ has become the highest point of $- f$ : a reflection in the $x$ -axis turns the graph upside down.

Changing the sign of the input

Now put the minus sign inside the bracket. Compare $y = f (x)$ with $y = f (- x)$ .

This time, the function does not work on the input you give it. It works on the negative of that input. At $x = 3$ , the new function works on $- 3$ , so it gives $f (- 3)$ . At $x = 1$ , it gives $f (- 1)$ .

$x$	$- 3$	$- 1$	$0$	$1$	$3$
$f (x)$	0	−4	−3	0	12
$f (- x)$	12	0	−3	−4	0

Look for outputs you recognise. The old $- 4$ at $x = - 1$ is now at $x = 1$ . The old 0 at $x = - 3$ is now at $x = 3$ . Every output appears again, at the input with the opposite sign.

This is the hard part of the lecture. Spend time here. Let us write the reason once, for any function $f$ .

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

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

Every point keeps its height and moves to the other side of the $y$ -axis, at the same distance. So the graph of $y = f (- x)$ is the reflection of the graph of $y = f (x)$ in the $y$ -axis.

The last lecture promised that its main sentence would come back here. Here it is. A change to the input moves the graph the opposite way, because the input must be changed back before the function sees the number it needs. For $f (x - 3)$ , the change was "subtract 3", so the input had to be 3 bigger. For $f (- x)$ , the change is "change the sign", and the only way to undo that is to change the sign again. So the point at $p$ moves to $- p$ .

That is it. That is the whole idea. A minus sign on the output flips every height. A minus sign on the input flips every position left to right.

A picture to hold in your mind

Think of the output as a height and the input as a position along a road. The minus sign outside says: "stay at your place on the road, but go as far below the road as you were above it". The minus sign inside says: "keep your height, but stand at the same distance on the other side of the start".

Reflections of graphs

$y = - f (x)$ is the reflection of $y = f (x)$ in the $x$ -axis. Every point $(p, q)$ moves to $(p, - q)$ .

$y = f (- x)$ is the reflection of $y = f (x)$ in the $y$ -axis. Every point $(p, q)$ moves to $(- p, q)$ .

$y = - f (- x)$ is both reflections together. Every point $(p, q)$ moves to $(- p, - q)$ , which is a half-turn (a rotation of 180°) about the origin O.

Now try both minus signs on five different graphs.

Graph builder — put a minus sign outside, inside, or both

Choose a graph. Then press the two buttons to put a minus sign outside the function, inside it, or both. Watch the marked point, and read the new equation.

Starting graph:

The graph of $y = f (- x)$ , where $f (x) = x^{2} + 2 x - 3$ , is the reflection of $y = f (x)$ in the $y$ -axis. Its equation is $y = x^{2} - 2 x - 3$ .

The graph of

y = f (x)

(ink) and its reflection

y = f (- x)

(teal). The lowest point moves from

(- 1, - 4)

(1, - 4)

, and the root

- 3

moves to

3

. The point

(0, - 3)

is on the mirror line, so it stays.

The same three changes for all five graphs in the graph builder:

y = x² + 2x − 3. Outside: y = −x² − 2x + 3. Inside: y = x² − 2x − 3. Both: y = −x² + 2x + 3.
y = √x. Outside: y = −√x. Inside: y = √(−x), with domain x ≤ 0. Both: y = −√(−x).
y = x³ − 4x. Outside: y = −x³ + 4x. Inside: the same, y = −x³ + 4x. Both: the original graph again.
y = x² − 4. Outside: y = −x² + 4. Inside: the original graph again. Both: y = −x² + 4.
y = 1/(x − 2). Outside: y = −1/(x − 2). Inside: y = −1/(x + 2), with the asymptote moved to x = −2. Both: y = 1/(x + 2).

What should you have seen? With the minus outside, every point moved straight up or down, across the $x$ -axis. With the minus inside, every point moved straight across the $y$ -axis. Two graphs behaved strangely. For $y = x^{2} - 4$ , the minus inside changed nothing. For $y = x^{3} - 4 x$ , the two minus signs gave the same graph. The rest of this lecture explains both.

Both minus signs: a half-turn

Put a minus sign in both places, $y = - f (- x)$ . The input changes sign, so $p$ becomes $- p$ . The output changes sign, so $q$ becomes $- q$ . The point $(p, q)$ moves to $(- p, - q)$ .

Now picture the two points. The origin O is exactly halfway between $(p, q)$ and $(- p, - q)$ . So you can reach one from the other by turning the plane through 180° about O. A reflection in one axis followed by a reflection in the other is the same as a half-turn.

The order of these two reflections does not matter: either order gives $(- p, - q)$ . A later lecture shows that this is not true for every pair of transformations.

Stop and think

Reflect the graph of $y = x^{3}$ in the $x$ -axis. Then, starting again, reflect it in the $y$ -axis. Which reflection gives which graph?

Show the answer

They give the same graph. In the $x$ -axis: $y = - x^{3}$ . In the $y$ -axis: $y = (- x)^{3} = - x^{3}$ . The equations are identical, so the graphs are identical. This does not break the rules. It is a fact about this curve. Because $(- x)^{3} = - x^{3}$ , changing the sign of the input has exactly the same effect as changing the sign of the output. Functions with this property have a name, and the next section gives it.

Graphs that match their own reflection

Some functions are unchanged by one of these moves. They have names.

A function is even if $f (- x) = f (x)$ for every $x$ in its domain. Reflecting its graph in the $y$ -axis gives back the same graph. So the graph is symmetric about the $y$ -axis. The classic example is $y = x^{2}$ , because $(- x)^{2} = x^{2}$ .

A function is odd if $f (- x) = - f (x)$ for every $x$ in its domain. Here the reflection in the $y$ -axis gives the same graph as the reflection in the $x$ -axis. Doing both gives back the original, so the graph is unchanged by a half-turn about O. The classic example is $y = x^{3}$ , because $(- x)^{3} = - x^{3}$ .

The graph of

y = x^{3} - 4 x

, an odd function. The point

(1, - 3)

and the point

(- 1, 3)

are both on the graph, and O is halfway between them. The same is true of

(2.5, 5.625)

and

(- 2.5, - 5.625)

. A half-turn about O swaps every such pair, so it leaves the graph unchanged.

The names come from powers of $x$ . An even power keeps its sign when $x$ changes sign: $(- x)^{4} = x^{4}$ . An odd power changes sign: $(- x)^{5} = - x^{5}$ . So a polynomial with only even powers is an even function, and a polynomial with only odd powers is an odd function. A constant counts as an even power, because $3 = 3 x^{0}$ .

Most functions are neither: they simply fail both tests. To decide, always work out $f (- x)$ and compare it with $f (x)$ and with $- f (x)$ . Sort the functions below with that test.

Sorter — even, odd, or neither?

Work out f(−x) for each function on paper. Then put each card in the right box.

Each function is tested by working out $f (- x)$ .

Function	f(−x)	Type
x⁴ − 3x²	x⁴ − 3x², the same as f(x)	even
x² + 5	x² + 5, the same as f(x)	even
3	3, the same as f(x)	even
x³ + x	−x³ − x, which is −f(x)	odd
1/x	−1/x, which is −f(x)	odd
5x	−5x, which is −f(x)	odd
x³ + 1	−x³ + 1, neither f(x) nor −f(x)	neither
x² + x	x² − x, neither f(x) nor −f(x)	neither
(x − 1)²	(x + 1)², neither f(x) nor −f(x)	neither

A third mirror you have already met

There is one more mirror line in this course. In the lecture on inverse functions, the graph of $f^{- 1}$ was the reflection of the graph of $f$ in the line $y = x$ . That reflection swaps the input and the output: $(p, q)$ moves to $(q, p)$ .

So each of the three mirrors is one simple change to a point: change the sign of $y$ , change the sign of $x$ , or swap them.

Worked examples

Worked Example 1 — Reflecting a parabola in the y-axis

$f (x) = x^{2} + 2 x - 3$ . Find the equation of $y = f (- x)$ , its lowest point and its roots.

Replace every $x$ with $(- x)$ , in a bracket:

f (- x) = (- x)^{2} + 2 (- x) - 3, so f (- x) = x^{2} - 2 x - 3 .

The $x^{2}$ term does not change, because a negative number squared is positive. The $2 x$ term changes sign. The constant does not change, because it has no $x$ in it.

Completing the square, $f (x) = (x + 1)^{2} - 4$ , so the lowest point of $f$ is $(- 1, - 4)$ . A reflection in the $y$ -axis moves it to $(1, - 4)$ . The roots of $f$ are $- 3$ and $1$ , so the new roots are $3$ and $- 1$ .

Check by factorising: $x^{2} - 2 x - 3 = (x - 3) (x + 1)$ , with roots $3$ and $- 1$ . ✓ Check the lowest point: $1 - 2 - 3 = - 4$ . ✓

Worked Example 2 — Tracking one point through three changes

The graph of $y = f (x)$ passes through $(3, - 2)$ . Find the matching point on (a) $y = - f (x)$ , (b) $y = f (- x)$ , (c) $y = - f (- x)$ .

(a) The minus is outside, so the output changes sign. $(3, - 2)$ moves to $(3, 2)$ .

(b) The minus is inside, so the position changes sign. $(3, - 2)$ moves to $(- 3, - 2)$ .

Check (b) by substituting. At $x = - 3$ , the new function gives $f (- (- 3)) = f (3) = - 2$ . ✓ The input $- 3$ is changed back to $3$ before $f$ sees it, which is exactly the reason for the reflection.

Worked Example 3 — Following an asymptote

Reflect the graph of $y = \frac{1}{x - 2}$ in the $y$ -axis. Find the new equation and its asymptote.

Replace $x$ with $(- x)$ :

y = \frac{1}{- x - 2}, and taking out the minus sign, y = - \frac{1}{x + 2} .

The old graph approaches the vertical line $x = 2$ , where the denominator is zero. A reflection in the $y$ -axis moves every position $p$ to $- p$ , so this line moves to $x = - 2$ . In the new equation, the denominator $x + 2$ is zero at $x = - 2$ . ✓

Check one point. $(3, 1)$ is on the old graph, because $\frac{1}{3 - 2} = 1$ . It should move to $(- 3, 1)$ . In the new equation: $- \frac{1}{- 3 + 2} = - \frac{1}{- 1} = 1$ . ✓

Worked Example 4 — Proving a function is odd

Show that $g (x) = x^{3} - 4 x$ is odd. Then decide whether $h (x) = x^{3} - 4 x + 1$ is odd, even or neither.

For $g$ : $g (- x) = (- x)^{3} - 4 (- x) = - x^{3} + 4 x$ . And $- g (x) = - (x^{3} - 4 x) = - x^{3} + 4 x$ . These are equal for every $x$ , so $g$ is odd.

See what this means for the roots. $g (x) = x (x - 2) (x + 2)$ , so the roots are $- 2$ , $0$ and $2$ . They come in pairs $\pm 2$ , with 0 in the middle. An odd function defined at 0 must have $g (0) = - g (0)$ , so $g (0) = 0$ . Its graph always passes through O.

For $h$ : $h (- x) = - x^{3} + 4 x + 1$ . Compare with $h (x) = x^{3} - 4 x + 1$ and with $- h (x) = - x^{3} + 4 x - 1$ . It matches neither, so $h$ is neither odd nor even.

Check: $h (1) = - 2$ and $h (- 1) = 4$ , which are neither equal nor opposites. ✓ The odd power $x^{3}$ does not make $h$ odd. The graph of $g$ has been moved up by 1, so its centre of half-turn symmetry is now $(0, 1)$ , not O.

Three classic mistakes

Trap 1 — Mixing up the two mirror lines

A minus sign outside changes the heights, so points move up or down, and the mirror line is the $x$ -axis. A minus sign inside changes the positions, so points move left or right, and the mirror line is the $y$ -axis. If you are unsure, take one point, such as $(2, 5)$ , and ask which of its coordinates changes sign.

Trap 2 — Changing the sign of only one term

$- f (x)$ means the negative of the whole output. For $f (x) = x^{2} + 2 x - 3$ , it is $- (x^{2} + 2 x - 3) = - x^{2} - 2 x + 3$ , with every term changed. Writing $- x^{2} + 2 x - 3$ changes only the first term, and it is a different graph. Always put $f (x)$ in a bracket first.

Trap 3 — Writing −x² for (−x)²

In $f (- x)$ , every $x$ is replaced by $(- x)$ , with its bracket. $(- x)^{2} = x^{2}$ , because the minus is squared too. But $- x^{2}$ means $- (x^{2})$ , which is negative. Without the bracket, $f (- x)$ for $f (x) = x^{2}$ would wrongly become $- x^{2}$ , the reflection in the $x$ -axis, not the $y$ -axis.

Try it yourself

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

$f (x) = x^{2} - 6 x + 5$ . Write the equations of $y = - f (x)$ and $y = f (- x)$ . Give the turning point of each graph, and say whether it is highest or lowest.

Hint 1
Find the turning point of $f$ first, by completing the square. Then move it.

Hint 2
$f (x) = (x - 3)^{2} - 4$ . For $- f (x)$ , change the sign of every term. For $f (- x)$ , replace $x$ with $(- x)$ .

Full answer
$f (x) = (x - 3)^{2} - 4$ , so its lowest point is $(3, - 4)$ . $- f (x) = - x^{2} + 6 x - 5$ . The reflection in the $x$ -axis moves $(3, - 4)$ to $(3, 4)$ , and turns the graph upside down, so $(3, 4)$ is a highest point. $f (- x) = (- x)^{2} - 6 (- x) + 5 = x^{2} + 6 x + 5$ . The reflection in the $y$ -axis moves $(3, - 4)$ to $(- 3, - 4)$ , still a lowest point. Check: $x^{2} + 6 x + 5 = (x + 3)^{2} - 4$ , lowest point $(- 3, - 4)$ . ✓
The graph of $y = f (x)$ has a highest point at $(2, 5)$ and crosses the $x$ -axis at $(- 1, 0)$ and $(4, 0)$ . Describe the graphs of $y = - f (x)$ and $y = f (- x)$ in the same way.

Hint 1
For each reflection, decide which coordinate changes sign. Then ask whether a highest point stays a highest point.

Hint 2
A reflection in the $x$ -axis turns the graph upside down. A reflection in the $y$ -axis does not.

Full answer
$y = - f (x)$ : every height changes sign. The highest point $(2, 5)$ becomes a lowest point at $(2, - 5)$ . The crossing points are on the mirror line, so they stay at $(- 1, 0)$ and $(4, 0)$ . $y = f (- x)$ : every position changes sign. The highest point moves to $(- 2, 5)$ , and it is still a highest point. The crossings move to $(1, 0)$ and $(- 4, 0)$ . Check one: $f (- x)$ at $x = - 4$ is $f (4) = 0$ . ✓
Decide whether each function is even, odd or neither. Show the working. (a) $x^{4} - 2 x^{2}$ (b) $x^{5} - x$ (c) $x^{3} + x^{2}$ (d) $\frac{3}{x}$

Hint 1
Work out $f (- x)$ , and compare it with $f (x)$ and with $- f (x)$ .

Hint 2
An even power of $- x$ is positive. An odd power of $- x$ is negative.

Full answer
(a) $(- x)^{4} - 2 (- x)^{2} = x^{4} - 2 x^{2} = f (x)$ , so it is even. (b) $(- x)^{5} - (- x) = - x^{5} + x = - (x^{5} - x) = - f (x)$ , so it is odd. (c) $(- x)^{3} + (- x)^{2} = - x^{3} + x^{2}$ . This is not $x^{3} + x^{2}$ , and not $- x^{3} - x^{2}$ , so it is neither. Check with $x = 1$ : $f (1) = 2$ and $f (- 1) = 0$ , which are neither equal nor opposites. ✓ (d) $\frac{3}{- x} = - \frac{3}{x} = - f (x)$ , so it is odd.
$f (x) = x^{2} - 4 x + 7$ . (a) Show that $f (2 + t) = f (2 - t)$ for every number $t$ . What does this tell you about the graph? (b) Is $f$ an even function? (c) Write $f$ as a translation of an even function.

Hint 1
Complete the square first. Then substitute $2 + t$ and $2 - t$ .

Hint 2
$f (x) = (x - 2)^{2} + 3$ . For (b), test one pair of inputs, $1$ and $- 1$ .

Full answer
(a) $f (x) = (x - 2)^{2} + 3$ . So $f (2 + t) = t^{2} + 3$ and $f (2 - t) = (- t)^{2} + 3 = t^{2} + 3$ . They are equal. Points at the same distance $t$ to the right and left of $x = 2$ have the same height, so the graph is symmetric about the vertical line $x = 2$ . (b) No. $f (1) = 4$ but $f (- 1) = 12$ , so $f (- x) \neq = f (x)$ . Its mirror line is $x = 2$ , not the $y$ -axis. (c) $g (x) = x^{2} + 3$ is even, because $(- x)^{2} + 3 = x^{2} + 3$ . And $f (x) = g (x - 2)$ , so the graph of $f$ is the graph of $g$ moved 2 to the right. The mirror line moved with it, from $x = 0$ to $x = 2$ . Check: $f (3.5) = 5.25$ and $f (0.5) = 5.25$ . ✓

The bigger idea

A reflection is a question about symmetry, and symmetry saves work. If you know a function is even, you only need to draw the right half of its graph. The left half is its mirror image. If it is odd, the left half is the right half turned through 180°. When you meet the graphs of sine and cosine, you will find that cosine is even and sine is odd, and much of their behaviour follows from that one fact.

Asking "what happens if I change the sign?" is one of the most useful questions in mathematics and physics. A pattern that survives a mirror or a half-turn tells you something about the rule that made it.

Key concepts in this lesson

The graph of $y = - f (x)$ is the reflection of $y = f (x)$ in the $x$ -axis, because every output changes sign: $(p, q)$ moves to $(p, - q)$ .
The graph of $y = f (- x)$ is the reflection of $y = f (x)$ in the $y$ -axis, because the input must change sign again to give the same output: $(p, q)$ moves to $(- p, q)$ .
The graph of $y = - f (- x)$ is a half-turn about the origin, which moves $(p, q)$ to $(- p, - q)$ .
A function is even if $f (- x) = f (x)$ , so its graph is symmetric about the $y$ -axis, and odd if $f (- x) = - f (x)$ , so its graph is unchanged by a half-turn about O.
To test a function, work out $f (- x)$ with every $x$ in a bracket, and compare it with $f (x)$ and $- f (x)$ ; most functions are neither even nor odd.