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Function Transformations and Composition · Solutions函数变换与复合 · 详解

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EASY易 MEDIUM中 HARD难 🇺🇸 US美 🇨🇦 ON安 🇨🇦 BC卑 🇨🇦 AB艾 SAT-style MCQSAT 风格选择题 AP-feeder FRQAP 衔接简答题 ON Provincial-style安大略省考风格 BC Provincial-style卑诗省考风格 Honors荣誉级
PART I  ·  SHORT RESPONSE · SOLUTIONS第一部分  ·  短答题 · 详解SAT MCQ + ON/BC short answer · 18 marksSAT 选择题 + 安/卑省考短答 · 共 18 分

Section A · Worked SolutionsA 部分 · 详细解答

Q1EASY易 🇺🇸 US美 SAT-style MCQSAT 风格选择题 §1 Shifts平移 · HSF-BF.B.3 [3 marks][3 分]

Describe $y = f(x + 4) - 5$ as a transformation of $y = f(x)$.将 $y = f(x + 4) - 5$ 描述为 $y = f(x)$ 的变换。

Answer:答案：  (D)  Shift left $4$ units and down $5$ units向左平移 $4$ 个单位，向下平移 $5$ 个单位

(a) Read master form $y = f(x - h) + k$从一般形式 $y = f(x - h) + k$ 读取 M1·A1·A1

The standard shift form is $y = f(x - h) + k$, where positive $h$ shifts right and positive $k$ shifts up. Rewrite $y = f(x + 4) - 5$ as $y = f(x - (-4)) + (-5)$, so $h = -4$ and $k = -5$. The horizontal shift is $h = -4$ (i.e. $4$ units left), and the vertical shift is $k = -5$ (i.e. $5$ units down). This matches option (D).标准平移形式为 $y = f(x - h) + k$，其中 $h$ 为正向右移、$k$ 为正向上移。把 $y = f(x + 4) - 5$ 改写为 $y = f(x - (-4)) + (-5)$，得 $h = -4$，$k = -5$。水平平移为 $h = -4$（即向左 $4$ 个单位），竖直平移为 $k = -5$（即向下 $5$ 个单位）。对应选项 (D)。
Why the wrong choices fail.干扰项分析。
  • (A) Right $4$, up $5$右 $4$，上 $5$: both signs flipped, the classic "the sign you see is the direction" mistake. Inside the bracket the sign flips; outside it does not.两个符号都搞反了，是典型的"看到什么符号就往哪边走"的错误。括号内符号要反向解读，括号外不需要。
  • (B) Right $4$, down $5$右 $4$，下 $5$: gets the vertical correct ($-5$ outside the bracket reads directly) but mis-reads the horizontal. The "$+4$ inside" feels like rightward, but solve $x + 4 = 0$ and you get $x = -4$, the parent's anchor point now sits at $x = -4$, so the graph moved left.竖直方向对（括号外的 $-5$ 直接读取）但水平方向错了。"括号内 $+4$"看着像向右，可解 $x + 4 = 0$ 得 $x = -4$，母函数的锚点现在落在 $x = -4$，所以图像向左移动。
  • (C) Left $4$, up $5$左 $4$，上 $5$: horizontal correct, vertical sign-flipped. The "$-5$ outside the function" is literally "subtract $5$ from every output", which lowers the graph.水平方向对、竖直方向符号搞反。"函数外的 $-5$"字面上就是"每个输出减 $5$"，会把图像压低。
Inside the bracket: opposite. Outside the bracket: direct.括号里：反向。括号外：直读。 This is the single-most-tested transformation rule in the entire unit. Memorise the diagnostic: to find $h$, ask "what value of $x$ makes the bracket zero?", that value is $h$. For $f(x + 4)$, the bracket is zero at $x = -4$, so $h = -4$ and the shift is leftward. For $f(x - 3)$, the bracket is zero at $x = 3$, so $h = +3$ and the shift is rightward. The "$- 5$ outside" needs no flipping, $k$ reads with its sign. The whole rest of §1–§4 builds on this single rule.这是整个单元最常考的变换规则。记住诊断方法：求 $h$ 就问"什么 $x$ 值使括号为零？"该值即 $h$。对 $f(x + 4)$，括号在 $x = -4$ 处为零，故 $h = -4$，向左移；对 $f(x - 3)$，括号在 $x = 3$ 处为零，故 $h = +3$，向右移。括号外的 "$-5$" 不需要变号，$k$ 直接带符号读出。§1–§4 后面所有内容都建立在这一条规则上。
Q2EASY易 🇺🇸 US美 SAT-style MCQSAT 风格选择题 §3 Reflections反射 · HSF-BF.B.3 [3 marks][3 分]

Image of $(-3, 5)$ under $y = f(-x)$.点 $(-3, 5)$ 在 $y = f(-x)$ 下的像。

Answer:答案：  (C)  $(3, 5)$

(a) Apply the $y$-axis reflection coordinate rule应用关于 $y$ 轴反射的坐标规则 M1·A1·A1

The transformation $y = f(-x)$ is a reflection over the $y$-axis: the coordinate rule (AB Math 30-1 indicator 5.1) is $(x, y) \to (-x, y)$. Applying it to $(-3, 5)$:变换 $y = f(-x)$ 是关于 $y$ 轴的反射：坐标规则（AB Math 30-1 指标 5.1）为 $(x, y) \to (-x, y)$。应用到 $(-3, 5)$： $$ (-3, 5) \;\to\; (-(-3), 5) \;=\; (3, 5). $$ This matches option (C).对应选项 (C)。
Why the wrong choices fail.干扰项分析。
  • (A) $(-3, -5)$: this is the $x$-axis reflection rule $(x, y) \to (x, -y)$, used for $y = -f(x)$, not $y = f(-x)$. Inside vs. outside, again.这是关于 $x$ 轴反射的规则 $(x, y) \to (x, -y)$，对应 $y = -f(x)$，不是 $y = f(-x)$。又是"括号内 vs. 括号外"的问题。
  • (B) $(3, -5)$: both coordinates negated, the origin reflection (composition of both reflections), which would correspond to $y = -f(-x)$.两个坐标都变号，是关于原点的反射（两种反射的复合），对应 $y = -f(-x)$。
  • (D) $(-5, -3)$: swaps the coordinates and negates them, this is a hybrid of the inverse rule $(x, y) \to (y, x)$ from §7 and a sign flip. None of those transformations is on offer here.交换坐标再变号，是 §7 中反函数规则 $(x, y) \to (y, x)$ 与变号的混合品。本题没有这种变换。
The minus sign sits where the transformation sits.负号在哪里，变换就在哪里。 Minus outside the function ($y = -f(x)$) flips the $y$-coordinate, because outputs get negated. Minus inside ($y = f(-x)$) flips the $x$-coordinate, because the input needed to produce a given output gets negated. This "inside / outside" lens unifies §1 shifts, §2 stretches, and §3 reflections, the parameter $a$ (outside) governs $y$, the parameter $b$ (inside) governs $x$. Once locked in, the SAT's "which point lies on the transformed graph" items become a five-second coordinate substitution.函数外的负号（$y = -f(x)$）翻转 $y$ 坐标，因为输出被取了负。函数内的负号（$y = f(-x)$）翻转 $x$ 坐标，因为产生某给定输出所需的输入被取了负。这种"内/外"的视角把 §1 平移、§2 伸缩、§3 反射统一起来：参数 $a$（外部）管 $y$、参数 $b$（内部）管 $x$。一旦锁定，SAT 上"哪一点在变换后图像上"这类题就只剩五秒钟的坐标代入。
Q3MEDIUM中 🇺🇸 US美 SAT-style MCQSAT 风格选择题 §5 Even / Odd / Neither偶 / 奇 / 非奇非偶 · HSF-BF.B.3 [3 marks][3 分]

Which function is odd?哪个函数是奇函数？

Answer:答案：  (B)  $f(x) = x^{3} - 4x$

(a) Apply the odd-function test $f(-x) = -f(x)$应用奇函数判别式 $f(-x) = -f(x)$ M1·A1·A1

For (B): $f(-x) = (-x)^{3} - 4(-x) = -x^{3} + 4x = -(x^{3} - 4x) = -f(x)$. $\checkmark$ So $f$ is odd.对选项 (B)：$f(-x) = (-x)^{3} - 4(-x) = -x^{3} + 4x = -(x^{3} - 4x) = -f(x)$。$\checkmark$ 故 $f$ 为奇函数。
Structural note: every term is an odd power of $x$ (the constant term is zero), which is the algebraic signature of an odd polynomial.结构上注意：每一项都是 $x$ 的奇次幂（常数项为零），这是奇多项式的代数特征。
Why the wrong choices fail.干扰项分析。
  • (A) $f(x) = x^{2} + 1$: $f(-x) = x^{2} + 1 = f(x)$, this is even, not odd. (Symmetric about the $y$-axis, not the origin.)$f(-x) = x^{2} + 1 = f(x)$，是偶函数而非奇函数。（关于 $y$ 轴对称，不是关于原点对称。）
  • (C) $f(x) = x^{3} + 2$: $f(-x) = -x^{3} + 2$, which equals neither $f(x)$ nor $-f(x)$. The constant term $+2$ kills the odd symmetry, an odd function must satisfy $f(0) = 0$.$f(-x) = -x^{3} + 2$，既不等于 $f(x)$ 也不等于 $-f(x)$。常数项 $+2$ 破坏了奇对称——奇函数必须满足 $f(0) = 0$。
  • (D) $f(x) = |x|$: $f(-x) = |-x| = |x| = f(x)$, this is even, not odd.$f(-x) = |-x| = |x| = f(x)$，是偶函数而非奇函数。
The fast diagnostic: check $f(0)$ and check every term's parity.快速诊断：先看 $f(0)$，再看每项的奇偶性。 For polynomial-type functions: an odd function must satisfy $f(0) = 0$ (substitute the equation $f(-0) = -f(0)$, so $f(0) = -f(0)$, hence $f(0) = 0$). Any non-zero constant term knocks out odd symmetry, that handles option (C) instantly. An even function's terms are all even powers of $x$ (constants count as $x^{0}$); an odd function's terms are all odd powers (with no constant). Mixed-parity terms make a function "neither". This three-check sequence, $f(0) = 0$, then term parity, then a full $f(-x)$ substitution as backup, takes about ten seconds per option.对多项式型函数：奇函数必满足 $f(0) = 0$（代入 $f(-0) = -f(0)$，故 $f(0) = -f(0)$，于是 $f(0) = 0$）。任何非零常数项都会破坏奇对称——(C) 立刻被淘汰。偶函数各项都是 $x$ 的偶次幂（常数算 $x^{0}$）；奇函数各项都是奇次幂（且无常数项）。混合奇偶则属于"非奇非偶"。三步诊断——$f(0) = 0$、各项奇偶、完整代入 $f(-x)$ 作为兜底——每个选项约十秒。
Q4MEDIUM中 🇨🇦 ON安 ON Provincial-style安大略省考风格 §2 Stretches (b-inside reversal)伸缩（$b$ 在内部反向作用） · MCR3U A1.5 [4 marks][4 分]

$(8, 3)$ on $y = f(x)$. (a) Image under $y = f(4 x)$ + name. (b) Image under $y = \tfrac{1}{2} f(x)$ + name.$(8, 3)$ 在 $y = f(x)$ 上。(a) $y = f(4 x)$ 下的像与变换名称。(b) $y = \tfrac{1}{2} f(x)$ 下的像与变换名称。

Answer:答案：  (a) $(2, 3)$, horizontal compression by factor $\tfrac{1}{4}$水平压缩 $\tfrac{1}{4}$ 倍  ·  (b) $(8, \tfrac{3}{2})$, vertical compression by factor $\tfrac{1}{2}$竖直压缩 $\tfrac{1}{2}$ 倍

(a) Horizontal stretch / compression by $b$由 $b$ 控制的水平伸 / 压 M1·A1

The transformation $y = f(b x)$ has the coordinate rule $(x, y) \to (\tfrac{x}{b}, y)$, $x$-coordinates are divided by $b$, the "$b$-inside reversal". Here $b = 4$, so the image of $(8, 3)$ is $(\tfrac{8}{4}, 3) = (2, 3)$. Because $|b| = 4 > 1$, this is a horizontal compression by factor $\tfrac{1}{4}$ (toward the $y$-axis), not a stretch.变换 $y = f(b x)$ 的坐标规则为 $(x, y) \to (\tfrac{x}{b}, y)$，即 $x$ 坐标被 $b$ 除——这正是"$b$ 在内部反向作用"。此处 $b = 4$，故 $(8, 3)$ 的像为 $(\tfrac{8}{4}, 3) = (2, 3)$。因 $|b| = 4 > 1$，这是水平压缩 $\tfrac{1}{4}$ 倍（向 $y$ 轴方向收缩），不是拉伸。

(b) Vertical stretch / compression by $a$由 $a$ 控制的竖直伸 / 压 M1·A1

The transformation $y = a f(x)$ has the coordinate rule $(x, y) \to (x, a y)$, $y$-coordinates are multiplied by $a$ (direct, not reversed). Here $a = \tfrac{1}{2}$, so the image of $(8, 3)$ is $(8, \tfrac{1}{2} \cdot 3) = (8, \tfrac{3}{2})$. Because $0 < |a| < 1$, this is a vertical compression by factor $\tfrac{1}{2}$ (toward the $x$-axis).变换 $y = a f(x)$ 的坐标规则为 $(x, y) \to (x, a y)$，即 $y$ 坐标乘以 $a$（直接作用，不反向）。此处 $a = \tfrac{1}{2}$，故 $(8, 3)$ 的像为 $(8, \tfrac{1}{2} \cdot 3) = (8, \tfrac{3}{2})$。因 $0 < |a| < 1$，这是竖直压缩 $\tfrac{1}{2}$ 倍（向 $x$ 轴方向收缩）。
$b$-inside reverses; $a$-outside does not.$b$ 在内部反向；$a$ 在外部不反向。 The single most common error on MCR3U transformation items is calling $y = f(4 x)$ a "horizontal stretch by $4$". It is the opposite, a compression by $\tfrac{1}{4}$, because the input $x$ gets divided by $b$ before being fed to $f$. To produce the output the parent had at input $X$, the new graph needs input $x = \tfrac{X}{b}$. Big $b$ shrinks the picture horizontally. The vertical parameter $a$ behaves naturally (no reversal): $y = 4 f(x)$ stretches by $4$, $y = \tfrac{1}{2} f(x)$ compresses by $\tfrac{1}{2}$. Lock the asymmetry, "inside reverses, outside doesn't", and the §4 master-form question becomes mechanical.MCR3U 变换题最常见的错误就是把 $y = f(4 x)$ 称为"水平拉伸 $4$ 倍"。恰恰相反——是 $\tfrac{1}{4}$ 的压缩，因为输入 $x$ 在送入 $f$ 之前被 $b$ 除。要复现母函数在输入 $X$ 处的输出，新图像需要输入 $x = \tfrac{X}{b}$。$b$ 越大，图像在水平方向越收缩。竖直方向的参数 $a$ 行为自然（不反向）：$y = 4 f(x)$ 拉伸 $4$ 倍，$y = \tfrac{1}{2} f(x)$ 压缩 $\tfrac{1}{2}$。把这条不对称——"内部反向、外部直读"——锁住，§4 一般形式题就纯属机械操作。
Q5MEDIUM中 🇨🇦 BC卑 🇨🇦 AB艾 BC Provincial-style卑诗省考风格 §1 + §3 Coordinate Rules坐标变换规则 · BC PC 12 / AB Math 30-1 RF 2.1, 5.1 [5 marks][5 分]

Transform $y = \sqrt{x}$. (a) Right $3$, down $1$. (b) Reflect over $x$-axis. (c) Reflect over $y$-axis.变换 $y = \sqrt{x}$。(a) 右 $3$、下 $1$。(b) 关于 $x$ 轴反射。(c) 关于 $y$ 轴反射。

Answer:答案：  (a) $y = \sqrt{x - 3} - 1$, image of $(4, 2)$ is $(7, 1)$$y = \sqrt{x - 3} - 1$，$(4, 2)$ 的像为 $(7, 1)$  ·  (b) $y = -\sqrt{x}$, image $(4, -2)$$y = -\sqrt{x}$，像 $(4, -2)$  ·  (c) $y = \sqrt{-x}$, image $(-4, 2)$$y = \sqrt{-x}$，像 $(-4, 2)$

(a) Shift right $3$, down $1$向右 $3$、向下 $1$ 平移 M1·A1

Right $h$, down $|k|$ replaces $x$ with $x - h$ inside and adds $k$ outside: image equation $y = \sqrt{x - 3} - 1$. Coordinate rule: $(x, y) \to (x + 3, y - 1)$. Image of $(4, 2)$: $(7, 1)$.向右 $h$、向下 $|k|$ 即把内部 $x$ 换为 $x - h$、外部加 $k$：变换后方程为 $y = \sqrt{x - 3} - 1$。坐标规则：$(x, y) \to (x + 3, y - 1)$。$(4, 2)$ 的像：$(7, 1)$。

(b) Reflect over the $x$-axis关于 $x$ 轴反射 M1·A1

Negate the output: image equation $y = -\sqrt{x}$. Coordinate rule (AB Math 30-1 indicator 5.1): $(x, y) \to (x, -y)$. Image of $(4, 2)$: $(4, -2)$.输出取负：方程为 $y = -\sqrt{x}$。坐标规则（AB Math 30-1 指标 5.1）：$(x, y) \to (x, -y)$。$(4, 2)$ 的像：$(4, -2)$。

(c) Reflect over the $y$-axis关于 $y$ 轴反射 A1

Negate the input: image equation $y = \sqrt{-x}$. Coordinate rule: $(x, y) \to (-x, y)$. Image of $(4, 2)$: $(-4, 2)$. (Note: the domain of the reflected graph is now $x \le 0$, not $x \ge 0$, the reflection moves the entire graph across the $y$-axis.)输入取负：方程为 $y = \sqrt{-x}$。坐标规则：$(x, y) \to (-x, y)$。$(4, 2)$ 的像：$(-4, 2)$。（注意：反射后图像的定义域变为 $x \le 0$ 而非 $x \ge 0$——反射把整条图像移到了 $y$ 轴另一侧。）
Three rules, four lines of work.三条规则，四行过程。 AB Math 30-1 indicator 5.1 spells out the trio that every transformation question composes: $(x, y) \to (x, -y)$ for $x$-axis, $(x, y) \to (-x, y)$ for $y$-axis, $(x, y) \to (y, x)$ for the $y = x$ line (which returns as the inverse rule in §7). Combine with shift rules $(x, y) \to (x + h, y + k)$ and stretch rules $(x, y) \to (\tfrac{x}{b}, a y)$ and you have the entire toolkit. The "image of $(4, 2)$" probe is the BC / AB provincial signature, markers want to see one explicit coordinate-rule application per transformation, not just the final equation.AB Math 30-1 指标 5.1 列明了所有变换都由以下三条规则组合而成：$(x, y) \to (x, -y)$（$x$ 轴）、$(x, y) \to (-x, y)$（$y$ 轴）、$(x, y) \to (y, x)$（$y = x$ 线，这在 §7 反函数规则中再次出现）。再加上平移规则 $(x, y) \to (x + h, y + k)$ 与伸缩规则 $(x, y) \to (\tfrac{x}{b}, a y)$，整套工具就齐了。"$(4, 2)$ 的像"这一问是 BC / AB 省考的标志写法——阅卷人希望每次变换都明确写出一次坐标规则的应用，而不是只给出最终方程。
PART II  ·  EXTENDED RESPONSE · SOLUTIONSAP-feeder FRQ + honors · 35 marks

Section B · Worked Solutions

Q6MEDIUM 🇺🇸 US AP-feeder FRQ §4 Master Form · HSF-BF.B.3 [8 marks]

$y = -2 f\bigl(\tfrac{1}{3}(x - 6)\bigr) + 4$. (a) Identify $a, b, h, k$. (b) Describe four transformations in order. (c) Image of $(3, 1)$.

Answer:  (a) $a = -2$, $b = \tfrac{1}{3}$, $h = 6$, $k = 4$  ·  (b) horizontal stretch by $3$, then shift right $6$; vertical stretch by $2$ with reflection over $x$-axis, then shift up $4$  ·  (c) $(15, 2)$

(a) Read $a, b, h, k$ from the master form A1·A1

Master form: $y = a f(b(x - h)) + k$. Matching: $a = -2$, $b = \tfrac{1}{3}$, the bracket $(x - 6)$ gives $h = 6$, and the constant outside gives $k = 4$.

(b) Order the four transformations M1·A1·A1·A1

Horizontal first (inside the function, applied to $x$ before evaluation):
  1. Horizontal stretch by factor $\tfrac{1}{|b|} = 3$ (since $b = \tfrac{1}{3}$, $|b| < 1$ produces a stretch, the inside-reversal rule from §2);
  2. Horizontal shift right by $h = 6$.
Vertical second (outside the function, applied after evaluation):
  1. Vertical stretch by factor $|a| = 2$ combined with reflection over the $x$-axis (since $a = -2 < 0$);
  2. Vertical shift up by $k = 4$.

(c) Image of $(3, 1)$ under the coordinate rule M1·A1

Combined rule: $(x, y) \to \bigl(\tfrac{x}{b} + h, \; a y + k\bigr) = \bigl(3 x + 6, \; -2 y + 4\bigr)$. Apply to $(3, 1)$: $$ \bigl(3(3) + 6, \; -2(1) + 4\bigr) \;=\; (15, 2). $$
HSVS: Horizontal Stretch, Horizontal Shift, Vertical Stretch, Vertical Shift, the lockstep order. The order of operations matters because stretches multiply (and reflections negate) while shifts add. Doing a shift before its matching stretch shifts the wrong point. The mnemonic order is "inside-out, multiply-before-add": apply the $b$-stretch, then the $h$-shift; apply the $a$-stretch, then the $k$-shift. AP-feeder courses (and IB Math AA HL Topic B5) test this order explicitly, naming all four transformations in the wrong sequence costs one A1 per swap. The coordinate-rule shortcut $\bigl(\tfrac{x}{b} + h, a y + k\bigr)$ is the entire HSVS sequence compressed into a single map.
Q7HARD 🇨🇦 ON ON Provincial-style §4 Master Form, b-Factor-Out · MCR3U A1.8 [9 marks]

$y = f(2 x - 10) + 3$. (a) Master form. (b) Order of operations. (c) Images of $(0, 0), (2, 4), (4, 0)$.

Answer:  (a) $y = f(2(x - 5)) + 3$, so $a = 1$, $b = 2$, $h = 5$, $k = 3$  ·  (c) $(5, 3), (6, 7), (7, 3)$

(a) Factor $b$ out of the bracket M1·A1·A1

Start from $y = f(2 x - 10) + 3$. Inside the function we have $2 x - 10 = 2(x - 5)$, factoring the leading $2$ out of the bracket. Master form is therefore: $$ y \;=\; f\bigl(2(x - 5)\bigr) + 3, $$ with $a = 1$, $b = 2$, $h = 5$, $k = 3$.

(b) Order of operations, with explicit reasoning M1·A1·A1

Horizontal (inside) first:
  1. Horizontal compression by factor $\tfrac{1}{|b|} = \tfrac{1}{2}$ (the inside-reversal: $|b| = 2 > 1$ produces a compression);
  2. Horizontal shift right by $h = 5$.
Why this order? The shift is read off the bracket after $b$ is factored out: the parent's anchor at $x = 0$ now lives at the value of $x$ that makes $2(x - 5) = 0$, namely $x = 5$. If you tried to apply the shift before the compression, you would shift the wrong point. The "factor first, read $h$ second" workflow is exactly what MCR3U A1.8 expects.
Vertical (outside) second: no vertical stretch ($a = 1$), no reflection; just shift up by $k = 3$.

(c) Image points via the coordinate rule M1·A1·A1

Coordinate rule: $(x, y) \to \bigl(\tfrac{x}{b} + h, \; y + k\bigr) = \bigl(\tfrac{x}{2} + 5, \; y + 3\bigr)$. Then:
  • $(0, 0) \to (\tfrac{0}{2} + 5, 0 + 3) = (5, 3)$.
  • $(2, 4) \to (\tfrac{2}{2} + 5, 4 + 3) = (6, 7)$.
  • $(4, 0) \to (\tfrac{4}{2} + 5, 0 + 3) = (7, 3)$.
If you skip the factor-out step, you read $h$ wrong by a factor of $b$. The most common MCR3U A1.8 error: looking at $y = f(2 x - 10) + 3$ and announcing "$h = 10$, shift right by $10$". Wrong, the shift is $5$, not $10$. The parent's anchor moved by the value that solves $2 x - 10 = 0$, which is $x = 5$. Provincial markers deduct sharply for this because it shows the student never internalised that $b$ multiplies everything inside the bracket, including any unfactored shift. Habit move: when you see $b \ne 1$ together with an unfactored bracket, factor $b$ out before doing anything else. This single discipline locks down every §4 problem you will ever see.
Q8MEDIUM 🇨🇦 BC 🇨🇦 AB BC Provincial-style §7 Inverses (swap-and-solve) · AB Math 30-1 RF 6.1–6.5 [9 marks]

$f(x) = 3 x - 5$, $g(x) = (x - 1)^{2} + 2$. (a) $f^{-1}$ + domain/range. (b) Verify $f(f^{-1}(x)) = x$. (c) Why $g$ has no inverse; restrict and invert.

Answer:  (a) $f^{-1}(x) = \tfrac{x + 5}{3}$, $D = R = \mathbb{R}$  ·  (c) restrict $x \ge 1$, then $g^{-1}(x) = 1 + \sqrt{x - 2}$, $D = [2, \infty)$, $R = [1, \infty)$

(a) Swap-and-solve for $f^{-1}$ M1·A1·A1

Start from $y = 3 x - 5$. Swap $x \leftrightarrow y$: $x = 3 y - 5$. Solve for $y$: $$ x + 5 \;=\; 3 y \;\Longrightarrow\; y \;=\; \frac{x + 5}{3}. $$ So $f^{-1}(x) = \tfrac{x + 5}{3}$. Domain: $\mathbb{R}$; range: $\mathbb{R}$. (Both the original and its inverse are linear with non-zero slope, hence defined on all reals.)

(b) Verify $f(f^{-1}(x)) = x$ M1·A1

$$ f(f^{-1}(x)) \;=\; f\bigl(\tfrac{x + 5}{3}\bigr) \;=\; 3 \cdot \tfrac{x + 5}{3} - 5 \;=\; (x + 5) - 5 \;=\; x. \;\checkmark $$

(c) Why $g$ fails the one-to-one test; restrict and invert R1·M1·A1·A1

$g(x) = (x - 1)^{2} + 2$ is a parabola with vertex $(1, 2)$, opening upward. It is not one-to-one on its natural domain $\mathbb{R}$: for example, $g(0) = 1 + 2 = 3$ and $g(2) = 1 + 2 = 3$, two distinct inputs sharing an output. Equivalently, the graph fails the horizontal-line test. So $g^{-1}$ does not exist as a function on $\mathbb{R}$.
Restrict to $x \ge 1$ (the right half of the parabola, including the vertex). On $[1, \infty)$, $g$ is strictly increasing, hence one-to-one. Swap-and-solve: $$ x \;=\; (y - 1)^{2} + 2 \;\Longrightarrow\; (y - 1)^{2} \;=\; x - 2 \;\Longrightarrow\; y - 1 \;=\; \pm \sqrt{x - 2}. $$ Because the restricted domain is $y \ge 1$ in the swap, take the $+$ branch: $y = 1 + \sqrt{x - 2}$. So $g^{-1}(x) = 1 + \sqrt{x - 2}$, with domain $[2, \infty)$ (the range of restricted $g$) and range $[1, \infty)$ (the restricted domain of $g$).
Domain of $f^{-1}$ = range of $f$; range of $f^{-1}$ = domain of $f$, always. This swap is the deepest fact in §7. When you restrict $g$'s domain to make it one-to-one, you also restrict the range of $g^{-1}$ to match (and vice versa). The choice of $\pm$ branch is dictated by the restricted domain: if you restricted to $x \le 1$ instead, you would pick the $-$ branch and get $g^{-1}(x) = 1 - \sqrt{x - 2}$. The vertex of the parabola is the natural restriction point because that is where the parabola changes monotonicity. AP graders (and the upcoming HSF-BF.B.4 items) test this explicit domain-range swap on the inverse, an unrestricted "$g^{-1}(x) = 1 \pm \sqrt{x - 2}$" answer loses the A1 because $\pm$ is not a function.
Q9HARDHonors 🇺🇸 US 🇨🇦 ON AP-feeder FRQ §6 Composition · HSF-BF.A.1.c (+) / MHF4U Strand C [9 marks]

$f(x) = \sqrt{x - 1}$, $g(x) = x^{2} - 5$. (a) $(f \circ g)(3), (g \circ f)(10)$. (b) $(f \circ g)(x)$ + domain. (c) $(g \circ f)(x)$ + domain. (d) Non-commutativity counter-example.

Answer:  (a) $(f \circ g)(3) = \sqrt{3}$, $(g \circ f)(10) = 4$  ·  (b) $(f \circ g)(x) = \sqrt{x^{2} - 6}$, $D = (-\infty, -\sqrt{6}] \cup [\sqrt{6}, \infty)$  ·  (c) $(g \circ f)(x) = x - 6$, $D = [1, \infty)$  ·  (d) e.g. $x = 5$: $(f \circ g)(5) = 2\sqrt{5}$, $(g \circ f)(5) = -1$

(a) Inner-then-outer evaluations M1·A1

$(f \circ g)(3) = f(g(3)) = f(3^{2} - 5) = f(4) = \sqrt{4 - 1} = \sqrt{3}$.
$(g \circ f)(10) = g(f(10)) = g(\sqrt{10 - 1}) = g(3) = 3^{2} - 5 = 4$.

(b) Formula and domain for $(f \circ g)(x)$ M1·A1·A1

$$ (f \circ g)(x) \;=\; f(g(x)) \;=\; f(x^{2} - 5) \;=\; \sqrt{(x^{2} - 5) - 1} \;=\; \sqrt{x^{2} - 6}. $$ Domain rule (inner output is a legal input to the outer): the inner $g(x) = x^{2} - 5$ accepts all reals, so the only constraint is the outer's domain, the radicand $x^{2} - 6 \ge 0$: $$ x^{2} \ge 6 \;\Longleftrightarrow\; |x| \ge \sqrt{6} \;\Longleftrightarrow\; x \in (-\infty, -\sqrt{6}] \cup [\sqrt{6}, \infty). $$

(c) Formula and domain for $(g \circ f)(x)$ M1·A1

$$ (g \circ f)(x) \;=\; g(f(x)) \;=\; g(\sqrt{x - 1}) \;=\; (\sqrt{x - 1})^{2} - 5 \;=\; (x - 1) - 5 \;=\; x - 6. $$ Domain: inherit the inner's domain $f$'s requirement $x - 1 \ge 0$, i.e. $x \ge 1$. (Important: even though the simplified formula $x - 6$ looks defined on $\mathbb{R}$, the composition is only valid where $f$ accepts $x$. The simplification cancelled out the radical, but the domain restriction survives.) So $D = [1, \infty)$.

(d) Non-commutativity counter-example R1·A1

Take $x = 5$ (which is in both domains: $5 \in [\sqrt{6}, \infty)$ for (b) since $\sqrt{6} \approx 2.45$, and $5 \in [1, \infty)$ for (c)). $$ (f \circ g)(5) \;=\; \sqrt{25 - 6} \;=\; \sqrt{19} \;\approx\; 4.36, $$ $$ (g \circ f)(5) \;=\; 5 - 6 \;=\; -1. $$ Since $\sqrt{19} \ne -1$, composition is not commutative.
Domain of a composition: inherit the inner's domain, then intersect with "inner output is in outer's domain". This two-step rule resolves the most common Honors-level slip. In part (c), the cancelled radical tempts students to claim the domain is all of $\mathbb{R}$, but composition cannot enlarge a function's domain: if $f$ rejects $x = 0$ then so does $g \circ f$. In part (b), the inner $g$ accepts everything, so the domain is purely the outer's constraint $x^{2} \ge 6$. The rule generalises: $\mathrm{dom}(f \circ g) = \{x \in \mathrm{dom}\, g : g(x) \in \mathrm{dom}\, f\}$. AP Calculus uses this every time the chain rule applies to a piecewise or radical inner, IB Math AA HL Topic E1 makes it explicit in the derivative-of-inverse formula. Composition is non-commutative for the same reason: order changes what is the "input" and what is the "wrap".
PART III  ·  MODELING / APPLIED · SOLUTIONSUniversal · 28 marks

Section C · Worked Solutions

Q10MEDIUM 🇺🇸 US AP-feeder FRQ §4 Master Form on $\sqrt{x}$ · HSF-BF.B.3 [9 marks]

$T(x) = 2 \sqrt{x - 3} + 1$ from parent $f(x) = \sqrt{x}$. (a) Master form params. (b) Order of transformations. (c) Domain + range with justification. (d) Image of $(0,0), (1,1), (4,2)$.

Answer:  (a) $a = 2$, $b = 1$, $h = 3$, $k = 1$  ·  (c) $D = [3, \infty)$, $R = [1, \infty)$  ·  (d) $(3, 1), (4, 3), (7, 5)$

(a) Read master-form parameters A1·A1

$T(x) = 2 \sqrt{x - 3} + 1 = 2 f(1 \cdot (x - 3)) + 1$. So $a = 2$, $b = 1$, $h = 3$, $k = 1$.

(b) Order of transformations A1·A1

No horizontal stretch ($b = 1$). Apply:
  1. Horizontal shift right by $h = 3$;
  2. Vertical stretch by factor $|a| = 2$ (no reflection since $a > 0$);
  3. Vertical shift up by $k = 1$.

(c) Domain and range, tracked through each transformation M1·A1·A1

Parent $f(x) = \sqrt{x}$ has $\mathrm{dom}\, f = [0, \infty)$ and $\mathrm{range}\, f = [0, \infty)$.
Horizontal shift right by $3$: domain becomes $[0 + 3, \infty) = [3, \infty)$; range unchanged.
Vertical stretch by $2$: domain unchanged $[3, \infty)$; range becomes $[2 \cdot 0, \infty) = [0, \infty)$.
Vertical shift up by $1$: domain unchanged $[3, \infty)$; range becomes $[0 + 1, \infty) = [1, \infty)$.
Final: $D = [3, \infty)$, $R = [1, \infty)$.

(d) Image points M1·A1

Coordinate rule: $(x, y) \to (x + 3, 2 y + 1)$.
  • $(0, 0) \to (3, 1)$.
  • $(1, 1) \to (4, 3)$.
  • $(4, 2) \to (7, 5)$.
Track domain and range through each transformation, do not derive them from the final equation. The temptation on $T(x) = 2 \sqrt{x - 3} + 1$ is to solve "$x - 3 \ge 0$" directly, which is fine for the domain but undersells the AP-feeder's HSF-IF.B.4 reading. The marker-rewarded approach is to follow the parent's $D$ and $R$ through each transformation: shifts add to the relevant interval, stretches scale it, reflections flip its sign. This makes the domain/range visible as "the parent's $[0, \infty)$ pushed right by $3$ and lifted by $1$", which is the same logic that classifies projectile-motion intervals in Unit 2 and that defines $\arcsin$'s restricted domain in Unit 8. The same tracking move generalises immediately to transformed exponentials, logs, and trig parents.
Q11MEDIUM 🇨🇦 ON ON Provincial-style §6 + §7 Composition + Inverse · MCR3U A1.4 / MHF4U C [9 marks]

$C(a) = 12 - 0.0065 a$ (°C at altitude $a$ m), $F(c) = \tfrac{9}{5} c + 32$. (a) $H(a) = (F \circ C)(a)$. (b) $H(2000)$. (c) $H^{-1}(F)$. (d) Altitude where $F = 32^{\circ}\mathrm{F}$.

Answer:  (a) $H(a) = 53.6 - 0.0117 a$  ·  (b) $H(2000) = 30.2^{\circ}\mathrm{F}$  ·  (c) $H^{-1}(F) = \tfrac{53.6 - F}{0.0117}$ m  ·  (d) $a \approx 1846$ m

(a) Build the composite $H = F \circ C$ M1·A1·A1

$$ H(a) \;=\; F(C(a)) \;=\; F(12 - 0.0065 a) \;=\; \tfrac{9}{5}(12 - 0.0065 a) + 32. $$ Expand: $\tfrac{9}{5} \cdot 12 = 21.6$ and $\tfrac{9}{5} \cdot 0.0065 = 0.0117$. So $$ H(a) \;=\; 21.6 - 0.0117 a + 32 \;=\; 53.6 - 0.0117 a \; (\text{degrees Fahrenheit}). $$

(b) Evaluate at $a = 2000$ m M1·A1

$$ H(2000) \;=\; 53.6 - 0.0117 \cdot 2000 \;=\; 53.6 - 23.4 \;=\; 30.2^{\circ}\mathrm{F}. $$

(c) Invert by swap-and-solve M1·A1·A1

Set $F = 53.6 - 0.0117 a$. Swap $a \leftrightarrow F$ (renaming output variable for clarity): $a = 53.6 - 0.0117 F$. Solve for $F$, which now denotes the input altitude... wait, let me re-set. Cleaner: from $F = 53.6 - 0.0117 a$, solve directly for $a$: $$ 0.0117 a \;=\; 53.6 - F \;\Longrightarrow\; a \;=\; \frac{53.6 - F}{0.0117}. $$ So $H^{-1}(F) = \dfrac{53.6 - F}{0.0117}$. In context: given a target Fahrenheit temperature $F$, $H^{-1}(F)$ outputs the altitude (in metres above base) at which the air sits at that temperature.

(d) Altitude where $F = 32^{\circ}\mathrm{F}$ A1

$$ H^{-1}(32) \;=\; \frac{53.6 - 32}{0.0117} \;=\; \frac{21.6}{0.0117} \;\approx\; 1846 \text{ m}. $$ Interpretation: the freezing point of water (in Fahrenheit terms) occurs about $1846$ m above the base of the resort.
Composition of linear maps is linear, and inversion swaps domain with range and slope with its reciprocal. Here both $C$ and $F$ are linear, so $H = F \circ C$ is linear, no surprise, but the slope is the product of the two component slopes ($\tfrac{9}{5} \times (-0.0065) = -0.0117$ per metre) and the intercept compounds through the inner. This is the engine behind unit conversion chains (metres $\to$ °C $\to$ °F) and is the prototype for the chain rule when both functions are non-linear. The inverse $H^{-1}$ inherits a reciprocal-style slope (in the linear case literally $-1 / 0.0117$ when written in $a$-as-function-of-$F$ form). MHF4U Strand C tests exactly this synthesis: build a real-world composition, evaluate forward, then invert to answer the reverse query.
Q12HARDHonors 🇨🇦 BC 🇺🇸 US BC Provincial-style §4 + §7 Synthesis on a Cubic · BC PC 12 / HSF-BF.B.4 [10 marks]

$T(x) = -2(x + 1)^{3} + 4$ from parent $f(x) = x^{3}$. (a) Master form. (b) Order. (c) Why $T$ stays one-to-one. (d) $T^{-1}(x)$ + domain/range. (e) Verify $T(T^{-1}(4))$.

Answer:  (a) $a = -2$, $b = 1$, $h = -1$, $k = 4$  ·  (d) $T^{-1}(x) = \sqrt[3]{\tfrac{4 - x}{2}} - 1$, $D = R = \mathbb{R}$  ·  (e) $T(T^{-1}(4)) = 4$

(a) Master-form parameters A1·A1

Rewrite $T(x) = -2(x - (-1))^{3} + 4 = -2 f(1 \cdot (x - (-1))) + 4$. So $a = -2$, $b = 1$, $h = -1$, $k = 4$.

(b) Order of transformations A1·A1

No horizontal stretch ($b = 1$). Apply:
  1. Horizontal shift left by $1$ (since $h = -1$);
  2. Vertical stretch by $|a| = 2$ combined with reflection over the $x$-axis (since $a = -2 < 0$);
  3. Vertical shift up by $k = 4$.

(c) Why $T$ remains one-to-one R1

The parent $f(x) = x^{3}$ is strictly monotonic (increasing) on $\mathbb{R}$, hence one-to-one. Each transformation in (b) preserves monotonicity up to sign: shifts do not change monotonicity, vertical stretches scale it (still monotonic), and the reflection $a < 0$ flips increasing to decreasing (still strictly monotonic). So $T$ is strictly decreasing on $\mathbb{R}$, hence one-to-one, and $T^{-1}$ exists on all of $\mathbb{R}$.

(d) Swap-and-solve for $T^{-1}$ M1·A1·A1·A1

Set $y = -2(x + 1)^{3} + 4$. Swap $x \leftrightarrow y$: $x = -2(y + 1)^{3} + 4$. Solve for $y$: $$ x - 4 \;=\; -2(y + 1)^{3} \;\Longrightarrow\; (y + 1)^{3} \;=\; \frac{4 - x}{2}. $$ Take the real cube root (defined for all reals): $$ y + 1 \;=\; \sqrt[3]{\tfrac{4 - x}{2}} \;\Longrightarrow\; y \;=\; \sqrt[3]{\tfrac{4 - x}{2}} - 1. $$ So $T^{-1}(x) = \sqrt[3]{\tfrac{4 - x}{2}} - 1$. Because $T$ is defined on all of $\mathbb{R}$ and its range is all of $\mathbb{R}$ (cube spans $\mathbb{R}$, the master form's transformations preserve that), $T^{-1}$ also has $D = R = \mathbb{R}$.

(e) Verification A1

$$ T^{-1}(4) \;=\; \sqrt[3]{\tfrac{4 - 4}{2}} - 1 \;=\; \sqrt[3]{0} - 1 \;=\; -1, $$ and $$ T(-1) \;=\; -2(-1 + 1)^{3} + 4 \;=\; -2(0) + 4 \;=\; 4. \;\checkmark $$ So $T(T^{-1}(4)) = 4$, confirming the inversion.
Odd parent + monotonicity-preserving transformations = guaranteed inverse. The cubic parent is the easy case for inverses precisely because $f(x) = x^{3}$ is one-to-one everywhere, unlike $f(x) = x^{2}$, which requires a domain restriction (recall Q8(c)). Any composition of shifts, non-zero stretches, and reflections applied to a one-to-one function stays one-to-one, the transformation toolkit cannot create or destroy monotonicity by itself. This is exactly why §7 invertibility questions on cubics and odd-degree polynomials never require a restriction, but quadratic and absolute-value parents always do. BC PC 12 and HSF-BF.B.4 reward students who explicitly cite the monotonicity-preservation argument before reaching for the swap-and-solve. The verification step in (e) is the audit move that catches sign errors in the swap, AP graders look for it.