Answer:
Domain: [tex]\displaystyle (-\infty,-3)\cup(-3,6)\cup(6,\infty)[/tex]
Range: [tex](-\infty,0)\cup(0,\frac{1}{9})\cup(\frac{1}{9},\infty)[/tex]
Step-by-step explanation:
Factor the numerator and denominator
[tex]\displaystyle f(x)=\frac{x-6}{x^2-3x-18}\\ \\f(x)=\frac{x-6}{(x-6)(x+3)}\\\\f(x)=\frac{1}{x+3}[/tex]
Because the factor [tex]x-6[/tex] exists in both the numerator and denominator, then there is a hole at [tex]x=6[/tex] because the function is not continuous there. Additionally, because the denominator cannot be 0, [tex]x\neq-3[/tex], and there will be a vertical asymptote at the line [tex]x=-3[/tex].
Thus, the domain of the function is [tex]\displaystyle (-\infty,-3)\cup(-3,6)\cup(6,\infty)[/tex].
Also, notice that because [tex]x\neq6[/tex], as the function gets closer to [tex]x=6[/tex], the function gets closer to [tex]\displaystyle f(6)=\frac{1}{6+3}=\frac{1}{9}[/tex], so [tex]\displaystyle y=\frac{1}{9}[/tex] is not included in the range.
Thus, the range is [tex](-\infty,0)\cup(0,\frac{1}{9})\cup(\frac{1}{9},\infty)[/tex].
