GCSE Maths: Graph transformations (Higher)
Part of the GCSE Maths functions and advanced graphs chapter (Higher). Chapter overview · whole chapter on one page · Download worksheet (PDF) · Answers (PDF)
Graph transformations
If you know the graph of y = f(x), you can sketch the graphs of many related functions without calculating anything new. Each point on the old graph moves to a new position, and the new graph is the old graph moved, flipped or stretched. This is called a transformation of the graph.
The rules are in the table. In each case, a point (x, y) on y = f(x) goes to the point shown.
| New function | What happens | Point (x, y) goes to |
|---|---|---|
| y = f(x) + a | translation up a units (down if a is negative) | (x, y + a) |
| y = f(x + a) | translation a units to the left (right if a is negative) | (x − a, y) |
| y = −f(x) | reflection in the x-axis | (x, −y) |
| y = f(−x) | reflection in the y-axis | (−x, y) |
| y = af(x) (extension) | stretch parallel to the y-axis, scale factor a | (x, ay) |
| y = f(ax) (extension) | stretch parallel to the x-axis, scale factor 1/a | (x/a, y) |
The graph below shows y = f(x) = x2 − 4x + 3 (dashed) and four transformed graphs (solid orange). Look at how the minimum point B(2, −1) moves each time.
Why does f(x + 2) move the graph to the left? The new function gives the same output as f but when the input is 2 smaller: for example the new function at x = 0 gives f(0 + 2) = f(2). So whatever f does at x = 2, the new graph does at x = 0: everything has shifted 2 to the left.
Everything moves together. Turning points, intercepts and asymptotes move with the graph. The reciprocal graph y = 1/x has asymptotes x = 0 and y = 0. The graph of y = 1/x + 2 has asymptotes x = 0 and y = 2, and the graph of y = 1/(x − 3) has asymptotes x = 3 and y = 0.
Completing the square. A quadratic written as y = (x + p)2 + q is the graph of y = x2 translated by p to the left and q up, so its turning point is (−p, q). This links to the work on completing the square in Chapter 14.
Key ideas
- f(x) + a: translate up a. f(x + a): translate a to the left. Together: the vector (−a, b) for f(x + a) + b.
- −f(x): reflect in the x-axis. f(−x): reflect in the y-axis.
- af(x): stretch in the y-direction by scale factor a. f(ax): stretch in the x-direction by scale factor 1/a.
- Changes outside the bracket affect y. Changes inside the bracket affect x (and work the opposite way).
- Turning points, roots and asymptotes move with the graph: transform the key points first.
Method. y = f(x) + 2 is a translation by the vector 0
2 (2 units up). Apply the rule to each key point.
- Identify the transformation. Look at what has been done to f(x): is it added outside the bracket, changed inside the bracket, or multiplied?y = f(x) + 2 is a translation by the vector 0
2 (2 units up) - Use the rule for this type. The graph moves up (or down) by the number added to f(x). Only the y-coordinates change: add the number to y.
- Apply the rule to each point. Work out the new coordinates.A (1, 0) → (1, 2)
B (2, −1) → (2, 1)
C (3, 0) → (3, 2)
D (0, 3) → (0, 5) - Sketch the new graph. Join the new points with a smooth curve of the same shape (flipped or stretched if the transformation does that). The old graph is shown dashed for comparison.
Check. Check one point: B′ = (2, 1). The new equation y = x2 − 4x + 5 gives y = 22 − 4 × 2 + 5 = 1 when x = 2. ✓
Watch out. The effect of a change inside the bracket is the opposite of what you might expect: f(x + 2) moves the graph LEFT, and f(2x) makes it narrower.
Method. y = f(x − 3) is a translation by the vector 3
0 (3 units to the right). Apply the rule to each key point.
- Identify the transformation. Look at what has been done to f(x): is it added outside the bracket, changed inside the bracket, or multiplied?y = f(x − 3) is a translation by the vector 3
0 (3 units to the right) - Use the rule for this type. Inside the bracket the change is the opposite way round to what you expect: y = f(x + a) moves the graph a units to the LEFT. Only the x-coordinates change: subtract a.
- Apply the rule to each point. Work out the new coordinates.A (1, 0) → (4, 0)
B (2, −1) → (5, −1)
C (3, 0) → (6, 0)
D (0, 3) → (3, 3) - Sketch the new graph. Join the new points with a smooth curve of the same shape (flipped or stretched if the transformation does that). The old graph is shown dashed for comparison.
Check. Check one point: B′ = (5, −1). The new equation y = x2 − 10x + 24 gives y = 52 − 10 × 5 + 24 = −1 when x = 5. ✓
Watch out. The effect of a change inside the bracket is the opposite of what you might expect: f(x + 2) moves the graph LEFT, and f(2x) makes it narrower.
Method. y = −f(x) is a reflection in the x-axis. Apply the rule to each key point.
- Identify the transformation. Look at what has been done to f(x): is it added outside the bracket, changed inside the bracket, or multiplied?y = −f(x) is a reflection in the x-axis
- Use the rule for this type. Minus f(x) turns the graph upside down (reflection in the x-axis). Only the y-coordinates change sign.
- Apply the rule to each point. Work out the new coordinates.A (0, 0) → (0, 0)
B (2, −4) → (2, 4)
C (3, 0) → (3, 0) - Sketch the new graph. Join the new points with a smooth curve of the same shape (flipped or stretched if the transformation does that). The old graph is shown dashed for comparison.
Check. Check one point: B′ = (2, 4). The new equation y = −x3 + 3x2 gives y = −23 + 3 × 22 = 4 when x = 2. ✓
Watch out. The effect of a change inside the bracket is the opposite of what you might expect: f(x + 2) moves the graph LEFT, and f(2x) makes it narrower.
Method. y = f(−x) is a reflection in the y-axis. Apply the rule to each key point.
- Identify the transformation. Look at what has been done to f(x): is it added outside the bracket, changed inside the bracket, or multiplied?y = f(−x) is a reflection in the y-axis
- Use the rule for this type. f(−x) flips the graph left to right (reflection in the y-axis). Only the x-coordinates change sign.
- Apply the rule to each point. Work out the new coordinates.A (0, 0) → (0, 0)
B (2, −4) → (−2, −4)
C (3, 0) → (−3, 0) - Sketch the new graph. Join the new points with a smooth curve of the same shape (flipped or stretched if the transformation does that). The old graph is shown dashed for comparison.
Check. Check one point: B′ = (−2, −4). The new equation y = −x3 − 3x2 gives y = −(−2)3 − 3 × (−2)2 = −4 when x = −2. ✓
Watch out. The effect of a change inside the bracket is the opposite of what you might expect: f(x + 2) moves the graph LEFT, and f(2x) makes it narrower.
Method. Write the new function using f, replace f by its rule, then simplify.
- Write the original function. f(x) = x2 − 4x
- Say what the new function is in terms of f. The new graph is y = f(x + 2).y = f(x + 2)
- Put the rule in. Replace every x in the rule by (x + 2).(x + 2)2 − 4(x + 2)
- Expand and simplify. Multiply out any brackets and collect like terms.y = x2 − 4
Check. Test with x = 3: the new rule gives 5; using the original rule with the changed input gives the same number. ✓
Watch out. Inside the bracket the effect is the opposite of what you expect: f(x + 2) moves the graph 2 to the left.
Method. Complete the square. The completed-square form is a transformation of y = x²: a translation left or right, then up or down.
- Halve the number in front of x. Completing the square: x2 + bx = (x + b/2)2 − (b/2)2.6 ÷ 2 = 3
- Write the bracket and subtract the square of the half. (x + 3)2 expands to x2 + 6x + 9, which has 9 too many, so subtract it.x2 + 6x = (x + 3)2 − 9
- Add the constant term. y = (x + 3)2 + 2
- Read the turning point. The graph of y = x2 has its minimum at (0, 0). Replacing x by x + 3 moves it 3 left, and adding 2 moves it 2 up.minimum at (−3, 2)
Check. Expand: (x + 3)2 + 2 = x2 + 6x + 11 = x2 + 6x + 11. ✓
Watch out. The x-coordinate of the turning point is the opposite sign to the number inside the bracket.
Watch out
A change inside the bracket does the opposite of what you expect: y = f(x + 2) moves 2 units to the left, not the right, and y = f(2x) is narrower, not wider. Also, −f(x) flips the graph over (y changes sign) but f(−x) flips it sideways (x changes sign). Only the coordinate that is affected changes.Exercise 25D Graph transformations
Non-calculatorMethod. y = f(x) + 3 is a translation by the vector 0
3 (3 units up). Apply the rule to each key point.
- Identify the transformation. Look at what has been done to f(x): is it added outside the bracket, changed inside the bracket, or multiplied?y = f(x) + 3 is a translation by the vector 0
3 (3 units up) - Use the rule for this type. The graph moves up (or down) by the number added to f(x). Only the y-coordinates change: add the number to y.
- Apply the rule to each point. Work out the new coordinates.A (1, 0) → (1, 3)
B (2, −1) → (2, 2)
C (3, 0) → (3, 3)
D (0, 3) → (0, 6)
Check. Check one point: B′ = (2, 2). The new equation y = x2 − 4x + 6 gives y = 22 − 4 × 2 + 6 = 2 when x = 2. ✓
Watch out. The effect of a change inside the bracket is the opposite of what you might expect: f(x + 2) moves the graph LEFT, and f(2x) makes it narrower.
Method. y = f(x − 2) is a translation by the vector 2
0 (2 units to the right). Apply the rule to each key point.
- Identify the transformation. Look at what has been done to f(x): is it added outside the bracket, changed inside the bracket, or multiplied?y = f(x − 2) is a translation by the vector 2
0 (2 units to the right) - Use the rule for this type. Inside the bracket the change is the opposite way round to what you expect: y = f(x + a) moves the graph a units to the LEFT. Only the x-coordinates change: subtract a.
- Apply the rule to each point. Work out the new coordinates.A (1, 0) → (3, 0)
B (2, −1) → (4, −1)
C (3, 0) → (5, 0)
D (0, 3) → (2, 3)
Check. Check one point: B′ = (4, −1). The new equation y = x2 − 8x + 15 gives y = 42 − 8 × 4 + 15 = −1 when x = 4. ✓
Watch out. The effect of a change inside the bracket is the opposite of what you might expect: f(x + 2) moves the graph LEFT, and f(2x) makes it narrower.
Method. y = −f(x) is a reflection in the x-axis. Apply the rule to each key point.
- Identify the transformation. Look at what has been done to f(x): is it added outside the bracket, changed inside the bracket, or multiplied?y = −f(x) is a reflection in the x-axis
- Use the rule for this type. Minus f(x) turns the graph upside down (reflection in the x-axis). Only the y-coordinates change sign.
- Apply the rule to each point. Work out the new coordinates.A (1, 0) → (1, 0)
B (2, −1) → (2, 1)
C (3, 0) → (3, 0)
D (0, 3) → (0, −3)
Check. Check one point: B′ = (2, 1). The new equation y = −x2 + 4x − 3 gives y = −22 + 4 × 2 − 3 = 1 when x = 2. ✓
Watch out. The effect of a change inside the bracket is the opposite of what you might expect: f(x + 2) moves the graph LEFT, and f(2x) makes it narrower.
Method. y = f(−x) is a reflection in the y-axis. Apply the rule to each key point.
- Identify the transformation. Look at what has been done to f(x): is it added outside the bracket, changed inside the bracket, or multiplied?y = f(−x) is a reflection in the y-axis
- Use the rule for this type. f(−x) flips the graph left to right (reflection in the y-axis). Only the x-coordinates change sign.
- Apply the rule to each point. Work out the new coordinates.A (1, 0) → (−1, 0)
B (2, −1) → (−2, −1)
C (3, 0) → (−3, 0)
D (0, 3) → (0, 3)
Check. Check one point: B′ = (−2, −1). The new equation y = x2 + 4x + 3 gives y = (−2)2 + 4 × (−2) + 3 = −1 when x = −2. ✓
Watch out. The effect of a change inside the bracket is the opposite of what you might expect: f(x + 2) moves the graph LEFT, and f(2x) makes it narrower.
Method. y = f(x) − 2 is a translation by the vector 0
−2 (2 units down). Apply the rule to each key point.
- Identify the transformation. Look at what has been done to f(x): is it added outside the bracket, changed inside the bracket, or multiplied?y = f(x) − 2 is a translation by the vector 0
−2 (2 units down) - Use the rule for this type. The graph moves up (or down) by the number added to f(x). Only the y-coordinates change: add the number to y.
- Apply the rule to each point. Work out the new coordinates.A (0, 0) → (0, −2)
B (2, −4) → (2, −6)
C (3, 0) → (3, −2)
Check. Check one point: B′ = (2, −6). The new equation y = x3 − 3x2 − 2 gives y = 23 − 3 × 22 − 2 = −6 when x = 2. ✓
Watch out. The effect of a change inside the bracket is the opposite of what you might expect: f(x + 2) moves the graph LEFT, and f(2x) makes it narrower.
Method. y = f(x + 1) is a translation by the vector −1
0 (1 unit to the left). Apply the rule to each key point.
- Identify the transformation. Look at what has been done to f(x): is it added outside the bracket, changed inside the bracket, or multiplied?y = f(x + 1) is a translation by the vector −1
0 (1 unit to the left) - Use the rule for this type. Inside the bracket the change is the opposite way round to what you expect: y = f(x + a) moves the graph a units to the LEFT. Only the x-coordinates change: subtract a.
- Apply the rule to each point. Work out the new coordinates.A (0, 0) → (−1, 0)
B (2, −4) → (1, −4)
C (3, 0) → (2, 0)
Check. Check one point: B′ = (1, −4). The new equation y = x3 − 3x − 2 gives y = 13 − 3 × 1 − 2 = −4 when x = 1. ✓
Watch out. The effect of a change inside the bracket is the opposite of what you might expect: f(x + 2) moves the graph LEFT, and f(2x) makes it narrower.
Method. y = 2f(x) is a stretch parallel to the y-axis, scale factor 2. Apply the rule to each key point.
- Identify the transformation. Look at what has been done to f(x): is it added outside the bracket, changed inside the bracket, or multiplied?y = 2f(x) is a stretch parallel to the y-axis, scale factor 2
- Use the rule for this type. Multiplying the whole function by a stretches the graph vertically (parallel to the y-axis). Multiply every y-coordinate by a.
- Apply the rule to each point. Work out the new coordinates.A (0, 0) → (0, 0)
B (2, −4) → (2, −8)
C (3, 0) → (3, 0)
Check. Check one point: B′ = (2, −8). The new equation y = 2x3 − 6x2 gives y = 2 × 23 − 6 × 22 = −8 when x = 2. ✓
Watch out. The effect of a change inside the bracket is the opposite of what you might expect: f(x + 2) moves the graph LEFT, and f(2x) makes it narrower.
Method. y = f(−x) is a reflection in the y-axis. Apply the rule to each key point.
- Identify the transformation. Look at what has been done to f(x): is it added outside the bracket, changed inside the bracket, or multiplied?y = f(−x) is a reflection in the y-axis
- Use the rule for this type. f(−x) flips the graph left to right (reflection in the y-axis). Only the x-coordinates change sign.
- Apply the rule to each point. Work out the new coordinates.A (0, 0) → (0, 0)
B (2, −4) → (−2, −4)
C (3, 0) → (−3, 0)
Check. Check one point: B′ = (−2, −4). The new equation y = −x3 − 3x2 gives y = −(−2)3 − 3 × (−2)2 = −4 when x = −2. ✓
Watch out. The effect of a change inside the bracket is the opposite of what you might expect: f(x + 2) moves the graph LEFT, and f(2x) makes it narrower.
Method. Write the new function using f, replace f by its rule, then simplify.
- Write the original function. f(x) = x2 − 4x
- Say what the new function is in terms of f. The new graph is y = f(x) + 3.y = f(x) + 3
- Put the rule in. Add (or subtract) the number to the whole function.(x2 − 4x) + 3
- Expand and simplify. Multiply out any brackets and collect like terms.y = x2 − 4x + 3
Check. Test with x = 3: the transformed curve should give 0 and the rule applied to the original function at the matching input agrees. ✓
Watch out. Inside the bracket the effect is the opposite of what you expect: f(x + 2) moves the graph 2 to the left.
Method. Write the new function using f, replace f by its rule, then simplify.
- Write the original function. f(x) = x2 − 4x
- Say what the new function is in terms of f. The new graph is y = f(x − 2).y = f(x − 2)
- Put the rule in. Replace every x in the rule by (x − 2).(x − 2)2 − 4(x − 2)
- Expand and simplify. Multiply out any brackets and collect like terms.y = x2 − 8x + 12
Check. Test with x = 3: the new rule gives −3; using the original rule with the changed input gives the same number. ✓
Watch out. Inside the bracket the effect is the opposite of what you expect: f(x + 2) moves the graph 2 to the left.
Method. Write the new function using f, replace f by its rule, then simplify.
- Write the original function. f(x) = x2 − 4x
- Say what the new function is in terms of f. The new graph is y = −f(x).y = −f(x)
- Put the rule in. Put a minus sign in front of the whole function: every sign in it changes.−(x2 − 4x)
- Expand and simplify. Multiply out any brackets and collect like terms.y = −x2 + 4x
Check. Test with x = 3: the transformed curve should give 3 and the rule applied to the original function at the matching input agrees. ✓
Watch out. Inside the bracket the effect is the opposite of what you expect: f(x + 2) moves the graph 2 to the left.
Method. Write the new function using f, replace f by its rule, then simplify.
- Write the original function. f(x) = x2 − 4x
- Say what the new function is in terms of f. The new graph is y = f(−x).y = f(−x)
- Put the rule in. Replace every x in the rule by (−x).(−x)2 − 4(−x)
- Expand and simplify. Multiply out any brackets and collect like terms.y = x2 + 4x
Check. Test with x = 3: the new rule gives 21; using the original rule with the changed input gives the same number. ✓
Watch out. Inside the bracket the effect is the opposite of what you expect: f(x + 2) moves the graph 2 to the left.
Method. Write the new function using f, replace f by its rule, then simplify.
- Write the original function. f(x) = x2 − 4x
- Say what the new function is in terms of f. The new graph is y = 2f(x).y = 2f(x)
- Put the rule in. Multiply the whole function by 2.2(x2 − 4x)
- Expand and simplify. Multiply out any brackets and collect like terms.y = 2x2 − 8x
Check. Test with x = 3: the transformed curve should give −6 and the rule applied to the original function at the matching input agrees. ✓
Watch out. Inside the bracket the effect is the opposite of what you expect: f(x + 2) moves the graph 2 to the left.
Method. A turning point is moved by a transformation in exactly the same way as any other point.
- Write the original minimum point. It is a point on the graph, so it is moved like any other point.B = (2, -1)
- Identify the transformation and use its rule. Adding a number to f(x) moves the whole graph up (or down). Every y-coordinate changes by that amount and every x-coordinate stays the same.y = f(x) + 4: (2, -1) → (2, 3)
- Write the new point. The shape has not been flipped, so it is still a minimum.minimum at (2, 3)
Check. The new point should lie on the new curve; for example for f(x) = x2 − 4x + 3 the new equation gives y = 3 at x = 2. ✓
Watch out. Reflecting in the x-axis turns a minimum into a maximum.
Method. A turning point is moved by a transformation in exactly the same way as any other point.
- Write the original minimum point. It is a point on the graph, so it is moved like any other point.B = (2, -1)
- Identify the transformation and use its rule. Adding a number inside the bracket moves the whole graph sideways. y = f(x + a) moves it a units to the left; y = f(x − a) moves it a units to the right. Every x-coordinate changes and every y-coordinate stays the same.y = f(x − 3): (2, -1) → (5, −1)
- Write the new point. The shape has not been flipped, so it is still a minimum.minimum at (5, −1)
Check. The new point should lie on the new curve; for example for f(x) = x2 − 4x + 3 the new equation gives y = −1 at x = 5. ✓
Watch out. Reflecting in the x-axis turns a minimum into a maximum.
Method. A turning point is moved by a transformation in exactly the same way as any other point.
- Write the original minimum point. It is a point on the graph, so it is moved like any other point.B = (2, -1)
- Identify the transformation and use its rule. Putting a minus sign in front of f(x) turns the graph upside down, a reflection in the x-axis. Each y-coordinate changes sign.y = −f(x): (2, -1) → (2, 1)
- Think about whether the minimum is still a minimum. Turning the graph upside down turns the lowest point into the highest point.The minimum becomes a maximum at (2, 1).
Check. The new point should lie on the new curve; for example for f(x) = x2 − 4x + 3 the new equation gives y = 1 at x = 2. ✓
Watch out. Reflecting in the x-axis turns a minimum into a maximum.
Method. A turning point is moved by a transformation in exactly the same way as any other point.
- Write the original minimum point. It is a point on the graph, so it is moved like any other point.B = (2, -1)
- Identify the transformation and use its rule. Replacing x by ax squeezes the graph towards the y-axis, parallel to the x-axis, by scale factor 1/a. Each x-coordinate is divided by a.y = f(2x): (2, -1) → (1, −1)
- Write the new point. The shape has not been flipped, so it is still a minimum.minimum at (1, −1)
Check. The new point should lie on the new curve; for example for f(x) = x2 − 4x + 3 the new equation gives y = −1 at x = 1. ✓
Watch out. Reflecting in the x-axis turns a minimum into a maximum.
Method. Roots are points on the x-axis, so transform the points (1, 0) and (3, 0).
- Roots are points on the x-axis. The graph of y = f(x) crosses the x-axis at the points (1, 0) and (3, 0).roots at x = 1 and x = 3
- Use the rule for this transformation on each point. Adding a number inside the bracket moves the whole graph sideways. y = f(x + a) moves it a units to the left; y = f(x − a) moves it a units to the right. Every x-coordinate changes and every y-coordinate stays the same.(1, 0) → (3, 0)
(3, 0) → (5, 0) - Write the roots. Only the x-values matter for a root.x = 3 and x = 5
Check. Plot the two new points on the x-axis and check they have moved as the rule says. ✓
Watch out. A translation up or down changes the roots; a reflection in the x-axis does not.
Method. Roots are points on the x-axis, so transform the points (1, 0) and (3, 0).
- Roots are points on the x-axis. The graph of y = f(x) crosses the x-axis at the points (1, 0) and (3, 0).roots at x = 1 and x = 3
- Use the rule for this transformation on each point. Replacing x by −x flips the graph left to right, a reflection in the y-axis. Each x-coordinate changes sign.(1, 0) → (−1, 0)
(3, 0) → (−3, 0) - Write the roots. Only the x-values matter for a root.x = −3 and x = −1
Check. Plot the two new points on the x-axis and check they have moved as the rule says. ✓
Watch out. A translation up or down changes the roots; a reflection in the x-axis does not.
Method. Roots are points on the x-axis, so transform the points (1, 0) and (3, 0). A reflection in the x-axis leaves points on the x-axis where they are, so the roots do not change.
- Roots are points on the x-axis. The graph of y = f(x) crosses the x-axis at the points (1, 0) and (3, 0).roots at x = 1 and x = 3
- Use the rule for this transformation on each point. Putting a minus sign in front of f(x) turns the graph upside down, a reflection in the x-axis. Each y-coordinate changes sign.(1, 0) → (1, 0)
(3, 0) → (3, 0) - Write the roots. Only the x-values matter for a root.x = 1 and x = 3
Check. Plot the two new points on the x-axis and check they have moved as the rule says. ✓
Watch out. A translation up or down changes the roots; a reflection in the x-axis does not.
Method. Roots are points on the x-axis, so transform the points (1, 0) and (3, 0).
- Roots are points on the x-axis. The graph of y = f(x) crosses the x-axis at the points (1, 0) and (3, 0).roots at x = 1 and x = 3
- Use the rule for this transformation on each point. Replacing x by ax squeezes the graph towards the y-axis, parallel to the x-axis, by scale factor 1/a. Each x-coordinate is divided by a.(1, 0) → (0.5, 0)
(3, 0) → (1.5, 0) - Write the roots. Only the x-values matter for a root.x = 0.5 and x = 1.5
Check. Plot the two new points on the x-axis and check they have moved as the rule says. ✓
Watch out. A translation up or down changes the roots; a reflection in the x-axis does not.
3 (3 units up); y = f(x) + 3
Method. Compare key points on the two graphs and decide which coordinate has changed.
- Compare the key points on the two graphs. Match each point on the solid curve to its image on the dashed or orange curve (A to A′, B to B′ ...).A (1, 0) → A′ (1, 3)
B (2, −1) → B′ (2, 2)
C (3, 0) → C′ (3, 3) - Decide which coordinate changed. If only y changes the change is outside the bracket; if only x changes it is inside the bracket.
- Write the transformation. y = f(x) + 3: a translation by the vector 0
3 (3 units up)
Check. Check one point of the dashed graph against the orange graph. ✓
Watch out. State the type of transformation and give its details (vector, axis of reflection).
0 (2 units to the right); y = f(x − 2)
Method. Compare key points on the two graphs and decide which coordinate has changed.
- Compare the key points on the two graphs. Match each point on the solid curve to its image on the dashed or orange curve (A to A′, B to B′ ...).A (0, 0) → A′ (2, 0)
B (2, −4) → B′ (4, −4)
C (3, 0) → C′ (5, 0) - Decide which coordinate changed. If only y changes the change is outside the bracket; if only x changes it is inside the bracket.
- Write the transformation. y = f(x − 2): a translation by the vector 2
0 (2 units to the right)
Check. Check one point of the dashed graph against the orange graph. ✓
Watch out. State the type of transformation and give its details (vector, axis of reflection).
Method. Compare key points on the two graphs and decide which coordinate has changed.
- Compare the key points on the two graphs. Match each point on the solid curve to its image on the dashed or orange curve (A to A′, B to B′ ...).A (1, 0) → A′ (1, 0)
B (2, −1) → B′ (2, 1)
C (3, 0) → C′ (3, 0) - Decide which coordinate changed. If only y changes the change is outside the bracket; if only x changes it is inside the bracket.
- Write the transformation. y = −f(x): a reflection in the x-axis
Check. Check one point of the dashed graph against the orange graph. ✓
Watch out. State the type of transformation and give its details (vector, axis of reflection).
Method. Compare key points on the two graphs and decide which coordinate has changed.
- Compare the key points on the two graphs. Match each point on the solid curve to its image on the dashed or orange curve (A to A′, B to B′ ...).A (0, 0) → A′ (0, 0)
B (2, −4) → B′ (−2, −4)
C (3, 0) → C′ (−3, 0) - Decide which coordinate changed. If only y changes the change is outside the bracket; if only x changes it is inside the bracket.
- Write the transformation. y = f(−x): a reflection in the y-axis
Check. Check one point of the dashed graph against the orange graph. ✓
Watch out. State the type of transformation and give its details (vector, axis of reflection).
Method. This adds 3 to every y-value, moving the graph up 3 and keeping its shape.
- Move the minimum. The dashed minimum is at (2, −1). Adding 3 to y givesminimum of the new graph at (2, 2)
- Find the graph with a minimum at (2, 2). Look at the diagram: the red curve has its lowest point at (2, 2).graph P
Check. At x = 0 the dashed graph is at 3, so the new graph is at 6, which matches P. ✓
Watch out. Check the lowest point to decide which graph it is.
Method. Subtracting 2 inside the bracket moves the graph 2 to the right.
- Move the minimum. The dashed minimum is at (2, −1). Moving 2 to the right givesminimum of the new graph at (4, −1)
- Find the graph with a minimum at (4, −1). The green curve has its lowest point at (4, −1).graph Q
Check. The new graph crosses the x-axis at 3 and 5, which matches Q. ✓
Watch out. f(x − 2) moves to the right, not the left.
Method. A minus sign outside turns the graph upside down about the x-axis.
- Move the minimum. The dashed minimum at (2, −1) becomes the highest point (2, 1).maximum of the new graph at (2, 1)
- Find the graph that opens downwards. The purple curve is upside down with its highest point at (2, 1).graph R
Check. The roots stay at 1 and 3, and R passes through those points. ✓
Watch out. Reflection in the x-axis keeps the roots in the same place.
Method. Use the rule for this transformation on the point (4, 7).
- Identify the transformation. Adding a number to f(x) moves the whole graph up (or down). Every y-coordinate changes by that amount and every x-coordinate stays the same.y = f(x) + 1 is a translation by the vector 0
1 (1 unit up) - Apply it to the point. Change only the coordinate that the rule says changes.(4, 7) → (4, 8)
Check. Reverse the rule from the new point and you get back to (4, 7). ✓
Watch out. Only one of the two coordinates changes in each of these transformations.
Method. Use the rule for this transformation on the point (4, 7).
- Identify the transformation. Adding a number inside the bracket moves the whole graph sideways. y = f(x + a) moves it a units to the left; y = f(x − a) moves it a units to the right. Every x-coordinate changes and every y-coordinate stays the same.y = f(x − 2) is a translation by the vector 2
0 (2 units to the right) - Apply it to the point. Change only the coordinate that the rule says changes.(4, 7) → (6, 7)
Check. Reverse the rule from the new point and you get back to (4, 7). ✓
Watch out. Only one of the two coordinates changes in each of these transformations.
Method. Use the rule for this transformation on the point (4, 7).
- Identify the transformation. Multiplying f(x) by a stretches the graph away from the x-axis, parallel to the y-axis. Each y-coordinate is multiplied by a.y = 2f(x) is a stretch parallel to the y-axis, scale factor 2
- Apply it to the point. Change only the coordinate that the rule says changes.(4, 7) → (4, 14)
Check. Reverse the rule from the new point and you get back to (4, 7). ✓
Watch out. Only one of the two coordinates changes in each of these transformations.
Method. Use the rule for this transformation on the point (4, 7).
- Identify the transformation. Replacing x by ax squeezes the graph towards the y-axis, parallel to the x-axis, by scale factor 1/a. Each x-coordinate is divided by a.y = f(½x) is a stretch parallel to the x-axis, scale factor 2
- Apply it to the point. Change only the coordinate that the rule says changes.(4, 7) → (8, 7)
Check. Reverse the rule from the new point and you get back to (4, 7). ✓
Watch out. Only one of the two coordinates changes in each of these transformations.
Method. Reciprocal graphs are translated like any others: the asymptotes move with the curve.
- Start with the asymptotes of y = 1/x. The reciprocal graph never touches the y-axis (x = 0) or the x-axis (y = 0).x = 0 and y = 0
- Write the new function in terms of f(x) = 1/x. The new graph is a translation of the old one.y = f(x) + 2
- Move each asymptote. The vertical asymptote moves with the sideways translation, and the horizontal asymptote moves with the up or down translation.x = 0 → x = 0
y = 0 → y = 2
Check. The function is not defined when x = 0 (the bottom of the fraction would be 0), and as x gets very large y gets close to 2. ✓
Watch out. x − 3 in the bracket moves the vertical asymptote to x = +3, not −3.
Method. Reciprocal graphs are translated like any others: the asymptotes move with the curve.
- Start with the asymptotes of y = 1/x. The reciprocal graph never touches the y-axis (x = 0) or the x-axis (y = 0).x = 0 and y = 0
- Write the new function in terms of f(x) = 1/x. The new graph is a translation of the old one.y = 1x − 3 = f(x − 3) + 0
- Move each asymptote. The vertical asymptote moves with the sideways translation, and the horizontal asymptote moves with the up or down translation.x = 0 → x = 3
y = 0 → y = 0
Check. The function is not defined when x = 3 (the bottom of the fraction would be 0), and as x gets very large y gets close to 0. ✓
Watch out. x − 3 in the bracket moves the vertical asymptote to x = +3, not −3.
Method. Reciprocal graphs are translated like any others: the asymptotes move with the curve.
- Start with the asymptotes of y = 1/x. The reciprocal graph never touches the y-axis (x = 0) or the x-axis (y = 0).x = 0 and y = 0
- Write the new function in terms of f(x) = 1/x. The new graph is a translation of the old one.y = 1x + 1 − 4 = f(x + 1) − 4
- Move each asymptote. The vertical asymptote moves with the sideways translation, and the horizontal asymptote moves with the up or down translation.x = 0 → x = -1
y = 0 → y = -4
Check. The function is not defined when x = -1 (the bottom of the fraction would be 0), and as x gets very large y gets close to -4. ✓
Watch out. x − 3 in the bracket moves the vertical asymptote to x = +3, not −3.
Method. Reciprocal graphs are translated like any others: the asymptotes move with the curve.
- Start with the asymptotes of y = 1/x. The reciprocal graph never touches the y-axis (x = 0) or the x-axis (y = 0).x = 0 and y = 0
- Write the new function in terms of f(x) = 1/x. The new graph is a translation of the old one.y = 1x − 2 + 1 = f(x − 2) + 1
- Move each asymptote. The vertical asymptote moves with the sideways translation, and the horizontal asymptote moves with the up or down translation.x = 0 → x = 2
y = 0 → y = 1
Check. The function is not defined when x = 2 (the bottom of the fraction would be 0), and as x gets very large y gets close to 1. ✓
Watch out. x − 3 in the bracket moves the vertical asymptote to x = +3, not −3.
Method. Start from y = 1/x, translate it 2 right and 1 up, and find where it meets the axes by substitution.
- Find the translation. 1/(x − 2) + 1 is f(x − 2) + 1 where f(x) = 1/x.translation by the vector (2, 1)
- Move the asymptotes. asymptotes: x = 2 and y = 1
- Find where it crosses the y-axis (x = 0). Substitute x = 0.y = 1/(0 − 2) + 1 = −0.5 + 1 = 0.5
- Find where it crosses the x-axis (y = 0). Solve 1/(x − 2) + 1 = 0.1/(x − 2) = −1, so x − 2 = −1 and x = 1
- Sketch. Draw the two branches of the curve, one in the top-right and one in the bottom-left of the asymptote cross.
Check. Both crossing points satisfy the equation: at (1, 0), 1/(1 − 2) + 1 = −1 + 1 = 0. ✓
Watch out. The graph has two separate branches and never crosses its asymptotes.
Method. Complete the square. The completed-square form is a transformation of y = x²: a translation left or right, then up or down.
- Halve the number in front of x. Completing the square: x2 + bx = (x + b/2)2 − (b/2)2.6 ÷ 2 = 3
- Write the bracket and subtract the square of the half. (x + 3)2 expands to x2 + 6x + 9, which has 9 too many, so subtract it.x2 + 6x = (x + 3)2 − 9
- Add the constant term. y = (x + 3)2 + 2
- Read the turning point. The graph of y = x2 has its minimum at (0, 0). Replacing x by x + 3 moves it 3 left, and adding 2 moves it 2 up.minimum at (−3, 2)
Check. Expand: (x + 3)2 + 2 = x2 + 6x + 11 = x2 + 6x + 11. ✓
Watch out. The x-coordinate of the turning point is the opposite sign to the number inside the bracket.
Method. Complete the square. The completed-square form is a transformation of y = x²: a translation left or right, then up or down.
- Halve the number in front of x. Completing the square: x2 + bx = (x + b/2)2 − (b/2)2.−4 ÷ 2 = −2
- Write the bracket and subtract the square of the half. (x + −2)2 expands to x2 + −4x + 4, which has 4 too many, so subtract it.x2 + −4x = (x + −2)2 − 4
- Add the constant term. y = (x + −2)2 + 5
- Read the turning point. The graph of y = x2 has its minimum at (0, 0). Replacing x by x + −2 moves it 2 right, and adding 5 moves it 5 up.minimum at (2, 5)
Check. Expand: (x + −2)2 + 5 = x2 + −4x + 9 = x2 + −4x + 9. ✓
Watch out. The x-coordinate of the turning point is the opposite sign to the number inside the bracket.
Method. Complete the square. The completed-square form is a transformation of y = x²: a translation left or right, then up or down.
- Halve the number in front of x. Completing the square: x2 + bx = (x + b/2)2 − (b/2)2.8 ÷ 2 = 4
- Write the bracket and subtract the square of the half. (x + 4)2 expands to x2 + 8x + 16, which has 16 too many, so subtract it.x2 + 8x = (x + 4)2 − 16
- Add the constant term. y = (x + 4)2 + 4
- Read the turning point. The graph of y = x2 has its minimum at (0, 0). Replacing x by x + 4 moves it 4 left, and adding 4 moves it 4 up.minimum at (−4, 4)
Check. Expand: (x + 4)2 + 4 = x2 + 8x + 20 = x2 + 8x + 20. ✓
Watch out. The x-coordinate of the turning point is the opposite sign to the number inside the bracket.
Method. Complete the square. The completed-square form is a transformation of y = x²: a translation left or right, then up or down.
- Halve the number in front of x. Completing the square: x2 + bx = (x + b/2)2 − (b/2)2.−2 ÷ 2 = −1
- Write the bracket and subtract the square of the half. (x + −1)2 expands to x2 + −2x + 1, which has 1 too many, so subtract it.x2 + −2x = (x + −1)2 − 1
- Add the constant term. y = (x + −1)2 − 4
- Read the turning point. The graph of y = x2 has its minimum at (0, 0). Replacing x by x + −1 moves it 1 right, and adding −4 moves it 4 down.minimum at (1, −4)
Check. Expand: (x + −1)2 − 4 = x2 + −2x + −3 = x2 + −2x + −3. ✓
Watch out. The x-coordinate of the turning point is the opposite sign to the number inside the bracket.
Method. Write both as f(x) transformations and read off the translation.
- Let f(x) = x2. f(x − 3) = (x − 3)2
- Spot the changes. The −3 inside the bracket is a translation 3 to the right (opposite to the sign). The +2 outside is a translation 2 up.3 right, 2 up
- Write it as a vector. A vector gives the move across (x) on top and up (y) below.translation by 3
2
Check. The turning point of y = x2 is (0, 0) and of the new graph is (3, 2): a move of 3 right and 2 up. ✓
Watch out. The sign inside the bracket is opposite to the direction of the move.
Method. Find a feature of the graph (such as the turning point) before and after the transformation and compare.
- Write both functions. f(x) = x2 and f(x + 3) = (x + 3)2
- Find the turning point (and root) of y = x2. It touches the x-axis when x2 = 0.x = 0
- Find where (x + 3)2 = 0. The bracket must be 0.x + 3 = 0, so x = −3
- Compare. The feature has moved from x = 0 to x = −3, which is 3 to the left.So f(x + 3) is a translation 3 units to the left. The reason is that the new function needs an input 3 smaller to produce the same output.
Check. At x = −3 the new function gives f(0) = 0, just as f gives 0 at x = 0. ✓
Watch out. “+3” in the bracket makes you reach the same value 3 earlier, so the graph moves left.
Method. Work out both equations and compare them (or compare one point).
- Find −f(x). Change every sign in f(x).−f(x) = −(x3 − 3x2) = −x3 + 3x2
- Find f(−x). Replace x by (−x): (−x)3 = −x3 and (−x)2 = x2.f(−x) = (−x)3 − 3(−x)2 = −x3 − 3x2
- Compare. The x2 terms have opposite signs, so the equations are different.−x3 + 3x2 ≠ −x3 − 3x2
- Confirm with a point. The point B(2, −4) goes to different places.B′ = (2, 4) for −f(x) but B′ = (−2, −4) for f(−x)
Check. Try x = 1: −f(1) = −(1 − 3) = 2 but f(−1) = −1 − 3 = −4. ✓
Watch out. −f(x) changes the sign of the whole output; f(−x) changes the sign of the input.
Method. Write the new function using f, replace f by its rule, then simplify.
- Write the original function. h(x) = −5x2 + 8x
- Say what the new function is in terms of f. The new graph is y = f(x − 1).y = f(x − 1)
- Put the rule in. Replace every x in the rule by (x − 1).−5(x − 1)2 + 8(x − 1)
- Expand and simplify. Multiply out any brackets and collect like terms.y = −5x2 + 18x − 13
Check. Test with x = 3: the new rule gives −4; using the original rule with the changed input gives the same number. ✓
Watch out. Inside the bracket the effect is the opposite of what you expect: f(x + 2) moves the graph 2 to the left.
Method. Use the rules: f(x + a) moves the point a to the left and + b moves it b up.
- Move the minimum. The new x-coordinate is the old one minus a (because the graph moves a to the left), and the new y is the old plus b.(3 − a, −2 + b) = (1, 5)
- Solve for a. Compare the x-coordinates.3 − a = 1, so a = 2
- Solve for b. Compare the y-coordinates.−2 + b = 5, so b = 7
Check. Start at (3, −2), move 2 to the left to (1, −2) and 7 up to (1, 5). ✓
Watch out. The sign of a works the opposite way round: f(x + 2) moves LEFT.
Method. Do the two transformations in order: first replace x by −x, then add 2.
- Name the starting function. f(x) = x2 − 4x + 3
- Reflect in the y-axis: replace x by (−x). Remember (−x)2 = x2.f(−x) = (−x)2 − 4(−x) + 3 = x2 + 4x + 3
- Translate 2 units up: add 2. y = x2 + 4x + 3 + 2 = x2 + 4x + 5
Check. The old minimum (2, −1) goes to (−2, −1) then to (−2, 1). The new equation gives (−2)2 + 4 × (−2) + 5 = 1 at x = −2. ✓
Watch out. Do the transformations in the order given.