Answer:
y = -2 (none of the options)
Step-by-step explanation:
Horizontal Asymptotes
A function f is said to have a horizontal asymptote y=a, if
[tex]\displaystyle \lim _{x\rightarrow -\infty }f(x)=a[/tex]
Or,
[tex]\displaystyle \lim _{x\rightarrow +\infty }f(x)=a[/tex]
The function provided in the question is
[tex]\displaystyle f(x)=\frac{-2x^2+3x+6}{x^2+1}[/tex]
We compute the limit
[tex]\displaystyle \lim _{x\rightarrow \infty }\frac{-2x^2+3x+6}{x^2+1}[/tex]
Dividing by [tex]x^2[/tex]
[tex]\displaystyle \lim _{x\rightarrow \infty }\frac{-2+3/x+6/x^2}{1+1/x^2}[/tex]
Knowing that
[tex]\displaystyle \lim _{x\rightarrow \infty }\frac{K}{x^n}=0,\ n\geqslant 1[/tex]
The limit results
[tex]\displaystyle \lim _{x\rightarrow \infty }\frac{-2+0+0}{1+0}=-2[/tex]
The only horizontal asymptote is [tex]y=-2[/tex]
None of the options presented is correct
