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zY#H)1_SZP77P&SCnE;x4RrGwv<$PHb|B7@jeyar;RYVMk($Ns&HJ{T@yJ{M9^Mo!# zI}n}laJO&3Pc~8lbHaSJ-#WTc9xBK=t5k-kL0~RVypBu9K$_==neyr&X^fq1UScnjDU7|--bLzIGCI%=@% zH^4|}O57O&4m8YHMgilpF-n5aNe$PG?&JJo&;m-MtWcg#%GveQ*Ex_*XSRHAzI2Dpz{W( zk8R_jTm?u7d`u?HetAWk7y}DDvQZb&*SbXa4Wg<6Z7IVjs&r+2;xQDCT8H`Z8=czG zn2UI3Q@B1G1;t-SFPYaq=#XLkfR^!sERZQvpLPd43>HodMfnwBTSEC;B@B(fYn>We zBOtx0UC_6Y$F33uW(CFZ!7;FY=EyD literal 0 HcmV?d00001 diff --git a/frontend/js/exam9/variants/v47.js b/frontend/js/exam9/variants/v47.js index 475b753..90e314b 100644 --- a/frontend/js/exam9/variants/v47.js +++ b/frontend/js/exam9/variants/v47.js @@ -110,21 +110,18 @@ $$\\dfrac{4 - x^2}{x^2 - 4} = \\dfrac{-(x^2 - 4)}{x^2 - 4} = -1$$ text: `График функции $f(x) = a(x-m)^2 + n$ изображён на рисунке. Используя график функции, найдите $a$, $m$ и $n$. Запишите формулу функции $y = f(x)$ в виде многочлена.`, + figure: `График параболы с вершиной (2; 2), ветви вверх`, sol: `Функция $f(x)=a(x-m)^2+n$ — парабола с вершиной в точке $(m;\\,n)$; знак $a$ определяет направление ветвей ($a>0$ — вверх, $a<0$ — вниз), а $|a|$ — «крутизну».

- Алгоритм по графику: -
    -
  • находим координаты вершины параболы — это $m$ (абсцисса) и $n$ (ордината);
  • -
  • берём любую другую точку $(x_0;\\,y_0)$ графика и подставляем в формулу: - $a = \\dfrac{y_0 - n}{(x_0 - m)^2}$;
  • -
  • раскрываем скобки: $f(x)=a(x-m)^2+n = ax^2 - 2am\\,x + (am^2+n)$.
  • -
- Типичный вариант (вершина $(1;\\,-4)$, ветви вниз, через точку $(0;-5)$): - $m=1$, $n=-4$, $a=\\dfrac{-5-(-4)}{(0-1)^2} = -1$. - $$f(x) = -(x-1)^2 - 4 = -x^2 + 2x - 1 - 4 = -x^2 + 2x - 5.$$ -
Ответ: $a$, $m$, $n$ снимаются с графика по вершине $(m;n)$ и контрольной точке; - $f(x)=ax^2-2am\\,x+(am^2+n)$. Например, при вершине $(1;-4)$ и $a=-1$: $f(x)=-x^2+2x-5$.
` + Шаг 1. Снимаем вершину параболы с графика: вершина находится в точке $(2;\\,2)$, значит $m=2$, $n=2$.
+ Шаг 2. Ветви направлены вверх — значит $a>0$. Берём вторую точку графика, например $(0;\\,6)$ (точка пересечения с осью ординат): + $$a = \\dfrac{y_0 - n}{(x_0 - m)^2} = \\dfrac{6 - 2}{(0 - 2)^2} = \\dfrac{4}{4} = 1.$$ + Шаг 3. Раскрываем скобки в форме $f(x)=a(x-m)^2+n$: + $$f(x) = (x-2)^2 + 2 = x^2 - 4x + 4 + 2 = x^2 - 4x + 6.$$ + Проверка по контрольным точкам графика:
+ $f(1)=1-4+6=3$, $f(3)=9-12+6=3$, $f(4)=16-16+6=6$ — совпадает с графиком. +
Ответ: $a=1$, $m=2$, $n=2$; $f(x)=x^2-4x+6$.
` }, { text: `Найдите сумму целых решений системы неравенств diff --git a/frontend/js/exam9/variants/v48.js b/frontend/js/exam9/variants/v48.js index f531cc3..396511d 100644 --- a/frontend/js/exam9/variants/v48.js +++ b/frontend/js/exam9/variants/v48.js @@ -95,20 +95,18 @@ $$\\dfrac{-6}{n+3} - \\dfrac{n-3}{n+3} = \\dfrac{-6 - (n - 3)}{n+3} = \\dfrac{-6 text: `График функции $f(x) = a(x-m)^2 + n$ изображён на рисунке. Используя график функции, найдите $a$, $m$ и $n$. Запишите формулу функции $y = f(x)$ в виде многочлена.`, + figure: `График параболы с вершиной (3; 1), ветви вверх`, sol: `Вершинная форма квадратичной функции: $f(x) = a(x - m)^2 + n$ — это парабола с вершиной в точке $(m;\\,n)$. Знак $a$ определяет направление ветвей ($a \\gt 0$ — вверх, $a \\lt 0$ — вниз). -
Алгоритм нахождения $a$, $m$, $n$ по графику: -
    -
  • Шаг 1. Находим координаты вершины параболы — это $m$ (абсцисса) и $n$ (ордината).
  • -
  • Шаг 2. Берём любую другую точку $(x_0;\\,y_0)$ графика и подставляем в формулу: $a = \\dfrac{y_0 - n}{(x_0 - m)^2}$.
  • -
  • Шаг 3. Раскрываем скобки по формуле квадрата разности $(x - m)^2 = x^2 - 2mx + m^2$: -$$f(x) = a(x - m)^2 + n = ax^2 - 2am\\,x + (am^2 + n)$$
  • -
-Типичный пример (вершина $(2;\\,1)$, ветви вверх, график проходит через $(0;\\,5)$): -
— $m = 2$, $n = 1$. -
— Для $a$: $a = \\dfrac{5 - 1}{(0 - 2)^2} = \\dfrac{4}{4} = 1$. -
— Формула в виде многочлена: -$$f(x) = (x - 2)^2 + 1 = x^2 - 4x + 4 + 1 = x^2 - 4x + 5$$ -
Ответ: читается с графика; $f(x) = ax^2 - 2am\\,x + (am^2 + n)$.
` +

+Шаг 1. Снимаем координаты вершины параболы с графика: вершина находится в точке $(3;\\,1)$, значит $m = 3$, $n = 1$. +

+Шаг 2. Ветви параболы направлены вверх, значит $a > 0$. Берём вторую точку графика, например $(1;\\,5)$: +$$a = \\dfrac{y_0 - n}{(x_0 - m)^2} = \\dfrac{5 - 1}{(1 - 3)^2} = \\dfrac{4}{4} = 1.$$ +
Шаг 3. Раскрываем скобки по формуле квадрата разности $(x - m)^2 = x^2 - 2mx + m^2$: +$$f(x) = (x - 3)^2 + 1 = x^2 - 6x + 9 + 1 = x^2 - 6x + 10.$$ +Проверка по контрольным точкам графика:
+$f(2)=4-12+10=2$, $f(4)=16-24+10=2$, $f(5)=25-30+10=5$ — совпадает с графиком. +
Ответ: $a=1$, $m=3$, $n=1$; $f(x) = x^2 - 6x + 10$.
` }, { text: `Найдите сумму целых решений системы неравенств