Graph Translations and Reflections
Higher
Move graphs by following what happens to their coordinates. Learn the four required translation/reflection rules, explain horizontal signs and turn a description into an equation.
Before you start: Functions, Linear Graphs, Quadratic Graphs and Graphical Solutions and Trigonometric Graphs. For geometric shapes, see Reflections, Rotations and Translations.
1. Follow inputs and outputs
The notation y = f(x) means that a rule takes an input x and produces an output y. A transformation moves every point of its graph, not just the intercepts. The same ideas can be expressed in words or in an ordinary equation.
| New equation | Movement | Point mapping |
|---|---|---|
| y = f(x) + a | a units up (down if a < 0) | (x, y) → (x, y + a) |
| y = f(x + a) | a units left (right if a < 0) | (x, y) → (x − a, y) |
| y = −f(x) | Reflect in the x-axis | (x, y) → (x, −y) |
| y = f(−x) | Reflect in the y-axis | (x, y) → (−x, y) |
An alteration outside f changes the output, so it acts vertically. An alteration inside f changes the input needed to get an old output, so it acts horizontally. In the graphs below, the original is solid blue and the image dashed orange. Both use the same axes and scales.
2. Vertical and horizontal translations
Worked example — y = f(x) + 2
Add 2 to every output. The x-coordinates stay fixed, so the graph moves two units up. The supplied graph has A(2, 3); its image is (2, 5). An endpoint (−2, 0) becomes (−2, 2).
Worked example — y = f(x + 2)
The old output 3 occurred when the old input was 2. The new input x must therefore satisfy x + 2 = 2, giving x = 0. So (2, 3) becomes (0, 3) and the graph moves two units left. Every x-coordinate decreases by 2; the heights stay fixed.
This explains the horizontal sign: f(x − 4) moves right 4 because the new input must be 4 larger to supply the old input. In words, “use the old rule on x minus four” places an old feature four units farther right.
3. The two reflections
Worked example — y = −f(x)
Negate the output. A(2, 3) becomes (2, −3), the same horizontal position below the axis. Points on the x-axis stay fixed. This is a reflection in the x-axis.
Worked example — y = f(−x)
Negate the input. To obtain the old input 2, the new x must be −2. Thus A(2, 3) becomes (−2, 3), with the same height. Points on the y-axis stay fixed. This is a reflection in the y-axis.
The transformations are different even when a symmetric graph can disguise a reflection. For y = x², reflecting in the y-axis leaves it unchanged because (−x)² = x². Use a labelled point or an asymmetric shape to check the rule.
4. Combine transformations and write equations
Worked example — y = f(x − 4) + 1
Move each point four units right and one unit up: (x, y) → (x + 4, y + 1). A(2, 3) becomes (6, 4). The supplied endpoints (−2, 0) and (3, 1) become (2, 1) and (7, 2).
Worked example — reflect, then shift
For y = −f(x) + 2, first reflect in the x-axis, then move up 2. A(2, 3) → (2, −3) → (2, −1). The mapping is (x, y) → (x, 2 − y).
Order matters here. Moving up 2 before reflecting would give −(f(x) + 2) = −f(x) − 2, sending A to (2, −5), a different result. Read the operations in the expression.
Worked example — write a quadratic after a shift
Move y = x² left 2 and down 1. Replace x by x + 2, then subtract 1: y = (x + 2)² − 1. Its vertex moves from (0, 0) to (−2, −1). At x = −2 the bracket is zero, which checks the direction.
Conversely, comparing y = (x − 3)² + 4 with y = x² tells you “right 3, up 4”. You can describe the movement without function notation.
5. Transform a trigonometric graph
Worked example — y = sin(x − 90°)
The input is reduced by 90°, so move the sine graph 90° right. The old maximum at (90°, 1) becomes (180°, 1). The old zero at (0°, 0) becomes (90°, 0).
The height range stays −1 to 1 and the period stays 360°. Translations and reflections change position or orientation, not the horizontal length of a cycle. For the triangle meaning of sine, use Right-Angled Trigonometry and Exact Values; for the original wave shapes, use Trigonometric Graphs.
Think Like a Mathematician
Can two different transformations produce the same picture? Yes: reflecting y = x² in the y-axis leaves it unchanged. That does not make the reflection rule wrong. Its symmetry means each image point was already on the graph. The supplied asymmetric piecewise graph makes the difference between x-axis and y-axis reflections visible.
Common Mistakes
Horizontal sign: f(x + a) moves left for positive a. Reflection axis: −f(x) changes heights; f(−x) changes horizontal positions. Whole graph: move all features and endpoints. Combined operations: read their order. Inverse confusion: −f(x) is not the inverse function f⁻¹(x).
Practice — Try It Yourself
Work on squared paper where a graph is requested. Try each question before opening its answer. Graph readings are estimates: give only the precision your scale supports.
1. Describe y = f(x) − 3 and map the point (2, 3).
Translate 3 down. Subtract 3 from the y-coordinate: (2, 3) → (2, 0).
2. Describe y = f(x − 5) and map (−1, 4). Explain the sign.
Translate 5 right: (−1, 4) → (4, 4). At the new x = 4, the expression x − 5 supplies the old input −1.
3. For a point (−2, 5), give its image under y = −f(x) and under y = f(−x). Name both reflections.
−f(x): (−2, −5), reflection in the x-axis. f(−x): (2, 5), reflection in the y-axis. Only the appropriate coordinate changes sign.
4. Sketch y = f(x + 1) − 2 using the supplied piecewise straight graph. Its vertices are (−2, 0), (0, 1), (2, 3), (3, 1).
Move left 1, down 2. Plot (−3, −2), (−1, −1), (1, 1), (2, −1) and join them in that order with straight segments, preserving the supplied shape. Its new domain is −3 ≤ x ≤ 2. Every vertex and endpoint must move.
5. Write the equation after y = x² moves right 3 and up 4. State the vertex.
y = (x − 3)² + 4, vertex (3, 4). Substituting x = 3 makes the square zero.
6. A student sketches y = f(x + 4) by moving right 4. Explain the error using an old point (2, 3).
The new input must satisfy x + 4 = 2, so x = −2. The point becomes (−2, 3): the graph should move left 4. The sign inside the function describes the changed input, not the direction to move the picture.
7. Map (2, 3) under y = −f(x) + 1. Describe the two operations in the right order.
Reflect in the x-axis, then move up 1: (2, 3) → (2, −3) → (2, −2). Equivalently, new y = 1 − old y.
8. Describe y = cos(x + 60°) relative to y = cos x. Where does the maximum (0°, 1) move?
Translate 60° left. The maximum moves to (−60°, 1). Equivalent maxima repeat every 360°, for example at (300°, 1).
9. A graph is reflected in the y-axis and then moved down 2. Write its equation and map (3, −1).
y = f(−x) − 2. First (3, −1) → (−3, −1), then → (−3, −3).
Frequently Asked Questions
Do I need a formula for the original graph?
No. Track the coordinates of a supplied shape, including its turning points, corners and endpoints.
Why does a plus inside mean left?
The input you write must be smaller to compensate for the added amount. If the old input is 2, x + 3 = 2 needs x = −1: the feature moves 3 left.
Are stretches included here?
This lesson covers translations and reflections. The four rules above preserve the shape; do not apply a geometric enlargement rule to them.