Предмет: Алгебра, автор: ata221

70 баллов! Прямо сейчас нужно! Главные мозги, профессоры, модераторы, магистры, светила наук, все-все-все, срочно! Пожалуйста, помогите.

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Автор ответа: NNNLLL54

0

$y=\dfrac{4x+2}{x+2}\\\\\\y\, (x+2)=4x+2\ \ ,\ \ \ yx+2y=4x+2\ \ ,\ \ \ yx-4x=2-2y\ \ ,\\\\\\x(y-4)=2-2y\ \ ,\ \ \ x=\dfrac{2-2y}{y-4}\ \ ,\ \ \ x=\dfrac{2(1-y)}{y-4}$

а) Обратная функция : $y=\dfrac{2(1-x)}{x-4}$ .

б) Асимптоты.

$y=\dfrac{2(1-x)}{x-4}\ \ ,\ \ y=\dfrac{2-2x}{x-4}\\\\\\\dfrac{2(1-x)}{x-4}=-2\cdot \dfrac{x-1}{x-4}=-2\cdot \dfrac{(x-4)+4-1}{x-4}=-2\cdot \Big(1+\dfrac{3}{x-4}\Big)=-2-\dfrac{6}{x-4}\\\\\\y=-2-\dfrac{6}{x-4}\ \ \ \Rightarrow \ \ \ y+2=-\dfrac{6}{x-4}$

Асимптоты - это прямые $y=-2$ и $x=4$ .

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