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Quadratic Functions and Equations · Solutions二次函数与方程 · 参考答案

Companion to the Practice Set · Mark-by-mark walkthroughs · SAT / AP-Feeder / ON / BC styles配套练习集答案 · 逐分讲解 · SAT / AP 衔接 / 安大略 / 卑诗风格

EASY易 MEDIUM中 HARD难 🇺🇸 US美 🇨🇦 ON安 🇨🇦 BC卑 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 · 参考答案SAT MCQ + ON/BC short answer · 22 marksSAT 选择题 + 安/卑省考短答 · 共 22 分

Section A · Worked SolutionsA 部分 · 详解

Q1EASY易 🇺🇸 US美 SAT-style MCQSAT 风格选择题 §1 Parabola Key Features抛物线关键特征 · HSF-IF.C.7a [3 marks][3 分]

Vertex of $y = (x - 3)^{2} - 4$.$y = (x - 3)^{2} - 4$ 的顶点。

Answer:答案：  (B)  $(3, -4)$

(a) Read the vertex off vertex form直接从顶点式读出顶点 M1·A1·A1

The equation is already in vertex form $y = a(x - h)^{2} + k$ with $a = 1$, $h = 3$, $k = -4$. The vertex is $(h, k) = (3, -4)$, which matches option (B).该方程已是顶点式 $y = a(x - h)^{2} + k$，其中 $a = 1$、$h = 3$、$k = -4$。顶点为 $(h, k) = (3, -4)$，与选项 (B) 一致。
Why the wrong choices fail.为何其他选项不对。
  • (A) $(-3, -4)$, sign-flip error: the student reads the $-3$ inside $(x - 3)$ as the $x$-coordinate of the vertex, instead of recognising that vertex form encodes $h$ as the value that zeroes the bracket.：符号错误。学生把 $(x - 3)$ 中的 $-3$ 当作顶点的 $x$ 坐标，而没有意识到顶点式中 $h$ 应取使括号为零的值。
  • (C) $(3, 4)$, correct $x$-coordinate but copies the magnitude $4$ without the sign; vertex form's $+k$ outside the bracket carries its sign directly.：$x$ 坐标正确，但抄写 $4$ 时漏掉了符号；顶点式中括号外的 $+k$ 直接携带其符号。
  • (D) $(-3, 4)$, both signs flipped at once; the "double negative" error from reading $(x - h)^{2} + k$ as if every sign were reversed.：两个符号都翻转了；"双重负号"错误，把 $(x - h)^{2} + k$ 中每个符号都颠倒着读。
The vertex is the $(h, k)$ that makes the squared bracket zero.顶点就是使平方括号为零的 $(h, k)$。 The fast diagnostic: ask "what value of $x$ makes $(x - 3)^{2}$ equal to zero?" The answer is $x = 3$, so $h = 3$, the sign in the bracket flips when you solve for $h$. The $y$-coordinate $k = -4$ sits outside the bracket and is read directly. This $(h, k)$-from-bracket move generalises to every transformed-graph question (Unit 2 §2.5 and downstream Unit 5 transformations), so lock it in.快速判断法：问"什么 $x$ 值使 $(x - 3)^{2}$ 为零？"答案是 $x = 3$，因此 $h = 3$——求解 $h$ 时括号内符号要翻转。$y$ 坐标 $k = -4$ 位于括号外，可直接读取。这种"从括号读 $(h, k)$"的方法适用于所有图像变换题（第 2 单元 §2.5 与后续第 5 单元的变换），务必掌握。
Q2EASY易 🇺🇸 US美 SAT-style MCQSAT 风格选择题 §3 Factoring (a = 1)因式分解（a = 1） · HSA-SSE.B.3a [3 marks][3 分]

Factor $x^{2} - 5x - 14$.将 $x^{2} - 5x - 14$ 因式分解。

Answer:答案：  (A)  $(x - 7)(x + 2)$

(a) Find two integers with product $-14$ and sum $-5$找两个整数，积为 $-14$、和为 $-5$ M1·A1

With $a = 1$, factoring $x^{2} + bx + c$ reduces to finding integers $p, q$ with $p \cdot q = c$ and $p + q = b$. Here $c = -14$, $b = -5$. Candidate pairs for product $-14$: $(1, -14), (-1, 14), (2, -7), (-2, 7)$. Sums: $-13, 13, -5, 5$. The pair $(2, -7)$ gives sum $-5$. $\checkmark$当 $a = 1$ 时，对 $x^{2} + bx + c$ 因式分解就是要找整数 $p, q$，使 $p \cdot q = c$ 且 $p + q = b$。此处 $c = -14$、$b = -5$。乘积为 $-14$ 的候选对：$(1, -14), (-1, 14), (2, -7), (-2, 7)$，对应的和为 $-13, 13, -5, 5$。$(2, -7)$ 给出和 $-5$。$\checkmark$

(b) Write the factorisation and check写出因式分解并验证 A1

So $x^{2} - 5x - 14 = (x + 2)(x - 7)$, which matches option (A). Verify by expansion: $(x - 7)(x + 2) = x^{2} + 2x - 7x - 14 = x^{2} - 5x - 14$. $\checkmark$因此 $x^{2} - 5x - 14 = (x + 2)(x - 7)$，与选项 (A) 一致。展开验证：$(x - 7)(x + 2) = x^{2} + 2x - 7x - 14 = x^{2} - 5x - 14$。$\checkmark$
Why the wrong choices fail.为何其他选项不对。
  • (B) $(x - 7)(x - 2)$, both factors negative gives product $+14$, not $-14$. Expansion check: $x^{2} - 9x + 14$.：两个因式都为负，乘积为 $+14$，而不是 $-14$。展开核对：$x^{2} - 9x + 14$。
  • (C) $(x + 7)(x - 2)$, correct magnitudes but sign-swapped; expansion gives $x^{2} + 5x - 14$ (wrong middle-term sign).：数值正确但符号互换；展开得 $x^{2} + 5x - 14$（中间项符号错）。
  • (D) $(x - 14)(x + 1)$, uses the $1 \cdot 14$ pair instead of $2 \cdot 7$; expansion gives $x^{2} - 13x - 14$.：使用了 $1 \cdot 14$ 这对而不是 $2 \cdot 7$；展开得 $x^{2} - 13x - 14$。
The sum tells you the signs; the product tells you the magnitudes.和决定符号，积决定数值。 For $x^{2} + bx + c$ with $c < 0$, the two factors must have opposite signs, one positive, one negative. With $c > 0$, both factors share the sign of $b$. Lock this sign rule down before listing pairs and you eliminate half the distractors instantly. Common trap: the SAT loves seeding the answer set with the same magnitudes in every wrong combination of signs.对 $x^{2} + bx + c$，若 $c < 0$，则两个因式必须异号（一正一负）；若 $c > 0$，两个因式都与 $b$ 同号。先记住这条符号规则再列因式对，能立即排除一半干扰项。常见陷阱：SAT 喜欢在答案集中把相同数值、不同符号的所有错误组合都摆出来。
Q3MEDIUM中 🇺🇸 US美 SAT-style MCQSAT 风格选择题 §4 Discriminant判别式 · HSA-REI.B.4 [3 marks][3 分]

Number of real solutions of $2x^{2} + 3x + 5 = 0$.$2x^{2} + 3x + 5 = 0$ 的实数解个数。

Answer:答案：  (C)  No real solutions无实数解

(a) Compute the discriminant计算判别式 M1·A1

With $a = 2$, $b = 3$, $c = 5$:代入 $a = 2$、$b = 3$、$c = 5$： $$ \Delta \;=\; b^{2} - 4ac \;=\; 3^{2} - 4(2)(5) \;=\; 9 - 40 \;=\; -31. $$

(b) Classify by sign of $\Delta$由 $\Delta$ 的符号判别 A1

Since $\Delta = -31 < 0$, the equation has no real solutions (the two roots are a complex-conjugate pair). This matches option (C).由于 $\Delta = -31 < 0$，方程无实数解（两个根为共轭复数对）。这与选项 (C) 一致。
Why the wrong choices fail.为何其他选项不对。
  • (A) Two distinct real solutions两个不同实数解, requires $\Delta > 0$; this would be the answer if $b^{2}$ exceeded $4ac$, but $9 < 40$.：要求 $\Delta > 0$；若 $b^{2}$ 大于 $4ac$ 才是该答案，但 $9 < 40$。
  • (B) One real solution (double root)一个实数解（重根）, requires $\Delta = 0$ exactly; here $\Delta = -31 \ne 0$.：要求 $\Delta = 0$；此处 $\Delta = -31 \ne 0$。
  • (D) Infinitely many real solutions无穷多实数解, impossible for a non-trivial polynomial equation; a quadratic equation has at most two roots over $\mathbb{C}$ by the fundamental theorem of algebra.：对于非平凡多项式方程不可能；由代数基本定理，二次方程在 $\mathbb{C}$ 上至多两个根。
$\sqrt{\text{negative}}$ is not real, do not "square-root through" a negative discriminant.负数的平方根不是实数；不要对负判别式"硬开平方"。 A common slip is to write $x = \dfrac{-3 \pm \sqrt{-31}}{4}$ and then claim $\sqrt{-31} = \pm \sqrt{31}$ (it is not, both signs of $\sqrt{31}$ are real, neither is $\sqrt{-31}$). When $\Delta < 0$, the quadratic formula produces $\pm i\sqrt{|\Delta|}/(2a)$, complex roots, not real ones. The SAT rewards students who answer "no real solutions" decisively after a five-second discriminant computation, rather than wading through the formula.常见错误是写 $x = \dfrac{-3 \pm \sqrt{-31}}{4}$ 然后声称 $\sqrt{-31} = \pm \sqrt{31}$（这是错的——$\sqrt{31}$ 的正负值都是实数，而 $\sqrt{-31}$ 都不是实数）。当 $\Delta < 0$ 时，求根公式给出 $\pm i\sqrt{|\Delta|}/(2a)$，是复根而非实根。SAT 奖励那些在五秒判别式计算后果断回答"无实数解"的学生，而不是硬套求根公式。
Q4MEDIUM中 🇨🇦 ON安 ON Provincial-style安大略省考风格 §2 Three Forms三种形式 · MPM2D Quadratic Relations二次关系 [6 marks][6 分]

$y = x^{2} - 8x + 11$. (a) Vertex form + vertex. (b) AOS + opening. (c) $y$-intercept.$y = x^{2} - 8x + 11$。(a) 顶点形式 + 顶点。(b) 对称轴 + 开口方向。(c) $y$ 轴截距。

Answer:答案：  (a) $y = (x - 4)^{2} - 5$, vertex $(4, -5)$，顶点 $(4, -5)$  ·  (b) $x = 4$, opens upward，开口向上  ·  (c) $(0, 11)$

(a) Complete the square配方法 M1·A1·A1

Take $x^{2} - 8x + 11$. Half of the $x$-coefficient is $-4$; its square is $16$. Add-and-subtract:取 $x^{2} - 8x + 11$。$x$ 系数的一半为 $-4$，其平方为 $16$。加一减一： $$ y \;=\; (x^{2} - 8x + 16) - 16 + 11 \;=\; (x - 4)^{2} - 5. $$ Vertex form: $y = (x - 4)^{2} - 5$, vertex $(h, k) = (4, -5)$.顶点形式：$y = (x - 4)^{2} - 5$，顶点 $(h, k) = (4, -5)$。

(b) Axis of symmetry and direction of opening对称轴与开口方向 A1·A1

Axis of symmetry: the vertical line through the vertex, $x = 4$.对称轴：过顶点的竖直线 $x = 4$。
Direction of opening: the leading coefficient $a = 1 > 0$, so the parabola opens upward, the vertex is a minimum.开口方向：首项系数 $a = 1 > 0$，所以抛物线开口向上，顶点为最小值点。

(c) $y$-intercept$y$ 轴截距 A1

Set $x = 0$ in the standard form: $y = 0 - 0 + 11 = 11$. The $y$-intercept is $(0, 11)$.在标准式中令 $x = 0$：$y = 0 - 0 + 11 = 11$。$y$ 轴截距为 $(0, 11)$。
Read three forms, three things; do not re-derive what is already visible.三种形式各读三件事；已显而易见的别再重推。 Standard form $y = ax^{2} + bx + c$ gives the $y$-intercept $(0, c)$ for free. Vertex form gives the vertex, the axis of symmetry, and the max/min for free. Factored form (when it exists) gives the $x$-intercepts for free. Provincial markers split marks across conversion (the M1) and reading (the A1s); students who only ever convert to one form and re-derive the rest leave easy A1s on the table. Train the eye to ask "which form gives me this feature without computation?" before reaching for the algebra.标准式 $y = ax^{2} + bx + c$ 免费给出 $y$ 轴截距 $(0, c)$。顶点式免费给出顶点、对称轴及最大/最小值。因式分解式（存在时）免费给出 $x$ 轴截距。省考阅卷在转换（M1 分）与读取（A1 分）之间分配分数；只会转换到一种形式再重推其余特征的学生白白丢分。养成习惯：动手算之前先问"哪种形式不用计算就能给我这个特征？"
Q5MEDIUM中 🇨🇦 BC卑 BC Provincial-style卑诗省考风格 §5 Completing the Square配方法 · BC PC11 quadratic functions二次函数 [7 marks][7 分]

$f(x) = 2x^{2} + 12x + 7$. (a) Vertex form by CTS. (b) Vertex, AOS, min. (c) Domain and range in interval notation.$f(x) = 2x^{2} + 12x + 7$。(a) 配方法化为顶点式。(b) 顶点、对称轴、最小值。(c) 用区间记法写出定义域与值域。

Answer:答案：  (a) $f(x) = 2(x + 3)^{2} - 11$  ·  (b) vertex $(-3, -11)$, AOS $x = -3$, min $-11$顶点 $(-3, -11)$，对称轴 $x = -3$，最小值 $-11$  ·  (c) $D = (-\infty, \infty)$, $R = [-11, \infty)$

(a) Complete the square with $a \ne 1$$a \ne 1$ 时的配方法 M1·A1·A1

Factor the leading coefficient out of the first two terms only:仅从前两项中提出首项系数： $$ f(x) \;=\; 2(x^{2} + 6x) + 7. $$ Half of $6$ is $3$; its square is $9$. Add-and-subtract $9$ inside the bracket:$6$ 的一半是 $3$，其平方为 $9$。在括号内加一减一 $9$： $$ f(x) \;=\; 2\bigl((x^{2} + 6x + 9) - 9\bigr) + 7 \;=\; 2(x + 3)^{2} - 18 + 7 \;=\; 2(x + 3)^{2} - 11. $$

(b) Vertex, axis of symmetry, minimum顶点、对称轴、最小值 A1·A1

From $f(x) = 2(x - (-3))^{2} + (-11)$: vertex $(-3, -11)$, axis of symmetry $x = -3$. Since $a = 2 > 0$ the parabola opens upward, so the vertex is a minimum: $f_{\min} = -11$ attained at $x = -3$.由 $f(x) = 2(x - (-3))^{2} + (-11)$：顶点 $(-3, -11)$，对称轴 $x = -3$。由 $a = 2 > 0$ 知抛物线开口向上，故顶点为最小值点：$f_{\min} = -11$，在 $x = -3$ 处取得。

(c) Domain and range定义域与值域 A1·A1

Every polynomial is defined for all real $x$: $D = (-\infty, \infty)$.任何多项式对所有实数 $x$ 都有定义：$D = (-\infty, \infty)$。
Since the minimum value is $-11$ and the parabola opens upward, the range is $R = [-11, \infty)$.由于最小值为 $-11$ 且抛物线开口向上，值域为 $R = [-11, \infty)$。
The "$- 9$ inside the bracket" is really "$- 18$ outside", track the multiplier."括号内的 $-9$" 实际上等同"括号外的 $-18$"——别忘了倍乘因子。 The most common BC PC11 sign error: writing $f(x) = 2(x + 3)^{2} - 9 + 7$ at the end, forgetting that the $-9$ sits inside a bracket multiplied by $2$. When you pull the $-9$ out, it becomes $-2 \cdot 9 = -18$, not $-9$. The audit move: re-expand your vertex form back to standard and confirm. Here $2(x + 3)^{2} - 11 = 2(x^{2} + 6x + 9) - 11 = 2x^{2} + 12x + 18 - 11 = 2x^{2} + 12x + 7$. $\checkmark$ The minute it takes to re-expand catches every sign slip.BC PC11 最常见的符号错误：最后写成 $f(x) = 2(x + 3)^{2} - 9 + 7$，忘记 $-9$ 位于乘 $2$ 的括号内。将 $-9$ 提出后变为 $-2 \cdot 9 = -18$，而非 $-9$。审查方法：把顶点式再展开回标准式核对。这里 $2(x + 3)^{2} - 11 = 2(x^{2} + 6x + 9) - 11 = 2x^{2} + 12x + 18 - 11 = 2x^{2} + 12x + 7$。$\checkmark$ 多花一分钟展开能抓住每个符号错。
PART II  ·  EXTENDED RESPONSE · SOLUTIONS第二部分  ·  简答题 · 参考答案AP-feeder FRQ + honors · 36 marksAP 衔接简答题 + 荣誉级 · 共 36 分

Section B · Worked SolutionsB 部分 · 详解

Q6MEDIUM中 🇺🇸 US美 AP-feeder FRQAP 衔接简答题 §2 Forms & Conversion三种形式与转换 · HSF-IF.C.8a [8 marks][8 分]

$g(x) = -2x^{2} + 8x + 10$. (a) Vertex form + max/min. (b) Factored form + $x$-intercepts. (c) $y$-intercept + AOS.$g(x) = -2x^{2} + 8x + 10$。(a) 顶点形式 + 最大/最小值。(b) 因式分解形式 + $x$ 轴截距。(c) $y$ 轴截距 + 对称轴。

Answer:答案：  (a) $g(x) = -2(x - 2)^{2} + 18$, vertex $(2, 18)$, maximum，顶点 $(2, 18)$，最大值  ·  (b) $g(x) = -2(x - 5)(x + 1)$, $x$-ints $x = 5, -1$，$x$ 轴截距 $x = 5, -1$  ·  (c) $y$-int $(0, 10)$, AOS $x = 2$$y$ 轴截距 $(0, 10)$，对称轴 $x = 2$

(a) Vertex form by completing the square用配方法化为顶点式 M1·A1·A1

Factor $-2$ from the first two terms only:仅从前两项中提出 $-2$： $$ g(x) \;=\; -2(x^{2} - 4x) + 10. $$ Half of $-4$ is $-2$; its square is $4$. Add-and-subtract $4$ inside the bracket:$-4$ 的一半是 $-2$，其平方为 $4$。在括号内加一减一 $4$： $$ g(x) \;=\; -2\bigl((x^{2} - 4x + 4) - 4\bigr) + 10 \;=\; -2(x - 2)^{2} + 8 + 10 \;=\; -2(x - 2)^{2} + 18. $$ Vertex $(2, 18)$. Since $a = -2 < 0$, the parabola opens downward, the vertex is a maximum.顶点 $(2, 18)$。由于 $a = -2 < 0$，抛物线开口向下，顶点为最大值点。

(b) Factored form by factoring out $-2$提出 $-2$ 化为因式分解形式 M1·A1·A1

$$ g(x) \;=\; -2x^{2} + 8x + 10 \;=\; -2(x^{2} - 4x - 5). $$ Factor $x^{2} - 4x - 5$: integers with product $-5$ and sum $-4$ are $-5$ and $1$, so $x^{2} - 4x - 5 = (x - 5)(x + 1)$. Hence对 $x^{2} - 4x - 5$ 因式分解：乘积为 $-5$、和为 $-4$ 的整数对为 $-5$ 与 $1$，所以 $x^{2} - 4x - 5 = (x - 5)(x + 1)$。因此 $$ g(x) \;=\; -2(x - 5)(x + 1). $$ The $x$-intercepts are the roots: $x = 5$ and $x = -1$, i.e. the points $(5, 0)$ and $(-1, 0)$.$x$ 轴截距即为根：$x = 5$ 与 $x = -1$，即点 $(5, 0)$ 与 $(-1, 0)$。

(c) $y$-intercept and axis of symmetry$y$ 轴截距与对称轴 A1·A1

$y$-intercept: $g(0) = -2(0) + 8(0) + 10 = 10$, so $(0, 10)$.$y$ 轴截距：$g(0) = -2(0) + 8(0) + 10 = 10$，故为 $(0, 10)$。
Axis of symmetry: the vertical line through the vertex, $x = 2$. (Cross-check using the midpoint of the $x$-intercepts: $\tfrac{5 + (-1)}{2} = 2$. $\checkmark$)对称轴：过顶点的竖直线 $x = 2$。（用 $x$ 轴截距的中点交叉验证：$\tfrac{5 + (-1)}{2} = 2$。$\checkmark$）
Three forms, three guaranteed cross-checks.三种形式 = 三条必然的交叉验证。 Once you have all three forms of the same parabola, every key feature is over-determined and you can audit your work in one line: vertex form's $h$ must equal the midpoint of factored form's roots; vertex form's $k$ must equal $g(h)$ computed from standard form; standard form's constant must equal $g(0)$ from any form. Here $h = 2 = \tfrac{5 + (-1)}{2}$, $k = 18 = -2(0)^{2} + 8(0) + 10 + \ldots$ wait, $g(2) = -2(4) + 16 + 10 = -8 + 16 + 10 = 18$. $\checkmark$ AP graders explicitly look for this cross-form consistency on HSF-IF.C.8a.一旦同一抛物线的三种形式都写出，每个关键特征都被冗余确定，可以一行就审查全部工作：顶点式的 $h$ 必须等于因式分解形式中根的中点；顶点式的 $k$ 必须等于由标准式计算的 $g(h)$；标准式的常数必须等于任意形式计算的 $g(0)$。此处 $h = 2 = \tfrac{5 + (-1)}{2}$，$g(2) = -2(4) + 16 + 10 = -8 + 16 + 10 = 18$。$\checkmark$ AP 阅卷在 HSF-IF.C.8a 上明确审查这种跨形式一致性。
Q7HARD难 🇨🇦 BC卑 BC Provincial-style卑诗省考风格 §3 Factoring (a ≠ 1)因式分解（a ≠ 1） · BC PC11 polynomial factoring多项式因式分解 [8 marks][8 分]

Solve by factoring. (a) $6x^{2} - 11x - 10 = 0$ (AC-method). (b) $4x^{2} - 25 = 0$ (DoS). (c) $9x^{2} - 12x + 4 = 0$ (perfect square).用因式分解法求解。(a) $6x^{2} - 11x - 10 = 0$（AC 法）。(b) $4x^{2} - 25 = 0$（平方差）。(c) $9x^{2} - 12x + 4 = 0$（完全平方）。

Answer:答案：  (a) $x = -\tfrac{2}{3}, \tfrac{5}{2}$  ·  (b) $x = \pm \tfrac{5}{2}$  ·  (c) $x = \tfrac{2}{3}$ (double root)（重根）

(a) AC-method on $6x^{2} - 11x - 10 = 0$对 $6x^{2} - 11x - 10 = 0$ 用 AC 法 M1·A1·A1

Compute $AC = 6 \cdot (-10) = -60$. Find integers with product $-60$ and sum $-11$. Candidates: $(-15, 4)$ gives sum $-11$. $\checkmark$计算 $AC = 6 \cdot (-10) = -60$。找乘积为 $-60$、和为 $-11$ 的整数对。候选：$(-15, 4)$ 给出和 $-11$。$\checkmark$
Split the middle term and group:拆分中间项并分组： $$ 6x^{2} - 15x + 4x - 10 \;=\; 3x(2x - 5) + 2(2x - 5) \;=\; (3x + 2)(2x - 5). $$ Setting each factor to zero: $3x + 2 = 0 \Rightarrow x = -\tfrac{2}{3}$; $2x - 5 = 0 \Rightarrow x = \tfrac{5}{2}$. Solution set $\bigl\{-\tfrac{2}{3}, \tfrac{5}{2}\bigr\}$.令每个因式为零：$3x + 2 = 0 \Rightarrow x = -\tfrac{2}{3}$；$2x - 5 = 0 \Rightarrow x = \tfrac{5}{2}$。解集 $\bigl\{-\tfrac{2}{3}, \tfrac{5}{2}\bigr\}$。

(b) Difference-of-squares pattern on $4x^{2} - 25 = 0$对 $4x^{2} - 25 = 0$ 应用平方差 M1·A1

Recognise $4x^{2} = (2x)^{2}$ and $25 = 5^{2}$, so this fits $A^{2} - B^{2} = (A - B)(A + B)$ with $A = 2x$, $B = 5$:识别 $4x^{2} = (2x)^{2}$ 与 $25 = 5^{2}$，故符合 $A^{2} - B^{2} = (A - B)(A + B)$，其中 $A = 2x$、$B = 5$： $$ 4x^{2} - 25 \;=\; (2x - 5)(2x + 5) \;=\; 0 \;\Longrightarrow\; x \;=\; \pm \tfrac{5}{2}. $$

(c) Perfect-square trinomial on $9x^{2} - 12x + 4 = 0$对 $9x^{2} - 12x + 4 = 0$ 应用完全平方三项式 M1·A1·A1

Test the perfect-square pattern $A^{2} - 2AB + B^{2} = (A - B)^{2}$ with $A^{2} = 9x^{2} \Rightarrow A = 3x$ and $B^{2} = 4 \Rightarrow B = 2$. Check the cross term: $-2AB = -2(3x)(2) = -12x$. $\checkmark$ So检验完全平方形式 $A^{2} - 2AB + B^{2} = (A - B)^{2}$，取 $A^{2} = 9x^{2} \Rightarrow A = 3x$、$B^{2} = 4 \Rightarrow B = 2$。核对交叉项：$-2AB = -2(3x)(2) = -12x$。$\checkmark$ 故 $$ 9x^{2} - 12x + 4 \;=\; (3x - 2)^{2} \;=\; 0 \;\Longrightarrow\; 3x - 2 \;=\; 0 \;\Longrightarrow\; x \;=\; \tfrac{2}{3}. $$ This is a double root (multiplicity 2), confirmed by the discriminant $\Delta = (-12)^{2} - 4(9)(4) = 144 - 144 = 0$.这是重根（重数 2），由判别式 $\Delta = (-12)^{2} - 4(9)(4) = 144 - 144 = 0$ 确认。
Spot the pattern before reaching for AC.动用 AC 法之前，先识别结构形式。 Difference of squares ($b = 0$ with $a, c$ both perfect squares of opposite sign) and perfect-square trinomial ($b^{2} = 4ac$ exactly) are the two patterns to scan for before setting up AC-method tables. Parts (b) and (c) are each one line of pattern-matching; running AC-method on them is six lines of wasted work. The structural diagnostic for a perfect square: check $b^{2} - 4ac$, if it's zero, you have $(A \pm B)^{2}$, where $A = \sqrt{a}\,x$ and $B = \sqrt{c}$, and the sign in the bracket matches the sign of $b$.平方差（$b = 0$ 且 $a, c$ 为符号相反的完全平方）与完全平方三项式（$b^{2} = 4ac$ 恰好成立）是 AC 法表格之前就该先扫描的两种形式。(b) 和 (c) 各只需一行形式识别；用 AC 法做要白写六行。完全平方的结构判断：检验 $b^{2} - 4ac$——若为零，即为 $(A \pm B)^{2}$，其中 $A = \sqrt{a}\,x$、$B = \sqrt{c}$，括号内符号与 $b$ 同号。
Q8MEDIUM中 🇨🇦 ON安 ON Provincial-style安大略省考风格 §4 Quadratic Formula求根公式 · MCR3U A2 [9 marks][9 分]

$3x^{2} - 7x - 2 = 0$. (a) $\Delta$. (b) Classify roots. (c) Solve via formula, exact radical. (d) $f(0)$ and graphical meaning.$3x^{2} - 7x - 2 = 0$。(a) $\Delta$。(b) 判别根的类型。(c) 用求根公式求解，给出精确根式。(d) $f(0)$ 及其图像意义。

Answer:答案：  (a) $\Delta = 73$  ·  (b) two distinct real roots两个不同实根  ·  (c) $x = \dfrac{7 \pm \sqrt{73}}{6}$  ·  (d) $f(0) = -2$, the $y$-intercept$f(0) = -2$，即 $y$ 轴截距

(a) Discriminant判别式 M1·A1

With $a = 3$, $b = -7$, $c = -2$:代入 $a = 3$、$b = -7$、$c = -2$： $$ \Delta \;=\; b^{2} - 4ac \;=\; (-7)^{2} - 4(3)(-2) \;=\; 49 + 24 \;=\; 73. $$

(b) Classify the roots判别根的类型 R1

$\Delta = 73 > 0$ and $73$ is not a perfect square, so the equation has two distinct real (irrational) roots.$\Delta = 73 > 0$ 且 $73$ 不是完全平方数，所以方程有两个不同的实数（无理）根。

(c) Solve via the quadratic formula用求根公式求解 M1·A1·A1·A1

$$ x \;=\; \frac{-b \pm \sqrt{\Delta}}{2a} \;=\; \frac{-(-7) \pm \sqrt{73}}{2(3)} \;=\; \frac{7 \pm \sqrt{73}}{6}. $$ Both roots: $x_{1} = \dfrac{7 + \sqrt{73}}{6}$ and $x_{2} = \dfrac{7 - \sqrt{73}}{6}$. $\sqrt{73}$ does not simplify ($73$ is prime), so this is the final exact form.两个根：$x_{1} = \dfrac{7 + \sqrt{73}}{6}$ 和 $x_{2} = \dfrac{7 - \sqrt{73}}{6}$。$\sqrt{73}$ 无法化简（$73$ 为素数），故此为最终精确形式。

(d) $f(0)$ and its meaning$f(0)$ 及其意义 A1·A1

$f(0) = 3(0)^{2} - 7(0) - 2 = -2$. Geometrically, $f(0)$ is the value of $f$ at $x = 0$, i.e. the height where the parabola crosses the $y$-axis. So the graph of $y = f(x)$ has $y$-intercept $(0, -2)$.$f(0) = 3(0)^{2} - 7(0) - 2 = -2$。从几何角度看，$f(0)$ 是 $f$ 在 $x = 0$ 处的值，即抛物线与 $y$ 轴相交的高度。所以 $y = f(x)$ 的图像有 $y$ 轴截距 $(0, -2)$。
Simplify the radical before stopping.收笔前请化简根式。 Provincial markers deduct an A1 for "unsimplified" radicals, answers like $\dfrac{7 \pm \sqrt{292}}{12}$ (the result of forgetting to reduce $\sqrt{4 \cdot 73}$ to $2\sqrt{73}$ and dividing through). Habit move: once the discriminant is computed, factor it into a perfect square times a square-free part: $\Delta = k^{2} \cdot m$ where $m$ is square-free, then $\sqrt{\Delta} = k\sqrt{m}$. Here $73$ is prime so $k = 1$. Always state explicitly "$\sqrt{73}$ is in lowest terms because $73$ is prime", markers want to see the awareness.省考阅卷会扣除"未化简"根式的 A1 分，比如答案写成 $\dfrac{7 \pm \sqrt{292}}{12}$（因为忘了把 $\sqrt{4 \cdot 73}$ 化为 $2\sqrt{73}$ 并约分）。养成习惯：算完判别式后，将其分解为完全平方乘以无平方因子部分：$\Delta = k^{2} \cdot m$，其中 $m$ 无平方因子，则 $\sqrt{\Delta} = k\sqrt{m}$。此处 $73$ 为素数所以 $k = 1$。一定要明确写"$\sqrt{73}$ 已是最简形式，因为 $73$ 为素数"——阅卷想看到这种自觉。
Q9HARD难Honors荣誉级 🇺🇸 US美 🇨🇦 BC卑 AP-feeder FRQAP 衔接简答题 §4 Complex Roots复根 · HSN-CN.A.7 (+) / BC PC11 [11 marks][11 分]

Family $x^{2} - 4x + c = 0$. (a) $\Delta(c)$. (b) $c$ for double root + the root. (c) Ranges of $c$ for two real, two complex. (d) $c = 13$: complex solutions, conjugate check. (e) Geometric meaning of no real roots.方程族 $x^{2} - 4x + c = 0$。(a) $\Delta(c)$。(b) 重根时的 $c$ 及其根。(c) 两实根、两复根时 $c$ 的取值范围。(d) $c = 13$：复数解与共轭验证。(e) 无实根的几何意义。

Answer:答案：  (a) $\Delta(c) = 16 - 4c$  ·  (b) $c = 4$, double root $x = 2$，重根 $x = 2$  ·  (c) two real: $c < 4$; two complex: $c > 4$两实根：$c < 4$；两复根：$c > 4$  ·  (d) $x = 2 \pm 3i$  ·  (e) graph lies entirely above $x$-axis图像完全位于 $x$ 轴上方

(a) Discriminant as a function of $c$将判别式表示为 $c$ 的函数 A1

With $a = 1$, $b = -4$, constant $c$:取 $a = 1$、$b = -4$、常数项 $c$： $$ \Delta(c) \;=\; (-4)^{2} - 4(1)(c) \;=\; 16 - 4c. $$

(b) Double-root condition重根条件 M1·A1

One real solution $\Leftrightarrow \Delta(c) = 0 \Leftrightarrow 16 - 4c = 0 \Leftrightarrow c = 4$. The root is $x = \dfrac{-b}{2a} = \dfrac{4}{2} = 2$. Sanity-check: $x^{2} - 4x + 4 = (x - 2)^{2}$. $\checkmark$一个实数解 $\Leftrightarrow \Delta(c) = 0 \Leftrightarrow 16 - 4c = 0 \Leftrightarrow c = 4$。根为 $x = \dfrac{-b}{2a} = \dfrac{4}{2} = 2$。核验：$x^{2} - 4x + 4 = (x - 2)^{2}$。$\checkmark$

(c) Discriminant ranges判别式的取值范围 M1·A1·A1

Two distinct real solutions $\Leftrightarrow \Delta(c) > 0 \Leftrightarrow 16 - 4c > 0 \Leftrightarrow c < 4$.两个不同实数解 $\Leftrightarrow \Delta(c) > 0 \Leftrightarrow 16 - 4c > 0 \Leftrightarrow c < 4$。
Two non-real complex solutions (conjugate pair) $\Leftrightarrow \Delta(c) < 0 \Leftrightarrow 16 - 4c < 0 \Leftrightarrow c > 4$.两个非实复数解（共轭对） $\Leftrightarrow \Delta(c) < 0 \Leftrightarrow 16 - 4c < 0 \Leftrightarrow c > 4$。
(Together with part (b), the three ranges $c < 4$, $c = 4$, $c > 4$ partition $\mathbb{R}$.)（连同 (b)，三段范围 $c < 4$、$c = 4$、$c > 4$ 划分了 $\mathbb{R}$。）

(d) Solve when $c = 13$; verify conjugacy$c = 13$ 时求解；验证共轭性 M1·A1·A1·R1

$\Delta(13) = 16 - 52 = -36$. Then$\Delta(13) = 16 - 52 = -36$。于是 $$ x \;=\; \frac{4 \pm \sqrt{-36}}{2} \;=\; \frac{4 \pm 6i}{2} \;=\; 2 \pm 3i. $$ So $x_{1} = 2 + 3i$ and $x_{2} = 2 - 3i$. Conjugacy check: the conjugate of $a + bi$ is $a - bi$, so $\overline{x_{1}} = \overline{2 + 3i} = 2 - 3i = x_{2}$. $\checkmark$ This is exactly the statement of HSN-CN.A.7: a real-coefficient polynomial whose roots include $a + bi$ must also have $a - bi$ as a root, because the conjugate of zero is zero.所以 $x_{1} = 2 + 3i$ 与 $x_{2} = 2 - 3i$。共轭性核验：$a + bi$ 的共轭为 $a - bi$，故 $\overline{x_{1}} = \overline{2 + 3i} = 2 - 3i = x_{2}$。$\checkmark$ 这正是 HSN-CN.A.7 的陈述：实系数多项式若以 $a + bi$ 为根，则 $a - bi$ 也必然是根，因为零的共轭为零。

(e) Geometric interpretation几何解释 A1

No real roots means the parabola $y = x^{2} - 4x + 13$ never meets the $x$-axis. Since $a = 1 > 0$ the parabola opens upward, and its minimum value $y = 13 - 4 = 9 > 0$ confirms the graph lies entirely above the $x$-axis.无实根意味着抛物线 $y = x^{2} - 4x + 13$ 不与 $x$ 轴相交。由 $a = 1 > 0$ 知抛物线开口向上，其最小值 $y = 13 - 4 = 9 > 0$ 确认图像完全位于 $x$ 轴上方。
Real coefficients lock complex roots into conjugate pairs, and the dividing line is $\Delta = 0$.实系数将复根锁为共轭对，分界线为 $\Delta = 0$。 The discriminant is a single real number whose sign sorts the three regimes: $\Delta > 0$, $\Delta = 0$, $\Delta < 0$. As $c$ slides through $4$ in this family, the two real roots merge at $x = 2$ (double root), then split apart vertically into the complex plane as $\pm i \sqrt{(c - 4)}$ added to the real part $\tfrac{-b}{2a} = 2$. This is the prototype for honors topics later: every real-coefficient polynomial of degree $n$ has its non-real complex roots in conjugate pairs (HSN-CN.A.7), which is why $\Delta < 0$ never produces a single complex root or three of them, only pairs.判别式是一个实数，其符号划分三种情形：$\Delta > 0$、$\Delta = 0$、$\Delta < 0$。当 $c$ 在这个方程族中越过 $4$ 时，两个实根在 $x = 2$ 处合并（重根），随后在复平面上竖直分裂为 $\pm i \sqrt{(c - 4)}$ 加上实部 $\tfrac{-b}{2a} = 2$。这是后续荣誉级主题的原型：任何 $n$ 次实系数多项式的非实复根都成共轭对（HSN-CN.A.7），这就是为什么 $\Delta < 0$ 不会产生单一复根或三个复根，而只能成对出现。
PART III  ·  MODELING / APPLIED · SOLUTIONS第三部分  ·  建模与应用 · 参考答案Universal · 32 marks通用题型 · 共 32 分

Section C · Modeling and ApplicationsC 部分 · 建模与应用

Q10MEDIUM中 🇺🇸 US美 AP-feeder FRQAP 衔接简答题 §6 Projectile Modeling投射运动建模 · HSF-IF.B.4 [10 marks][10 分]

Rocket: $h(t) = -4.9 t^{2} + 30 t + 2$ (m, s). (a) $h(0)$ + meaning. (b) Time + height at max. (c) Time hits ground. (d) Practical domain/range + monotone intervals.火箭：$h(t) = -4.9 t^{2} + 30 t + 2$（米、秒）。(a) $h(0)$ 及其含义。(b) 最大高度的时刻与高度。(c) 落地的时刻。(d) 实际定义域/值域 + 单调区间。

Answer:答案：  (a) $h(0) = 2$ m (launch platform height)$h(0) = 2$ m（发射平台高度）  ·  (b) $t \approx 3.1$ s, $h_{\max} \approx 47.9$ m，$h_{\max} \approx 47.9$ m  ·  (c) $t \approx 6.2$ s  ·  (d) $D = [0, 6.2]$, $R = [0, 47.9]$; increasing on $[0, 3.1]$, decreasing on $[3.1, 6.2]$$D = [0, 6.2]$，$R = [0, 47.9]$；在 $[0, 3.1]$ 上递增，在 $[3.1, 6.2]$ 上递减

(a) Initial height初始高度 A1·A1

$h(0) = -4.9(0)^{2} + 30(0) + 2 = 2$ m. In context: the rocket is launched from a platform $2$ metres above the ground, so $h(0) = 2$ is the launch height (the constant term encodes the starting altitude).$h(0) = -4.9(0)^{2} + 30(0) + 2 = 2$ m。结合情境：火箭从距地面 $2$ m 高的平台发射，故 $h(0) = 2$ 即发射高度（常数项编码起始高度）。

(b) Time and height at maximum最大值时刻与高度 M1·A1·A1

The vertex of $h(t) = at^{2} + bt + c$ is at $t = -\dfrac{b}{2a}$. With $a = -4.9$, $b = 30$:$h(t) = at^{2} + bt + c$ 的顶点在 $t = -\dfrac{b}{2a}$。代入 $a = -4.9$、$b = 30$： $$ t^{*} \;=\; -\frac{30}{2(-4.9)} \;=\; \frac{30}{9.8} \;\approx\; 3.0612\ldots \;\approx\; 3.1 \;\text{s}. $$ Maximum height:最大高度： $$ h(t^{*}) \;=\; -4.9(3.0612)^{2} + 30(3.0612) + 2 \;\approx\; -45.918 + 91.837 + 2 \;\approx\; 47.918 \;\approx\; 47.9 \;\text{m}. $$ (Cross-check via $k = c - \tfrac{b^{2}}{4a} = 2 - \tfrac{900}{-19.6} = 2 + 45.918 = 47.918$. $\checkmark$)（用 $k = c - \tfrac{b^{2}}{4a} = 2 - \tfrac{900}{-19.6} = 2 + 45.918 = 47.918$ 交叉验证。$\checkmark$）

(c) Time the rocket hits the ground火箭落地的时刻 M1·A1·A1

Solve $h(t) = 0$ for $t > 0$:求解 $h(t) = 0$，取 $t > 0$： $$ -4.9 t^{2} + 30 t + 2 \;=\; 0 \;\Longrightarrow\; t \;=\; \frac{-30 \pm \sqrt{900 + 39.2}}{-9.8} \;=\; \frac{-30 \pm \sqrt{939.2}}{-9.8}. $$ $\sqrt{939.2} \approx 30.646$. The two algebraic roots are$\sqrt{939.2} \approx 30.646$。两个代数根为 $$ t \;\approx\; \frac{-30 - 30.646}{-9.8} \;\approx\; 6.188 \quad \text{and} \quad t \;\approx\; \frac{-30 + 30.646}{-9.8} \;\approx\; -0.066. $$ Reject the negative root (before launch). So $t \approx 6.2$ s.舍去负根（早于发射）。所以 $t \approx 6.2$ s。

(d) Practical domain and range; monotone intervals实际定义域与值域；单调区间 A1·A1

Practical domain: $t \in [0, 6.2]$, from launch through impact.实际定义域：$t \in [0, 6.2]$，从发射到落地。
Practical range: $h \in [0, 47.9]$, from ground level (at impact) up to the peak. Note the platform height $2$ is not the minimum of the practical range; the rocket also reaches ground level $0$ on the way down.实际值域：$h \in [0, 47.9]$，从地面（落地时）到峰值。注意平台高度 $2$ 不是实际值域的最小值；火箭下降途中也到达地面 $0$。
Increasing on $[0, 3.1]$ (ascent); decreasing on $[3.1, 6.2]$ (descent). This matches the HSF-IF.B.4 reading: the function rises until the vertex, then falls.在 $[0, 3.1]$ 上递增（上升段）；在 $[3.1, 6.2]$ 上递减（下降段）。这与 HSF-IF.B.4 的解读一致：函数上升至顶点后下降。
Algebraic domain $\ne$ practical domain, and the negative root is the giveaway.代数定义域 $\ne$ 实际定义域，负根就是提示。 The algebraic domain of $h$ is all of $\mathbb{R}$, but the model is only valid where the situation makes sense: $t \ge 0$ (no time-travel) and $h(t) \ge 0$ (rocket above ground). The negative algebraic root $t \approx -0.066$ is a tell that the model would have predicted "ground crossing in the past", physically meaningless, but a useful sanity check that the algebra is right. AP graders explicitly reward students who state the practical domain in interval notation and name a reason ("rocket cannot exist before launch" or "rocket cannot pass through the ground"); writing only $t \ge 0$ misses an A1.$h$ 的代数定义域是全部 $\mathbb{R}$，但模型只在情境合理处有效：$t \ge 0$（不能时间倒流）且 $h(t) \ge 0$（火箭在地面之上）。负的代数根 $t \approx -0.066$ 是模型预测"过去落地"的提示——物理上无意义，但用于验证代数正确性。AP 阅卷明确奖励用区间记法且写出理由（"火箭在发射前不存在"或"火箭不能穿越地面"）的学生；只写 $t \ge 0$ 会丢一个 A1。
Q11MEDIUM中 🇨🇦 ON安 ON Provincial-style安大略省考风格 §6 Revenue Modeling营收建模 · MCR3U A2.5 [11 marks][11 分]

Bakery: base $200$ croissants at CA\$3.00; each CA\$0.10 hike drops sales by $5$. Let $x =$ number of \$0.10 hikes. (a) $P(x), Q(x)$. (b) $R(x)$ expanded. (c) Optimal $x$ + max revenue. (d) Optimal price + quantity. (e) Restriction on $x$.烘焙店：基准每日售出 $200$ 个，单价 CA\$3.00；每涨价 CA\$0.10 销量下降 $5$ 个。设 $x =$ 涨价 \$0.10 的次数。(a) $P(x)$、$Q(x)$。(b) $R(x)$ 展开。(c) 最优 $x$ 与最大营收。(d) 最优单价与销量。(e) $x$ 的取值限制。

Answer:答案：  (a) $P(x) = 3 + 0.10 x$, $Q(x) = 200 - 5 x$  ·  (b) $R(x) = -0.5 x^{2} + 5 x + 600$  ·  (c) $x^{*} = 5$, $R_{\max} = \$612.50$  ·  (d) price $\$3.50$, quantity $175$单价 $\$3.50$，销量 $175$  ·  (e) $0 \le x \le 40$, $x \in \mathbb{Z}$

(a) Linear price and quantity线性单价与销量 A1·A1

Each \$0.10 hike adds \$0.10 to the price: $P(x) = 3.00 + 0.10 x$ (in CA\$).每次 \$0.10 涨价使单价增加 \$0.10：$P(x) = 3.00 + 0.10 x$（单位 CA\$）。
Each hike drops sales by $5$: $Q(x) = 200 - 5 x$ (in croissants).每次涨价销量下降 $5$ 个：$Q(x) = 200 - 5 x$（牛角包数）。

(b) Revenue in expanded standard form营收的标准展开式 M1·A1

$$ R(x) \;=\; P(x) \cdot Q(x) \;=\; (3 + 0.10 x)(200 - 5 x). $$ Expand:展开： $$ R(x) \;=\; 600 - 15 x + 20 x - 0.5 x^{2} \;=\; -0.5 x^{2} + 5 x + 600. $$

(c) Maximise $R(x)$使 $R(x)$ 最大 M1·A1·A1·A1

$R$ is a downward-opening parabola ($a = -0.5 < 0$), so the maximum is at the vertex:$R$ 是开口向下的抛物线（$a = -0.5 < 0$），最大值在顶点处取得： $$ x^{*} \;=\; -\frac{b}{2a} \;=\; -\frac{5}{2(-0.5)} \;=\; \frac{5}{1} \;=\; 5. $$ Maximum revenue:最大营收： $$ R(5) \;=\; -0.5(25) + 5(5) + 600 \;=\; -12.5 + 25 + 600 \;=\; \$612.50. $$

(d) Optimal price and quantity最优单价与销量 A1·A1

Price: $P(5) = 3.00 + 0.10(5) = \$3.50$ per croissant.单价：$P(5) = 3.00 + 0.10(5) = \$3.50$ 每个牛角包。
Quantity: $Q(5) = 200 - 5(5) = 175$ croissants/day.销量：$Q(5) = 200 - 5(5) = 175$ 个/日。
(Cross-check: $\$3.50 \times 175 = \$612.50$. $\checkmark$)（交叉验证：$\$3.50 \times 175 = \$612.50$。$\checkmark$）

(e) Restriction on $x$$x$ 的取值限制 A1

Two restrictions follow from the situation. First, $Q(x) \ge 0 \Leftrightarrow 200 - 5 x \ge 0 \Leftrightarrow x \le 40$: at $40$ hikes the bakery sells zero croissants, beyond that the model would predict negative sales. Second, $x \ge 0$ (number of hikes cannot be negative) and $x$ must be a non-negative integer (\$0.10 hikes are discrete). Hence $x \in \{0, 1, 2, \ldots, 40\}$, i.e. $0 \le x \le 40$ with $x \in \mathbb{Z}$.由情境得出两条限制。首先，$Q(x) \ge 0 \Leftrightarrow 200 - 5 x \ge 0 \Leftrightarrow x \le 40$：涨价 $40$ 次后烘焙店销量为零，再涨模型会预测负销量。其次，$x \ge 0$（涨价次数不能为负），且 $x$ 必须为非负整数（\$0.10 涨价是离散的）。所以 $x \in \{0, 1, 2, \ldots, 40\}$，即 $0 \le x \le 40$ 且 $x \in \mathbb{Z}$。
Optimum-of-product-of-linears: always a parabola, always halfway between the zeros.线性乘积的最优值：恒为抛物线，恒位于两零点之间。 When revenue is the product of a linear price and a linear quantity, $R(x) = (a + bx)(c + dx)$, the result is a quadratic whose roots are $x = -a/b$ and $x = -c/d$, the price-zero and the quantity-zero. The maximum sits at the midpoint of those two roots: $x^{*} = \tfrac{1}{2}(-a/b - c/d)$. Here zeros are at $x = -30$ (price hits zero) and $x = 40$ (quantity hits zero); midpoint is $\tfrac{-30 + 40}{2} = 5$. That midpoint shortcut catches arithmetic errors faster than re-running $-b/(2a)$, and provincial markers reward the student who explicitly cites it.当营收是线性单价与线性销量之积 $R(x) = (a + bx)(c + dx)$ 时，结果为二次函数，根为 $x = -a/b$（单价零点）与 $x = -c/d$（销量零点）。最大值在这两根的中点：$x^{*} = \tfrac{1}{2}(-a/b - c/d)$。此处零点为 $x = -30$（单价为零）与 $x = 40$（销量为零）；中点为 $\tfrac{-30 + 40}{2} = 5$。这个中点捷径比重算 $-b/(2a)$ 更能快速捕捉算术错误，省考阅卷会奖励明确引用该法的学生。
Q12HARD难Honors荣誉级 🇨🇦 BC卑 🇺🇸 US美 BC Provincial-style卑诗省考风格 §7 Quadratic Inequality二次不等式 · BC PC11 inequalities不等式 [11 marks][11 分]

$2x^{2} - 5x - 12 < 0$. (a) Factor + roots. (b) Sign chart. (c) Solution set ($< 0$). (d) Solution set ($\ge 0$). (e) Park profit context $\Pi(x) = -2x^{2} + 5x + 12$, profitable prices $x \ge 0$.$2x^{2} - 5x - 12 < 0$。(a) 因式分解 + 根。(b) 符号表。(c) ($< 0$) 解集。(d) ($\ge 0$) 解集。(e) 公园利润情境 $\Pi(x) = -2x^{2} + 5x + 12$，$x \ge 0$ 时盈利价格区间。

Answer:答案：  (a) $(2x + 3)(x - 4)$, roots $x = -\tfrac{3}{2}, 4$，根 $x = -\tfrac{3}{2}, 4$  ·  (c) $\bigl(-\tfrac{3}{2}, 4\bigr)$  ·  (d) $\bigl(-\infty, -\tfrac{3}{2}\bigr] \cup [4, \infty)$  ·  (e) $0 \le x < 4$ (dollars)$0 \le x < 4$（美元）

(a) Factor and roots因式分解与根 M1·A1·A1

AC-method on $2x^{2} - 5x - 12$: $AC = 2 \cdot (-12) = -24$. Integers with product $-24$ and sum $-5$: $(-8, 3)$. Split and group:对 $2x^{2} - 5x - 12$ 用 AC 法：$AC = 2 \cdot (-12) = -24$。乘积 $-24$、和 $-5$ 的整数：$(-8, 3)$。拆分并分组： $$ 2x^{2} - 8x + 3x - 12 \;=\; 2x(x - 4) + 3(x - 4) \;=\; (2x + 3)(x - 4). $$ Roots of $2x^{2} - 5x - 12 = 0$: $x = -\tfrac{3}{2}$ and $x = 4$.$2x^{2} - 5x - 12 = 0$ 的根：$x = -\tfrac{3}{2}$ 与 $x = 4$。

(b) Sign chart符号表 M1·A1·A1

The two roots partition the real line into three intervals: $\bigl(-\infty, -\tfrac{3}{2}\bigr)$, $\bigl(-\tfrac{3}{2}, 4\bigr)$, $(4, \infty)$. Test one point from each, using the factored form $f(x) = (2x + 3)(x - 4)$:两个根将实轴分成三个区间：$\bigl(-\infty, -\tfrac{3}{2}\bigr)$、$\bigl(-\tfrac{3}{2}, 4\bigr)$、$(4, \infty)$。在每个区间取一个测试点，利用因式分解形式 $f(x) = (2x + 3)(x - 4)$：
  • $x = -2$: $(2(-2) + 3)(-2 - 4) = (-1)(-6) = +6 > 0$. Sign: $+$符号：$+$.
  • $x = 0$: $(3)(-4) = -12 < 0$. Sign: $-$符号：$-$.
  • $x = 5$: $(13)(1) = +13 > 0$. Sign: $+$符号：$+$.
Sign chart (left to right): $\;+ \;\big|\; -\tfrac{3}{2} \;\big|\; - \;\big|\; 4 \;\big|\; +$. Since $a = 2 > 0$ the parabola opens upward, so the signs $+, -, +$ are exactly what symmetry predicts.符号表（从左到右）：$\;+ \;\big|\; -\tfrac{3}{2} \;\big|\; - \;\big|\; 4 \;\big|\; +$。由 $a = 2 > 0$ 知抛物线开口向上，所以符号 $+, -, +$ 正是对称性所预测的。

(c) Solution set of $2x^{2} - 5x - 12 < 0$$2x^{2} - 5x - 12 < 0$ 的解集 M1·A1

Strictly negative on $\bigl(-\tfrac{3}{2}, 4\bigr)$ from the sign chart. Solution set: $\bigl(-\tfrac{3}{2}, 4\bigr)$ (open interval, endpoints excluded because the inequality is strict).由符号表可知严格为负在 $\bigl(-\tfrac{3}{2}, 4\bigr)$。解集：$\bigl(-\tfrac{3}{2}, 4\bigr)$（开区间，端点不包含，因不等式严格）。

(d) Solution set of $2x^{2} - 5x - 12 \ge 0$$2x^{2} - 5x - 12 \ge 0$ 的解集 A1

Non-strict $\ge 0$: take the $+$ regions plus the two roots. Solution set: $\bigl(-\infty, -\tfrac{3}{2}\bigr] \cup [4, \infty)$ (closed at the roots).非严格 $\ge 0$：取 $+$ 区域外加两个根。解集：$\bigl(-\infty, -\tfrac{3}{2}\bigr] \cup [4, \infty)$（根处闭合）。

(e) Park profit context公园利润情境 M1·A1

Notice $\Pi(x) = -2x^{2} + 5x + 12 = -(2x^{2} - 5x - 12)$, so $\Pi(x) > 0 \Leftrightarrow 2x^{2} - 5x - 12 < 0 \Leftrightarrow x \in \bigl(-\tfrac{3}{2}, 4\bigr)$ by part (c). Combined with the situational restriction $x \ge 0$ (admission price in dollars cannot be negative), the profitable price range is $\;0 \le x < 4$. The park earns strictly positive profit when daily admission is set somewhere between \$0 and \$4 (excluding \$4 itself, where profit is exactly zero).注意 $\Pi(x) = -2x^{2} + 5x + 12 = -(2x^{2} - 5x - 12)$，所以由 (c) 得 $\Pi(x) > 0 \Leftrightarrow 2x^{2} - 5x - 12 < 0 \Leftrightarrow x \in \bigl(-\tfrac{3}{2}, 4\bigr)$。结合情境约束 $x \ge 0$（美元票价不能为负），盈利价格范围为 $\;0 \le x < 4$。当日票价定在 \$0 与 \$4 之间（不含 \$4 本身，此处利润恰为零）时，公园获得严格正利润。
Sign chart $\Rightarrow$ instant relative inequality, no re-work.符号表 $\Rightarrow$ 立得所有相关不等式，无需重做。 Once the sign chart is drawn, every related inequality, $< 0$, $\le 0$, $> 0$, $\ge 0$, reads off directly without recomputing. Strict vs. non-strict only changes whether the root endpoints are included. The "$\ge 0$" solution is the complement of the "$< 0$" solution (as a subset of $\mathbb{R}$); the union of the two answers is all of $\mathbb{R}$, with the roots counted in whichever side has $\ge$ or $\le$. The deeper move in (e): negating $\Pi(x) = -f(x)$ flips the sign chart wholesale ($+ \leftrightarrow -$), so "$\Pi > 0$" $=$ "$f < 0$", the algebra you already did transfers directly, no second sign chart needed. BC provincial markers love this kind of structural reuse.一旦符号表画出，所有相关不等式（$< 0$、$\le 0$、$> 0$、$\ge 0$）都可直接读出，无需重算。严格与非严格的差别仅在于是否包含根端点。"$\ge 0$" 解集是"$< 0$" 解集在 $\mathbb{R}$ 中的补集；两个答案的并集是整个 $\mathbb{R}$，根归入带 $\ge$ 或 $\le$ 的那一侧。(e) 中的更深一招：对 $\Pi(x) = -f(x)$ 取负使整张符号表翻转（$+ \leftrightarrow -$），所以 "$\Pi > 0$" $=$ "$f < 0$"——已做的代数直接转用，无需再做第二张符号表。BC 省考阅卷喜爱这种结构重用。