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OxfordAQA · International GCSE · Mathematics Topik 2 3:56 Narasi bahasa Inggris · Subtitle bahasa Inggris + 中文 disematkan langsung
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Unit 3.2 is the biggest unit, 35 percent of each Extension paper, and there is no formulae sheet.
3.2 单元是最大的单元，占每份拓展卷的 35%，而且没有公式表。
Recall the quadratic formula and completed square, the line equations, the derivative of k x to the n, and the two nth-term forms.
要记住求根公式和配方形式、直线方程、kxⁿ 的导数，以及两种通项形式。
Make a the subject of a x plus 3 equals 5xt minus 2a.
把 ax + 3 = 5xt − 2a 中的 a 变为主元。
Gather every a term on one side and everything else on the other, then take a out as a factor and divide by x plus 2.
把所有含 a 的项移到一边，其余移到另一边，再提出公因式 a，除以 x + 2。
Factorise top and bottom, and the whole factor x plus 3 cancels.
分子分母因式分解，整个因式 x + 3 约去。
To subtract, use the common denominator x plus 2 times x minus 1, and put the second numerator in a bracket so the minus sign reaches both terms: 2x minus 5 on top.
相减时用公分母 (x + 2)(x − 1)，并把第二个分子加上括号，使负号作用于两项：分子是 2x − 5。
Prove n squared plus n plus 9 is odd for every integer.
证明对每个整数 n，n² + n + 9 都是奇数。
n squared plus n is n times n plus 1: two consecutive integers, one of them even, so the product is even.
n² + n = n(n + 1)：两个连续整数，其中一个是偶数，所以乘积是偶数。
Even plus 9 is odd.
偶数加 9 是奇数。
f g of x is x minus 5 squared, plus 1.
fg(x) = (x − 5)² + 1。
Setting it to 5 leaves x minus 5 squared equal to 4, so x is 3 or 7.
令它等于 5，得 (x − 5)² = 4，所以 x = 3 或 7。
For an inverse, swap and rearrange: f of x equals 3x has inverse x over 3.
求反函数时交换并整理：f(x) = 3x 的反函数是 x/3。
Through 1, 7 and 2, 9 the gradient is 2 over 1, which is 2, and the line through them is y equals 2x plus 5.
过 (1, 7) 和 (2, 9) 的斜率是 2/1 = 2，所以过这两点的直线是 y = 2x + 5。
A perpendicular line has gradient minus a half, since the gradients multiply to minus 1.
垂直的直线斜率为 −½，因为两斜率之积为 −1。
Completing the square rewrites x squared minus 6x plus 5 as x minus 3 squared, minus 4.
配方把 x² − 6x + 5 写成 (x − 3)² − 4。
The square is never negative, so the parabola has its minimum at 3, minus 4, on the line x equals 3, with roots 1 and 5.
平方项永不为负，所以抛物线在 (3, −4) 取最小值，对称轴是 x = 3，根为 1 和 5。
The cubic y equals x cubed minus 12x plus 5 has a maximum at minus 2, 21 and a minimum at 2, minus 11, where the gradient is zero.
三次曲线 y = x³ − 12x + 5 在 (−2, 21) 有极大值，在 (2, −11) 有极小值，这些点斜率为零。
At x equals 3 the gradient is 15, so the tangent there has gradient 15.
x = 3 处斜率为 15，所以那里的切线斜率为 15。
Set the derivative 3x squared minus 12 to zero: x is plus or minus 2.
令导数 3x² − 12 = 0：x = ±2。
The gradient changes from plus to minus at minus 2, a maximum, and minus to plus at 2, a minimum.
在 −2 处斜率由正变负，是极大值；在 2 处由负变正，是极小值。
At x equals 3 the gradient is 15, giving y equals 15x minus 49.
x = 3 处斜率为 15，得 y = 15x − 49。
Critical values first; the parabola's sign does the rest.
先求临界值，抛物线符号定区间。
Factorise x squared minus x minus 6 to find the critical values, minus 2 and 3.
把 x² − x − 6 因式分解，得到临界值 −2 和 3。
The parabola opens upward, so it is above the axis outside them: x below minus 2 or above 3.
抛物线开口向上，所以在两临界值之外位于 x 轴上方：x < −2 或 x > 3。
5, 9, 17, 29, 45: first differences 4, 8, 12, 16, and a constant second difference of 4, so a is 2.
5, 9, 17, 29, 45：一阶差 4, 8, 12, 16，二阶差恒为 4，所以 a = 2。
Subtract 2n squared and the residuals 3, 1, minus 1, minus 3, minus 5 are linear: minus 2n plus 5.
减去 2n²，余下 3, 1, −1, −3, −5 是线性的：−2n + 5。
The nth term passes through every point.
通项经过每一个点。
Check yourself.
自我检测。
9x squared minus 25 is a difference of two squares.
9x² − 25 是平方差。
In f g, apply g first.
fg 中先作用 g。
A line perpendicular to gradient 2 has gradient minus a half.
与斜率 2 垂直的直线斜率为 −½。
Stationary points solve dy by dx equals 0.
驻点满足 dy/dx = 0。
And a is half the second difference.
a 是二阶差的一半。