One rule covers every transformed graph: y = a·f(b(x − h)) + k. Read what each letter does, move a couple of points, and the equation writes itself.
Read the graph
The dashed graph is y = |x|. Which equation matches the solid graph?
The V opens downward, so there is a minus in front. The corner is at (1, 3): right 1 is (x − 1) inside, up 3 is + 3 outside. From the corner, one step right drops the graph 2, to (2, 1), so it is stretched by 2: y = −2|x − 1| + 3. B has the corner at x = −1. C drops only ½ per step — a wider V. D opens upward.
y = a·f(b(x − h)) + k
| Letter | Where | What it does |
|---|---|---|
| k | outside, added | up k (as written) |
| h | inside, subtracted | right h (sign reversed) |
| a | outside, multiplied | vertical stretch by |a|; flip if a < 0 |
| b | inside, multiplied | every x divided by b; flip if b < 0 |
Outside letters act on y and do what they say. Inside letters act on x and do the opposite.
y = 2x²: (1, 1) → (1, 2)
every height doubled — looks narrower
y = ½x²: (2, 4) → (2, 2)
every height halved — looks wider
"Narrower" and "wider" are how it looks; what actually happened is a vertical stretch or shrink. The x-coordinates never moved.
√x gives 2 at x = 4
the point (4, 2)
√4x gives 2 when 4x = 4
so x = 1
(4, 2) → (1, 2)
x divided by 4, not multiplied
Just as f(x − 3) moves right, f(4x) shrinks toward the y-axis: the input reaches the old value 4 times sooner. Multiplying the x-coordinate by 4 to get (16, 2) is the usual wrong answer.
A fraction inside stretches instead. On y = √x3, the height 2 needs x3 = 4, so x = 12: (4, 2) → (12, 2). Dividing by b = ⅓ is multiplying by 3 — and every x-intercept moves the same way, so a zero at x = −1 becomes a zero at x = −3.
| Equation | Mirror | (x, y) becomes |
|---|---|---|
| y = −f(x) | x-axis | (x, −y) |
| y = f(−x) | y-axis | (−x, y) |
| y = −f(−x) | both: the origin | (−x, −y) |
For an even function such as x² or |x|, f(−x) = f(x), so the y-axis flip changes nothing.
A minus inside, with a shift. The point (4, 1) is on y = f(x). Where does it go on y = f(−x + 3)?
−x + 3 = 4
the inside must equal the old input
−x = 1 → x = −1
− 3, then change both signs
(4, 1) → (−1, 1)
the inside never touches y
check: f(−(−1) + 3) = f(4) = 1 ✓
the new point reaches back to (4, 1)
What is different: factored, the inside is −(x − 3): flip over the y-axis (4 → −4), then right 3 (−4 → −1).
Reading f(−x + 3) as "flip, then left 3" sends 4 to −4 − 3 = −7. Factor out the minus first, or solve "inside = old input"; both give −1.
A negative fraction in front. The point (6, −4) is on y = f(x). Where does it go on y = −½f(x) + 2?
x: stays 6
nothing inside
y: −½(−4) + 2 = 2 + 2 = 4
flip and halve, then up 2
(6, −4) → (6, 4)
the image
What is different: a = −½ does two jobs at once — a flip and a shrink — and both happen before the + 2.
A square root that runs left. Find the domain and range of y = √2 − x + 1.
2 − x ≥ 0 → x ≤ 2
the inside of a root cannot be negative
√−(x − 2): flip over the y-axis, right 2
the curve starts at (2, 1) and runs left
domain x ≤ 2, range y ≥ 1
the + 1 lifts the lowest height
What is different: a minus inside flips the domain (x ≤ 2, not x ≥ 2) but leaves the range alone, because it acts on x only.
y = 3f(x + 1) − 5
start from (4, −2)
x: 4 − 1 = 3
+1 inside moves LEFT 1
y: 3(−2) − 5 = −11
multiply first, then subtract
= (3, −11)
the image
Subtracting 5 before multiplying gives 3(−2 − 5) = −21 and the point (3, −21). The equation says 3 times f, then minus 5 — so the stretch comes first, the shift last.
When the inside mixes a squeeze and a shift, factor first: f(3x + 6) = f(3(x + 2)). The point (6, 1) on y = f(x) then moves like this:
3(x + 2) = 6
the inside must equal the old input
x + 2 = 2, x = 0
squeeze by 3 (6 → 2), then left 2 (2 → 0)
(6, 1) → (0, 1)
the inside never touches y
Squeezing by 3 and then moving left 6 (the number you can see) sends 6 to 2 − 6 = −4 — wrong. After factoring, the shift is 2. Solving "inside = old input" avoids the question of order altogether.
Transformation items give a point to move, an equation to match to a described sequence, a graph to match, or a range to find. Treat x and y separately — outside acts on y as written (stretch, then shift); inside acts on x backwards (factor first, then undo) — and check by putting your new point into the new equation.
y = √x: x ≥ 0, y ≥ 0
the starting curve
y = −2√x − 1 + 4
right 1, flip, stretch 2, up 4
domain x ≥ 1
the start moved right 1
range y ≤ 4
the flip turned ≥ into ≤, the shift moved 0 to 4
Move the intercepts
The graph of y = f(x) crosses the x-axis at x = −4 and x = 6. Where does y = f(x2) cross it?
f is zero when its input is −4 or 6, so x2 = −4 or x2 = 6, giving x = −8 and x = 12. Dividing inside stretches the graph sideways by 2. A halves the intercepts — that would be f(2x). C forgets the change, and D is the flip f(−x).
