Найти точки экстремума и интервалы монотонности функций

$y = \frac{x^2+3}{x-1}; x \neq 1\\y' =\frac{2x(x-1) - x^2+3}{(x-1)^2} = \frac{2x^2-2x-x^2-3}{(x-1)^2} = \frac{x^2-2x-3}{(x-1)^2}\\y' = 0\\x^2-2x-3 = 0\\D = 4+12 = 16\\x_{1/2} = \frac{2\frac+-\sqrt{16}}2; x_1 = -1; x_2 = 3$

↓ ↓ ↑ ↑ y

________|________|_______|______

- -1 - 1 + 3 + y'

y ↓ (-∞; 1)

y ↑ (1; ∞)