Describe the vertical asymptotes and holes for the graph of y=x-4/x^2+3x+2,

You have a vertical asymptote at those points where the denominator is equal to zero, you have holes at those points where both the nominator and the denominator are equal to zero at the same time.

Therefore:
x - 4 = 0  
x = 4

x² + 3x + 2 = 0
(x + 2)(x + 1) = 0
x = -2 and x = -1

There are no points in which both numerator and denominator are equal to zero, therefore there are no holes, while x = -2 and x = -1 are vertical asymptotes.

I’m pretty sure the answer is a