a·f(b(x − c)) + d, including the two counterintuitive ones (c and b)You know y = x² cold. So you also know — right now, without plotting a single point — the graphs of:
y = x² + 3 y = (x − 5)² y = −2x² y = (3x)²
(up 3; right 5; flipped and stretched tall; squeezed thin.) One parent graph plus four dials. The entire transformation game is learning what each dial does — especially the two that move backwards from what they look like.
g(x) = a · f( b(x − c) ) + d
| Dial | Effect | Direction intuition |
|---|---|---|
d |
vertical translation by d | honest: +3 means up 3 |
c |
horizontal translation by c | backwards-looking: (x − 5) means right 5 |
a |
vertical dilation by factor |a|; if a < 0, reflection over the x-axis | honest: outputs multiply by a |
b |
horizontal dilation by factor 1/|b|; if b < 0, reflection over the y-axis | backwards: b = 3 squeezes to one-third width |
Why the horizontal dials act backwards: c and b operate on the input before f sees it. For g(x) = f(x − 5) to produce what f produced at 2, x must be 7 — everything happens 5 units later, i.e., the graph slides right. For g(x) = f(3x), x = 2 already delivers what f had at 6 — features arrive at one-third the x-value: compression.
Outputs transform honestly; inputs transform inversely. That one sentence generates the whole table.
A point (p, q) on f lands on g = a·f(b(x − c)) + d at:
new x: solve b(x − c) = p → x = p/b + c new y: a·q + d
Horizontal changes touch only x; vertical changes touch only y. When a multi-transformation question gets confusing, track two or three landmark points and let the graph follow.
Within the template, a safe order: dilations/reflections first, then translations (apply b and a before c and d). The template's own structure b(x − c) bakes this in — if the expression arrives as f(3x − 6), factor it first: f(3(x − 2)) — compress by 1/3, then shift right 2. Unfactored reading ("left 6 then compress"?) produces wrong answers; factoring ends the debate.
x → x/b + c.y → ay + d. If a < 0, the range flips (the max becomes the min) — re-order the endpoints.Example: f has domain [2, 6] and range [0, 4]. Then g(x) = −2f(x − 1) + 3 has domain [3, 7] (shift right 1) and range: y → −2y + 3 sends 0 → 3 and 4 → −5 → range [−5, 3].
[GRAPH: Parent f(x) = |x|-like V shape with vertex at origin, alongside g(x) = −2f(x − 1) + 3: vertex translated to (1, 3), arms opening downward twice as steep. Landmark points labeled: (0,0)→(1,3); (2,2)→(3,−1). Dashed arrows showing right-1, stretch-and-flip, up-3.]
Problem: Describe how the graph of g(x) = −3f(x + 2) − 4 is obtained from the graph of f.
Solution: Working from the template: c = −2 (shift left 2), a = −3 (vertical stretch by 3 and reflection over the x-axis), d = −4 (shift down 4). Order: shift left 2, stretch/reflect vertically, shift down 4. (Any order that applies the vertical translation after the vertical stretch/reflection is correct.)
Interpretation: "+2 inside" = left, the input dial moving backwards, as always.
Problem: The point (6, 1) is on the graph of f. Find the corresponding point on g(x) = f(3x − 6).
Strategy: Factor the inside; then solve for the new x.
Solution: g(x) = f(3(x − 2)). The input to f equals 6 when 3(x − 2) = 6 → x − 2 = 2 → x = 4. Output unchanged (no a or d). New point: (4, 1).
Interpretation: Solving b(x − c) = p never lies, no matter how tangled the inside expression looks.
Problem: f has domain [−1, 3] and range [2, 8]. Find the domain and range of g(x) = −f(2x) + 5.
Solution: - Domain: x/2 maps... solve 2x ∈ [−1, 3] → x ∈ [−1/2, 3/2]. Domain [−0.5, 1.5]. - Range: y → −y + 5: 2 → 3 and 8 → −3. Flip re-orders: range [−3, 3].
Interpretation: The −1 multiplier turned the old max (8) into the new min (−3). Always re-sort after a negative a.
Problem: Show that y = 4x² can be described as a transformation of y = x² in two different ways, and verify they agree at a point.
Solution: - Vertical: a = 4 → vertical stretch by factor 4. - Horizontal: 4x² = (2x)² → b = 2 → horizontal compression by factor 1/2.
Check with the parent point (1, 1): vertical stretch sends it to (1, 4) — and indeed 4(1)² = 4 ✓. Horizontal compression sends it to (1/2, 1) — and 4(1/2)² = 1 ✓. Both transformed graphs contain the correct transformed points; the two descriptions generate the same parabola.
Interpretation: Dilations of power functions can masquerade as each other. The exam exploits this with "which transformation…?" items where two choices look different but only one matches the algebra given.
g(x) = f(x − 3) + 2 is the graph of f translated1. (B). −3 inside → right 3; +2 outside → up 2. (A) misreads the inside sign; (D) swaps the two dials.
g(x) = f(2x) is2. (C). b = 2 compresses horizontally to 1/2 width — inputs reach their old values twice as fast. (A) reads b backwards; (B) confuses inside with outside.
g(x) = −f(x) is the graph of f reflected over3. (A). Minus outside flips outputs: x-axis mirror. (B) needs f(−x); (C) needs both.
y = (x + 4)² − 1 is4. (D). Inside zero at x = −4; then down 1: vertex (−4, −1). (A) misreads the inside sign; (C) misses the −1.
g(x) = f(x − 1) is5. (B). Shift right 1: [2 + 1, 6 + 1] = [3, 7]. (A) shifts left; (C) forgets horizontal changes touch the domain.
g(x) = 3f(x) − 2 is6. (A). y → 3y − 2 sends 0 → −2 and 4 → 10: [−2, 10]. (C) applies the −2 before the 3; (D) adds instead.
g(x) = f(−x)?7. (C). g(−3) = f(−(−3)) = f(3) = 5 → point (−3, 5). (D) requires f(−3), which is unknown; (A)/(B) flip the output — but only the input is negated.
y = f(3(x − 2)), the corresponding point is8. (A). Solve 3(x − 2) = 6 → x = 4; y unchanged → (4, 1). (C) multiplies instead of solving (18 = 3·6); (B) divides then forgets the shift; (D) adds 2 to 6 directly.
9. (B). f(3(−x)) = f(−3x) = f(3x) since f is even ✓. (A) breaks symmetry by moving the mirror line; (C) adds an odd piece; (D) multiplies by an odd function, producing an odd product.
y = 2f(−x) + 1 from the graph of f:10. (D). b = −1: y-axis reflection; a = 2: vertical stretch; d = +1: up 1, applied last. (A) reflects over the wrong axis; (B) stretches after translating (wrong: 2(f(−x) + 1) = 2f(−x) + 2 ≠ 2f(−x) + 1); (C) reflecting last is harmless here, but it also stretches after shifting — same flaw as (B).
y = 4x² can be obtained from y = x² by a horizontal11. (A). 4x² = (2x)²: b = 2 → horizontal compression by 1/2. (C) confuses the vertical factor 4 with the horizontal one; (B)/(D) read the dilation backwards.
12. (FRQ-style, 6 points) (i) [2 pts] Domain: solve 2x ∈ [0, 8] → x ∈ [0, 4]. Range: y → −(1/2)y + 3 sends −2 → 4 and 6 → 0; re-order → [0, 4]. (ii) [2 pts] New x: 2x = 3 → x = 1.5; new y: −(1/2)(6) + 3 = 0 → point (1.5, 0). The negative a flips the graph vertically, so f's local maximum becomes a local minimum of g. (iii) [2 pts] f increasing on (0, 3) → f(2x) increasing on (0, 1.5) (compression relocates the interval) → multiplying by −1/2 reverses monotonicity → g is decreasing on (0, 1.5). Reason: a negative vertical factor turns increasing into decreasing; the horizontal compression by 1/2 maps the interval (0, 3) to (0, 1.5).
12. (FRQ-style) 🚫 The function f has domain [0, 8] and range [−2, 6], with a local maximum at (3, 6). Let g(x) = −(1/2)f(2x) + 3.
(i) Find the domain and range of g.
(ii) Find the coordinates of the point on g corresponding to f's local maximum (3, 6), and state whether it is a local maximum or a local minimum of g.
(iii) If f is increasing on (0, 3), on what interval is g decreasing, and why?
The quadratic function p(x) = 2x² − 12x + 11.
(a) (i) Write p in vertex form by completing the square. (ii) State the vertex and the range of p.
(b) Describe the graph of p as a sequence of transformations applied to y = x² (name each transformation and its order).
(c) The function w(x) = p(x + 3) − 11. (i) Write w(x) in simplest vertex form. (ii) Find the zeros of w in exact form.
(a) [2 pts] (i) [1 pt] p(x) = 2(x² − 6x) + 11 = 2(x² − 6x + 9) − 18 + 11 = 2(x − 3)² − 7. (ii) [1 pt] Vertex (3, −7); a = 2 > 0 opens up → range [−7, ∞).
(b) [2 pts] Starting from y = x²: vertical stretch by factor 2, then translate right 3 and down 7. [1 pt for the correct dilation, 1 pt for both translations in a valid order — translations after the stretch]
(c) [2 pts] (i) [1 pt] w(x) = 2((x + 3) − 3)² − 7 − 11 = 2x² − 18. Vertex form: 2(x − 0)² − 18, vertex (0, −18). (ii) [1 pt] 2x² − 18 = 0 → x² = 9 → x = ±3.
1. (B). −3 inside → right 3; +2 outside → up 2. (A) misreads the inside sign; (D) swaps the two dials.
2. (C). b = 2 compresses horizontally to 1/2 width — inputs reach their old values twice as fast. (A) reads b backwards; (B) confuses inside with outside.
3. (A). Minus outside flips outputs: x-axis mirror. (B) needs f(−x); (C) needs both.
4. (D). Inside zero at x = −4; then down 1: vertex (−4, −1). (A) misreads the inside sign; (C) misses the −1.
5. (B). Shift right 1: [2 + 1, 6 + 1] = [3, 7]. (A) shifts left; (C) forgets horizontal changes touch the domain.
6. (A). y → 3y − 2 sends 0 → −2 and 4 → 10: [−2, 10]. (C) applies the −2 before the 3; (D) adds instead.
7. (C). g(−3) = f(−(−3)) = f(3) = 5 → point (−3, 5). (D) requires f(−3), which is unknown; (A)/(B) flip the output — but only the input is negated.
8. (A). Solve 3(x − 2) = 6 → x = 4; y unchanged → (4, 1). (C) multiplies instead of solving (18 = 3·6); (B) divides then forgets the shift; (D) adds 2 to 6 directly.
9. (B). f(3(−x)) = f(−3x) = f(3x) since f is even ✓. (A) breaks symmetry by moving the mirror line; (C) adds an odd piece; (D) multiplies by an odd function, producing an odd product.
10. (D). b = −1: y-axis reflection; a = 2: vertical stretch; d = +1: up 1, applied last. (A) reflects over the wrong axis; (B) stretches after translating (wrong: 2(f(−x) + 1) = 2f(−x) + 2 ≠ 2f(−x) + 1); (C) reflecting last is harmless here, but it also stretches after shifting — same flaw as (B).
11. (A). 4x² = (2x)²: b = 2 → horizontal compression by 1/2. (C) confuses the vertical factor 4 with the horizontal one; (B)/(D) read the dilation backwards.
12. (FRQ-style, 6 points) (i) [2 pts] Domain: solve 2x ∈ [0, 8] → x ∈ [0, 4]. Range: y → −(1/2)y + 3 sends −2 → 4 and 6 → 0; re-order → [0, 4]. (ii) [2 pts] New x: 2x = 3 → x = 1.5; new y: −(1/2)(6) + 3 = 0 → point (1.5, 0). The negative a flips the graph vertically, so f's local maximum becomes a local minimum of g. (iii) [2 pts] f increasing on (0, 3) → f(2x) increasing on (0, 1.5) (compression relocates the interval) → multiplying by −1/2 reverses monotonicity → g is decreasing on (0, 1.5). Reason: a negative vertical factor turns increasing into decreasing; the horizontal compression by 1/2 maps the interval (0, 3) to (0, 1.5).
🎯 Exam tip: When a transformation MC offers "left" vs. "right" or "stretch" vs. "compress" as its trap pair, don't argue with your memory — evaluate one landmark point both ways. Thirty seconds of arithmetic beats any mnemonic under exam pressure.