State the vertical, horizontal asymptotes and zeros of the rational function, f(x) = x2+3x+2 x2+5x+4 . why is there no zero at x = –1?

1 ) x² + 5 x + 4 = 0
x² + 4 x + x + 4 = 0
x  ( x + 4 ) + ( x + 4 ) = 0
( x + 1 ) ( x + 4 ) = 0
Vertical Asymptotes are:
x = - 1 and x = - 4
2 ) Horizontal Asymptote:
lim ( x → ∞ ) ( x² + 3 x + 2 ) / ( x² + 5 x + 4 ) = ( L` Hospitals Rule )
= lim ( x → ∞ ) ( 2 x + 3 ) / ( 2 x + 5 ) = 
= 2 / 2 = 1
H. A. :  y = 1
3 ) Zeros:
x² + 3 x + 2 = 0
x² + 2 x + x + 2 = 0
x ( x + 2 ) + ( x + 2 ) = 0
( x + 1 ) ( x + 2 ) = 0
x1 = - 1 , x2 = - 2
4 ) There is no zero at x = - 1 ( only x = - 2 )
because: f ( - 1 ) = 0 / 0   ( not defined ) 
This function is defined for:
x ∈ R \ { - 4, - 1 }

Many things that seem threatening in the dark become welcoming when we shine light on them

Ithink is a. true, but not sure