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 }