01 · Translations
P(−4, 3) lies on y = f(x). Find its image on f(x + 5) − 2.
Hint
Solve x + 5 = −4, then subtract 2 from the old output.
Worked solution
x = −9, y = 3 − 2 = 1
P′ = (−9, 1)
Understand · explore · practise
Graph transformations
Learn translations, stretches and reflections of y = f(x). Map points, roots and asymptotes, combine transformations and find unknown parameters.
Before you startFunction notation, coordinates and basic graph shapes
01 / Vertical shifts
Changing the output moves a point up or down.
If (p, q) lies on y = f(x), then q = f(p). Adding k to the output gives f(p) + k = q + k.
y = f(x) + k
(p, q) → (p, q + k)
Positive k moves every point up; negative k moves it down. The translation vector has horizontal component 0 and vertical component k.
For f(x) = x² − 4, the minimum (0, −4) moves to (0, −1) under f(x) + 3. The new roots solve x² − 1 = 0, so they are −1 and 1. Old roots generally stop being roots after a vertical shift.
Rearrange the y sideWorked example
y − 5 = f(x)
Add 5 to isolate y.
y = f(x) + 5
Translate up 5, not down 5.
If P = (−3, 2), then P′ = (−3, 7)
The input coordinate does not change.
02 / Horizontal shifts
Ask which new input gives the old one.
For y = f(x − h), the old input p occurs when x − h = p. Therefore the new input is p + h.
y = f(x − h)
(p, q) → (p + h, q)
For example, f(x − 3) moves the graph right 3; f(x + 3) moves it left 3. The sign appears reversed because you solve for the new input.
Substitute into every occurrence of x. If f(x) = x(x − 4), then f(x + 2) = (x + 2)(x − 2) = x² − 4. Its roots move from 0 and 4 to −2 and 2.
Its new y-intercept is f(2) = −4. The image of an old y-intercept need not remain on the y-axis.
Watch the input condition move a point
Pause, replay or seek freely. The notes explain the same idea and stay in view.
03 / Stretches
Inside multipliers divide horizontal coordinates.
For positive scale constants a and b:
y = af(x): (p, q) → (p, aq)
y = f(bx): (p, q) → (p/b, q)
2f(x) doubles vertical distances from the x-axis. f(2x) halves horizontal distances from the y-axis. A scale factor between 0 and 1 is a compression, still described as a stretch by that factor.
For f(x/3), the inside multiplier is 1/3, so the horizontal scale factor is 3. Vertical scaling preserves roots when a ≠ 0; horizontal scaling moves each root using the same input rule.
Compare two stretchesWorked example
f(x) = 4 − x²
Roots ±2; maximum (0, 4).
f(2x) = 4 − 4x²
Roots ±1; maximum (0, 4).
2f(x) = 8 − 2x²
Roots ±2; maximum (0, 8).
3y = f(x) ⇒ y = f(x)/3
Vertical factor 1/3. A point (6, −9) maps to (6, −3).
04 / Reflections
Track which coordinate changes sign.
Negating the output reflects the graph in the x-axis. Negating the input reflects it in the y-axis.
y = −f(x): (p, q) → (p, −q)
y = f(−x): (p, q) → (−p, q)
For f(x) = x² + x, f(−x) = x² − x, while −f(x) = −x² − x. These are different functions.
Some graphs already have symmetry: x² looks unchanged under f(−x). For x³, both reflections produce −x³. A symmetric example can hide the distinction, so use the coordinate rule.
A negative multiplier combines a reflection with a positive stretch. For example, −2f(x) maps (p, q) to (p, −2q).
05 / Explore points
You can transform a graph without knowing its formula.
The blue graph joins four known points with straight segments. Choose a transformation and compare each image in the table.
The input range of the original graph is −2 ≤ x ≤ 4; its output range is −1 ≤ y ≤ 3. Both endpoints are included. Translate, stretch or reflect these sets with the points, reversing the order of endpoints after a negative scale.
For an unfamiliar smooth curve, use the same method: map labelled points, turning points and asymptotes, then preserve the transformed overall shape. A negative vertical scale swaps maxima and minima.
A transformed point is not automatically an axis intercept. You can give an intercept only if its new coordinate is zero or the equation supplies enough information to find it.
y = f(x − 2)Blue → gold
| Point | Original | Image |
|---|---|---|
| A | (-2, 1) | (0, 1) |
| B | (0, 3) | (2, 3) |
| C | (2, -1) | (4, -1) |
| D | (4, 1) | (6, 1) |
Every x-coordinate increases by 2; y stays unchanged. Domain: 0 ≤ x ≤ 6. Range: −1 ≤ y ≤ 3.
06 / Combine changes
Solve the inside expression before changing the output.
A useful general form, with a ≠ 0 and b ≠ 0, is:
y = a f(b(x − h)) + k
(p, q) → (p/b + h, aq + k)
Horizontal: stretch by 1/|b|, reflect in the y-axis if b < 0, then translate by h. Vertical: stretch by |a|, reflect in the x-axis if a < 0, then translate by k.
For f(2x − 4), solve 2x − 4 = p. The image input is p/2 + 2. Equivalently, write f(2(x − 2)): compress horizontally by 1/2, then move right 2.
Order matters on the same coordinate. Moving right 2 before compressing by 1/2 gives p/2 + 1. Likewise, 2f(x) + 3 is not the same as 2(f(x) + 3).
Why exclude zero scale factors?
With a = 0, all defined outputs collapse to k, so roots, range and asymptotes need separate treatment. With b = 0, the input to f is always zero: the expression is constant if f(0) exists, and undefined otherwise. The point-mapping rule p/b cannot be used.
One point, several operationsWorked example
P = (6, −2)
g(x) = −3f(2x − 4) + 5
Find the new input before the output.
2x − 4 = 6 ⇒ x = 5
The horizontal image is 5.
y = −3(−2) + 5 = 11
The output changes independently.
P′ = (5, 11)
Check g(5) = −3f(6) + 5 = 11.
07 / Asymptotes
Transform the lines as well as the points.
Under y = a f(b(x − h)) + k, with a and b non-zero, a vertical asymptote x = r moves to x = r/b + h. A horizontal asymptote y = s moves to y = as + k.
f(x) = 3/x
g(x) = −2f(2(x − 1)) + 3
g(x) = −3/(x − 1) + 3
The new asymptotes are x = 1 and y = 3. Its intercepts are (2, 0) and (0, 6), found by solving the new equation. The domain excludes 1; the range excludes 3.
If you know only a sketch and some labelled points, there may not be enough information to find a new intercept exactly. State the coordinates you can justify; do not invent an unlabelled value.
y = −3/(x − 1) + 3Move asymptotes
y = −3/(x − 1) + 3 has vertical asymptote x = 1 and horizontal asymptote y = 3. Intercepts: (2,0) and (0,6).
08 / Find parameters
Use the condition at the transformed point.
Suppose f(x) = (x + 2)(x − 1)(x − 4). The graph y = f(x + c) passes through (3, 0) when 3 + c is one of the original roots.
3 + c = −2, 1 or 4
c = −5, −2 or 1
For y = f(ax) to pass through (2, 0), solve 2a = −2, 1 or 4. Thus a = −1, 1/2 or 2. A zero coordinate may give no constraint or an impossible condition, so substitute rather than divide blindly.
One point does not identify a unique transformation
Mapping (4, 2) to (2, 2) could be a translation left 2, y = f(x + 2), or a horizontal compression by 1/2, y = f(2x). A second point or a specified transformation type is needed to distinguish them.
Recover a squared-reciprocal shiftWorked example
y = 9/(x + c)² − 1
The graph passes through (0, 0)
Substitute x = y = 0, keeping c ≠ 0.
0 = 9/c² − 1 ⇒ c² = 9
There are two possible shifts.
c = −3 or c = 3
Both satisfy the domain restriction at the origin.
09 / Your turn
Write the coordinate rule before calculating.
Each answer includes the input equation or output operation that justifies the mapping.
02 · A mixed scale
Q(6, −4) lies on y = f(x). Find its image on y = −f(x/3)/2.
Hint
The inside factor is 1/3.
Worked solution
x/3 = 6 ⇒ x = 18
y = −(−4)/2 = 2
Q′ = (18, 2)
03 · Rearrange first
R(2, 7) lies on y = f(x). Find its image on 2y + 3 = f(x).
Hint
Isolate y.
Worked solution
y = (f(x) − 3)/2
y = (7 − 3)/2 = 2
R′ = (2, 2)
04 · Roots and a maximum
f(x) = 9 − x². State the roots and turning point of y = f(3x), and of y = −2f(x).
Hint
Use the original roots ±3 and maximum (0, 9).
Worked solution
For f(3x) = 9 − 9x²: roots ±1, maximum (0, 9). For −2f(x) = 2x² − 18: roots ±3 and a minimum at (0, −18); it has no maximum over the reals.
05 · Inside order
S(−2, 5) lies on y = f(x). Find its image on y = 3f(2x + 6) − 4.
Hint
Solve 2x + 6 = −2.
Worked solution
x = −4, y = 3(5) − 4 = 11
S′ = (−4, 11)
06 · Move asymptotes
A graph has vertical asymptote x = −2, horizontal asymptote y = 4 and point (0, 1). Give their images under y = −2f(3(x − 1)) + 5.
Hint
Use x′ = x/3 + 1 and y′ = −2y + 5.
Worked solution
Vertical: x = 1/3
Horizontal: y = −3
Point: (1, 3)
The point (1, 3) is not a y-intercept. The given information does not identify the new y-intercept.
07 · More than one parameter
f(x) = (x + 3)(x − 2)². Find all c such that y = f(x − c) passes through (1, 0).
Hint
1 − c must equal an original root.
Worked solution
1 − c = −3 or 2
c = 4 or −1
The repeated root 2 gives only one parameter value.
08 · Domain and range
f has domain [−2, 4] and range [−1, 3]. Find the domain and range of y = −2f(2(x − 1)) + 1.
Hint
Transform all allowed inputs and outputs, reversing endpoint order for the negative scale.
Worked solution
−2 ≤ 2(x − 1) ≤ 4
0 ≤ x ≤ 3
Outputs run from −2(3) + 1 = −5 to −2(−1) + 1 = 3. Domain [0, 3]; range [−5, 3].
09 · Identify two possibilities
T(8, −6) maps to (4, −3). Give one transformation using only a translation and one using only stretches.
Hint
A translation adds coordinate differences; stretches multiply distances from the axes.
Worked solution
Translation: y = f(x + 4) + 3
Stretches: y = ½f(2x)
Both map the stated point correctly but generally produce different graphs.
10 · Extension: shared roots
f(x) = (x² − 1)(x² − 9). Find all k for which f(x) and f(x − k) have exactly three roots in common.
Hint
List the roots and shift that set by k.
Worked solution
The original roots are {−3, −1, 1, 3}; the shifted roots are {k − 3, k − 1, k + 1, k + 3}. A common root forces k to be a difference of two original roots: 0, ±2, ±4 or ±6. At k = 0 all four coincide; ±4 gives two common roots; ±6 gives one. Exactly three occur for k = 2 or k = −2.
10 / Recap
Map inputs and outputs separately.
- Outside additions move vertically; inside additions require solving for the new input.
- Outside multipliers scale y; inside multipliers scale x by their reciprocal.
- −f(x) reflects in the x-axis; f(−x) reflects in the y-axis.
- Rearrange into y = … before reading a transformation.
- Move asymptotes and domain restrictions too.
- Use labelled information; do not assume a transformed point is still an intercept.
Section 1 of 10 · Vertical shifts