Предмет: Алгебра, автор: 77788864

Изобразите на плоскости множество точек, заданных неравенством
у>4х+1​.

Ответы

Автор ответа: Аноним
2

Графиком функции y = 4x + 1 является прямая, проходящая через две точки : (0;1), (-1;-3).

Теперь перейдем к неравенству : y > 4x + 1. Выберем контрольную точку (-1;1) и подставляем ее координаты в неравенство

1 > -4 + 1

1 > -3 - ВЕРНО

Поэтому множество точек на плоскости будет левая часть графика прямой y = 4x + 1 (где выбрали контрольную точку).

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