Предмет: Математика, автор: timuruteev1337

Прямая y=4x+8 параллельна касательной к графику функции y=x²-5x+7. Найдите абциссу точки
П.С. Я сейчас в 10 классе , и решил начать готовиться к егэ. В интернете нет нормального,понятного объяснения,как решать эту задачу! Пожалуйста,помогите.

Ответы

Автор ответа: szaikina350

Ответ:

уравнение касательной y–yo=y'(xo)(x–xo)

y'=2x–5

угловые коэфф–ты прямой и касательной д. б. равны, тогда

y'(xo)=4

4=2xo–5 ⇒ xo=4,5

timuruteev1337: мне бы объяснить

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