Describe the vertical asymptote(s) and hole(s) for the graph: y=(x-4)/(x^2+3x+2)

The vertical asymptote(s) are the vales of x, for which x^2 + 3x + 2 = 0
x^2 + x + 2x + 2 = 0
x(x + 1) + 2(x + 1) = 0
(x + 2)(x + 1) = 0
x = -2 and x = -1
The vertical asymptotes are the lines x = -2 and x = -1.

y = (x - 4)/(x^2 + 3x + 2) = (x - 2)(x + 2)/(x + 1)(x + 2) = (x - 2)/(x + 1)
At x + 2 = 0
x = -2
y = (-2 - 2)/(-2 + 1) = -4/-1 = 4
The hole is at point (-2, 4)

hope it is correct if not i'm srry i tried to