I missed one calculus class so I don't know how to solve this... I've attached a picture of how my teacher solved a similar problem. However, it's a little confusing for me. Would appreciate if you could help!

how my teacher solved a similar problem-

Find all vertical and horizontal asymptotes of f (x ) =; (x - 3) (x + 2)(x2 - 4x+ 3) The function is continuous when ( 2c + 2 ) ( 20 2 -40c + 3 ) * 0 . 15 ( 2C+ 2 ) ( 20 2 - 42+ 3 ) = 0 then 20= - 2 / or 2 2 - 42 + 3 = 0 ( 20-3) ( 20 - 1) = 0 , 50 ) 20 = 3 / or May have Vertical asymptotes at these valves . Now lim = - Do . denom . is the So x=- 2 is a VA . lim 2 C -D 3 + 5 ( 2 - lim - DC-0 3+ ( 2C+2) ( x - 1) = 10 lim 2- 5 (0) = lim DC -D 3 ( 2C + 2 ) ( 20 - 1 ) = 10 So me= 3 is a removable discontinuity , not an asymptote . lim 20-0 1+ 8 ( 20) = = = DO . So xewl is also a VA .2. (10 marks) Find all vertical and horizontal asymptotes of 5x2 - 3x - 2 f (2) = 0c 2 - 1 Show all steps and explain your answer fully