Adding outside moves the graph up or down.
y = f(x) + k
up k units when k > 0
y = f(x) − k
down k units
Every output gets k added, so every point climbs by the same k. The shape does not change at all — only its height.
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- Grade 7-12 Math Review
- Transformations of Functions
Transformations of Functions
Change the equation a little and the whole graph moves, flips or stretches — and every change follows a rule you can read off the equation.
Write the new function
The graph of f(x) = x² is shifted 3 units right and 2 units up. Which function gives the new graph?
Up 2 is added outside: + 2. Right 3 goes inside with the opposite sign: (x − 3). So g(x) = (x − 3)² + 2, with its vertex at (3, 2). A moves the graph LEFT 3 — the plus sign inside is the classic slip. C swaps the two moves, putting the vertex at (2, 3). D is not a shift at all: it changes the shape of the formula without moving the vertex to (3, 2).
Changing x inside moves it sideways — against the sign.
y = (x − 3)²
lowest point where x − 3 = 0
x = 3
so the vertex moved RIGHT to (3, 0)
Minus inside means right
It is tempting to read y = (x − 3)² as "3 to the left" because of the minus sign. Ask instead which x makes the inside zero: x = 3, on the right. In the same way, the blue curve has x + 2 inside, which is zero at x = −2, so its vertex sits on the left.
A minus sign flips it — where you put it decides which way.
| Equation | Flip | (4, 2) becomes |
|---|---|---|
| y = −f(x) | over the x-axis | (4, −2) |
| y = f(−x) | over the y-axis | (−4, 2) |
A minus outside changes every y. A minus inside changes every x.
Multiplying stretches or shrinks it.
| Equation | What happens | Why |
|---|---|---|
| y = a·f(x), a > 1 | vertical stretch — narrower | every y is multiplied by a |
| y = a·f(x), 0 < a < 1 | vertical shrink — wider | every y gets smaller |
| y = f(bx), b > 1 | horizontal shrink | every x is DIVIDED by b |
The point (1, 1) on y = x² becomes (1, 2) on y = 2x². The x stayed; only the height doubled.
Put them together by moving a few points.
y = −2(x + 1)² + 3
(x, y) → (x − 1, −2y + 3)
left 1; stretch by 2, flip, then up 3
(0, 0) → (−1, 3)
the vertex
(1, 1) → (0, 1)
−2(1) + 3 = 1
(−1, 1) → (−2, 1)
check: −2(−2 + 1)² + 3 = 1 ✓
Stretch before you shift up
The y-values are multiplied by −2 first and then raised by 3. Adding 3 first gives −2(1 + 3) = −8 for the point (1, 1), which puts it far below the real graph. The equation says −2 times the square, plus 3 — follow the order of operations.
Variations: a point under a squeeze, two flips at once, a factor hiding inside.
A horizontal squeeze on a point: (6, −4) is on y = f(x). Where is it on y = f(3x)?
3x = 6
the inside must still feed 6 to f
x = 2
divide the x by 3
(2, −4)
y is untouched
f(3 · 2) = f(6) = −4 ✓
check
A number multiplying x inside divides the x-coordinate. It does not multiply it.
Both flips at once: (5, 2) is on y = f(x). Where is it on y = −f(−x)?
f(−x): (5, 2) → (−5, 2)
the inside minus flips x
−f(−x): (−5, 2) → (−5, −2)
the outside minus flips y
Two flips together turn the graph half a turn about the origin: both signs change.
A factor hiding inside: where is the vertex of y = (2x − 6)²?
2x − 6 = 0
the inside is zero at the vertex
x = 3
not 6
(2x − 6)² = (2(x − 3))²
factor out the 2 to see the shift
vertex (3, 0)
right 3
Read the shift only after the x stands alone in the bracket. Reading "right 6" from 2x − 6 is the trap.
The domain and range move with the graph.
y = √x
domain x ≥ 0, range y ≥ 0
y = √x + 4 − 1
left 4, down 1
domain x ≥ −4
the starting point moved left 4
range y ≥ −1
and down 1
With a flip: y = −√x + 3 + 4
domain x ≥ −3
the inside must be 0 or more
√x + 3 ≥ 0
a root is never negative
−√x + 3 ≤ 0
the minus turns it 0 or less
range y ≤ 4
adding 4 makes it 4 or less
The flip turns the range over
Copying the shift and writing y ≥ 4 forgets the minus. The curve starts at (−3, 4) and runs DOWN from there: at x = 1, y = −√4 + 4 = 2, which is below 4.
On the exam
Periodical exams usually give an equation and ask for the move (up, left, flip, stretch), the new vertex, or the domain and range; entrance tests turn it round and give the moves, asking for the equation or where one point lands. The fastest route is to treat x and y separately: inside the bracket acts on x against the sign, outside acts on y as written. For a 10-second check, put the new point back into the new equation.
Now you try
Practice problem
Move the point
The point (2, 6) is on the graph of y = f(x). Which point must be on the graph of
y = ½·f(x − 4) + 1?
Inside, x − 4 moves the point right 4: x = 2 + 4 = 6. Outside, halve the y and then add 1: ½(6) + 1 = 4. The point is (6, 4). A moves left instead of right. B doubles instead of halving. D adds 1 before halving, ½(6 + 1) = 3.5 — the order trap from step 5.
Hint: handle x and y separately. The x only sees what is inside the brackets; the y sees everything outside, multiplication first.