This lesson is the core of the exam's biggest category, Representations of Functions: Symbolic, Graphical and Tabular (30%). Graph-matching items appear in both sections; in Section 2 there is no calculator, so the skills here must run by hand.
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 across the x-axis | honest: outputs multiply by a |
| b | horizontal dilation by factor 1/|b|; if b < 0, reflection across 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, so 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 the graph of f lands on g(x) = a·f(b(x − c)) + d at:
new x: solve b(x − c) = p new y: a·q + d
Horizontal changes touch only x; vertical changes touch only y. When a multi-step transformation item gets confusing, track one or two landmark points and let the graph follow. This is also the fastest safeguard against the left/right and stretch/compress traps: thirty seconds of arithmetic beats any mnemonic.
Factor before reading. If the inside arrives unfactored — g(x) = f(2x + 6) — factor first: f(2(x + 3)). Compress by 1/2, then shift left 3. Reading "left 6, then compress" off the unfactored form gives wrong answers; better yet, skip the debate entirely and solve 2x + 6 = p for the new x.
Within the template, a safe order: dilations and reflections first, then translations. The structure b(x − c) bakes this in. Reversing the order changes the function: applying "up 1" before "stretch by 2" turns f into 2(f(x) + 1) = 2f(x) + 2, not 2f(x) + 1.
Example: f has domain [−2, 4] and range [1, 6]. For g(x) = −f(x) + 2: the domain is untouched (no horizontal dials), still [−2, 4]; the range maps through y → −y + 2, sending 1 → 1 and 6 → −4, giving [−4, 1] after re-ordering.
The 30% category lives on translating among three views of one function:
Graph → equation. For "Which of the following could be an equation of the function graphed above?", read off anchor features and eliminate: 1. Locate the vertex/turning point or intercepts — this pins c and d. 2. Check opening direction / end behavior — this pins the sign of a. 3. Confirm with one more plotted point — this pins the magnitude of a.
[GRAPH: A parabola in the xy-plane with vertex at (3, −2), opening upward, passing through (5, 2) and (1, 2). Matching equation: y = (x − 3)² − 2 — vertex gives the right-3, down-2 translation of y = x²; the point (5, 2) confirms a = 1 since (5−3)² − 2 = 2.]
The answer choices always include the sign-flipped vertex (x + 3 vs. x − 3), the flipped d, and the reflected a — each eliminated by one anchor feature.
Table → equation. Test each candidate formula against two or three rows — a single mismatch kills a candidate. Prioritize easy rows: x = 0 first, then symmetric pairs like x = ±2, which also reveal symmetry at a glance.
| x | −2 | −1 | 0 | 1 | 2 |
|---|---|---|---|---|---|
| g(x) | 7 | 4 | 3 | 4 | 7 |
Here g(0) = 3 eliminates any candidate with g(0) ≠ 3 instantly, and the mirror-image rows (g(−x) = g(x)) announce y-axis symmetry — so the formula must be even: g(x) = x² + 3 fits every row.
Dilations of power functions can masquerade as each other: y = 9x² is y = x² stretched vertically by 9 — and also, since 9x² = (3x)², compressed horizontally by factor 1/3. Both descriptions produce the same parabola; verify with the parent point (1, 1), which lands at (1, 9) under the vertical stretch and at (1/3, 1) under the compression — and both image points satisfy y = 9x². Exam items exploit this by offering two "different-looking" transformation descriptions where only one matches the algebra given — or, as here, by asking you to recognize that both are correct.
1. A — Correct: −4 inside the parentheses moves the graph right 4 (inputs transform inversely); −2 outside moves it down 2 (outputs transform honestly). B) reads (x − 4) as "left 4" — the classic inside-sign misread. C) reads −2 outside as "up." D) makes both misreads at once. E) swaps the two dials' magnitudes. Fix: inside the parentheses, the sign lies (x − c means right c); outside, it tells the truth.
2. D — Correct: only vertical dials appear, so x stays 2 and y maps through 3(5) − 4 = 11. A) transforms x as well, but a and d never touch inputs. B) computes 3(5 − 4), applying the translation before the stretch. C) adds 4 instead of subtracting: 3(5) + 4. E) transforms x and mangles y. Fix: vertical operations act on y only, in template order — multiply by a first, then add d.
3. B — Correct: solve 2x + 6 = 4 → x = −1; no a or d, so y stays −3: (−1, −3). Check: g(−1) = f(2(−1) + 6) = f(4) = −3 ✓. A) applies "divide by 2, then subtract 6" (4/2 − 6 = −4) — transformations applied in the wrong order to the point. C) makes a sign error: (4 + 6)/2 = 5. D) feeds 4 forward through the inside (2·4 + 6 = 14) instead of solving backwards. E) has the right x but flips y, though nothing acts on outputs here. Fix: set the entire inside expression equal to the old x-coordinate and solve — the new input is whatever makes the inside produce p.
4. A — Correct: no horizontal dials, so the domain stays [−2, 4]; the range maps through y → −y + 2, sending 1 → 1 and 6 → −4, and re-ordering gives [−4, 1]. B) adds 2 but ignores the reflection (1 → 3, 6 → 8). C) swaps domain and range — that's inverse-function behavior, not transformation behavior. D) reflects but forgets the +2 (and mis-orders). E) transforms the domain even though −f(x) + 2 contains no b or c. Fix: map both range endpoints through y → ay + d, then re-sort — a negative a always flips which endpoint is the minimum.
5. E — Correct: vertex (3, −2) forces the form a(x − 3)² − 2, opening upward means a > 0, and (5, 2) gives a(2)² − 2 = 2 → a = 1: y = (x − 3)² − 2. A) puts the vertex at (−3, −2) — the inside-sign misread. B) puts the vertex at (3, 2), misreading d. C) opens downward, contradicting the graph. D) misreads both vertex coordinates. Fix: match vertex first (it pins c and d), then opening direction (sign of a), then confirm with one plotted point.
| x | −2 | −1 | 0 | 1 | 2 |
|---|---|---|---|---|---|
| g(x) | 7 | 4 | 3 | 4 | 7 |
Which of the following could define g?
6. A — Correct: x² + 3 reproduces every row: 7, 4, 3, 4, 7; the mirror-image table announces an even function, and g(0) = 3 seals it. B) (x + 3)² gives g(0) = 9 ≠ 3 (it happens to match at x = −1 — one matching row is not a match). C) x² − 3 gives g(0) = −3, confusing "up 3" with "down 3." D) 3x² gives g(0) = 0, confusing the vertical stretch with the vertical shift. E) (x − 3)² gives g(0) = 9 and breaks the symmetry. Fix: test candidates at x = 0 first, then one symmetric pair — a formula must match every row, not just one.
7. C — Correct: b = 3 compresses horizontally to 1/3 of the original width — inputs reach their old values three times as fast. A) reads the horizontal dial forwards instead of backwards. B) confuses inside multiplication with outside multiplication (that would be 3f(x)). D) applies the reciprocal to the wrong axis. E) confuses dilation with translation. Fix: a constant multiplying x inside acts inversely — b = 3 squeezes; b = 1/3 stretches.
8. E — Correct: g(−5) = f(−(−5)) + 1 = f(5) + 1 = 5, so (−5, 5) is on g. A) keeps x = 5, but f(5) is only reachable through the input −5 once x is negated. B) flips the output's sign — b = −1 negates inputs, never outputs — and botches the +1. C) negates x correctly but drops the +1. D) leaves x alone and flips y, reversing both roles. Fix: f(−x) reflects across the y-axis (inputs negate); +1 then lifts outputs — handle the two dials on their own axes.
9. B — Correct: y-axis symmetry is the definition of an even function: g(−x) = g(x). A) is odd symmetry (origin), the wrong mirror. C) even functions need not vanish at 0 (x² + 3 doesn't). D) even functions are never one-to-one on a domain containing ±x pairs — symmetric inputs share outputs. E) symmetry says nothing about sign (x² − 4 is even and negative near 0). Fix: y-axis mirror ⇔ g(−x) = g(x); origin rotation ⇔ g(−x) = −g(x) — attach each symmetry to its equation.
10. D — Correct: |x − 1| is smallest (zero) at x = 1, and there g(1) = 2(0) + 3 = 3: minimum point (1, 3). A) misreads (x − 1) as a left shift. B) treats +3 as a downward shift. C) makes both misreads. E) confuses the stretch factor 2 with the horizontal shift. Fix: for a·|x − c| + d with a > 0, the minimum sits exactly at (c, d) — inside constant gives x, outside constant gives y.
11. C — Correct: g(x) = f(3x) is defined when 3x lands in [0, 12], i.e., 0 ≤ x ≤ 4. A) multiplies the domain by 3 instead of solving (the backwards-dial trap). B) assumes horizontal dilations leave the domain alone. D) treats b = 3 as a shift right 3. E) treats it as a shift left 3. Fix: to transform a domain, require the whole inside expression to lie in the original domain and solve the inequality for x.
12. E — Correct: 9x² = (3x)², so stretching y = x² vertically by 9 and compressing it horizontally by 1/3 yield the same parabola — check (1, 1): it maps to (1, 9) and to (1/3, 1), and both satisfy y = 9x². A) overgeneralizes; for power functions the two kinds of dilation can coincide. B) compares the factors numerically, but factors on different axes aren't comparable that way. C) misstates the equivalence — the horizontal counterpart is a compression by 1/3, not a stretch by 9. D) invents a partial coincidence; the graphs agree everywhere, since (3x)² = 9x² for all x. Fix: two transformation descriptions match exactly when their algebraic results are identical — rewrite both symbolically and compare formulas, not labels.
1. A — Correct: −4 inside the parentheses moves the graph right 4 (inputs transform inversely); −2 outside moves it down 2 (outputs transform honestly). B) reads (x − 4) as "left 4" — the classic inside-sign misread. C) reads −2 outside as "up." D) makes both misreads at once. E) swaps the two dials' magnitudes. Fix: inside the parentheses, the sign lies (x − c means right c); outside, it tells the truth.
2. D — Correct: only vertical dials appear, so x stays 2 and y maps through 3(5) − 4 = 11. A) transforms x as well, but a and d never touch inputs. B) computes 3(5 − 4), applying the translation before the stretch. C) adds 4 instead of subtracting: 3(5) + 4. E) transforms x and mangles y. Fix: vertical operations act on y only, in template order — multiply by a first, then add d.
3. B — Correct: solve 2x + 6 = 4 → x = −1; no a or d, so y stays −3: (−1, −3). Check: g(−1) = f(2(−1) + 6) = f(4) = −3 ✓. A) applies "divide by 2, then subtract 6" (4/2 − 6 = −4) — transformations applied in the wrong order to the point. C) makes a sign error: (4 + 6)/2 = 5. D) feeds 4 forward through the inside (2·4 + 6 = 14) instead of solving backwards. E) has the right x but flips y, though nothing acts on outputs here. Fix: set the entire inside expression equal to the old x-coordinate and solve — the new input is whatever makes the inside produce p.
4. A — Correct: no horizontal dials, so the domain stays [−2, 4]; the range maps through y → −y + 2, sending 1 → 1 and 6 → −4, and re-ordering gives [−4, 1]. B) adds 2 but ignores the reflection (1 → 3, 6 → 8). C) swaps domain and range — that's inverse-function behavior, not transformation behavior. D) reflects but forgets the +2 (and mis-orders). E) transforms the domain even though −f(x) + 2 contains no b or c. Fix: map both range endpoints through y → ay + d, then re-sort — a negative a always flips which endpoint is the minimum.
5. E — Correct: vertex (3, −2) forces the form a(x − 3)² − 2, opening upward means a > 0, and (5, 2) gives a(2)² − 2 = 2 → a = 1: y = (x − 3)² − 2. A) puts the vertex at (−3, −2) — the inside-sign misread. B) puts the vertex at (3, 2), misreading d. C) opens downward, contradicting the graph. D) misreads both vertex coordinates. Fix: match vertex first (it pins c and d), then opening direction (sign of a), then confirm with one plotted point.
6. A — Correct: x² + 3 reproduces every row: 7, 4, 3, 4, 7; the mirror-image table announces an even function, and g(0) = 3 seals it. B) (x + 3)² gives g(0) = 9 ≠ 3 (it happens to match at x = −1 — one matching row is not a match). C) x² − 3 gives g(0) = −3, confusing "up 3" with "down 3." D) 3x² gives g(0) = 0, confusing the vertical stretch with the vertical shift. E) (x − 3)² gives g(0) = 9 and breaks the symmetry. Fix: test candidates at x = 0 first, then one symmetric pair — a formula must match every row, not just one.
7. C — Correct: b = 3 compresses horizontally to 1/3 of the original width — inputs reach their old values three times as fast. A) reads the horizontal dial forwards instead of backwards. B) confuses inside multiplication with outside multiplication (that would be 3f(x)). D) applies the reciprocal to the wrong axis. E) confuses dilation with translation. Fix: a constant multiplying x inside acts inversely — b = 3 squeezes; b = 1/3 stretches.
8. E — Correct: g(−5) = f(−(−5)) + 1 = f(5) + 1 = 5, so (−5, 5) is on g. A) keeps x = 5, but f(5) is only reachable through the input −5 once x is negated. B) flips the output's sign — b = −1 negates inputs, never outputs — and botches the +1. C) negates x correctly but drops the +1. D) leaves x alone and flips y, reversing both roles. Fix: f(−x) reflects across the y-axis (inputs negate); +1 then lifts outputs — handle the two dials on their own axes.
9. B — Correct: y-axis symmetry is the definition of an even function: g(−x) = g(x). A) is odd symmetry (origin), the wrong mirror. C) even functions need not vanish at 0 (x² + 3 doesn't). D) even functions are never one-to-one on a domain containing ±x pairs — symmetric inputs share outputs. E) symmetry says nothing about sign (x² − 4 is even and negative near 0). Fix: y-axis mirror ⇔ g(−x) = g(x); origin rotation ⇔ g(−x) = −g(x) — attach each symmetry to its equation.
10. D — Correct: |x − 1| is smallest (zero) at x = 1, and there g(1) = 2(0) + 3 = 3: minimum point (1, 3). A) misreads (x − 1) as a left shift. B) treats +3 as a downward shift. C) makes both misreads. E) confuses the stretch factor 2 with the horizontal shift. Fix: for a·|x − c| + d with a > 0, the minimum sits exactly at (c, d) — inside constant gives x, outside constant gives y.
11. C — Correct: g(x) = f(3x) is defined when 3x lands in [0, 12], i.e., 0 ≤ x ≤ 4. A) multiplies the domain by 3 instead of solving (the backwards-dial trap). B) assumes horizontal dilations leave the domain alone. D) treats b = 3 as a shift right 3. E) treats it as a shift left 3. Fix: to transform a domain, require the whole inside expression to lie in the original domain and solve the inequality for x.
12. E — Correct: 9x² = (3x)², so stretching y = x² vertically by 9 and compressing it horizontally by 1/3 yield the same parabola — check (1, 1): it maps to (1, 9) and to (1/3, 1), and both satisfy y = 9x². A) overgeneralizes; for power functions the two kinds of dilation can coincide. B) compares the factors numerically, but factors on different axes aren't comparable that way. C) misstates the equivalence — the horizontal counterpart is a compression by 1/3, not a stretch by 9. D) invents a partial coincidence; the graphs agree everywhere, since (3x)² = 9x² for all x. Fix: two transformation descriptions match exactly when their algebraic results are identical — rewrite both symbolically and compare formulas, not labels.