Question

`f(x) = 2 - (3)/(x+1)`

Domain:
V.A:
Roots:
Y-int:
H.A:
Holes:
O.A:

Also, draw on the graph.

Answer 1

1.
`x\in (-\infty, -1)\cup (-1,\infty)`

2.
`x=-1`

3.
`x=(1)/(2)`

4.
`(0,-1)`

5.
`y=2`

6. There are no holes for this function.

7. There are no oblique asymptotes.

Explanation:

Consider the function
`y=2-(3)/(x+1).` The denominator of the fraction includes the expression
`x+1.` Since the denominator cannot be equal to 0, then

`x+1\\eq 0,\\ \\x\\eq -1.`

Thus, the range is
`x\in (-\infty, -1)\cup (-1,\infty)`

and line
`x=-1` is a vertical asymptote.

The roots of the function are:

`f(x)=0\Rightarrow 2-(3)/(x+1)=0,\\ \\(2x+2-3)/(x+1)=0\Rightarrow 2x+2-3=0,\\ \\2x-1=0,\\ \\\2x=1,\\ \\x=(1)/(2).`

When
`x=0,`
`f(0)=2-(3)/(0+1),\\ \\f(0)=2-3=-1.`

Point (0,-1) is y-intercept.

The line
`y=2` is a horizontal asymptote.

There are no holes for this function.

There are no oblique asymptotes.