Answer:
1. [tex]x\in (-\infty, -1)\cup (-1,\infty)[/tex]
2. [tex]x=-1[/tex]
3. [tex]x=\dfrac{1}{2}[/tex]
4. [tex](0,-1)[/tex]
5. [tex]y=2[/tex]
6. There are no holes for this function.
7. There are no oblique asymptotes.
Step-by-step explanation:
Consider the function [tex]y=2-\dfrac{3}{x+1}.[/tex] The denominator of the fraction includes the expression [tex]x+1.[/tex] Since the denominator cannot be equal to 0, then
[tex]x+1\neq 0,\\ \\x\neq -1.[/tex]
Thus, the range is [tex]x\in (-\infty, -1)\cup (-1,\infty)[/tex]
and line [tex]x=-1[/tex] is a vertical asymptote.
The roots of the function are:
[tex]f(x)=0\Rightarrow 2-\dfrac{3}{x+1}=0,\\ \\\dfrac{2x+2-3}{x+1}=0\Rightarrow 2x+2-3=0,\\ \\2x-1=0,\\ \\\2x=1,\\ \\x=\dfrac{1}{2}.[/tex]
When [tex]x=0,[/tex] [tex]f(0)=2-\dfrac{3}{0+1},\\ \\f(0)=2-3=-1.[/tex]
Point (0,-1) is y-intercept.
The line [tex]y=2[/tex] is a horizontal asymptote.
There are no holes for this function.
There are no oblique asymptotes.
