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Ответ:
ludamishina26
05.03.2023 19:51


1) у= х2-3х+2

парабола, ветви вверх

2) В(х;у) - вершина

х=3/2 =1,5   у= 2,25-4,5+2 = -0,25   В(1,5; -0,25) - вершина

3) х2-3х+2 = 0

Д= 9-8 = 1

х(1) = (3+1) / 2 = 2

х(2)  = (3-1)/ 2 = 1

y=0 при  х=1, х=2

4) у>0 при х∈(-∞; 1) U (2; +∞)

    у< 0 при х∈(1; 2)

5) для построения чертим координатную плоскость, отмечаем стрелками положительные направления по каждой оси (вверх и вправо),подписываем их (х и у) , отмечаем  начало координат (О) и единичные отрезки*

(*) удобнее взять ед отрезок в 2 клетки,

на координатной плоскости отмечаем вершину В,  через нее вертикально проводим пунктирную линию  - ось симметрии параболы,

ставим нули функции точки   (1; 0) и (2; 0)

далее отмечаем точки х=0 у= 2,  и симметрично   х=3 у= 2

соединяем плавной линией точки.  Подписываем график. Всё!


   


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Ответ:
daryaavdeeva01
05.03.2023 19:51
Y = - x² - 3x + 1 - квадратичная функция. Графиком этой функции является парабола, ветви направлены вниз. Вершину параболы будем искать следующим образом:

m = -b/2a = - (-3)/2*(-1) = -1,5 - координата абсциссы.

Подставим теперь в функцию

y = - (-1.5)² - 3 * (-1.5) + 1 = 3,25

(-1.5; 3.25) - координаты вершины параболы.

у = х+5 - линейная функция. Графиком линейной функции является прямая, которая проходит через точки (0;5), (-5;0)

Графики пересекаются в точке (-2;3), где x=-2 и y=3 - решения системы уравнения
На координатной плоскости построены графики функций y=-x^2-3x+1 и y=x+5 . используя эти графики реши
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