Для функции у =3x^2+x-4/x-3 найти: а) уравнение касательной к графику в точке хо = 0; б) уравнение наклонной асимптоты графика; в) точки экстремума.

а) Для нахождения уравнения касательной к графику в точке х0 = 0 нужно найти производную функции y = 3x^2 + x - 4/x - 3 и подставить значение х = 0.

y' = (6x + 1)(x - 3) + 4/(x - 3)^2

Подставляем х = 0:

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4/9 y' = -3 + 4/9 y' = -3 + 4/9 y' = -3 + 4/9 y' = -3 + 4/9 y' = -3 + 4/9 y' = -3