Given
[tex]f(x)=\frac{3}{x-3}+1[/tex]
The vertical asymptotes are given by the value of x which makes the denominator equal to zero. In our case,
[tex]\begin{gathered} x-3=0 \\ \Rightarrow x=3 \end{gathered}[/tex]
Thus, the vertical asymptote is x=3.
As for the horizontal asymptote, notice that
[tex]\lim _{x\to\infty}f(x)=\lim _{x\to\infty}(\frac{3}{x-3}+1)=3\lim _{x\to\infty}\frac{1}{x-3}+1=0+1=1[/tex]
Similarly,
[tex]\lim _{x\to-\infty}f(x)=3\lim _{x\to-\infty}\frac{1}{x-3}+1=3\cdot0+1=1[/tex]
Therefore, the horizontal asymptote is y=1.
As for the domain of the function, the only points that are not included in the domain are those that cause a vertical asymptote. Hence,
[tex]\text{domain}(f(x))=\mleft\lbrace x\in\R|x\ne3\mright\rbrace[/tex]
Similarly, since y=1 is a horizontal asymptote,
[tex]\text{range}(f(x))=\mleft\lbrace y\in\R|y\ne1\mright\rbrace[/tex]
Thus, the domain is all real numbers except 3, and the range is all real numbers except 1.
