Answer:
Step-by-step explanation:
Domain: the values that the x cannot be (in this case, the values that make the denominator equal to 0)
x²-5x+6=0
(x-2)(x-3)=0
x≠ 2, 3
in interval notation: (-∞,2)∪(2,3)∪(3,∞)
We just determined that x≠2 or 3 which means that the vertical aysmptotes are
x= 2,3
To find the holes graph both the numerator and denominator then see if anything cancels
2x²+x= x(2x+1)
x²-5x+6= (x-2)(x-3)
because nothing more to do here there are no holes
