Reflecting a graph across the axes
How do you turn a graph upside down, or make it face the other way, by changing only its rule?
Where you are 7 in Transformations of functions
When you reach for it
You already have two ways to move a graph without changing its shape: add a number to the output and the graph slides up or down, replace x by x - h and it slides sideways (shifting a graph). Both keep the graph the right way up. Sometimes the graph you need has the right shape but is upside down, or faces the other way: a parabola that opens downward, a square root curve that runs to the left. That is a reflection, and a single minus sign in the rule makes it. Where the minus sign goes decides which way the graph flips.
You also reach for it in reverse, when a graph in front of you looks like a parent graph turned over and you want to name what was done to it.
The procedure
Flipping the outputs: -f(x)
Start with a function f and build g(x) = -f(x). For each input, g does whatever f does and then takes the opposite of the answer. The inputs are untouched and every output changes sign, so if (a, b) is on the graph of f, then (a, -b) is on the graph of g:
A point above the horizontal axis lands the same distance below it, at the same x, and a point below lands above. That is the reflection of a point across the x-axis, done to every point of the graph at once. So the graph of y = -f(x) is the graph of f reflected across the x-axis, with the axis as the mirror.
Two things never change. A point on the axis has b = 0, and -0 = 0, so the zeros of f stay where they are. And the inputs are the same, so the domain is the same. The range is what flips: every output is replaced by its opposite.
Flipping the inputs: f(-x)
Now build h(x) = f(-x). This time the minus sign acts first: h takes its input, changes its sign, and hands the result to f. To find where the point (a, b) of f goes, ask which input makes h hand a to f. The answer is -a, because
So (-a, b) is on the graph of h. The output stays and the input changes sign:
Every point moves straight across the vertical axis to the same distance on the other side, so the graph of y = f(-x) is the graph of f reflected across the y-axis. The y-intercept, where a = 0, stays put. The range is unchanged and the domain flips.
This is the same reversal you met with shifts, where f(x - 3) moves right although it shows a minus sign. A change made inside, to the input, has to be undone to see where the points go.
The figure shows both flips of \displaystyle y = \sqrt{x} at once. The dashed curve is \displaystyle \sqrt{x}. Below it is \displaystyle -\sqrt{x}, each point dropped straight down through the x-axis. To its left is \displaystyle \sqrt{-x}, each point carried straight across the y-axis.
Both at once
The two signs do not interfere. -f(-x) changes the sign of the input and then of the output, so (a, b) \mapsto (-a, -b): a reflection across one axis followed by a reflection across the other, which together turn the graph a half turn about the origin.
Naming a reflection from a graph
Compare matching points of the parent and the new graph. If each point (a, b) of the parent has the partner (a, -b), the new graph is y = -f(x). If the partner is (-a, b), it is y = f(-x). If it is (-a, -b), it is y = -f(-x). A quick first look: a graph that has gone upside down points to the outside sign, and one that now faces the other way points to the inside sign.
Sometimes a flip is invisible, or two flips look the same. Reflecting y = x^2 across the y-axis changes nothing, because (-x)^2 = x^2: the parabola is already its own mirror image, the property studied in Symmetry about the vertical axis and the origin. For y = x^3 the two flips give the same graph (Example 3). Then more than one rule names the graph, and each is correct.
Worked examples
Turning the square root curve upside down
Graph \displaystyle g(x) = -\sqrt{x} from the parent graph \displaystyle y = \sqrt{x}, and give its domain and range.
SolutionThe minus sign is outside the square root, so it acts on the outputs: g(x) = -f(x) with \displaystyle f(x) = \sqrt{x}. Each parent point (a, b) goes to (a, -b). Use the parent points with whole-number outputs:
| x | 0 | 1 | 4 | 9 |
|---|---|---|---|---|
| \displaystyle \sqrt{x} | 0 | 1 | 2 | 3 |
| \displaystyle -\sqrt{x} | 0 | -1 | -2 | -3 |
Plot the new points and join them in the parent’s shape. The curve starts at the origin and runs to the right, falling more and more slowly: the lower curve in the figure above. The starting point (0, 0) did not move, since it is on the mirror.
The inputs are the parent’s inputs, so the domain is still x \ge 0. The parent’s outputs are all y \ge 0, and their opposites are all y \le 0, so the range is y \le 0.
Making the square root curve face left
Graph \displaystyle h(x) = \sqrt{-x}, give its domain and range, and find h(-9).
SolutionNow the minus sign is inside, on the input: h(x) = f(-x) with \displaystyle f(x) = \sqrt{x}. Each parent point (a, b) goes to (-a, b):
| x | 0 | -1 | -4 | -9 |
|---|---|---|---|---|
| \displaystyle \sqrt{-x} | 0 | 1 | 2 | 3 |
Check one entry with the rule: \displaystyle h(-4) = \sqrt{-(-4)} = \sqrt{4} = 2. The curve starts at the origin and runs to the left, rising: the upper left curve in the figure above.
The square root needs -x \ge 0, which means x \le 0, so the domain is x \le 0. A negative input is exactly what \displaystyle \sqrt{-x} wants, since then -x is positive. The outputs are the parent’s outputs, so the range is still y \ge 0.
Finally \displaystyle h(-9) = \sqrt{9} = 3, the height the parent has at x = 9.
Turning the cube over
Graph g(x) = -x^3 by reflecting y = x^3 across the x-axis. Then reflect y = x^3 across the y-axis instead, and compare.
SolutionBy the order of operations the power is taken first and the minus sign applies to its result, so -x^3 means -(x^3). The sign is outside and acts on outputs: each point (a, a^3) goes to (a, -a^3).
| x | -2 | -1 | 0 | 1 | 2 |
|---|---|---|---|---|---|
| x^3 | -8 | -1 | 0 | 1 | 8 |
| -x^3 | 8 | 1 | 0 | -1 | -8 |
The parent cubic rises from lower left to upper right through the origin. The new graph falls from upper left to lower right.
For the other flip, replace x by -x:
Three negative factors give a negative product, so (-x)^3 and -x^3 are the same function, and the two reflections give the same graph. The table shows it too: the -x^3 row is the x^3 row written backwards, which is what a left to right flip does to a table whose inputs sit symmetrically about 0. So y = -x^3 may be named either way.
Naming the reflection of 1/x
The graph in the figure is a reflection of \displaystyle y = \frac{1}{x}. Name the reflection and write the rule.
The parent \displaystyle y = \frac{1}{x} has its branches in the first and third quadrants, through (1, 1) and (-1, -1). This graph has them in the second and fourth, through (1, -1) and (-1, 1). Compare matching points: (1, 1) became (1, -1) and \displaystyle \left(2, \frac{1}{2}\right) became \displaystyle \left(2, -\frac{1}{2}\right). Same input, opposite output: a reflection across the x-axis, so the rule is
Check a marked point: at x = -1, \displaystyle -\frac{1}{-1} = 1, so (-1, 1) is on it.
Reflecting across the y-axis instead gives \displaystyle \frac{1}{-x}, which equals \displaystyle -\frac{1}{x} for every x \ne 0, so that name is just as correct, as with the cube. Either flip carries each axis onto itself, so the branches still approach the axes without touching them: the domain is still every x \ne 0 and the range every y \ne 0.
A mirror-image agility course
Two practice rings for dog agility share a wall. Berenike measures x in meters east of the center line and y in meters north of the shared wall. Her dog’s path, y = p(x), passes the five obstacles at these points:
| x | -6 | -2 | 0 | 3 | 8 |
|---|---|---|---|---|---|
| p(x) | 5 | 1 | 2 | 7 | 4 |
(a) Dariusz sets the course in the same ring with east and west swapped. Write the rule for his path and its table.
(b) Octavia’s ring lies south of the wall, and she sets the course reflected across the wall, with negative y meaning meters south. Write the rule and the table.
Solution(a) Swapping east and west replaces each x by its opposite and keeps y: (a, b) \mapsto (-a, b), a reflection across the center line, which is the y-axis. So Dariusz’s path is q(x) = p(-x):
| x | 6 | 2 | 0 | -3 | -8 |
|---|---|---|---|---|---|
| q(x) | 5 | 1 | 2 | 7 | 4 |
Check with the rule: q(-3) = p(3) = 7.
(b) Reflecting across the wall, the x-axis, keeps x and changes the sign of y. Octavia’s path is r(x) = -p(x):
| x | -6 | -2 | 0 | 3 | 8 |
|---|---|---|---|---|---|
| r(x) | -5 | -1 | -2 | -7 | -4 |
Each obstacle is as far south of the wall as Berenike’s is north of it.
The new tables hold only the reflected points. q(4) cannot be found, because it needs p(-4), which was never measured: a reflection moves known points but cannot supply new ones.
A flip joined to a shift
Graph y = -(x - 2)^2 from y = x^2. Give the vertex, the direction it opens and the range.
SolutionRead the rule from the inside out. Replacing x by x - 2 turns x^2 into (x - 2)^2, the parabola shifted 2 to the right, with vertex (2, 0). The minus sign in front then acts on the whole output (x - 2)^2, reflecting that shifted parabola across the x-axis.
| x | 0 | 1 | 2 | 3 | 4 |
|---|---|---|---|---|---|
| (x - 2)^2 | 4 | 1 | 0 | 1 | 4 |
| -(x - 2)^2 | -4 | -1 | 0 | -1 | -4 |
The vertex (2, 0) lies on the mirror, so it stays. Every other point drops below the axis, so the parabola opens downward and the vertex is now its highest point. A square is never negative, so -(x - 2)^2 \le 0 for every x, with 0 only at x = 2: the range is y \le 0.
In vertex form, a minus sign in front of the bracket is exactly this reflection.
Boundaries
The mirror is an axis. -f(x) reflects across the x-axis, the line y = 0, and f(-x) across the y-axis, the line x = 0. A mirror along another line, such as y = 3, needs a shift as well, which is the business of Combining transformations and the general form. Reflecting across the line y = x swaps inputs and outputs and gives the graph of an inverse, which need not be a function.
The sign must reach the whole output, or every input. The reflection of \displaystyle y = \sqrt{x} + 2 across the x-axis is \displaystyle -(\sqrt{x} + 2) = -\sqrt{x} - 2, not \displaystyle -\sqrt{x} + 2. In f(-x), every x in the rule is replaced, including any inside a bracket or a root.
Each flip changes one of domain and range. -f(x) keeps the domain and negates the range; f(-x) keeps the range and negates the domain.
A flip can leave a graph unchanged, as f(-x) does to y = x^2, and two different flips can give the same graph, as for y = x^3 and \displaystyle y = \frac{1}{x}. Naming a reflection from a graph then has more than one right answer.
Where this goes next
Stretches and compressions change a graph’s size, and a negative factor is a stretch with this flip attached. Combining transformations and the general form gathers every move into one rule, a f(b(x - h)) + k, where a negative a reflects across the x-axis and a negative b across the y-axis; Graphing by transforming a parent graph draws such graphs and Finding the rule of a transformed graph or table works backwards from them. Asking whether f(-x) equals f(x), or equals -f(x), is the test for even and odd functions.
Exercises
Write the rule
For each function, write -f(x) and f(-x), and simplify.
- f(x) = x^2 + 1
- f(x) = 2x + 3
- \displaystyle f(x) = \sqrt{x - 1}. Give the domain of each new function too.
- \displaystyle f(x) = \frac{1}{x + 3}
Points and values
- The point (4, -7) is on the graph of f. Name the matching point on the graph of y = -f(x), of y = f(-x) and of y = -f(-x).
- f(2) = 5 and f(-2) = 1. With g(x) = -f(x) and h(x) = f(-x), find g(2), g(-2), h(2) and h(-2).
- The graph of f crosses the x-axis at x = -3 and x = 6, and the y-axis at y = 4. Where does the graph of -f(x) cross each axis? Where does the graph of f(-x)?
Tables
Use this table for exercises 8 and 9.
| x | -4 | -1 | 0 | 2 | 6 |
|---|---|---|---|---|---|
| f(x) | 3 | -2 | 5 | 0 | -1 |
- Make a table for g(x) = -f(x) and one for h(x) = f(-x).
- Find h(-2), h(4) and g(-1). Explain why h(3) cannot be found.
Graphs, domain and range
- Describe the graph of y = -|x|: its shape, its highest or lowest point, and its range.
- A function f has domain -2 \le x \le 5 and range 1 \le y \le 6. Give the domain and range of -f(x) and of f(-x).
- Give the domain and range of \displaystyle y = -\sqrt{x - 3}, and three points on its graph.
- Simplify (-x)^2 and -x^2. Which one is y = x^2 reflected across the x-axis, and what does the other show about reflecting y = x^2 across the y-axis?
Name the reflection
- The graph of f passes through (1, 3), (2, 5) and (4, 9). A reflection of it passes through (-1, 3), (-2, 5) and (-4, 9). Write the reflection in terms of f.
- Another reflection of the same graph passes through (1, -3), (2, -5) and (4, -9). Write it in terms of f.
Each graph in the figure is \displaystyle y = \sqrt{x} or y = x^2 after one or two reflections.
- Name the reflections that give graph A, and write its rule.
- Name the reflection that gives graph B, and write its rule.
An agility course
- Valerian’s course uses the same grid as Example 5. His dog’s path is the line \displaystyle y = \frac{x}{2} + 4, passing the tunnel at (6, 7) and a jump at (-2, 3). Ismena sets the same course with east and west swapped. Where are her tunnel and jump, and what is the rule for her dog’s path?
Reasoning
- Explain why the minus sign in -f(x) moves points up and down, while the one in f(-x) moves them left and right.
- Explain why reflecting a graph across the x-axis twice gives back the original graph. Is the same true for the y-axis?
- Explain why the graph of -f(x) has the same x-intercepts as the graph of f, and the graph of f(-x) the same y-intercept.
- Ismena says the graph of y = -f(x) always lies below the x-axis. Test her claim with f(x) = x - 3, and say what is true instead.
Answers
- -f(x) = -x^2 - 1 and f(-x) = (-x)^2 + 1 = x^2 + 1, which is f itself: this parabola is its own mirror image across the y-axis.
- -f(x) = -2x - 3 and f(-x) = 2(-x) + 3 = -2x + 3.
- \displaystyle -f(x) = -\sqrt{x - 1}, domain x \ge 1. \displaystyle f(-x) = \sqrt{-x - 1}, domain x \le -1, since -x - 1 \ge 0 means x \le -1.
- \displaystyle -f(x) = -\frac{1}{x + 3} and \displaystyle f(-x) = \frac{1}{-x + 3} = \frac{1}{3 - x}.
- (4, 7), (-4, -7) and (-4, 7). The outside sign changes the second coordinate, the inside sign the first.
- g(2) = -5, g(-2) = -1, h(2) = f(-2) = 1 and h(-2) = f(2) = 5.
- -f(x) crosses the x-axis at x = -3 and x = 6 (zeros do not move) and the y-axis at y = -4. f(-x) crosses the x-axis at x = 3 and x = -6 and the y-axis at y = 4.
- g: the inputs -4, -1, 0, 2, 6 give -3, 2, -5, 0, 1. h: the inputs 4, 1, 0, -2, -6 give 3, -2, 5, 0, -1.
- h(-2) = f(2) = 0, h(4) = f(-4) = 3 and g(-1) = -f(-1) = 2. h(3) = f(-3), and the table does not give f(-3).
- An upside-down V whose corner (0, 0) is its highest point. The left arm rises to the corner with slope 1 and the right arm falls with slope -1. Range y \le 0.
- -f(x): domain -2 \le x \le 5, range -6 \le y \le -1. f(-x): domain -5 \le x \le 2, range 1 \le y \le 6.
- Domain x \ge 3, range y \le 0. Points (3, 0), (4, -1) and (7, -2): the curve starts at (3, 0) and falls to the right.
- (-x)^2 = x^2 and -x^2 = -(x^2). y = -x^2 is the reflection across the x-axis, opening downward. (-x)^2 = x^2 shows the reflection across the y-axis leaves y = x^2 unchanged.
- y = f(-x): each point (a, b) became (-a, b).
- y = -f(x): each point (a, b) became (a, -b).
- Graph A is \displaystyle y = \sqrt{x} reflected across both axes: \displaystyle y = -\sqrt{-x}. The parent’s (4, 2) is at (-4, -2), both coordinates negated.
- Graph B is y = x^2 reflected across the x-axis: y = -x^2. The parent’s (2, 4) is at (2, -4). Reflecting across both axes gives the same graph, since -(-x)^2 = -x^2.
- Tunnel (-6, 7), jump (2, 3), path \displaystyle y = -\frac{x}{2} + 4, found by replacing x with -x in \displaystyle \frac{x}{2} + 4.
- In -f(x) the input reaches f unchanged and only the answer is negated, so each point keeps its x and its height changes sign. In f(-x) the answer is untouched, and the input that gives f’s old height b is now -a instead of a, so each point keeps its height and moves to the opposite x.
- Two reflections across the x-axis give -(-f(x)) = f(x), the original function. Yes: f(-(-x)) = f(x) as well.
- An x-intercept is a point (a, 0), and (a, -0) = (a, 0), so it is on -f(x) too. The y-intercept of f(-x) is at f(-0) = f(0), the same height as for f.
- Here -f(x) = 3 - x, which is 3 at x = 0, above the axis, so the claim is false. What is true: -f(x) is below the axis exactly where f(x) is above it, here for x > 3.