📚 New SAT Math: An In-Depth Look at the Newly Added Topics | 新SAT数学新增知识点深度解析
The redesigned SAT Math section introduces several topics that were not tested on the old exam. Understanding these new concepts is essential for achieving a top score, as they appear in both the calculator and no-calculator portions. This guide breaks down each key addition with clear explanations, examples, and bilingual support to help students master the material.
改革后的新SAT数学部分引入了若干在旧考试中不曾出现的知识点。掌握这些新增概念对于取得高分至关重要，因为它们同时出现在可用计算器与不可用计算器的部分中。本文将对每个关键新增内容进行详细解析，提供清晰的解释与示例，并以双语形式帮助学生彻底吃透这些内容。
1. Complex Numbers (i = √-1) | 复数（虚数单位 i）
A complex number takes the form a + bi, where a and b are real numbers and i is the imaginary unit defined as i = √(-1). The most fundamental property is i² = -1. SAT questions often ask you to add, subtract, or multiply complex numbers, or to rationalise a denominator by multiplying by the complex conjugate.
复数写作 a + bi 的形式，其中 a 和 b 为实数，i 是虚数单位，定义为 i = √(-1)。最基本的性质是 i² = -1。SAT 考题经常要求你对复数进行加、减、乘运算，或者通过乘以其共轭复数来将分母实数化。
For example, to simplify (2 + 3i)(1 – i), use FOIL: 2·1 + 2·(-i) + 3i·1 + 3i·(-i) = 2 – 2i + 3i – 3i². Since i² = -1, -3i² = +3, giving 5 + i.
例如，化简 (2 + 3i)(1 – i) 需要先用乘法展开：2·1 + 2·(-i) + 3i·1 + 3i·(-i) = 2 – 2i + 3i – 3i²。因为 i² = -1，-3i² = +3，结果为 5 + i。
The complex conjugate of a + bi is a – bi. Multiplying a complex number by its conjugate always yields a real number: (a + bi)(a – bi) = a² + b².
a + bi 的共轭复数是 a – bi。将复数与其共轭相乘总能得到一个实数：(a + bi)(a – bi) = a² + b²。
2. Radian Measure and the Unit Circle | 弧度制与单位圆
The SAT now expects you to work comfortably with radians. Radian measure relates an angle to the arc length on a circle: π radians = 180°. Converting between degrees and radians is straightforward: multiply degrees by π/180, or radians by 180/π.
新SAT要求考生能够熟练运用弧度制。弧度制通过圆的弧长来度量角：π 弧度 = 180°。度与弧度的转换很简单：度数乘以 π/180 得到弧度，弧度乘以 180/π 得到度数。
The unit circle is a circle of radius 1 centred at the origin. Points on the unit circle have coordinates (cos θ, sin θ) for an angle θ measured from the positive x-axis. Common angles like π/6, π/4, π/3 appear frequently and you should memorise their sine and cosine values.
单位圆是以原点为圆心、半径为 1 的圆。对于从正 x 轴开始度量的角 θ，单位圆上点的坐标为 (cos θ, sin θ)。诸如 π/6、π/4、π/3 等常见角经常出现，你应当熟记它们的正弦与余弦值。
3. Basic Trigonometry: Sine, Cosine, Tangent | 基础三角函数：正弦、余弦、正切
Sine, cosine, and tangent are now explicitly tested. In a right triangle, sin θ = opposite/hypotenuse, cos θ = adjacent/hypotenuse, and tan θ = opposite/adjacent. Beyond right-triangle definitions, you need to understand the functions on the coordinate plane and their relationships, such as the Pythagorean identity: sin²θ + cos²θ = 1.
正弦、余弦和正切现在被明确地考查。在直角三角形中，sin θ = 对边/斜边，cos θ = 邻边/斜边，tan θ = 对边/邻边。除了直角三角形定义外，你还需要理解这些函数在坐标平面上的意义及其相互关系，例如毕达哥拉斯恒等式：sin²θ + cos²θ = 1。
You may be asked to find the sine of an angle given its cosine, or to solve simple trigonometric equations. Recognising the graphs of y = sin x and y = cos x is also helpful for certain problems.
题目可能会要求你在已知一个角的余弦时求其正弦，或者解简单的三角方程。能识别 y = sin x 和 y = cos x 的图像对于某些问题也有帮助。
4. Quadratic Functions: Vertex Form and Completing the Square | 二次函数：顶点形式与配方法
The vertex form of a quadratic function is y = a(x – h)² + k, where (h, k) is the vertex of the parabola. The SAT often requires you to rewrite a standard form quadratic (y = ax² + bx + c) into vertex form by completing the square. This skill also helps you find the maximum or minimum value of the function.
二次函数的顶点形式为 y = a(x – h)² + k，其中 (h, k) 是抛物线的顶点。SAT 经常要求你通过配方法将标准形式的二次函数 (y = ax² + bx + c) 改写为顶点形式。这项技能也能帮你找到该函数的最大值或最小值。
To complete the square for y = x² + 6x + 5, take half of 6 (which is 3), square it (9), and add and subtract 9 inside the expression: y = (x² + 6x + 9) – 9 + 5 = (x + 3)² – 4. The vertex is (-3, -4).
对 y = x² + 6x + 5 进行配方：取 6 的一半（3），将其平方（9），并在式子中加上再减去 9：y = (x² + 6x + 9) – 9 + 5 = (x + 3)² – 4。顶点为 (-3, -4)。
5. Polynomial Division and the Remainder Theorem | 多项式除法与余数定理
Long division of polynomials and the Remainder Theorem are new to the SAT. If a polynomial f(x) is divided by (x – a), the remainder is f(a). If f(a) = 0, then (x – a) is a factor. This provides a quick way to evaluate a polynomial at a given point or to factor it completely.
多项式长除法和余数定理是 SAT 的新增内容。若多项式 f(x) 除以 (x – a)，所得余数为 f(a)。若 f(a) = 0，则 (x – a) 是一个因式。这提供了一种快速求多项式在某点取值或将其完全因式分解的方法。
For instance, given f(x) = 2x³ – 3x² + 4x – 5, the remainder when divided by (x – 2) is f(2) = 2(8) – 3(4) + 8 – 5 = 16 – 12 + 8 – 5 = 7. No lengthy division is needed.
例如，已知 f(x) = 2x³ – 3x² + 4x – 5，除以 (x – 2) 的余数为 f(2) = 2(8) – 3(4) + 8 – 5 = 16 – 12 + 8 – 5 = 7。无需进行冗长的除法即可得到。
6. Standard Deviation in Statistics | 统计学中的标准差
Standard deviation measures how spread out a data set is. A small standard deviation means the data points cluster closely around the mean; a large one indicates wide variability. You do not need to compute it by hand—SAT questions test your understanding of its meaning and how changes in data affect it.
标准差衡量一组数据的离散程度。标准差小意味着数据点紧密聚集在均值附近；标准差大则表示数据变异性大。你不需要手动计算——SAT 题目考查的是你对标准差含义的理解以及数据变化会如何影响它。
For example, adding a constant to every data point shifts all values equally, so the standard deviation remains unchanged. Multiplying by a constant, however, scales the spread and multiplies the standard deviation by the absolute value of that constant.
例如，给每个数据点加上一个常数会同等程度平移所有值，因此标准差保持不变。而乘以一个常数则会缩放分布范围，使标准差乘以该常数的绝对值。
7. Scatterplots, Correlation, and Lines of Best Fit | 散点图、相关性及最佳拟合线
The new SAT emphasises interpreting scatterplots and linear models. You need to draw or identify a line of best fit, use it to make predictions, and understand the meaning of slope and intercept within the context. The correlation coefficient r describes the strength and direction of a linear relationship, ranging from -1 to 1.
新SAT强调散点图与线性模型的解读。你需要画出或识别一条最佳拟合线，用它进行预测，并理解斜率和截距在具体情境中的含义。相关系数 r 描述线性关系的强度与方向，取值范围为 -1 到 1。
A common question: ‘The line of best fit for the data is y = 2.3x + 15. What does the slope 2.3 represent?’ It means that for every one-unit increase in x, the predicted y increases by 2.3 units.
常见问题：“数据的最佳拟合线为 y = 2.3x + 15，斜率 2.3 代表什么？”它意味着 x 每增加一个单位，预测的 y 值增加 2.3 个单位。
8. Margin of Error and Statistical Inference | 误差范围与统计推断
Questions about margin of error appear when survey results are reported. The margin of error indicates the range within which the true population parameter is likely to fall. The SAT expects you to draw reasonable conclusions from a study, considering the margin of error, sample size, and randomness.
有关误差范围的问题常见于调查报告。误差范围指出了真实总体参数可能落入的区间。SAT 要求你据此得出合理的结论，并考虑误差范围、样本量和随机性。
For instance, if a poll shows 54% support for a proposal with a margin of error of ±3%, the true support could be anywhere between 51% and 57%. You cannot definitively claim that a majority supports it if the lower end dips below 50%.
例如，一项民调显示有 54% 的人支持某提案，误差范围为 ±3%，那么真实支持率可能在 51% 到 57% 之间。如果下限低于 50%，你便无法断言多数人支持该提案。
9. Circle Equations (x – h)² + (y – k)² = r² | 圆的方程
Circle equations now appear regularly. The standard form gives the centre (h, k) and radius r. You must be able to complete the square for both x and y terms to convert a general form equation like x² + y² + Ax + By + C = 0 into standard form. Once in standard form, extracting the centre and radius is straightforward.
圆的方程如今经常出现。标准形式显示了圆心 (h, k) 和半径 r。你必须能够通过对 x 和 y 项都进行配方，将诸如 x² + y² + Ax + By + C = 0 的一般形式方程转化为标准形式。化为标准形式后，提取圆心和半径就很简单了。
For example, x² + y² – 6x + 4y – 3 = 0 can be rewritten as (x – 3)² + (y + 2)² = 16, so the circle has centre (3, -2) and radius 4.
例如，x² + y² – 6x + 4y – 3 = 0 可改写为 (x – 3)² + (y + 2)² = 16，因此该圆的圆心为 (3, -2)，半径为 4。
10. Exponential Growth and Decay Models | 指数增长与衰减模型
Exponential functions are modelled as f(t) = a · (1 ± r)^t, where a is the initial amount, r is the growth or decay rate per time period, and t is time. You must interpret the meaning of each component in real-world contexts, such as population growth, radioactive decay, or compound interest.
指数函数模型通常写作 f(t) = a · (1 ± r)^t，其中 a 为初始值，r 为每个时间周期的增长或衰减率，t 为时间。你需要解读每个组成部分在现实情境中的含义，例如人口增长、放射性衰变或复利计算。
The SAT may also require you to compare linear and exponential growth and explain why an exponential model is appropriate (e.g., constant percentage change rather than constant absolute change). The natural base e appears occasionally, but understanding its meaning as an irrational constant approximately equal to 2.718 is enough.
SAT 也可能要求你比较线性与指数增长，并解释为何指数模型更为合适（例如，恒定百分比变化而非恒定绝对变化）。自然底数 e 偶尔会出现，但只需明白它是一个约等于 2.718 的无理常数即可。
11. Solving Linear-Quadratic Systems | 线性与二次方程组的求解
A linear-quadratic system consists of one linear equation and one quadratic equation, such as y = 2x + 1 and y = x² – x – 3. Solutions represent the intersection points of a line and a parabola. The substitution method is usually the most efficient: replace y from the linear equation into the quadratic.
线性与二次方程组由一条线性方程和一条二次方程组成，例如 y = 2x + 1 和 y = x² – x – 3。方程组的解代表直线与抛物线的交点。代入法通常最为高效：将线性方程中的 y 代入二次方程中。
Solving x² – x – 3 = 2x + 1 gives x² – 3x – 4 = 0, which factors as (x – 4)(x + 1) = 0. The solutions are x = 4 and x = -1, and substituting back yields the full coordinates. This topic also tests your ability to verify solutions and interpret the number of intersections.
解 x² – x – 3 = 2x + 1 得 x² – 3x – 4 = 0，分解为 (x – 4)(x + 1) = 0。解为 x = 4 和 x = -1，代回即可得到完整坐标。该考点也考验你验证解以及解读交点个数的能力。
12. Data Interpretation and Nonlinear Relationships | 数据解读与非线性关系
Beyond linear models, you must recognise when a data set is better modelled by a quadratic or exponential function. For example, if data shows a constant second difference, a quadratic model fits; if it grows by a constant factor, an exponential model fits. You should also interpret graphs that are curved and identify key features such as maxima, minima, and asymptotes.
除了线性模型外，你还需要识别何时数据集更适合用二次或指数函数来建模。例如，若数据呈现恒定的二阶差分，则二次模型适用；若数据以恒定倍数增长，则指数模型适用。你还需解读弯曲图像，并识别最大值、最小值及渐近线等关键特征。
Problems may present a table of values and ask which equation best represents the data. Check the patterns: a quadratic will have symmetric values around the vertex, while exponential data will have a constant ratio between consecutive y-values when x increases by a fixed amount.
题目可能给出一个数值表格，并询问哪个方程最能代表这些数据。检查模式：二次数据在顶点附近呈对称分布，而指数数据在 x 增加固定量时，连续的 y 值之间将保持恒定比值。
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